# NCERT Solutions Class 12 Mathematics Chapter 10

Vector Algebra is one of the important concepts utilised in mathematics and physics, and NCERT solutions for class 12 chapter 10 elaborate on it. We do come across various questions in our daily lives, such as the height of a tree or, to put it in another way, how hard should a ball be hit to score a goal? The only thing that one can say in response to such questions is 'Magnitude' and such magnitudes are called scalars. However, if we also need to determine the direction in which a tree is growing or in what direction one must hit the ball to reach the goal, we must use another quantity known as a Vector.

Therefore, a vector is defined as a quantity with both direction and magnitude. We need scalars such as: length, mass, time, distance, speed, area, volume as well as vectors like displacement, velocity, acceleration, force, weight, both for Mathematics and Physical Science. NCERT Solutions for Class 12 Mathematics Chapter 10 demonstrates solutions with the use of vectors.

The problems in the NCERT Solutions for Class 12 Mathematics chapter 10 are based on real-life scenarios that help students understand the topic easily. The lesson begins with an overview of some basic vector concepts and builds on them to thoroughly explain more complex subjects. The best thing about the NCERT solutions for Class 12 Mathematics Chapter 10 vector algebra is that it conveys tough sections in vernacular and easy language so that it could easily be understood by students with different intellect levels.

### Key Topics Covered in NCERT Solutions Class 12 Mathematics Chapter 10

The topics discussed in NCERT Solutions Class 12 Mathematics Chapter 10 are equally essential because they explain various sections of vectors, such as direction cosines, dot product, cross product, section formula, and associated properties. Furthermore, all topics are interconnected, implying that students cannot move on to the next section without first mastering the prior one. Therefore, studying each and every topic is necessary.

List of NCERT Solutions Class 12 Mathematics Chapter 10 Exercises

The word 'Vector' comes from 'Vectus', a Latin word, which means "to carry." Modern vector theory began in 1800 when Caspar Wessel (1745-1818) and Jean Robert Argand (1768-1822) demonstrated how to use a directed line segment in a coordinate plane to give the geometric interpretation of a complex number. Mathematicians conducted additional research, culminating in the topic as we know it today. Below are links to NCERT solutions class 12 Mathematics chapter 10, offering highly engaging facts and recommendations to assist students in studying the subject matter with ease.

There are various lessons in both Mathematics and Physics that need students to have a solid understanding of vectors. As a result, you must regularly review the solutions in these links and take notes on all the procedures and formulas stated in them. NCERT Solutions for Class 12 Mathematics Chapter 10 Vector Algebra for each practice question is provided in the links below.

Class 12 Mathematics Chapter 10 Ex 10.1 - 5 Questions

Class 12 Mathematics Chapter 10 Ex 10.2 - 19 Questions

Class 12 Mathematics Chapter 10 Ex 10.3 - 18 Questions

Class 12 Mathematics Chapter 10 Ex 10.4 - 12 Questions

Class 12 Mathematics Chapter 10 Miscellaneous Ex - 19 Questions

Total Questions: There are 63 sums in Chapter 10 of Class 12 Mathematics Vector Algebra, divided into 20 simple computational questions, 37 medium-level problems, and 6 difficult problems.

NCERT Solutions Class 12 Mathematics Chapter 10 Formula List

All the formulas students need to understand this chapter are used in NCERT Solutions for class 12 Mathematics chapter 10 to help execute arithmetic operations on vectors. The same laws that apply to whole numbers or scalars do not apply to vectors, which have their direction. As a result, the computations for vectors alter when this element is taken into consideration.

Several geometrical implications of vectors are discussed in this chapter, which contains additional formulas and processes some of which are mentioned below.

• Let S and T be two given vectors; then the dot product is represented by S.T = |S| |T| cos θ.
• If θ = 0°, meaning both S and T are in the same direction, then T = |S| |T|.
• If θ = 90°, meaning S and T are orthogonal then T = 0.
• If we have two vectors S = (S1, S2, S3 ….. Sn) and T = (T1, T2, T3 ….. Tn) then the dot product is given as S.T = (S1T1 + S2T2 + S3T3 ….. SnTn)

Students should also create a chart on important formulae so that they can be accessed easily whenever needed.

NCERT Mathematics Syllabus

Class 12 NCERT Mathematics Syllabus

Term - 1

 Unit Name Chapter Name Relations and Function Relations and Functions Inverse Trigonometric Functions Algebra Matrices Determinants Calculus Continuity and Differentiability Application of Derivatives Linear Programming Linear Programming

Term - 2

 Unit Name Chapter Name Calculus Integrals Application of Integrals Differential Equations Vectors and Three-Dimensional Geometry Vector Algebra Three Dimensional Geometry Probability Probability

Experts at Extramarks create NCERT Solutions to help students understand concepts more easily and precisely. NCERT Solutions provide detailed step-by-step explanations of problems found in textbooks. Solutions are available for various classes mentioned in the below links-

• NCERT Solutions class 1
• NCERT Solutions class 2
• NCERT Solutions class 3
• NCERT Solutions class 4
• NCERT Solutions class 5
• NCERT Solutions class 6
• NCERT Solutions class 7
• NCERT Solutions class 8
• NCERT Solutions class 9
• NCERT Solutions class 10
• NCERT Solutions class 11
• NCERT Solutions class 12

NCERT Mathematics Exam Pattern

 Duration of Marks 3 hours 15 minutes Marks for Internal 20 marks Marks for Theory 80 marks Total Number of Questions 38 Questions Very short answer question 20 Questions Short answer questions 7 Questions Long Answer Questions (4 marks each) 7 Questions Long Answer Questions (6 marks each) 4 Questions

NCERT Exemplar Class 12 Mathematics

NCERT Exemplar Class 12 Mathematics are available and can be accessed from NCERT official website. NCERT Exemplar Class 12 Mathematics Chapter 10 consists of problems with their solutions to help students prepare for their final exams. These Exemplar questions are a little more complex, and cover all the important concepts in Class 12 Mathematics Chapter 10. Students will fully understand all the concepts covered in chapter 10 by practising these NCERT Exemplars for Mathematics Class 12.

Introduction to vectors, types of vectors, addition and multiplication of vectors, vector components, section formula, scalar product, projection of a vector on a line, and vector product are all covered here. Each question in these materials is connected to topics covered in the CBSE Class 12 syllabus.

### Key Features of NCERT Solutions Class 12 Mathematics Chapter 10

NCERT Solutions for Class 12 Mathematics are not only easy to comprehend, but they also include essential practice questions based on the recent CBSE syllabus.

The following are the important aspects to consider:

• You would be better positioned to clear your doubts because it provides in-depth knowledge in simple language.
• It simplifies and categorises every concept in the topic, making it easier to grasp.
• It offers multiple concepts for approaching the question paper and is based on CBSE pattern.
• These answers are a result of extensive research and are designed to lay the groundwork for your future studies.

Q.1

$\mathrm{Find}|\stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{b}}|,\mathrm{if}\stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}-7\stackrel{^}{\mathrm{j}}+7\stackrel{^}{\text{k}}\mathrm{and}\stackrel{\to }{\mathrm{b}}=3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}.$

Ans.

$\begin{array}{l}\mathrm{Given}\text{vectors are:}\stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}-7\stackrel{^}{\mathrm{j}}+7\stackrel{^}{\mathrm{k}}\mathrm{and}\stackrel{\to }{\mathrm{b}}=3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\\ \mathrm{Then},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{b}}=|\begin{array}{ccc}\stackrel{^}{\mathrm{i}}& \stackrel{^}{\mathrm{j}}& \stackrel{^}{\mathrm{k}}\\ 1& -7& 7\\ 3& -2& 2\end{array}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(-14+14\right)\stackrel{^}{\mathrm{i}}+\left(2-21\right)\stackrel{^}{\mathrm{j}}+\left(-2+21\right)\stackrel{^}{\mathrm{k}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-19\stackrel{^}{\mathrm{j}}+19\stackrel{^}{\mathrm{k}}\\ \therefore |\stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{b}}|=|-19\stackrel{^}{\mathrm{j}}+19\stackrel{^}{\mathrm{k}}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\left(-19\right)}^{2}+{\left(19\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\sqrt{19}\end{array}$

Q.2

$\begin{array}{l}\mathrm{Find}\mathrm{a}\mathrm{unit}\mathrm{vector}\mathrm{perpendicular}\mathrm{to}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{vector}\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}\\ \mathrm{and}\stackrel{\to }{\mathrm{a}}-\stackrel{\to }{\mathrm{b}},\mathrm{where}\stackrel{\to }{\mathrm{a}}=3\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\text{k}}\mathrm{and}\stackrel{\to }{\mathrm{b}}=\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-2\stackrel{^}{\mathrm{k}}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\text{given vectors are}\stackrel{\to }{\mathrm{a}}=3\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\mathrm{and}\stackrel{\to }{\mathrm{b}}=\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-2\stackrel{^}{\mathrm{k}}.\\ \mathrm{Then},\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}=3\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}+\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-2\stackrel{^}{\mathrm{k}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}-\stackrel{\to }{\mathrm{b}}=3\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}-\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+4\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ \left(\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}\right)×\left(\stackrel{\to }{\mathrm{a}}-\stackrel{\to }{\mathrm{b}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=|\begin{array}{ccc}\stackrel{^}{\mathrm{i}}& \stackrel{^}{\mathrm{j}}& \stackrel{^}{\mathrm{k}}\\ 4& 4& 0\\ 2& 0& 4\end{array}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(16-0\right)\stackrel{^}{\mathrm{i}}-\left(16-0\right)\stackrel{^}{\mathrm{j}}+\left(0-8\right)\stackrel{^}{\mathrm{k}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=16\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-16\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}-8\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ |\left(\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}\right)×\left(\stackrel{\to }{\mathrm{a}}-\stackrel{\to }{\mathrm{b}}\right)|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=|16\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-16\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}-8\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\left(16\right)}^{2}+{\left(-16\right)}^{2}+{\left(-8\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=24\\ \text{Hence, the unit vector perpendicular to each of the vectors}\\ \left(\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}\right)\text{\hspace{0.17em}\hspace{0.17em}and}\left(\stackrel{\to }{\mathrm{a}}-\stackrel{\to }{\mathrm{b}}\right)\text{is given by the relation,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=±\frac{\left(\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}\right)×\left(\stackrel{\to }{\mathrm{a}}-\stackrel{\to }{\mathrm{b}}\right)}{|\left(\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}\right)×\left(\stackrel{\to }{\mathrm{a}}-\stackrel{\to }{\mathrm{b}}\right)|}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=±\frac{16\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-16\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}-8\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}}{24}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=±\frac{2\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}-\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}}{3}\end{array}$

Q.3

$\begin{array}{l}\mathrm{If}\mathrm{a}\mathrm{unit}\mathrm{vector}\stackrel{\to }{\mathrm{a}}\mathrm{makes}\mathrm{angles}\frac{\mathrm{\pi }}{3}\mathrm{with}\stackrel{^}{\mathrm{i}},\frac{\mathrm{\pi }}{4}\mathrm{with}\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{}\mathrm{and}\mathrm{an}\\ \mathrm{acute}\mathrm{angle}\mathrm{\theta }\mathrm{with}\stackrel{^}{\text{k}},\mathrm{then}\mathrm{find}\mathrm{\theta }\mathrm{and}\mathrm{hence},\mathrm{the}\mathrm{components}\mathrm{of}\stackrel{\to }{\mathrm{a}}.\end{array}$

Ans.

$\begin{array}{l}\text{Let unit vector}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}={\text{a}}_{\text{1}}\stackrel{^}{\mathrm{i}}+{\text{a}}_{\text{2}}\stackrel{^}{\mathrm{j}}+{\text{a}}_{\text{3}}\stackrel{^}{\mathrm{k}}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}|\stackrel{\to }{\mathrm{a}}|=1.\\ \text{It is given that\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}\text{makes angles}\frac{\mathrm{\pi }}{3}\text{\hspace{0.17em}\hspace{0.17em}with}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}},\text{\hspace{0.17em}}\frac{\mathrm{\pi }}{4}\text{\hspace{0.17em}\hspace{0.17em}with}\stackrel{^}{\mathrm{j}}\text{and an acute}\\ \text{angle}\mathrm{\theta }\text{with\hspace{0.17em}}\stackrel{^}{\mathrm{k}}.\\ \text{Then, we have:}\\ \text{cos}\frac{\mathrm{\pi }}{3}=\frac{{\mathrm{a}}_{1}}{|\stackrel{\to }{\mathrm{a}}|}⇒\frac{1}{2}=\frac{{\mathrm{a}}_{1}}{1}\\ ⇒{\mathrm{a}}_{1}=\frac{1}{2}\\ \text{cos}\frac{\mathrm{\pi }}{4}=\frac{{\mathrm{a}}_{2}}{|\stackrel{\to }{\mathrm{a}}|}⇒\frac{1}{\sqrt{2}}=\frac{{\mathrm{a}}_{2}}{1}\\ ⇒{\mathrm{a}}_{2}=\frac{1}{\sqrt{2}}\\ \text{cos}\mathrm{\theta }=\frac{{\mathrm{a}}_{3}}{|\stackrel{\to }{\mathrm{a}}|}⇒\text{cos}\mathrm{\theta }=\frac{{\mathrm{a}}_{3}}{1}\\ ⇒{\mathrm{a}}_{3}=\text{cos}\mathrm{\theta }\\ \mathrm{Since},\text{\hspace{0.17em}}|\stackrel{\to }{\mathrm{a}}|=1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\sqrt{{{\mathrm{a}}_{1}}^{2}+{{\mathrm{a}}_{2}}^{2}+{{\mathrm{a}}_{3}}^{2}}=1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{{\mathrm{a}}_{1}}^{2}+{{\mathrm{a}}_{2}}^{2}+{{\mathrm{a}}_{3}}^{2}=1\\ {\left(\frac{1}{2}\right)}^{2}+{\left(\frac{1}{\sqrt{2}}\right)}^{2}+{\text{cos}}^{\text{2}}\mathrm{\theta }=1\\ {\text{\hspace{0.17em}\hspace{0.17em}cos}}^{\text{2}}\mathrm{\theta }=1-\frac{1}{4}-\frac{1}{2}\\ {\text{\hspace{0.17em}\hspace{0.17em}cos}}^{\text{2}}\mathrm{\theta }=\frac{1}{4}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}cos}\mathrm{\theta }=\frac{1}{2}=\mathrm{cos}\frac{\mathrm{\pi }}{3}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\theta }=\frac{\mathrm{\pi }}{3}\\ \mathrm{Thus},\text{}\mathrm{\theta }=\frac{\mathrm{\pi }}{3}\text{and and the components of}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}\hspace{0.17em}are\hspace{0.17em}}\left(\frac{1}{2},\frac{1}{\sqrt{2}},\frac{1}{2}\right).\end{array}$

Q.4

$\mathrm{Show}\mathrm{that}\left(\stackrel{\to }{\mathrm{a}}-\stackrel{\to }{\mathrm{b}}\right)×\left(\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}\right)=2\left(\stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{b}}\right)$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\left(\stackrel{\to }{\mathrm{a}}-\stackrel{\to }{\mathrm{b}}\right)×\left(\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}\right)\\ \text{\hspace{0.17em}}=\stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{b}}-\stackrel{\to }{\mathrm{b}}×\stackrel{\to }{\mathrm{a}}-\stackrel{\to }{\mathrm{b}}×\stackrel{\to }{\mathrm{b}}\\ \text{\hspace{0.17em}}=0+\stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{b}}-0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[âˆµ\stackrel{\to }{\mathrm{b}}×\stackrel{\to }{\mathrm{a}}=\stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{b}}\right]\\ \text{\hspace{0.17em}}=2\left(\stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{b}}\right)=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.5 Represent graphically a displacement of 40 km, 30° east of north.

Ans.

$\begin{array}{l}\text{Here},\text{\hspace{0.17em}}\mathrm{vector}\text{}\stackrel{\to }{\mathrm{OP}}\text{represents the displacement of 40 km, 30°}\\ \text{East of North.}\end{array}$

Q.6 Classify the following measures as scalars and vectors.

(i) 10 kg
(ii) 2 meters north-west
(iii) 40°
(iv) 40 watt
(v) 10-19 coulomb
(vi) 20 m/s2

Ans.

(i) 10 kg is a scalar quantity because it involves only magnitude.
(ii) 2 meters north-west is a vector quantity as it involves both magnitude and direction.
(iii) 40° is a scalar quantity as it involves only magnitude.
(iv) 40 watts is a scalar quantity as it involves only magnitude.
(v) 10–19 coulomb is a scalar quantity as it involves only magnitude.
(vi) 20 m/s2 is a vector quantity as it involves magnitude as well as direction

Q.7 Classify the following as scalar and vector quantities.

(i) time period
(ii) distance
(iii) force
(iv) velocity
(v) work done

Ans.

(i) Time period is a scalar quantity as it has only magnitude.
(ii) Distance is a scalar quantity as it has only magnitude.
(iii) Force is a vector quantity as it has both magnitude and direction.
(iv) Velocity is a vector quantity as it has both magnitude as well as direction.
(v) Work done is a scalar quantity as it has only magnitude.

Q.8 nIn fig.(a square), identify the following vectors.
(i) Coinitial
(ii) Equal
(iii) Collinear but not equal

Ans.

$\begin{array}{l}\text{(i) Vectors}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}\hspace{0.17em}and}\stackrel{\to }{\mathrm{d}}\text{\hspace{0.17em}are coinitial because they have the same\hspace{0.17em}initial point.}\\ \text{(ii) Vectors}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}and}\stackrel{\to }{\mathrm{d}}\text{\hspace{0.17em}\hspace{0.17em}are equal because they have the same magnitude and direction.}\\ \text{(iii) Vectors}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}\hspace{0.17em}and}\stackrel{\to }{\mathrm{c}}\text{\hspace{0.17em}are collinear but not equal. This is because although they are parallel,}\\ \mathrm{their}\mathrm{directions}\mathrm{are}\mathrm{not}\mathrm{the}\mathrm{same}\text{.}\end{array}$

Q.9

$\text{Answer the following as true or false.}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\begin{array}{l}\left(\mathrm{i}\right) \stackrel{\to }{\mathrm{a}} \mathrm{and} -\stackrel{\to }{\mathrm{a}}\mathrm{ }\mathrm{are}\mathrm{collinear}.\\ \left(\mathrm{ii}\right)\mathrm{Two}\mathrm{collinear}\mathrm{vectors}\mathrm{are}\mathrm{always}\mathrm{equal}\mathrm{inmagnitude}.\\ \left(\mathrm{iii}\right)\mathrm{Two}\mathrm{vectors}\mathrm{having}\mathrm{same}\mathrm{magnitude}\mathrm{are}\mathrm{collinear}.\\ \left(\mathrm{iv}\right)\mathrm{Two}\mathrm{collinear}\mathrm{vectors}\mathrm{having}\mathrm{the}\mathrm{same}\mathrm{magnitude}\\ \mathrm{are}\mathrm{equal}.\end{array}$

Ans.

(i) True. Since, both vectors are parallel to same line.
(ii) False.
Collinear vectors are those vectors that are parallel to the same line, irrespective of their magnitudes and directions.
(iii) False.
Collinear vectors are those vectors that are parallel to the same line, irrespective of their magnitudes and directions.
(iv) False.
Collinear vectors are those vectors that are parallel to the same line, irrespective of their magnitudes and directions.

Q.10

$\mathrm{Find}\mathrm{\lambda }\mathrm{and}\mathrm{\mu }\mathrm{if}\left(2\stackrel{^}{\mathrm{i}}+6\stackrel{^}{\mathrm{j}}+27\stackrel{^}{\text{k}}\right)×\left(\stackrel{^}{\mathrm{i}}+\mathrm{\lambda }\stackrel{^}{\mathrm{j}}+\mathrm{\mu }\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\right)=0.$

Ans.

$\begin{array}{l}\left(2\stackrel{^}{\mathrm{i}}+6\stackrel{^}{\mathrm{j}}+27\stackrel{^}{\mathrm{k}}\right)×\left(\stackrel{^}{\mathrm{i}}+\mathrm{\lambda }\stackrel{^}{\mathrm{j}}+\mathrm{\mu }\mathrm{ }\stackrel{^}{\mathrm{k}}\right)=0\\ ⇒|\begin{array}{ccc}\stackrel{^}{\mathrm{i}}& \stackrel{^}{\mathrm{j}}& \stackrel{^}{\mathrm{k}}\\ 2& 6& 27\\ 1& \mathrm{\lambda }& \mathrm{\mu }\end{array}|=0\\ \left(6\mathrm{\mu }-27\mathrm{\lambda }\right)\stackrel{^}{\mathrm{i}}-\left(2\mathrm{\mu }-27\right)\stackrel{^}{\mathrm{j}}+\left(2\mathrm{\lambda }-6\right)\mathrm{ }\stackrel{^}{\mathrm{k}}=0\stackrel{^}{\mathrm{i}}+0\stackrel{^}{\mathrm{j}}+0\stackrel{^}{\mathrm{k}}\\ \mathrm{On}\mathrm{comparing}\mathrm{the}\mathrm{corresponding}\mathrm{components},\mathrm{we}\mathrm{have}:\\ 6\mathrm{\mu }-27\mathrm{\lambda }=0, 2\mathrm{\mu }-27=0\mathrm{and}2\mathrm{\lambda }-6=0\\ ⇒\mathrm{\mu }=\frac{27}{2}\mathrm{and}\mathrm{\lambda }=3\end{array}$

Q.11

$\begin{array}{l}\text{Compute the magnitude of the following vectors:}\\ \stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}};\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=2\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}-7\stackrel{^}{\mathrm{j}}-3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}};\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}=\frac{1}{\sqrt{3}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+\frac{1}{\sqrt{3}}\stackrel{^}{\mathrm{j}}-\frac{1}{\sqrt{3}}\stackrel{^}{\mathrm{k}};\end{array}$

Ans.

$\begin{array}{l}\text{The given vectors are:}\\ \stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}};\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=2\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}-7\stackrel{^}{\mathrm{j}}-3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}};\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}=\frac{1}{\sqrt{3}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+\frac{1}{\sqrt{3}}\stackrel{^}{\mathrm{j}}-\frac{1}{\sqrt{3}}\stackrel{^}{\mathrm{k}}\\ \text{Magnitude of}\stackrel{\to }{\mathrm{a}}=|\stackrel{\to }{\mathrm{a}}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\left(1\right)}^{2}+{\left(1\right)}^{2}+{\left(1\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{3}\\ \text{Magnitude of}\stackrel{\to }{\mathrm{b}}=|\stackrel{\to }{\mathrm{b}}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\left(2\right)}^{2}+{\left(-7\right)}^{2}+{\left(-3\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{62}\\ \text{Magnitude of}\stackrel{\to }{\mathrm{c}}=|\stackrel{\to }{\mathrm{c}}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\left(\frac{1}{\sqrt{3}}\right)}^{2}+{\left(\frac{1}{\sqrt{3}}\right)}^{2}+{\left(\frac{1}{\sqrt{3}}\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{\frac{3}{3}}=1\end{array}$

Q.12 Write two different vectors having same magnitude.

Ans.

$\begin{array}{l}\begin{array}{l}\text{Two different vectors are:}\end{array}\\ \begin{array}{l}\stackrel{\to }{\mathrm{a}}\text{â€„}=\stackrel{^}{\mathrm{i}}\text{â€„}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}};\text{â€„}\stackrel{\to }{\mathrm{b}}\text{â€„}=\text{â€„}2\stackrel{^}{\mathrm{i}}\text{â€„}+3\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\end{array}\\ \begin{array}{l}\text{Then}\end{array}\\ \text{Magnitudeâ€„ofâ€„}\stackrel{\to }{\mathrm{a}}\text{â€„=â€„}\left|\stackrel{^}{\mathrm{i}}\text{â€„}+\text{â€„}2\stackrel{^}{\mathrm{j}}+\text{â€„}3\stackrel{^}{\mathrm{k}}\text{â€„}\right|\\ \begin{array}{l}\text{=â€„}\sqrt{{\text{1}}^{\text{2}}{\text{+2}}^{\text{2}}{\text{+3}}^{\text{2}}}\end{array}\\ \begin{array}{l}\text{=}\sqrt{\text{1+4+9}}\end{array}\\ \begin{array}{l}\text{=â€„}\sqrt{\text{14}}\end{array}\\ \begin{array}{l}\text{Magnitude â€„ofâ€„}\stackrel{\to }{\mathrm{b}}=\left|2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right|\end{array}\\ \begin{array}{l}\text{=}\sqrt{{\text{2}}^{\text{2}}{\text{+3}}^{\text{2}}{\text{+1}}^{\text{2}}}\end{array}\\ \begin{array}{l}\text{=}\sqrt{\text{4+9+1}}\end{array}\\ \begin{array}{l}\text{=}\sqrt{\text{14}}\end{array}\\ \begin{array}{l}\text{Hence,}\stackrel{\to }{\mathrm{a}}\text{â€„andâ€„}\stackrel{\to }{\mathrm{b}}\text{â€„areâ€„twoâ€„differentâ€„vectorsâ€„havingâ€„the â€„same}\end{array}\\ \begin{array}{l}\text{magnitude.}\end{array}\\ \begin{array}{l}\text{The vectors are different because they have different directions.}\end{array}\end{array}$

Q.13 Write two different vectors having same direction.

Ans.

$\begin{array}{l}\begin{array}{l}\text{Two different vectors are:}\end{array}\\ \begin{array}{l}\stackrel{\to }{\mathrm{a}}=2\stackrel{^}{\mathrm{i}}\text{â€„}+2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}};\text{â€„}\stackrel{\to }{\mathrm{b}}=\stackrel{^}{\mathrm{i}}\text{â€„}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\end{array}\\ \begin{array}{l}\text{Then,}\end{array}\\ \begin{array}{l}\text{Magnitudeâ€„ofâ€„}\stackrel{\to }{\text{a}}\end{array}\text{â€„}=\text{â€„}\begin{array}{l}\left|2\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right|\end{array}\\ \begin{array}{l}=\text{â€„}\sqrt{{2}^{2}+{2}^{2}+{2}^{2}}\end{array}\\ \begin{array}{l}=\text{â€„}\sqrt{4+4+4}\end{array}\\ \begin{array}{l}=\text{â€„}\sqrt{12}\end{array}\\ =\text{â€„}\begin{array}{l}2\sqrt{3}\end{array}\\ \text{Direction cosinesâ€„ofâ€„}\stackrel{\to }{\mathrm{a}}\text{â€„}\mathrm{are}:\\ \begin{array}{l}\frac{2}{2\sqrt{3}},\frac{2}{2\sqrt{3}},\frac{2}{2\sqrt{3}}\text{â€„}\mathrm{i}.\mathrm{e}.,\text{â€„}\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\end{array}\\ \begin{array}{l}\text{Magnitudeâ€„of}\stackrel{\to }{\mathrm{b}}=\left|\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right|\end{array}\\ \begin{array}{l}=\sqrt{{1}^{2}+{1}^{2}+{1}^{2}}\end{array}\\ \begin{array}{l}=\sqrt{1+1+1}\end{array}\\ \begin{array}{l}=\sqrt{3}\end{array}\\ \begin{array}{l}\text{Direction â€„cosinesâ€„ofâ€„}\stackrel{\to }{\mathrm{b}}\text{â€„are:}\end{array}\\ \begin{array}{l}\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\end{array}\\ \begin{array}{l}\text{Direction â€„cosinesâ€„ofâ€„}\stackrel{\to }{\mathrm{a}}\text{â€„}\mathrm{and}\text{â€„}\stackrel{\to }{\mathrm{b}}\text{â€„areâ€„same.â€„Hence,â€„these â€„two}\end{array}\\ \begin{array}{l}\text{vectorsâ€„haveâ€„sameâ€„direction.}\end{array}\end{array}$

Q.14

$\begin{array}{l}\text{Find the values of}\mathrm{x}\text{and}\mathrm{y}\text{so that the vectors\hspace{0.17em}\hspace{0.17em}}2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}\text{and}\\ \mathrm{x}\stackrel{^}{\mathrm{i}}+\mathrm{y}\stackrel{^}{\mathrm{j}}\text{are equal.}\end{array}$

Ans.

$\begin{array}{l}\text{Since,}\\ \text{\hspace{0.17em}\hspace{0.17em}}2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}=\mathrm{x}\stackrel{^}{\mathrm{i}}+\mathrm{y}\stackrel{^}{\mathrm{j}}\\ \text{So, on comparing coefficients of}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}and\hspace{0.17em}}\stackrel{^}{\mathrm{j}},\text{we get}\\ \mathrm{x}=2\text{and}\mathrm{y}=3.\end{array}$

Q.15 Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (– 5, 7).

Ans.

$\begin{array}{l}\text{The vector with the initial point P (2, 1) and terminal point}\\ \text{Q (-5, 7) can be given by,}\\ \stackrel{\to }{\mathrm{PQ}}=\left(-5-2\right)\stackrel{^}{\mathrm{i}}+\left(7-1\right)\stackrel{^}{\mathrm{j}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-7\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+6\stackrel{^}{\mathrm{j}}\\ \text{Hence, the required scalar components are -7 and 6 while the}\\ \text{vector components are\hspace{0.17em}\hspace{0.17em}}-7\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}6\stackrel{^}{\mathrm{j}}.\end{array}$

Q.16

$\begin{array}{l}\mathrm{Given}\mathrm{that}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=0\mathrm{and}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=0.\mathrm{What}\mathrm{can}\mathrm{you}\mathrm{conclude}\\ \mathrm{about}\mathrm{the}\mathrm{vectors}\stackrel{\to }{\mathrm{a}}\mathrm{and}\text{â€‹}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}?\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=0\\ \mathrm{Either}\text{}|\stackrel{\to }{\mathrm{a}}|=0\text{or}|\stackrel{\to }{\mathrm{b}}|=0,\\ \text{or}\stackrel{\to }{\mathrm{a}}\perp \stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}\left(\mathrm{if}\text{}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\text{are non-zero.}\right)\\ \left(\mathrm{ii}\right)\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=0\\ \mathrm{Either}\text{}|\stackrel{\to }{\mathrm{a}}|=0\text{or}|\stackrel{\to }{\mathrm{b}}|=0,\text{}\\ \text{or}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}\hspace{0.17em}}âˆ¥\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}\left(\mathrm{if}\text{}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\text{are non-zero.}\right)\\ \text{But,}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}\hspace{0.17em}and}\stackrel{\to }{\mathrm{b}}\text{cannot be perpendicular and parallel simultaneously.}\\ \text{Hence,\hspace{0.17em}â€‹}|\stackrel{\to }{\mathrm{a}}|=0\text{or}|\stackrel{\to }{\mathrm{b}}|=0.\end{array}$

Q.17

$\begin{array}{l}\text{Find the sum of the vectors}\stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=-2\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}+5\stackrel{^}{\mathrm{k}}\\ \text{and}\stackrel{\to }{\mathrm{c}}=\stackrel{^}{\mathrm{i}}-6\stackrel{^}{\mathrm{j}}-7\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\text{\hspace{0.17em}}.\end{array}$

Ans.

$\begin{array}{l}\text{The given vectors are:}\\ \stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=-2\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}+5\stackrel{^}{\mathrm{k}}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{and}\stackrel{\to }{\mathrm{c}}=\stackrel{^}{\mathrm{i}}-6\stackrel{^}{\mathrm{j}}-7\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\text{\hspace{0.17em}}\\ \text{Then,}\\ \stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}+\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}=\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}+\left(-2\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}+5\stackrel{^}{\mathrm{k}}\right)+\stackrel{^}{\mathrm{i}}-6\stackrel{^}{\mathrm{j}}-7\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ =-\text{\hspace{0.17em}}4\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\end{array}$

Q.18

$\begin{array}{l}\text{Find the unit vector in the direction of the vector\hspace{0.17em}}\\ \stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}.\end{array}$

Ans.

$\begin{array}{l}\text{The unit vector}\stackrel{^}{\mathrm{a}}\text{\hspace{0.17em}\hspace{0.17em}in the direction of vector\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\text{\hspace{0.17em}\hspace{0.17em}is given}\\ \text{by}\stackrel{^}{\mathrm{a}}=\frac{\stackrel{\to }{\mathrm{a}}}{|\stackrel{\to }{\mathrm{a}}|}\\ =\frac{\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}}{|\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}|}\\ =\frac{\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}}{\sqrt{{1}^{2}+{1}^{2}+{2}^{2}}}\\ =\frac{\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}}{\sqrt{6}}\\ =\frac{\stackrel{^}{\mathrm{i}}}{\sqrt{6}}+\frac{\stackrel{^}{\mathrm{j}}}{\sqrt{6}}+\frac{2}{\sqrt{6}}\stackrel{^}{\mathrm{k}}\end{array}$

Q.19

$\begin{array}{l}\text{Find the unit vector in the direction of the vector\hspace{0.17em}}\stackrel{\to }{\mathrm{PQ}},\text{}\\ \text{where P and Q are the points}\left(\text{1,2,3}\right)\text{and}\left(\text{4,5,6}\right)\\ \text{respectively.}\end{array}$

Ans.

$\begin{array}{l}\text{Here,â€„}\stackrel{\to }{\text{OP}}=\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\text{and}\stackrel{\to }{\mathrm{OQ}}=4\stackrel{^}{\mathrm{i}}+5\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}\\ \text{So,â€„\hspace{0.17em}}\stackrel{\to }{\mathrm{PQ}}=\stackrel{\to }{\mathrm{OQ}}-\stackrel{\to }{\mathrm{OP}}\\ =\left(4\stackrel{^}{\mathrm{i}}+5\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}\right)-\left(\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\\ \text{Unitâ€„vectorâ€„ofâ€„\hspace{0.17em}}\stackrel{\to }{\mathrm{PQ}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{PQ}}}{\left|\stackrel{\to }{\mathrm{PQ}}\right|}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}}{\left|3\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\right|}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}}{\sqrt{{3}^{2}+{3}^{2}+{3}^{2}}}\\ \text{\hspace{0.17em}â€„â€„\hspace{0.17em}\hspace{0.17em}}=\frac{3\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}}{3\sqrt{3}}\\ \text{\hspace{0.17em}\hspace{0.17em}â€„â€„\hspace{0.17em}}=\frac{\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}}{\sqrt{3}}\\ \text{\hspace{0.17em}â€„â€„\hspace{0.17em}\hspace{0.17em}}=\frac{\stackrel{^}{\mathrm{i}}}{\sqrt{3}}+\frac{\stackrel{^}{\mathrm{j}}}{\sqrt{3}}+\frac{\stackrel{^}{\mathrm{k}}}{\sqrt{3}}\end{array}$

Q.20

$\begin{array}{l}\text{For given vectors,\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}=2\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\text{and}\stackrel{\to }{\mathrm{b}}=-\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}},\text{find the}\\ \text{unit vector in the direction of the vector}\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}.\end{array}$

Ans.

$\begin{array}{l}\text{The\hspace{0.17em}\hspace{0.17em}given vectors are:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}=2\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\text{and}\stackrel{\to }{\mathrm{b}}=-\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\\ \text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}+\stackrel{\to }{\mathrm{b}}=2\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}+\left(-\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{k}}\\ |\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}+\stackrel{\to }{\mathrm{b}}|=|\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{k}}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{1}^{2}+{1}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{2}\\ \therefore \text{Unit vector of \hspace{0.17em}}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}+\stackrel{\to }{\mathrm{b}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}+\stackrel{\to }{\mathrm{b}}}{|\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}+\stackrel{\to }{\mathrm{b}}|}=\frac{\stackrel{^}{\mathrm{i}}}{\sqrt{2}}+\frac{\stackrel{^}{\mathrm{k}}}{\sqrt{2}}\end{array}$

Q.21

$\begin{array}{l}\mathrm{Find}\mathrm{a}\mathrm{vector}\mathrm{in}\mathrm{the}\mathrm{direction}\mathrm{of}\mathrm{vector}5\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\mathrm{which}\mathrm{has}\\ \mathrm{magnitude}8\mathrm{units}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{unit}\mathrm{vector}\mathrm{of}5\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}=\frac{5\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}}{|5\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}|}\\ =\frac{5\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}}{\sqrt{{5}^{2}+{\left(-1\right)}^{2}+{2}^{2}}}\\ =\frac{5\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}}{\sqrt{25+1+4}}\\ =\frac{5\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}}{\sqrt{30}}\\ \mathrm{Thus},\mathrm{the}\mathrm{vector}\mathrm{in}\mathrm{the}\mathrm{direction}\mathrm{of}\left(5\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)\mathrm{which}\mathrm{has}\\ \mathrm{magnitude}8\mathrm{units}\mathrm{is}:\\ 8\stackrel{^}{\mathrm{a}}=8×\frac{5\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}}{\sqrt{30}}\\ =\frac{40\mathrm{ }\stackrel{^}{\mathrm{i}}-8\stackrel{^}{\mathrm{j}}+16\mathrm{ }\stackrel{^}{\mathrm{k}}}{\sqrt{30}}\\ =\frac{40}{\sqrt{30}}\mathrm{ }\stackrel{^}{\mathrm{i}}-\frac{8}{\sqrt{30}}\stackrel{^}{\mathrm{j}}+\frac{16}{\sqrt{30}}\mathrm{ }\stackrel{^}{\mathrm{k}}\end{array}$

Q.22

$\begin{array}{l}\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{vectors}\mathrm{}2\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\mathrm{and}-4\stackrel{^}{\mathrm{i}}+6\stackrel{^}{\mathrm{j}}-8\stackrel{^}{\mathrm{k}} \mathrm{are}\\ \mathrm{collinear}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{vectors}\mathrm{are}\\ \stackrel{\to }{\mathrm{a}}=2\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\\ \stackrel{\to }{\mathrm{b}}=-4\stackrel{^}{\mathrm{i}}+6\stackrel{^}{\mathrm{j}}–8\stackrel{^}{\mathrm{k}}\\ =-2\left(2\stackrel{^}{\mathrm{i}}–3\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\right)\\ \stackrel{\to }{\mathrm{b}}\mathrm{ }=-2\stackrel{\to }{\mathrm{a}},\mathrm{which}\mathrm{is}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\stackrel{\to }{\mathrm{a}}=\mathrm{\lambda }\stackrel{\to }{\mathrm{b}},\\ \mathrm{where},\mathrm{\lambda }=-2\\ \mathrm{Hence},\mathrm{the}\mathrm{given}\mathrm{vectors}\mathrm{are}\mathrm{collinear}.\end{array}$

Q.23

$\text{Find the direction cosines of the vector}\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}.$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{vector} \mathrm{is}\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}.\\ |\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}|=\sqrt{{1}^{2}+{2}^{2}+{3}^{2}}\\ =\sqrt{1+4+9}\\ =\sqrt{14}\\ \mathrm{So},\mathrm{the}\mathrm{direction}\mathrm{cosines}\mathrm{of}\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}} \mathrm{are} \left(\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}\right).\end{array}$

Q.24 Find the direction cosines of the vector joining the points A(1, 2, –3) and B(–1, – 2, 1) directed A to B.

Ans.

$\begin{array}{l}\text{The given points are}\mathrm{A}\left(1,\text{}2,–3\right)\text{and}\mathrm{B}\left(–1,–2,1\right).\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{AB}}=\left(-1-1\right)\stackrel{^}{\mathrm{i}}+\left(-2-2\right)\stackrel{^}{\mathrm{j}}+\left(1+3\right)\stackrel{^}{\mathrm{k}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-2\stackrel{^}{\mathrm{i}}-4\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\\ \therefore |\stackrel{\to }{\mathrm{AB}}|=\sqrt{{\left(-2\right)}^{2}+{\left(-4\right)}^{2}+{4}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{4+16+16}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{36}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=6\\ \text{Hence, the direction cosines of \hspace{0.17em}}\stackrel{\to }{\mathrm{AB}}\text{\hspace{0.17em}\hspace{0.17em}are\hspace{0.17em}\hspace{0.17em}}\left(\frac{-2}{6},\frac{-4}{6},\frac{4}{6}\right)=\left(\frac{-1}{3},\frac{-2}{3},\frac{2}{3}\right).\end{array}$

Q.25

$\begin{array}{l}\text{Show that the vector}\stackrel{^}{\text{i}}+\stackrel{^}{\text{j}}+\stackrel{^}{\text{k}}\text{is equally inclined to the axes}\\ \text{OX, OY and OZ.}\end{array}$

Ans.

$\begin{array}{l}\text{The given vector is}\stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}},\text{then}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}|\stackrel{\to }{\mathrm{a}}|=|\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}|\\ =\sqrt{{1}^{2}+{1}^{2}+{1}^{2}}\\ =\sqrt{3}\\ \text{The direction cosines of}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}are}\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right).\\ \text{Let}\mathrm{\alpha },\text{\hspace{0.17em}}\mathrm{\beta }\text{and}\mathrm{\gamma }\text{be the angles formed by}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}with the positive}\\ \text{directions of x, y, and z axes.}\\ \text{Then, we have cos}\mathrm{\alpha }=\frac{1}{\sqrt{3}},\text{cos}\mathrm{\beta }=\frac{1}{\sqrt{3}}\text{,\hspace{0.17em}cos}\mathrm{\gamma }=\frac{1}{\sqrt{3}}.\\ \text{Hence, the given vector is equally inclined to axes OX,}\\ \text{OY, and OZ.}\end{array}$

Q.26

$\begin{array}{l}\mathrm{Let}\mathrm{the}\mathrm{vectors}\stackrel{\to }{\mathrm{a}},\text{\hspace{0.17em}}\stackrel{\to }{\text{\hspace{0.17em}}\mathrm{b}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}\text{}\mathrm{be}\mathrm{given}\mathrm{as}{\mathrm{a}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{a}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{a}}_{3}\stackrel{^}{\text{k}},\text{\hspace{0.17em}}\\ {\mathrm{b}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{b}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{b}}_{3}\stackrel{^}{\mathrm{k}},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{c}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{c}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{c}}_{3}\stackrel{^}{\mathrm{k}}.\mathrm{Then}\mathrm{show}\mathrm{that}\\ \stackrel{\to }{\mathrm{a}}×\left(\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}\right)=\stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{c}}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\text{given vectors are:}{\mathrm{a}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{a}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{a}}_{3}\stackrel{^}{\mathrm{k}},\text{\hspace{0.17em}}{\mathrm{b}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{b}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{b}}_{3}\stackrel{^}{\mathrm{k}},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{c}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{c}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{c}}_{3}\stackrel{^}{\mathrm{k}}.\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\stackrel{\to }{\mathrm{a}}×\left(\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}\right)\\ =\left({\mathrm{a}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{a}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{a}}_{3}\stackrel{^}{\mathrm{k}}\right)×\left({\mathrm{b}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{b}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{b}}_{3}\stackrel{^}{\mathrm{k}}+\text{\hspace{0.17em}}{\mathrm{c}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{c}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{c}}_{3}\stackrel{^}{\mathrm{k}}\right)\\ =\left({\mathrm{a}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{a}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{a}}_{3}\stackrel{^}{\mathrm{k}}\right)×\left\{\left({\mathrm{b}}_{1}+{\mathrm{c}}_{1}\right)\stackrel{^}{\mathrm{i}}+\left({\mathrm{b}}_{2}+{\mathrm{c}}_{2}\right)\stackrel{^}{\mathrm{j}}+\left({\mathrm{b}}_{3}+{\mathrm{c}}_{3}\right)\stackrel{^}{\mathrm{k}}\right\}\\ =|\begin{array}{ccc}\stackrel{^}{\mathrm{i}}& \stackrel{^}{\mathrm{j}}& \stackrel{^}{\mathrm{k}}\\ {\mathrm{a}}_{1}& {\mathrm{a}}_{2}& {\mathrm{a}}_{3}\\ \left({\mathrm{b}}_{1}+{\mathrm{c}}_{1}\right)& \left({\mathrm{b}}_{2}+{\mathrm{c}}_{2}\right)& \left({\mathrm{b}}_{3}+{\mathrm{c}}_{3}\right)\end{array}|\\ =\left\{{\mathrm{a}}_{2}\left({\mathrm{b}}_{3}+{\mathrm{c}}_{3}\right)-{\mathrm{a}}_{3}\left({\mathrm{b}}_{2}+{\mathrm{c}}_{2}\right)\right\}\stackrel{^}{\mathrm{i}}+\left\{{\mathrm{a}}_{1}\left({\mathrm{b}}_{3}+{\mathrm{c}}_{3}\right)-{\mathrm{a}}_{3}\left({\mathrm{b}}_{1}+{\mathrm{c}}_{1}\right)\right\}\stackrel{^}{\mathrm{j}}\\ +\left\{{\mathrm{a}}_{1}\left({\mathrm{b}}_{2}+{\mathrm{c}}_{2}\right)-{\mathrm{a}}_{2}\left({\mathrm{b}}_{3}+{\mathrm{c}}_{3}\right)\right\}\stackrel{^}{\mathrm{k}}\\ =\left\{{\mathrm{a}}_{2}{\mathrm{b}}_{3}+{\mathrm{a}}_{2}{\mathrm{c}}_{3}-{\mathrm{a}}_{3}{\mathrm{b}}_{2}-{\mathrm{a}}_{3}{\mathrm{c}}_{2}\right\}\stackrel{^}{\mathrm{i}}+\left\{{\mathrm{a}}_{1}{\mathrm{b}}_{3}+{\mathrm{a}}_{1}{\mathrm{c}}_{3}-{\mathrm{a}}_{3}{\mathrm{b}}_{1}-{\mathrm{a}}_{3}{\mathrm{c}}_{1}\right\}\stackrel{^}{\mathrm{j}}\\ +\left\{{\mathrm{a}}_{1}{\mathrm{b}}_{2}+{\mathrm{a}}_{1}{\mathrm{c}}_{2}-{\mathrm{a}}_{2}{\mathrm{b}}_{3}-{\mathrm{a}}_{2}{\mathrm{c}}_{3}\right\}\stackrel{^}{\mathrm{k}}\text{\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{b}}=\left({\mathrm{a}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{a}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{a}}_{3}\stackrel{^}{\mathrm{k}}\right)×\left({\mathrm{b}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{b}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{b}}_{3}\stackrel{^}{\mathrm{k}}\right)\\ =|\begin{array}{ccc}\stackrel{^}{\mathrm{i}}& \stackrel{^}{\mathrm{j}}& \stackrel{^}{\mathrm{k}}\\ {\mathrm{a}}_{1}& {\mathrm{a}}_{2}& {\mathrm{a}}_{3}\\ {\mathrm{b}}_{1}& {\mathrm{b}}_{2}& {\mathrm{b}}_{3}\end{array}|\\ =\left({\mathrm{a}}_{2}{\mathrm{b}}_{3}-{\mathrm{a}}_{3}{\mathrm{b}}_{2}\right)\stackrel{^}{\mathrm{i}}+\left({\mathrm{a}}_{1}{\mathrm{b}}_{3}-{\mathrm{a}}_{3}{\mathrm{b}}_{1}\right)\stackrel{^}{\mathrm{j}}+\left({\mathrm{a}}_{1}{\mathrm{b}}_{2}-{\mathrm{a}}_{2}{\mathrm{b}}_{1}\right)\stackrel{^}{\mathrm{k}}\\ \stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{c}}=\left({\mathrm{a}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{a}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{a}}_{3}\stackrel{^}{\mathrm{k}}\right)×\left({\mathrm{c}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{c}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{c}}_{3}\stackrel{^}{\mathrm{k}}\right)\\ =|\begin{array}{ccc}\stackrel{^}{\mathrm{i}}& \stackrel{^}{\mathrm{j}}& \stackrel{^}{\mathrm{k}}\\ {\mathrm{a}}_{1}& {\mathrm{a}}_{2}& {\mathrm{a}}_{3}\\ {\mathrm{c}}_{1}& {\mathrm{c}}_{2}& {\mathrm{c}}_{3}\end{array}|\\ =\left({\mathrm{a}}_{2}{\mathrm{b}}_{3}-{\mathrm{a}}_{3}{\mathrm{b}}_{2}\right)\stackrel{^}{\mathrm{i}}+\left({\mathrm{a}}_{1}{\mathrm{b}}_{3}-{\mathrm{a}}_{3}{\mathrm{b}}_{1}\right)\stackrel{^}{\mathrm{j}}+\left({\mathrm{a}}_{1}{\mathrm{b}}_{2}-{\mathrm{a}}_{2}{\mathrm{b}}_{1}\right)\stackrel{^}{\mathrm{k}}\\ \stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{c}}\\ =\left({\mathrm{a}}_{2}{\mathrm{b}}_{3}-{\mathrm{a}}_{3}{\mathrm{b}}_{2}\right)\stackrel{^}{\mathrm{i}}+\left({\mathrm{a}}_{1}{\mathrm{b}}_{3}-{\mathrm{a}}_{3}{\mathrm{b}}_{1}\right)\stackrel{^}{\mathrm{j}}+\left({\mathrm{a}}_{1}{\mathrm{b}}_{2}-{\mathrm{a}}_{2}{\mathrm{b}}_{1}\right)\stackrel{^}{\mathrm{k}}+\left({\mathrm{a}}_{2}{\mathrm{b}}_{3}-{\mathrm{a}}_{3}{\mathrm{b}}_{2}\right)\stackrel{^}{\mathrm{i}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\left({\mathrm{a}}_{1}{\mathrm{b}}_{3}-{\mathrm{a}}_{3}{\mathrm{b}}_{1}\right)\stackrel{^}{\mathrm{j}}+\left({\mathrm{a}}_{1}{\mathrm{b}}_{2}-{\mathrm{a}}_{2}{\mathrm{b}}_{1}\right)\stackrel{^}{\mathrm{k}}\\ =\left({\mathrm{a}}_{2}{\mathrm{b}}_{3}-{\mathrm{a}}_{3}{\mathrm{b}}_{2}+{\mathrm{a}}_{2}{\mathrm{b}}_{3}-{\mathrm{a}}_{3}{\mathrm{b}}_{2}\right)\stackrel{^}{\mathrm{i}}+\left({\mathrm{a}}_{1}{\mathrm{b}}_{3}-{\mathrm{a}}_{3}{\mathrm{b}}_{1}+{\mathrm{a}}_{1}{\mathrm{b}}_{3}-{\mathrm{a}}_{3}{\mathrm{b}}_{1}\right)\stackrel{^}{\mathrm{j}}\\ +\left({\mathrm{a}}_{1}{\mathrm{b}}_{2}-{\mathrm{a}}_{2}{\mathrm{b}}_{1}+{\mathrm{a}}_{1}{\mathrm{b}}_{2}-{\mathrm{a}}_{2}{\mathrm{b}}_{1}\right)\stackrel{^}{\mathrm{k}}\\ =\left({\mathrm{a}}_{2}{\mathrm{b}}_{3}+{\mathrm{a}}_{2}{\mathrm{b}}_{3}-{\mathrm{a}}_{3}{\mathrm{b}}_{2}-{\mathrm{a}}_{3}{\mathrm{b}}_{2}\right)\stackrel{^}{\mathrm{i}}+\left({\mathrm{a}}_{1}{\mathrm{b}}_{3}+{\mathrm{a}}_{1}{\mathrm{b}}_{3}-{\mathrm{a}}_{3}{\mathrm{b}}_{1}-{\mathrm{a}}_{3}{\mathrm{b}}_{1}\right)\stackrel{^}{\mathrm{j}}\\ +\left({\mathrm{a}}_{1}{\mathrm{b}}_{2}+{\mathrm{a}}_{1}{\mathrm{b}}_{2}-{\mathrm{a}}_{2}{\mathrm{b}}_{1}-{\mathrm{a}}_{2}{\mathrm{b}}_{1}\right)\stackrel{^}{\mathrm{k}}\text{\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we have}\\ \text{\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}×\left(\stackrel{\to }{\mathrm{b}}×\stackrel{\to }{\mathrm{c}}\right)=\stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{c}}.\text{\hspace{0.17em}\hspace{0.17em}}\\ \text{\hspace{0.17em}Hence, the given result is proved.}\end{array}$

Q.27

$\begin{array}{l}\text{Find the position vector of a point R which divides the}\\ \text{line joining two points P and Q whose position vectors}\\ \text{are\hspace{0.17em}\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}–\stackrel{^}{\mathrm{k}}\text{and}–\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\text{\hspace{0.17em}}\mathrm{k}}\text{respectively, in the ration 2:1}\\ \left(\text{i}\right)\text{internally\hspace{0.17em}}\\ \left(\text{ii}\right)\text{\hspace{0.17em}externally}\end{array}$

Ans.

$\begin{array}{l}\text{The given vectors are}\stackrel{\to }{\mathrm{OP}}=\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\text{and}\stackrel{\to }{\mathrm{OQ}}=-\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\text{\hspace{0.17em}}\mathrm{k}}.\\ \text{The position vector of point R dividing the line segment joining}\\ \text{two points\hspace{0.17em}P and Q in the ratio 2: 1 is given by:}\\ \left(\text{i}\right)\text{Internally:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{OR}}=\frac{2\left(-\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\text{\hspace{0.17em}}\mathrm{k}}\right)+1\left(\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\right)}{2+1}\\ =\frac{-\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+4\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}}{3}\\ =-\frac{1}{3}\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+\frac{4}{3}\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}+\frac{1}{3}\stackrel{^}{\mathrm{k}}\\ \left(\text{ii}\right)\text{The position vector of point R which divides the line joining}\\ \text{two points P and Q externally in the ratio 2:1 is given by,}\\ \text{\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{OR}}=\frac{2\left(-\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\text{\hspace{0.17em}}\mathrm{k}}\right)-1\left(\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\right)}{2-1}\\ =\frac{-3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+0\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}+3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}}{1}\\ =-3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\end{array}$

Q.28 Find the position vector of the mid-point of the vector joining the points P(2, 3, 4) and Q(4, 1, –2).

Ans.

$\begin{array}{l}\text{The position vector of mid-point R of the vector joining points}\\ \text{P (2, 3, 4) and Q (4, 1,-2) is given by,}\\ \text{\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{OR}}=\frac{\left(2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\right)+\left(4\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}-2\stackrel{^}{\mathrm{k}}\right)}{2}\\ =\frac{6\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}}{2}\\ =3\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\end{array}$

Q.29

$\begin{array}{l}\text{Show that the points A, B and C with position vectors,}\\ \stackrel{\to }{\mathrm{a}}=3\stackrel{^}{\mathrm{i}}-4\stackrel{^}{\mathrm{j}}-4\stackrel{^}{\mathrm{k}},\stackrel{\to }{\mathrm{b}}=2\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\text{and}\stackrel{\to }{\mathrm{c}}=\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}-5\stackrel{^}{\mathrm{k}}\text{respectively}\\ \text{form the vertices of a right angled triangle.}\end{array}$

Ans.

$\begin{array}{l}\text{Position vectors of points A, B, and C are respectively given as:}\\ \stackrel{\to }{\mathrm{a}}=3\stackrel{^}{\mathrm{i}}-4\stackrel{^}{\mathrm{j}}-4\stackrel{^}{\mathrm{k}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=2\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\text{\hspace{0.17em}and}\stackrel{\to }{\mathrm{c}}=\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}-5\stackrel{^}{\mathrm{k}}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{AB}}=\stackrel{\to }{\mathrm{b}}-\stackrel{\to }{\mathrm{a}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(2\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)-\left(3\stackrel{^}{\mathrm{i}}-4\stackrel{^}{\mathrm{j}}-4\stackrel{^}{\mathrm{k}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+5\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{BC}}=\stackrel{\to }{\mathrm{c}}-\stackrel{\to }{\mathrm{b}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}-5\stackrel{^}{\mathrm{k}}\right)-\left(2\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}-6\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{CA}}=\stackrel{\to }{\mathrm{a}}-\stackrel{\to }{\mathrm{c}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(3\stackrel{^}{\mathrm{i}}-4\stackrel{^}{\mathrm{j}}-4\stackrel{^}{\mathrm{k}}\right)-\left(\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}-5\stackrel{^}{\mathrm{k}}\right)\end{array}$

$\begin{array}{c}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ \text{So,\hspace{0.17em}}{|\stackrel{\to }{\mathrm{AB}}|}^{2}={|-\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+5\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}|}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=35\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{|\stackrel{\to }{\mathrm{BC}}|}^{2}={|-\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}-6\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}|}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=41\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{|\stackrel{\to }{\mathrm{CA}}|}^{2}={|2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}|}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=6\\ {|\stackrel{\to }{\mathrm{AB}}|}^{2}=\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{|\stackrel{\to }{\mathrm{BC}}|}^{2}+{|\stackrel{\to }{\mathrm{CA}}|}^{2}\\ =36+5=41\end{array}$

$\text{Hence, ABC is a right-angled triangle}$

Q.30n

$\begin{array}{l}\mathrm{If}\mathrm{either}\stackrel{\to }{\mathrm{a}}=\stackrel{\to }{0}\mathrm{or}\stackrel{\to }{\mathrm{b}}=\stackrel{\to }{0},\mathrm{then}\stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{b}}=\stackrel{\to }{0}.\mathrm{Is}\mathrm{the}\mathrm{converse}\mathrm{true}?\\ \mathrm{Justify}\mathrm{your}\mathrm{answer}\mathrm{with}\mathrm{an}\mathrm{example}.\end{array}$

Ans.

$\begin{array}{l}\text{Let us consider any parallel non-zero vectors so that\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{b}}=0\\ \mathrm{Let}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}=4\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+6\stackrel{^}{\mathrm{j}}+8\stackrel{^}{\mathrm{k}}\text{and}\stackrel{\to }{\mathrm{b}}=2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\\ \mathrm{Then},\\ \stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{b}}=|\begin{array}{ccc}\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}& \stackrel{^}{\mathrm{j}}& \stackrel{^}{\mathrm{k}}\\ 4& 6& 8\\ 2& 3& 4\end{array}|\\ =\left(24-24\right)\stackrel{^}{\mathrm{i}}+\left(16-16\right)\stackrel{^}{\mathrm{j}}+\left(12-12\right)\stackrel{^}{\mathrm{k}}\\ =0\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+0\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}+0\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ =\stackrel{\to }{0}\\ \mathrm{And}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}|\stackrel{\to }{\mathrm{a}}|=|4\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+6\stackrel{^}{\mathrm{j}}+8\stackrel{^}{\mathrm{k}}|\\ =\sqrt{{4}^{2}+{6}^{2}+{8}^{2}}\\ =\sqrt{16+36+64}\\ =\sqrt{116}\\ \therefore \text{\hspace{0.17em}}|\stackrel{\to }{\mathrm{a}}|\ne 0\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}|\stackrel{\to }{\mathrm{b}}|=|2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}|\\ =\sqrt{{2}^{2}+{3}^{2}+{4}^{2}}\\ =\sqrt{4+9+16}\\ =\sqrt{29}\\ \therefore \text{\hspace{0.17em}}|\stackrel{\to }{\mathrm{b}}|\ne 0\\ \text{Hence, the converse of the given statement need not be true.}\end{array}$

Q.31 Find the area of the triangle with vertices A (1, 1, 2), B (2, 3, 5) and C (1, 5, 5).

Ans.

$\begin{array}{l}\text{The vertices of triangle ABC are given as A (1, 1, 2), B (2, 3, 5)}\\ \text{and C (1, 5, 5).}\\ \text{The adjacent sides\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{AB}}\text{and}\stackrel{\to }{\mathrm{BC}}\text{\hspace{0.17em}\hspace{0.17em}of}\mathrm{\Delta }\text{ABC are given as:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{AB}}=\left(2-1\right)\stackrel{^}{\mathrm{i}}+\left(3-1\right)\stackrel{^}{\mathrm{j}}+\left(5-2\right)\stackrel{^}{\mathrm{k}}\\ =\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{BC}}=\left(1-2\right)\stackrel{^}{\mathrm{i}}+\left(5-3\right)\stackrel{^}{\mathrm{j}}+\left(5-5\right)\stackrel{^}{\mathrm{k}}\\ =-\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}\\ \therefore \stackrel{\to }{\mathrm{AB}}×\stackrel{\to }{\mathrm{BC}}\\ =\frac{1}{2}|\begin{array}{ccc}\stackrel{^}{\mathrm{i}}& \stackrel{^}{\mathrm{j}}& \stackrel{^}{\mathrm{k}}\\ 1& 2& 3\\ -1& 2& 0\end{array}|\\ =\frac{1}{2}\left\{\left(0-6\right)\stackrel{^}{\mathrm{i}}-\left(0+3\right)\stackrel{^}{\mathrm{j}}+\left(2+2\right)\stackrel{^}{\mathrm{k}}\right\}\\ =\frac{1}{2}\left(-6\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\right)\\ \mathrm{The}\text{area}\mathrm{\Delta ABC}\\ =\frac{1}{2}|\stackrel{\to }{\mathrm{AB}}×\stackrel{\to }{\mathrm{BC}}|\\ =\frac{1}{2}|-\text{\hspace{0.17em}}6\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}|\\ =\frac{1}{2}\sqrt{{\left(-\text{\hspace{0.17em}}6\right)}^{2}+{\left(-3\right)}^{2}+{4}^{2}}\\ =\frac{1}{2}\sqrt{36+9+16}\\ =\frac{1}{2}\sqrt{61}\text{square units}\\ \text{Thus, the area of}\mathrm{\Delta }\text{ABC is}\frac{1}{2}\sqrt{61}\text{square units}.\end{array}$

Q.32

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{parallelogram}\mathrm{whose}\mathrm{adjacents}\mathrm{sides}\\ \mathrm{are}\mathrm{determined}\mathrm{by}\mathrm{the}\mathrm{vectors}\stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\mathrm{and}\stackrel{\to }{\mathrm{b}}=2\stackrel{^}{\mathrm{i}}-7\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}.\end{array}$

Ans.

$\begin{array}{l}\text{The area of the parallelogram whose adjacent sides are\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}\text{and}\stackrel{\to }{\mathrm{b}}\\ \mathrm{is}\text{}|\stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{b}}|.\\ \text{Adjacent sides are in vector form, which are as follows:}\\ \stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\text{and}\\ \stackrel{\to }{\mathrm{b}}=2\stackrel{^}{\mathrm{i}}-7\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\\ \therefore \stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{b}}=|\begin{array}{ccc}\stackrel{^}{\mathrm{i}}& \stackrel{^}{\mathrm{j}}& \stackrel{^}{\mathrm{k}}\\ 1& -1& 3\\ 2& -7& 1\end{array}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(-1+21\right)\stackrel{^}{\mathrm{i}}+\left(1-6\right)\stackrel{^}{\mathrm{j}}+\left(-7+2\right)\stackrel{^}{\mathrm{k}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=20\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-5\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}-5\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ \text{\hspace{0.17em}}|\stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{b}}|=|20\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-5\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}-5\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{20}^{2}+{\left(-5\right)}^{2}+{\left(-5\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{400+25+25}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{450}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=15\sqrt{2}\\ \text{Thus, the area of the given parallelogram is 15}\sqrt{2}\text{\hspace{0.17em}square units.}\end{array}$

Q.33

$\begin{array}{l}\mathrm{Let}\mathrm{the}\mathrm{vectors}\stackrel{\to }{\mathrm{a}}\mathrm{and}\stackrel{\to }{\mathrm{b}}\mathrm{be}\mathrm{such}\mathrm{that}|\stackrel{\to }{\mathrm{a}}|=3\mathrm{and}|\stackrel{\to }{\mathrm{b}}|=\frac{\sqrt{3}}{2},\mathrm{then}\\ \stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{b}}\mathrm{is}\mathrm{a}\mathrm{unit}\mathrm{vector},\mathrm{if}\mathrm{the}\mathrm{angle}\mathrm{between}\stackrel{\to }{\mathrm{a}}\mathrm{and}\stackrel{\to }{\mathrm{b}}\mathrm{is}\\ \left(\mathrm{A}\right)\frac{\mathrm{\pi }}{6}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \left(\mathrm{B}\right)\frac{\mathrm{\pi }}{4}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \left(\mathrm{C}\right)\frac{\mathrm{\pi }}{3}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \left(\mathrm{D}\right)\frac{\mathrm{\pi }}{2}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Given}:\text{\hspace{0.17em}\hspace{0.17em}}|\stackrel{\to }{\mathrm{a}}|=3\text{and}|\stackrel{\to }{\mathrm{b}}|=\frac{\sqrt{2}}{3}\\ \mathrm{Since},\text{}\stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{b}}=|\stackrel{\to }{\mathrm{a}}||\stackrel{\to }{\mathrm{b}}|\mathrm{sin\theta }\text{\hspace{0.17em}}\stackrel{^}{\mathrm{n}},\text{where \hspace{0.17em}}\stackrel{^}{\mathrm{n}}\text{is a unit vector perpendicular}\\ \text{to both}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}and}\stackrel{\to }{\mathrm{b}}\text{â€‹ and}\mathrm{\theta }\text{â€‹ is the angle between}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}and}\stackrel{\to }{\mathrm{b}}\text{â€‹}.\\ \mathrm{Since},\text{}\stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{b}}\text{is a unit vector, so}|\stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{b}}|=1.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}|\stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{b}}|=1\\ ⇒||\stackrel{\to }{\mathrm{a}}||\stackrel{\to }{\mathrm{b}}|\mathrm{sin\theta }\text{\hspace{0.17em}}\stackrel{^}{\mathrm{n}}|=1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}|\stackrel{\to }{\mathrm{a}}||\stackrel{\to }{\mathrm{b}}||\mathrm{sin\theta }|\text{\hspace{0.17em}}=1\\ \text{\hspace{0.17em}\hspace{0.17em}}3×\frac{\sqrt{2}}{3}×\mathrm{sin\theta }=1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{sin\theta }=\frac{1}{\sqrt{2}}=\mathrm{sin}\frac{\mathrm{\pi }}{4}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\theta }=\frac{\mathrm{\pi }}{4}\\ \text{Hence,}\stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}is a unit vector if the angle between}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}and}\stackrel{\to }{\mathrm{b}}\text{is\hspace{0.17em}}\frac{\mathrm{\pi }}{4}.\\ \text{The correct answer is B.}\end{array}$

Q.34

$\begin{array}{l}\mathrm{Area}\mathrm{of}\mathrm{a}\mathrm{rectangle}\mathrm{having}\mathrm{vertices}\mathrm{A},\mathrm{B},\mathrm{C}\mathrm{and}\mathrm{D}\mathrm{with}\\ \mathrm{position}\mathrm{vector}-\stackrel{^}{\mathrm{i}}+\frac{1}{2}\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\text{k}},\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+\frac{1}{2}\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}},\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-\frac{1}{2}\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\\ \mathrm{and}-\stackrel{^}{\mathrm{i}}-\frac{1}{2}\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\text{\hspace{0.17em}}\mathrm{respectively}\mathrm{is}\\ \left(\mathrm{A}\right)\frac{1}{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \left(\mathrm{B}\right)1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \left(\mathrm{C}\right)2\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \left(\mathrm{D}\right)4\\ \text{\hspace{0.17em}}\end{array}$

Ans.

$\begin{array}{l}\text{The position vectors of vertices A, B, C, and D of rectangle}\\ \text{ABCD are given as:}\\ \stackrel{\to }{\mathrm{OA}}=-\stackrel{^}{\mathrm{i}}+\frac{1}{2}\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}},\text{}\stackrel{\to }{\mathrm{OB}}=\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+\frac{1}{2}\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{OC}}=\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-\frac{1}{2}\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\text{and}\\ \stackrel{\to }{\mathrm{OD}}=-\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-\frac{1}{2}\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\\ \text{The adjacent sides}\stackrel{\to }{\mathrm{AB}}\text{\hspace{0.17em}and}\stackrel{\to }{\mathrm{BC}}\text{\hspace{0.17em}of the given rectangle are given as:}\\ \stackrel{\to }{\mathrm{AB}}=\left(1+1\right)\stackrel{^}{\mathrm{i}}+\left(\frac{1}{2}-\frac{1}{2}\right)\stackrel{^}{\mathrm{j}}+\left(4-4\right)\stackrel{^}{\mathrm{k}}=2\stackrel{^}{\mathrm{i}}\\ \stackrel{\to }{\mathrm{BC}}=\left(1-1\right)\stackrel{^}{\mathrm{i}}+\left(-\frac{1}{2}-\frac{1}{2}\right)\stackrel{^}{\mathrm{j}}+\left(4-4\right)\stackrel{^}{\mathrm{k}}=-\stackrel{^}{\mathrm{j}}\\ \therefore \stackrel{\to }{\mathrm{AB}}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{BC}}\\ =|\begin{array}{ccc}\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}& \stackrel{^}{\mathrm{j}}& \stackrel{^}{\mathrm{k}}\\ 2& 0& 0\\ 0& -1& 0\end{array}|\\ =\left(-2-0\right)\stackrel{^}{\mathrm{k}}\\ =-2\stackrel{^}{\mathrm{k}}\\ |\stackrel{\to }{\mathrm{AB}}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{BC}}|\\ =|-2\stackrel{^}{\mathrm{k}}|\\ =2\\ \text{Now, it is known that the area of a parallelogram whose adjacent}\\ \text{sides are\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}\text{and}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{is}\text{}|\stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{b}}|.\\ \text{Hence, the area of the given rectangle is\hspace{0.17em}}|\stackrel{\to }{\mathrm{AB}}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{BC}}|=2\text{square units.}\\ \text{The correct answer is C.}\end{array}$

Q.35

$\mathrm{Find}\left[\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}\right]\mathrm{if}\stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}-2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+3\text{\hspace{0.17em}}\stackrel{^}{\text{k}},\stackrel{\to }{\mathrm{b}}=2\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}-3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}=3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}-2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}.$

Ans.

$\begin{array}{l}\left[\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}\right]=|\begin{array}{ccc}1& -2& 3\\ 2& -3& 1\\ 3& 1& -2\end{array}|\\ \text{\hspace{0.17em}\hspace{0.17em}}=1|\begin{array}{cc}-3& 1\\ 1& -2\end{array}|+2|\begin{array}{cc}2& 1\\ 3& -2\end{array}|+3|\begin{array}{cc}2& -3\\ 3& 1\end{array}|\\ \text{\hspace{0.17em}\hspace{0.17em}}=1\left(6-1\right)+2\left(-4-3\right)+3\left(2+9\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=1\left(5\right)+2\left(-7\right)+3\left(11\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=5-14+33\\ \text{\hspace{0.17em}\hspace{0.17em}}=24\\ \therefore \left[\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}\right]\text{\hspace{0.17em}}=24\end{array}$

Q.36

$\mathrm{In}\mathrm{triangle},\mathrm{which}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{is}\mathrm{not}\mathrm{true}:\phantom{\rule{0ex}{0ex}}\begin{array}{l}\left(\mathrm{A}\right)\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{AB}}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{BC}}+\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{CA}}=0\\ \left(\mathrm{B}\right)\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{AB}}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{BC}}–\mathrm{AC}=0\\ \left(\mathrm{C}\right)\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{AB}}+\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{BC}}–\stackrel{\to }{\mathrm{CA}}=0\\ \left(\mathrm{D}\right)\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{AB}}\text{\hspace{0.17em}}–\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{CB}}+\stackrel{\to }{\mathrm{CA}}=0\end{array}$

Ans.

$\begin{array}{l}\text{On applying the triangle law of addition in the given triangle,}\\ \text{we have:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{AB}}+\stackrel{\to }{\mathrm{BC}}=\stackrel{\to }{\mathrm{AC}}...\left(\text{i}\right)\\ ⇒\stackrel{\to }{\mathrm{AB}}+\stackrel{\to }{\mathrm{BC}}=-\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{CA}}\\ ⇒\stackrel{\to }{\mathrm{AB}}+\stackrel{\to }{\mathrm{BC}}+\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{CA}}=\stackrel{\to }{0}...\left(\text{ii}\right)\\ \therefore \text{The equation given in alternative A is true.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{AB}}+\stackrel{\to }{\mathrm{BC}}=\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{AC}}\\ ⇒\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{AB}}+\stackrel{\to }{\mathrm{BC}}-\stackrel{\to }{\mathrm{AC}}=\stackrel{\to }{0}\\ \therefore \text{The equation given in alternative B is true.}\\ \text{From equation}\left(\text{ii}\right)\text{, we have:}\\ \stackrel{\to }{\mathrm{AB}}-\stackrel{\to }{\mathrm{CB}}+\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{AC}}=\stackrel{\to }{0}\\ \therefore \text{The equation given in alternative D is true.}\\ \text{Now, the equation given in alternative C,}\\ \stackrel{\to }{\mathrm{AB}}+\stackrel{\to }{\mathrm{BC}}-\stackrel{\to }{\mathrm{CA}}=\stackrel{\to }{0}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{AB}}+\stackrel{\to }{\mathrm{BC}}=\stackrel{\to }{\mathrm{CA}}\text{\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{iii}\right)\\ \text{From equation}\left(\text{i}\right)\text{and}\left(\text{iii}\right)\text{,â€‹\hspace{0.17em}we have}\end{array}$

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{CA}}=\stackrel{\to }{\mathrm{AC}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\stackrel{\to }{\mathrm{AC}}=\stackrel{\to }{\mathrm{AC}}\\ \text{\hspace{0.17em}}⇒\stackrel{\to }{\mathrm{AC}}+\stackrel{\to }{\mathrm{AC}}=\stackrel{\to }{0}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\stackrel{\to }{\mathrm{AC}}=\stackrel{\to }{0}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{AC}}=\stackrel{\to }{0},\text{which is not true.}\\ \text{Hence, the equation given in alternative C is incorrect.}\\ \text{The correct answer is C.}\end{array}$

Q.37

$\begin{array}{l}\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{vectors}\stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}-2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+3\text{\hspace{0.17em}}\stackrel{^}{\text{k}},\stackrel{\to }{\mathrm{b}}=-2\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}-4\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}},\text{\hspace{0.17em}}\\ \stackrel{\to }{\mathrm{c}}=\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}-\mathbf{3}\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+5\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\text{}\mathrm{are}\mathrm{coplanar}.\end{array}$

Ans.

$\begin{array}{l}\left[\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}\right]=|\begin{array}{ccc}1& -2& 3\\ -2& 3& -\text{\hspace{0.17em}}4\\ 1& -3& 5\end{array}|\\ \text{\hspace{0.17em}\hspace{0.17em}}=1|\begin{array}{cc}3& -4\\ -3& 5\end{array}|+2|\begin{array}{cc}-2& -4\\ 1& 5\end{array}|+3|\begin{array}{cc}-2& 3\\ 1& -3\end{array}|\\ \text{\hspace{0.17em}\hspace{0.17em}}=1\left(15-12\right)+2\left(-10+4\right)+3\left(6-3\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=1\left(3\right)+2\left(-6\right)+3\left(3\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=3-12+9\\ \text{\hspace{0.17em}}=0\\ \mathrm{Since},\text{}\left[\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}\right]=0.\text{So, given vectors are coplanar.}\end{array}$

Q.38

$\begin{array}{l}\mathrm{Find}\text{\hspace{0.17em}}\mathrm{\lambda }\mathrm{if}\mathrm{the}\mathrm{vectors}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}-\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\stackrel{^}{\text{k}},3\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+\mathrm{\lambda }\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}-3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\text{}\mathrm{are}\\ \mathrm{coplanar}.\end{array}$

Ans.

$\begin{array}{l}\left[\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}\right]=|\begin{array}{ccc}1& -1& 1\\ 3& 1& 2\\ 1& \mathrm{\lambda }& -3\end{array}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}\right]\text{\hspace{0.17em}\hspace{0.17em}}=1|\begin{array}{cc}1& 2\\ \mathrm{\lambda }& -3\end{array}|-\left(-1\right)|\begin{array}{cc}3& 2\\ 1& -3\end{array}|+1|\begin{array}{cc}3& 1\\ 1& \mathrm{\lambda }\end{array}|\\ \text{\hspace{0.17em}\hspace{0.17em}}=1\left(-3-2\mathrm{\lambda }\right)+1\left(-9-2\right)+1\left(3\mathrm{\lambda }-1\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=-3-2\mathrm{\lambda }-11+3\mathrm{\lambda }-1\\ \text{\hspace{0.17em}\hspace{0.17em}}=\mathrm{\lambda }-15\\ \mathrm{Since},\text{given vectors are coplanar. So,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}\right]=0\\ ⇒\mathrm{\lambda }-15=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\lambda }=15\end{array}$

Q.39

$\begin{array}{l}\mathrm{Let}\stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\stackrel{^}{\text{k}},\stackrel{\to }{\mathrm{b}}=\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}={\mathrm{c}}_{1}\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+{\mathrm{c}}_{2}\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+{\mathrm{c}}_{3}\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}.\mathrm{Then}\\ \left(\mathrm{a}\right)\mathrm{If}{\mathrm{c}}_{1}=1\mathrm{and}{\mathrm{c}}_{2}=2,\mathrm{find}{\mathrm{c}}_{3}\mathrm{which}\mathrm{makes}\stackrel{\to }{\mathrm{a}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}\mathrm{coplanar}.\\ \left(\mathrm{b}\right)\mathrm{If}{\mathrm{c}}_{1}=-1\mathrm{and}{\mathrm{c}}_{3}=1,\text{\hspace{0.17em}}\mathrm{show}\mathrm{that}\mathrm{no}\mathrm{value}\mathrm{of}{\mathrm{c}}_{1}\mathrm{can}\\ \mathrm{make}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}\text{\hspace{0.17em}}\mathrm{coplanar}.\\ \end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{Given}:\stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}},\stackrel{\to }{\mathrm{b}}=\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}\mathrm{and}\\ \text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}={\mathrm{c}}_{1}\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+{\mathrm{c}}_{2}\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+{\mathrm{c}}_{3}\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ \mathrm{Putting}{\text{c}}_{\text{1}}=1{\text{and c}}_{\text{2}}=2,\text{we get}\\ \text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}=1\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+{\mathrm{c}}_{3}\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ \mathrm{Since},\text{}\stackrel{\to }{\mathrm{a}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}\mathrm{are}\mathrm{coplanar}.\mathrm{Then},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}\right]=|\begin{array}{ccc}1& 1& 1\\ 1& 0& 0\\ 1& 2& {\mathrm{c}}_{3}\end{array}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}\right]\text{\hspace{0.17em}\hspace{0.17em}}=1|\begin{array}{cc}0& 0\\ 2& {\mathrm{c}}_{3}\end{array}|-1|\begin{array}{cc}1& 0\\ 1& {\mathrm{c}}_{3}\end{array}|+1|\begin{array}{cc}1& 0\\ 1& 2\end{array}|\\ \text{\hspace{0.17em}\hspace{0.17em}}=1\left(0\right)-1\left({\mathrm{c}}_{3}-0\right)+1\left(2-0\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=0-{\mathrm{c}}_{3}+2\\ \mathrm{Since},\text{given vectors are coplanar. So,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}\right]=0\\ ⇒-{\mathrm{c}}_{3}+2=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{c}}_{3}=2\\ \left(\mathrm{ii}\right)\mathrm{Given}:\stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}},\stackrel{\to }{\mathrm{b}}=\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}\mathrm{and}\\ \text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}={\mathrm{c}}_{1}\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+{\mathrm{c}}_{2}\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+{\mathrm{c}}_{3}\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ \mathrm{Putting}{\text{c}}_{\text{2}}=-1{\text{and c}}_{\text{3}}=1,\text{we get}\\ \text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}={\mathrm{c}}_{1}\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}-1\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+1\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ \mathrm{Since},\text{}\stackrel{\to }{\mathrm{a}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}\mathrm{are}\mathrm{coplanar}.\mathrm{Then},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}\right]=|\begin{array}{ccc}1& 1& 1\\ 1& 0& 0\\ {\mathrm{c}}_{1}& -1& 1\end{array}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}\right]\text{\hspace{0.17em}\hspace{0.17em}}=1|\begin{array}{cc}0& 0\\ -1& 1\end{array}|-1|\begin{array}{cc}1& 0\\ {\mathrm{c}}_{1}& 1\end{array}|+1|\begin{array}{cc}1& 0\\ {\mathrm{c}}_{1}& -1\end{array}|\\ \text{\hspace{0.17em}\hspace{0.17em}}=1\left(0\right)-1\left(1-0\right)+1\left(-1-0\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=0-1-1\\ =-2\\ \mathrm{Since},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}\right]\ne 0\\ \mathrm{So},\text{there will be no effect on the value of}\left[\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}\right]\text{by any}\\ {\text{value of c}}_{\text{1}}{\text{.Thus, no value of c}}_{\text{1}}\text{can make}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}},\stackrel{\to }{\mathrm{b}}\text{and \hspace{0.17em}}\stackrel{\to }{\mathrm{c}}\text{coplanar.}\end{array}$

Q.40

$\begin{array}{l}\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{four}\mathrm{points}\mathrm{with}\mathrm{position}\mathrm{vectors}\\ 4\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+8\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+12\text{\hspace{0.17em}}\stackrel{^}{\text{k}},2\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+4\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+6\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}},\text{\hspace{0.17em}}3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+5\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+4\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}5\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+8\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+5\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ \mathrm{are}\mathrm{coplanar}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{the}\mathrm{four}\mathrm{points}\mathrm{with}\mathrm{position}\mathrm{vectors}\mathrm{be}\\ \mathrm{OA}=4\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+8\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+12\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}},\mathrm{OB}=2\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+4\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+6\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}},\text{\hspace{0.17em}}\mathrm{OC}=3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+5\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+4\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ \mathrm{and}\mathrm{OD}=5\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+8\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+5\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}},\text{then}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{AB}}=\stackrel{\to }{\mathrm{OB}}-\stackrel{\to }{\mathrm{OA}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(2\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+4\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+6\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\right)-\left(4\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+8\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+12\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-2\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}-4\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}-6\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{AC}}=\stackrel{\to }{\mathrm{OC}}-\stackrel{\to }{\mathrm{OA}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+5\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+4\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\right)-\left(4\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+8\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+12\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}-3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}-8\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{AD}}=\stackrel{\to }{\mathrm{OD}}-\stackrel{\to }{\mathrm{OA}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(5\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+8\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+5\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\right)-\left(4\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+8\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+12\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+0\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}-7\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ \mathrm{Thus},\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}}\left[\stackrel{\to }{\mathrm{AB}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{AC}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{AD}}\right]=|\begin{array}{ccc}-2& -4& -6\\ -1& -3& -8\\ 1& 0& -7\end{array}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-2|\begin{array}{cc}-3& -8\\ 0& -7\end{array}|-\left(-4\right)|\begin{array}{cc}-1& -8\\ \text{\hspace{0.17em}\hspace{0.17em}}1& -7\end{array}|+\left(-6\right)|\begin{array}{cc}-1& -3\\ \text{\hspace{0.17em}\hspace{0.17em}}1& \text{\hspace{0.17em}\hspace{0.17em}}0\end{array}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-2\left(21-0\right)+4\left(7+8\right)-6\left(0+3\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-42+60-18\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=0\\ \mathrm{Thus},\text{}\mathrm{the}\mathrm{given}\mathrm{four}\text{points}\mathrm{with}\mathrm{position}\mathrm{vectors}\text{are coplanar.}\end{array}$

Q.41

$\begin{array}{l}\text{If}\stackrel{\to }{\mathrm{a}}\text{}\stackrel{\to }{\mathrm{b}}\text{are two collinear vectors, then which of the following}\\ \text{are incorrect:}\\ \left(\text{A}\right)\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=\mathrm{\lambda }\stackrel{\to }{\mathrm{a}},\text{for some scalar}\mathrm{\lambda }\\ \left(\text{B}\right)\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}=±\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\\ \left(\text{C}\right)\text{the respective components of}\stackrel{\to }{\mathrm{a}}\text{and}\stackrel{\to }{\mathrm{b}}\text{are proportional}\\ \left(\text{D}\right)\text{both the vectors}\stackrel{\to }{\mathrm{a}}\text{and}\stackrel{\to }{\mathrm{b}}\text{have same direction, but}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}different magnitudes.}\end{array}$

Ans.

$\begin{array}{l}\text{If}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}\hspace{0.17em}and}\stackrel{\to }{\mathrm{b}}\text{are two collinear vectors, then they are parallel.}\\ \text{Therefore, we have:}\\ \text{\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=\mathrm{\lambda }\stackrel{\to }{\mathrm{a}}\left(\mathrm{\lambda }\text{is a scalar quantities.}\right)\\ \text{If\hspace{0.17em}}\mathrm{\lambda }=±1,\text{then \hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=±1\stackrel{\to }{\mathrm{a}}.\\ \text{If\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}={\mathrm{a}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{a}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{a}}_{3}\stackrel{^}{\mathrm{k}}\text{and}\stackrel{\to }{\mathrm{b}}={\mathrm{b}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{b}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{b}}_{3}\stackrel{^}{\mathrm{k}},\text{\hspace{0.17em}then}\\ \text{\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=\mathrm{\lambda }\stackrel{\to }{\mathrm{a}}⇒{\mathrm{b}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{b}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{b}}_{3}\stackrel{^}{\mathrm{k}}=\mathrm{\lambda }\left({\mathrm{a}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{a}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{a}}_{3}\stackrel{^}{\mathrm{k}}\right)\end{array}$

$\begin{array}{l}⇒{\mathrm{b}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{b}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{b}}_{3}\stackrel{^}{\mathrm{k}}=\left({\mathrm{\lambda a}}_{1}\right)\stackrel{^}{\mathrm{i}}+\left({\mathrm{\lambda a}}_{2}\right)\stackrel{^}{\mathrm{j}}+\left({\mathrm{\lambda a}}_{3}\right)\stackrel{^}{\mathrm{k}}\\ ⇒{\mathrm{b}}_{1}={\mathrm{\lambda a}}_{1},{\mathrm{b}}_{2}={\mathrm{\lambda a}}_{2},{\mathrm{b}}_{3}={\mathrm{\lambda a}}_{3}\\ ⇒\frac{{\mathrm{b}}_{1}}{{\mathrm{a}}_{1}}=\frac{{\mathrm{b}}_{2}}{{\mathrm{a}}_{2}}=\frac{{\mathrm{b}}_{3}}{{\mathrm{a}}_{3}}=\mathrm{\lambda }\\ \text{Thus, the respective components of}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}\hspace{0.17em}and}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}\hspace{0.17em}are proportional.}\\ \text{However, vectors}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}\hspace{0.17em}and}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}\hspace{0.17em}can have different directions.}\\ \text{Hence, the statement given in D is incorrect.}\\ \text{The correct answer is D.}\end{array}$

Q.42

$\begin{array}{l}\mathrm{Find}\mathrm{x}\mathrm{such}\mathrm{that}\mathrm{the}\mathrm{four}\mathrm{points}\mathrm{A}\left(3,2,1\right),\mathrm{B}\left(4,\mathrm{x},5\right),\\ \mathrm{C}\left(4,2,-2\right)\mathrm{and}\mathrm{D}\left(6,5,-1\right)\mathrm{are}\mathrm{coplanar}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{the}\mathrm{four}\mathrm{points}\mathrm{with}\mathrm{position}\mathrm{vectors}\mathrm{be}\\ \mathrm{OA}=3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}},\mathrm{OB}=4\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+\mathrm{x}\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+5\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}},\text{\hspace{0.17em}}\mathrm{OC}=4\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}-2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ \mathrm{andOD}=6\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+5\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}-1\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}},\text{then}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{AB}}=\stackrel{\to }{\mathrm{OB}}-\stackrel{\to }{\mathrm{OA}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(4\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+\mathrm{x}\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+5\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\right)-\left(3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+\left(\mathrm{x}-2\right)\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+4\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{AC}}=\stackrel{\to }{\mathrm{OC}}-\stackrel{\to }{\mathrm{OA}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(4\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}-2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\right)-\left(4\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+8\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+12\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=0\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}-6\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}-14\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{AD}}=\stackrel{\to }{\mathrm{OD}}-\stackrel{\to }{\mathrm{OA}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(6\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+5\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}-1\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\right)-\left(4\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+8\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}+12\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}-3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}-13\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ \mathrm{Since},\text{given vectors are coplanar. So,}\\ \text{\hspace{0.17em}}\left[\stackrel{\to }{\mathrm{AB}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{AC}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{AD}}\right]=0\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\stackrel{\to }{\mathrm{AB}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{AC}}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{AD}}\right]=|\begin{array}{ccc}1& \mathrm{x}-2& 4\\ 0& -6& -14\\ 2& -3& -13\end{array}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1|\begin{array}{cc}-6& -14\\ -3& -13\end{array}|-\left(\mathrm{x}-2\right)|\begin{array}{cc}0& -14\\ 2& -13\end{array}|+4|\begin{array}{cc}0& -6\\ 2& -3\end{array}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1\left(78-42\right)-\left(\mathrm{x}-2\right)\left(0+28\right)+4\left(0+12\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=36-28\text{\hspace{0.17em}}\mathrm{x}+56+48\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=140-28\text{\hspace{0.17em}}\mathrm{x}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}140-28\text{\hspace{0.17em}}\mathrm{x}=0\\ ⇒\mathrm{x}=\frac{140}{28}=5\\ \mathrm{Thus},\text{the value of x is 5.}\end{array}$

Q.43

$\begin{array}{l}\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{vectors}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}},\stackrel{\to }{\mathrm{b}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}\mathrm{are}\mathrm{coplanar}\mathrm{if}\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}}+\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}},\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}}+\stackrel{\to }{\mathrm{c}\text{\hspace{0.17em}}}\text{\hspace{0.17em}}\\ \mathrm{and}\stackrel{\to }{\mathrm{c}\text{\hspace{0.17em}}}+\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}}\text{\hspace{0.17em}}\mathrm{are}\mathrm{coplanar}.\end{array}$

Ans.

$\begin{array}{l}\left[\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}}+\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}}+\stackrel{\to }{\mathrm{c}\text{\hspace{0.17em}}},\stackrel{\to }{\mathrm{c}\text{\hspace{0.17em}}}+\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}}\right]=\left(\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}}+\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}}\right).\left(\left(\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}}+\stackrel{\to }{\mathrm{c}\text{\hspace{0.17em}}}\right)×\left(\stackrel{\to }{\mathrm{c}\text{\hspace{0.17em}}}+\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}}\right)\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=\left(\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}}+\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}}\right).\left(\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}}×\stackrel{\to }{\mathrm{c}\text{\hspace{0.17em}}}+\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}}×\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}}+\stackrel{\to }{\mathrm{c}\text{\hspace{0.17em}}}×\stackrel{\to }{\mathrm{c}\text{\hspace{0.17em}}}+\stackrel{\to }{\mathrm{c}\text{\hspace{0.17em}}}×\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=\left(\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}}+\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}}\right).\left(\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}}×\stackrel{\to }{\mathrm{c}\text{\hspace{0.17em}}}+\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}}×\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}}+0+\stackrel{\to }{\mathrm{c}\text{\hspace{0.17em}}}×\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}.}\left(\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}}×\stackrel{\to }{\mathrm{c}\text{\hspace{0.17em}}}\right)\text{\hspace{0.17em}}+\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}}.\left(\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}}×\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}}\right)\text{\hspace{0.17em}}+\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}}\left(\stackrel{\to }{\mathrm{c}\text{\hspace{0.17em}}}×\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}}\right)+\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}.}\left(\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}}×\stackrel{\to }{\mathrm{c}\text{\hspace{0.17em}}}\right)\text{\hspace{0.17em}}\\ +\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}}.\left(\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}}×\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}}\right)\text{\hspace{0.17em}}+\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}}.\left(\stackrel{\to }{\mathrm{c}\text{\hspace{0.17em}}}×\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=\left[\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}\text{\hspace{0.17em}}}\right]+\left[\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}}\right]+\left[\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}}\right]+\left[\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}\text{\hspace{0.17em}}}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\left[\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}}\right]+\left[\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}}=\left[\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}\text{\hspace{0.17em}}}\right]+0+0+0\text{\hspace{0.17em}}+0+\left[\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}\text{\hspace{0.17em}}}\right]\\ \left[âˆµ\left[\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}}\right]=\left[\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}}\right]=\left[\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}\text{\hspace{0.17em}}}\right]=\left[\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}}\right]=0\right]\\ \text{\hspace{0.17em}\hspace{0.17em}}=2\left[\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}\text{\hspace{0.17em}}}\right]\\ \mathrm{Since},\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}}+\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}},\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}}+\stackrel{\to }{\mathrm{c}\text{\hspace{0.17em}}}\text{\hspace{0.17em}}\mathrm{and}\stackrel{\to }{\mathrm{c}\text{\hspace{0.17em}}}+\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}}\text{\hspace{0.17em}}\mathrm{are}\mathrm{coplanar}.\text{}\mathrm{So},\\ \left[\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}}+\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}}+\stackrel{\to }{\mathrm{c}\text{\hspace{0.17em}}},\stackrel{\to }{\mathrm{c}\text{\hspace{0.17em}}}+\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}}\right]=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}2\left[\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}\text{\hspace{0.17em}}}\right]=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\stackrel{\to }{\mathrm{a}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}\text{\hspace{0.17em}}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}\text{\hspace{0.17em}}}\right]=0\\ \mathrm{So},\text{}\mathrm{the}\mathrm{vectors}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}},\stackrel{\to }{\mathrm{b}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}\mathrm{are}\mathrm{coplanar}.\end{array}$

Q.44 Write down a unit vector in XY-plane, making an angle of 30° with the positive direction in X-axis.

Ans.

$\begin{array}{l}\text{If is}\stackrel{\to }{\mathrm{r}}\text{\hspace{0.17em}a unit vector in the XY-plane, then\hspace{0.17em}}\stackrel{\to }{\mathrm{r}}\text{\hspace{0.17em}}=\mathrm{cos\theta }\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+\mathrm{sin\theta }\stackrel{^}{\mathrm{j}}.\\ \text{Here,}\mathrm{\theta }\text{is the angle made by the unit vector with the}\\ \text{positive direction of the x-axis.}\\ \text{Therefore, for}\mathrm{\theta }\text{= 30°:}\\ \text{\hspace{0.17em}}\stackrel{\to }{\mathrm{r}}\text{\hspace{0.17em}}=\mathrm{cos}\text{30°\hspace{0.17em}\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+\mathrm{sin}\text{30°\hspace{0.17em}}\stackrel{^}{\mathrm{j}}=\frac{\sqrt{3}}{2}\stackrel{^}{\mathrm{i}}+\frac{1}{2}\stackrel{^}{\mathrm{j}}\\ \text{Thus, the required unit vector is\hspace{0.17em}}\frac{\sqrt{3}}{2}\stackrel{^}{\mathrm{i}}+\frac{1}{2}\stackrel{^}{\mathrm{j}}.\end{array}$

Q.45 Find the scalar components and magnitude of the vector joining the points P(x1, y1, z1) and Q(x2, y2, z2).

Ans.

$\begin{array}{l}\text{The vector joining the points\hspace{0.17em}}\mathrm{P}\left({\mathrm{x}}_{1},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}_{1},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{z}}_{1}\right)\text{and}\mathrm{Q}\left({\mathrm{x}}_{2},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}_{2},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{z}}_{2}\right)\text{is:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{PQ}}=\left({\mathrm{x}}_{2}-{\mathrm{x}}_{1}\right)\stackrel{^}{\mathrm{i}}+\left({\mathrm{y}}_{2}-{\mathrm{y}}_{1}\right)\stackrel{^}{\mathrm{j}}+\left({\mathrm{z}}_{2}-{\mathrm{z}}_{1}\right)\stackrel{^}{\mathrm{k}}\\ |\stackrel{\to }{\mathrm{PQ}}|=|\left({\mathrm{x}}_{2}-{\mathrm{x}}_{1}\right)\stackrel{^}{\mathrm{i}}+\left({\mathrm{y}}_{2}-{\mathrm{y}}_{1}\right)\stackrel{^}{\mathrm{j}}+\left({\mathrm{z}}_{2}-{\mathrm{z}}_{1}\right)\stackrel{^}{\mathrm{k}}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\left({\mathrm{x}}_{2}-{\mathrm{x}}_{1}\right)}^{2}+{\left({\mathrm{y}}_{2}-{\mathrm{y}}_{1}\right)}^{2}+{\left({\mathrm{z}}_{2}-{\mathrm{z}}_{1}\right)}^{2}}\\ \text{Hence, the scalar components and the magnitude of the vector}\\ \text{joining the given points are respectively}\\ \left\{\left({\mathrm{x}}_{2}-{\mathrm{x}}_{1}\right),\text{\hspace{0.17em}\hspace{0.17em}}\left({\mathrm{y}}_{2}-{\mathrm{y}}_{1}\right),\text{\hspace{0.17em}\hspace{0.17em}}\left({\mathrm{z}}_{2}-{\mathrm{z}}_{1}\right)\right\}\text{and}\sqrt{{\left({\mathrm{x}}_{2}-{\mathrm{x}}_{1}\right)}^{2}+{\left({\mathrm{y}}_{2}-{\mathrm{y}}_{1}\right)}^{2}+{\left({\mathrm{z}}_{2}-{\mathrm{z}}_{1}\right)}^{2}}.\end{array}$

Q.46 A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of north and stops. Determine the girl’s displacement from her initial point of departure.

Ans.

Let O and B be the initial and final positions of the girl respectively. Then, the girl’s position can be shown as:

$\begin{array}{l}\text{Now, we have:}\\ \text{\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{OA}}=-\text{\hspace{0.17em}}4\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{AB}}=\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}|\stackrel{\to }{\mathrm{AB}}|\mathrm{cos}60\mathrm{°}+\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}|\stackrel{\to }{\mathrm{AB}}|\text{sin60°}\\ =\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\left(3\right)\frac{1}{2}+\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\left(3\right)\text{}\frac{\sqrt{3}}{2}\\ =\frac{3}{2}\stackrel{^}{\mathrm{i}}+\frac{3\sqrt{3}}{2}\stackrel{^}{\mathrm{j}}\\ \text{By the triangle law of vector addition, we have:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{OB}}=\stackrel{\to }{\mathrm{OA}}+\stackrel{\to }{\mathrm{AB}}\\ =-\text{\hspace{0.17em}}4\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+\frac{3}{2}\stackrel{^}{\mathrm{i}}+\frac{3\sqrt{3}}{2}\stackrel{^}{\mathrm{j}}\\ =\left(-\text{\hspace{0.17em}}4+\frac{3}{2}\right)\stackrel{^}{\mathrm{i}}+\frac{3\sqrt{3}}{2}\stackrel{^}{\mathrm{j}}\\ =-\frac{5}{2}\stackrel{^}{\mathrm{i}}+\frac{3\sqrt{3}}{2}\stackrel{^}{\mathrm{j}}\\ \text{Hence, the girl’s displacement from her initial point of departure}\\ \text{is\hspace{0.17em}\hspace{0.17em}}-\frac{5}{2}\stackrel{^}{\mathrm{i}}+\frac{3\sqrt{3}}{2}\stackrel{^}{\mathrm{j}}.\end{array}$

Q.47

$\begin{array}{l}\mathrm{If}\stackrel{\to }{\mathrm{a}}=\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}},\mathrm{then}\mathrm{is}\mathrm{it}\mathrm{true}\mathrm{that}|\stackrel{\to }{\mathrm{a}}|=|\stackrel{\to }{\mathrm{b}}|+|\stackrel{\to }{\mathrm{c}}|?\mathrm{Justify}\mathrm{your}\\ \mathrm{answer}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{In}\text{}\mathrm{\Delta ABC},\text{\hspace{0.17em}}\mathrm{let}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{CB}}=\stackrel{\to }{\mathrm{a}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{CA}}=\stackrel{\to }{\mathrm{b}}\text{and}\stackrel{\to }{\mathrm{AB}}=\stackrel{\to }{\mathrm{c}}.\\ \text{Now, by the triangle law of vector addition, we have\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}=\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}.\\ \mathrm{Since},\text{sum of any two sides of a triangle is always greater than third side.}\\ \therefore |\stackrel{\to }{\mathrm{a}}|<|\stackrel{\to }{\mathrm{b}}|+|\stackrel{\to }{\mathrm{c}}|\\ \text{Hence, it is not true that\hspace{0.17em}}|\stackrel{\to }{\mathrm{a}}|=|\stackrel{\to }{\mathrm{b}}|+|\stackrel{\to }{\mathrm{c}}|.\end{array}$

Q.48

$\mathrm{Find}\text{}\mathrm{the}\text{}\mathrm{value}\text{}\mathrm{of}\text{}\mathrm{x}\text{}\mathrm{for}\text{}\mathrm{which}\text{}\mathrm{x}\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\text{k}}\right)\mathrm{is}\mathrm{a}\mathrm{unit}\mathrm{vector}.$

Ans.

$\begin{array}{l}\mathrm{x}\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)\mathrm{is}\mathrm{a}\mathrm{unit}\mathrm{vector},\text{}\\ \text{so\hspace{0.17em}\hspace{0.17em}}|\mathrm{x}\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)|=1\\ ⇒\mathrm{x}|\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)|=1\\ \text{\hspace{0.17em}}\mathrm{x}\sqrt{{1}^{2}+{1}^{2}+{1}^{2}}=1\\ ⇒\mathrm{x}\sqrt{3}=1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=±\frac{1}{\sqrt{3}}\\ \text{Thus, the required value of x is\hspace{0.17em}}±\frac{1}{\sqrt{3}}.\end{array}$

Q.49

$\begin{array}{l}\mathrm{Find}\mathrm{a}\mathrm{vector}\mathrm{of}\mathrm{magnitude}5\mathrm{units},\mathrm{and}\mathrm{parallel}\mathrm{to}\mathrm{the}\\ \mathrm{resultant}\mathrm{of}\mathrm{the}\mathrm{vectors}\stackrel{\to }{\mathrm{a}}=2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\mathrm{and}\stackrel{\to }{\mathrm{b}}=\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\text{given}\mathrm{vectorsare}\\ \stackrel{\to }{\mathrm{a}}=2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\mathrm{and}\stackrel{\to }{\mathrm{b}}=\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\\ \mathrm{Let}\text{}\stackrel{\to }{\mathrm{c}}\text{be the resultant vectors of}\stackrel{\to }{\mathrm{a}}\text{and}\stackrel{\to }{\mathrm{b}}.\\ \mathrm{Then},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}=\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}+\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\\ =3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}\\ \therefore |\stackrel{\to }{\mathrm{c}}|=|3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}|\\ =\sqrt{{3}^{2}+{1}^{2}}\\ =\sqrt{10}\\ \mathrm{So},\text{\hspace{0.17em}}\stackrel{^}{\mathrm{c}}=\frac{3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}}{\sqrt{10}}\\ \text{Thus, the vector of magnitude 5 units and parallel to the}\\ \text{resultant of vectors\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}\text{and}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}\mathrm{is}\\ ±5.\stackrel{^}{\mathrm{c}}=±5.\text{\hspace{0.17em}}\frac{3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}}{\sqrt{10}}=±\text{\hspace{0.17em}}\frac{3\sqrt{10}}{2}\stackrel{^}{\mathrm{i}}±\frac{\sqrt{10}}{2}\stackrel{^}{\mathrm{j}}\end{array}$

Q.50

$\begin{array}{l}\text{Find the angle between two vectors}\stackrel{\to }{\mathrm{a}}\text{and}\stackrel{\to }{\mathrm{b}}\text{with}\\ \text{magnitude}\sqrt{\text{3}}\text{and\hspace{0.17em}\hspace{0.17em}2, respectively having}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=\sqrt{\text{6}}.\end{array}$

Ans.

$\begin{array}{l}\text{It is given that,}\\ |\stackrel{\to }{\mathrm{a}}|=\sqrt{3},|\stackrel{\to }{\mathrm{b}}|=2\text{and}\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}=\sqrt{6}\\ \text{Now, we know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}=|\stackrel{\to }{\mathrm{a}}||\stackrel{\to }{\mathrm{b}}|\mathrm{cos\theta }\\ ⇒\sqrt{6}=\sqrt{3}.2\mathrm{cos\theta }\\ ⇒\mathrm{cos\theta }=\frac{\sqrt{6}}{\sqrt{3}.2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{\sqrt{2}}\end{array}$

$\begin{array}{l}⇒\mathrm{cos\theta }=\mathrm{cos}\frac{\mathrm{\pi }}{4}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\theta }=\frac{\mathrm{\pi }}{4}\\ \text{Hence, the angle between the given vectors\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}\text{and}\stackrel{\to }{\mathrm{b}}\text{is}\frac{\mathrm{\pi }}{4}.\end{array}$

Q.51

$\text{Find the angle between the vectors}\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\text{and}3\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}.$

Ans.

$\begin{array}{l}\text{The given vectors are\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\text{and}\stackrel{\to }{\mathrm{b}}=3\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}|\stackrel{\to }{\mathrm{a}}|=|\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}|\\ =\sqrt{{1}^{2}+{\left(-2\right)}^{2}+{3}^{2}}\\ =\sqrt{1+4+9}\\ =\sqrt{14}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}|\stackrel{\to }{\mathrm{b}}|=|3\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}|\\ =\sqrt{{3}^{2}+{\left(-2\right)}^{2}+{1}^{2}}\\ =\sqrt{9+4+1}\\ =\sqrt{14}\\ \text{Now,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}=\left(\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\right).\left(3\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)\\ =3+4+3=10\end{array}$

$\begin{array}{l}\text{Also, we know that}\\ \text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}=|\stackrel{\to }{\mathrm{a}}|.|\stackrel{\to }{\mathrm{b}}|\mathrm{cos\theta }\\ 10=\sqrt{14}.\sqrt{14}\mathrm{cos\theta }\\ \mathrm{cos\theta }=\frac{10}{14}\\ =\frac{5}{7}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\theta }={\mathrm{cos}}^{-1}\left(\frac{5}{7}\right)\end{array}$

Q.52

$\text{Find the projection of the vector}\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}\text{on the vector}\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}.$

Ans.

$\begin{array}{l}\text{Let}\stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}\text{and}\stackrel{\to }{\mathrm{b}}=\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}\\ \text{Now, projection of vector}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}\hspace{0.17em}on}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}\hspace{0.17em}is given by,}\\ \frac{1}{|\stackrel{\to }{\mathrm{b}}|}\left(\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}\right)=\frac{1}{|\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}|}\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}\right).\left(\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{\sqrt{{1}^{2}+{1}^{2}}}\left(1-1\right)=0\\ \text{Hence, the projection of vector}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}on\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\text{is 0.}\end{array}$

Q.53

$\begin{array}{l}\text{Find the projection of the vector}\stackrel{^}{\mathrm{i}}+\mathbf{3}\stackrel{^}{\mathrm{j}}+7\stackrel{^}{k}\text{on the vector}\\ 7\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+8\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}.\end{array}$

Ans.

$\begin{array}{l}\text{Let}\stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+7\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\text{and}\stackrel{\to }{\mathrm{b}}=7\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+8\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ \text{Now, projection of vector}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}\hspace{0.17em}on}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}\hspace{0.17em}is given by,}\\ \frac{1}{|\stackrel{\to }{\mathrm{b}}|}\left(\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}\right)=\frac{1}{|7\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+8\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}|}\left(\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+7\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\right).\left(7\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+8\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{\sqrt{{7}^{2}+{\left(-1\right)}^{2}+{8}^{2}}}\left(7-3+56\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{\sqrt{49+1+64}}\left(60\right)=\frac{1}{\sqrt{114}}\left(60\right)\\ \text{Hence, the projection of vector}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}on\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\text{is}\frac{\text{60}}{\sqrt{114}}\text{.}\end{array}$

Q.54

$\begin{array}{l}\text{Show that each of the given three vectors is a unit vector:}\\ \frac{1}{7}\left(2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}\right),\text{\hspace{0.17em}\hspace{0.17em}}\frac{1}{7}\left(3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-6\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right),\text{\hspace{0.17em}}\frac{1}{7}\left(6\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-3\stackrel{^}{\mathrm{k}}\right)\\ \text{Also, show that they are mutually perpendicular to each}\\ \text{other.}\end{array}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}=\frac{1}{7}\left(2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=\frac{1}{7}\left(3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-6\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}=\frac{1}{7}\left(6\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-3\stackrel{^}{\mathrm{k}}\right)\\ ⇒\text{\hspace{0.17em}}|\stackrel{\to }{\mathrm{a}}|=|\frac{1}{7}\left(2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}\right)|\end{array}$

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}}=\frac{1}{7}\sqrt{{2}^{2}+{3}^{2}+{6}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{1}{7}\sqrt{4+9+36}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{7}{7}=1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}|\stackrel{\to }{\mathrm{b}}|=|\frac{1}{7}\left(3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-6\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)\text{\hspace{0.17em}}|\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{1}{7}\sqrt{{3}^{2}+{\left(-6\right)}^{2}+{2}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{1}{7}\sqrt{9+36+4}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{7}{7}=1\\ \text{and}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}|\stackrel{\to }{\mathrm{c}}|=|\frac{1}{7}\left(6\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-3\stackrel{^}{\mathrm{k}}\right)\text{\hspace{0.17em}}|\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{1}{7}\sqrt{{6}^{2}+{2}^{2}+{\left(-3\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{1}{7}\sqrt{36+4+9}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{7}{7}=1\\ \text{Thus, each of the given three vectors is a unit vector.}\\ \text{Now,}\\ \text{\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=\frac{1}{7}\left(2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}\right).\frac{1}{7}\left(3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-6\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)\\ =\frac{1}{49}\left(6-18+12\right)=0\end{array}$

$\begin{array}{l}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}=\frac{1}{7}\left(3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-6\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right).\frac{1}{7}\left(6\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-3\stackrel{^}{\mathrm{k}}\right)\\ =\frac{1}{49}\left(18-12-6\right)=0\\ \stackrel{\to }{\mathrm{c}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}=\frac{1}{7}\left(6\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-3\stackrel{^}{\mathrm{k}}\right).\frac{1}{7}\left(2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}\right)\\ =\frac{1}{49}\left(12+6-18\right)=0\\ \text{Hence, the given three vectors are mutually perpendicular}\\ \text{to each other.}\end{array}$

Q.55

$\text{Find}\left|\stackrel{\to }{\text{a}}\right|\text{and}\left|\stackrel{\to }{\text{b}}\right|\text{, if}\left(\stackrel{\to }{\text{a}}+\stackrel{\to }{\text{b}}\right)\text{.}\left(\stackrel{\to }{\text{a}}–\stackrel{\to }{\text{b}}\right)=\text{8 and}\left|\stackrel{\to }{\text{a}}\right|=\text{8}\left|\stackrel{\to }{\text{b}}\right|\text{.}$

Ans.

$\begin{array}{l}\text{We have,}\\ \left(\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}\right).\left(\stackrel{\to }{\mathrm{a}}-\stackrel{\to }{\mathrm{b}}\right)=8\text{and}|\stackrel{\to }{\mathrm{a}}|=8|\stackrel{\to }{\mathrm{b}}|\\ \text{So,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}\right).\left(\stackrel{\to }{\mathrm{a}}-\stackrel{\to }{\mathrm{b}}\right)=8\\ ⇒\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{a}}-\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}-\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=8\\ ⇒{|\stackrel{\to }{\mathrm{a}}|}^{2}-{|\stackrel{\to }{\mathrm{b}}|}^{2}=8\left[âˆµ\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(8|\stackrel{\to }{\mathrm{b}}|\right)}^{2}-{|\stackrel{\to }{\mathrm{b}}|}^{2}=8\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}64{|\stackrel{\to }{\mathrm{b}}|}^{2}-{|\stackrel{\to }{\mathrm{b}}|}^{2}=8\\ \end{array}$

$\begin{array}{l}⇒\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\stackrel{\to }{b}|=\sqrt{\frac{8}{63}}=\frac{2\sqrt{2}}{3\sqrt{7}}\\ \therefore \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\stackrel{\to }{a}|=8|\stackrel{\to }{b}|\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}=8×\frac{2\sqrt{2}}{3\sqrt{7}}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}=\frac{16\sqrt{2}}{3\sqrt{7}}\end{array}$

Q.56

$\text{Evaluate the product}\left(3\stackrel{\to }{\mathrm{a}}-5\stackrel{\to }{\mathrm{b}}\right).\left(2\stackrel{\to }{\mathrm{a}}+7\stackrel{\to }{\mathrm{b}}\right).$

Ans.

$\begin{array}{l}\left(3\stackrel{\to }{\mathrm{a}}-5\stackrel{\to }{\mathrm{b}}\right).\left(2\stackrel{\to }{\mathrm{a}}+7\stackrel{\to }{\mathrm{b}}\right)=6\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}+21\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}-10\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}-35\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=6{|\stackrel{\to }{\mathrm{a}}|}^{2}+21\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}-10\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}-35\text{\hspace{0.17em}}{|\stackrel{\to }{\mathrm{b}}|}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=6{|\stackrel{\to }{\mathrm{a}}|}^{2}+11\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}-35\text{\hspace{0.17em}}{|\stackrel{\to }{\mathrm{b}}|}^{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[âˆµ\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}\right]\end{array}$

Q.57

$\begin{array}{l}\text{Find the magnitude of the vectors}\stackrel{\to }{\text{a}}\text{and}\stackrel{\to }{\text{b}}\text{, having the same}\\ \text{magnitude and such that the angle between them is 60°}\\ \text{and their scalar product is}\frac{1}{2}\text{.}\end{array}$

Ans.

$\begin{array}{l}\text{Let}\mathrm{\theta }\text{be the angle between the vectors}\stackrel{\to }{\text{a}}\text{and}\stackrel{\to }{\text{b}}\text{.\hspace{0.17em}Then,}\\ \stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}.\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=|\stackrel{\to }{\mathrm{a}}||\stackrel{\to }{\mathrm{b}}|\mathrm{cos\theta }\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{2}=|\stackrel{\to }{\mathrm{a}}||\stackrel{\to }{\mathrm{a}}|\mathrm{cos}60°\left[âˆµ|\stackrel{\to }{\mathrm{a}}|=|\stackrel{\to }{\mathrm{b}}|\right]\end{array}$

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{2}={|\stackrel{\to }{\mathrm{a}}|}^{2}×\frac{1}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1={|\stackrel{\to }{\mathrm{a}}|}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}|\stackrel{\to }{\mathrm{a}}|=1\\ \text{So,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}|\stackrel{\to }{\mathrm{b}}|=|\stackrel{\to }{\mathrm{a}}|=1\end{array}$

Q.58

$\text{Find}|\stackrel{\to }{\mathrm{x}}|,\text{if for a unit vector}\stackrel{\to }{\mathrm{a}},\left(\stackrel{\to }{\mathrm{x}}-\stackrel{\to }{\mathrm{a}}\right).\left(\stackrel{\to }{\mathrm{x}}+\stackrel{\to }{\mathrm{a}}\right)=12.$

Ans.

$\begin{array}{l}\text{Since,}\\ \left(\stackrel{\to }{\mathrm{x}}-\stackrel{\to }{\mathrm{a}}\right).\left(\stackrel{\to }{\mathrm{x}}+\stackrel{\to }{\mathrm{a}}\right)=12\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{x}}.\stackrel{\to }{\mathrm{x}}+\stackrel{\to }{\mathrm{x}}.\stackrel{\to }{\mathrm{a}}-\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{x}}-\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}=12\\ \text{\hspace{0.17em}\hspace{0.17em}}{|\stackrel{\to }{\mathrm{x}}|}^{2}-{|\stackrel{\to }{\mathrm{a}}|}^{2}=12\left[âˆµ\stackrel{\to }{\mathrm{x}}.\stackrel{\to }{\mathrm{a}}=\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{x}}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}{|\stackrel{\to }{\mathrm{x}}|}^{2}-{1}^{2}=12\left[\stackrel{\to }{\mathrm{a}}\text{is a unit vector so,}|\stackrel{\to }{\mathrm{a}}|=1\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{|\stackrel{\to }{\mathrm{x}}|}^{2}=12+1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}|\stackrel{\to }{\mathrm{x}}|=\sqrt{13}\end{array}$

Q.59

$\begin{array}{l}\mathrm{If}\stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\text{k}},\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=2\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\mathrm{and}\stackrel{\to }{\mathrm{c}}=\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}},\mathrm{find}\mathrm{a}\mathrm{unit}\\ \mathrm{vector}\mathrm{parallel}\mathrm{to}\mathrm{the}\mathrm{vector}2\stackrel{\to }{\mathrm{a}}-\stackrel{\to }{\mathrm{b}}+3\stackrel{\to }{\mathrm{c}}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\text{given vectors are:}\\ \stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}},\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=2\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\mathrm{and}\stackrel{\to }{\mathrm{c}}=\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\\ \mathrm{So},\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}}2\stackrel{\to }{\mathrm{a}}-\stackrel{\to }{\mathrm{b}}+3\stackrel{\to }{\mathrm{c}}=2\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)-\left(2\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\right)+3\left(\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}-2\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}-3\stackrel{^}{\mathrm{k}}+3\stackrel{^}{\mathrm{i}}-6\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\\ |2\stackrel{\to }{\mathrm{a}}-\stackrel{\to }{\mathrm{b}}+3\stackrel{\to }{\mathrm{c}}|=|3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{3}^{2}+{\left(-3\right)}^{2}+{2}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{9+9+4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{22}\\ \text{Thus, the unit vector along\hspace{0.17em}}\left(2\stackrel{\to }{\mathrm{a}}-\stackrel{\to }{\mathrm{b}}+3\stackrel{\to }{\mathrm{c}}\right)\text{\hspace{0.17em}}\mathrm{is}\\ \frac{2\stackrel{\to }{\mathrm{a}}-\stackrel{\to }{\mathrm{b}}+3\stackrel{\to }{\mathrm{c}}}{|2\stackrel{\to }{\mathrm{a}}-\stackrel{\to }{\mathrm{b}}+3\stackrel{\to }{\mathrm{c}}|}=\frac{3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}}{\sqrt{22}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3}{\sqrt{22}}\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-\frac{3}{\sqrt{22}}\stackrel{^}{\mathrm{j}}+\frac{2}{\sqrt{22}}\stackrel{^}{\mathrm{k}}\end{array}$

Q.60

$\begin{array}{l}\text{If}\stackrel{\to }{\mathrm{a}}=2\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}},\stackrel{\to }{\mathrm{b}}=-\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\text{and}\stackrel{\to }{\mathrm{c}}=3\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}\text{are such that}\\ \stackrel{\to }{\mathrm{a}}+\mathrm{\lambda }\stackrel{\to }{\mathrm{b}}\text{is perpendicular to}\stackrel{\to }{\mathrm{c}},\text{then find the value of}\mathrm{\lambda }.\end{array}$

Ans.

$\begin{array}{l}\text{The given vectors are:}\\ \stackrel{\to }{\mathrm{a}}=2\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}},\stackrel{\to }{\mathrm{b}}=-\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\text{and}\stackrel{\to }{\mathrm{c}}=3\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}\end{array}$

$\begin{array}{l}\therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}+\mathrm{\lambda }\stackrel{\to }{\mathrm{b}}=\left(2\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\right)+\mathrm{\lambda }\left(-\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(2-\mathrm{\lambda }\right)\stackrel{^}{\mathrm{i}}+\left(2+2\mathrm{\lambda }\right)\stackrel{^}{\mathrm{j}}+\left(3+\mathrm{\lambda }\right)\stackrel{^}{\mathrm{k}}\\ \text{Since,\hspace{0.17em}}\left(\stackrel{\to }{\mathrm{a}}+\mathrm{\lambda }\stackrel{\to }{\mathrm{b}}\right)\text{is perpendicular to}\stackrel{\to }{\mathrm{c}}\text{, so}\\ \left(\stackrel{\to }{\mathrm{a}}+\mathrm{\lambda }\stackrel{\to }{\mathrm{b}}\right).\stackrel{\to }{\mathrm{c}}=0\\ ⇒\left\{\left(2-\mathrm{\lambda }\right)\stackrel{^}{\mathrm{i}}+\left(2+2\mathrm{\lambda }\right)\stackrel{^}{\mathrm{j}}+\left(3+\mathrm{\lambda }\right)\stackrel{^}{\mathrm{k}}\right\}.\left(3\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}\right)=0\\ ⇒3\left(2-\mathrm{\lambda }\right)+\left(2+2\mathrm{\lambda }\right)=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}6-3\mathrm{\lambda }+2+2\mathrm{\lambda }=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}8-\mathrm{\lambda }=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\lambda }=8\\ \text{Therefore, the required value of}\mathrm{\lambda }\text{is 8.}\end{array}$

Q.61 Show that the points A(1,–2,–8), B(5, 0, –2) and C(11, 3, 7) are collinear, and find the ratio in which B divides AC.

Ans.

$\begin{array}{l}\text{The given points are A (1,}-\text{2,}-\text{8), B (5, 0,}-\text{2) and C(11, 3, 7).}\\ \therefore \stackrel{\to }{\mathrm{AB}}=\left(5-1\right)\stackrel{^}{\mathrm{i}}+\left(0+2\right)\stackrel{^}{\mathrm{j}}+\left(-2+8\right)\stackrel{^}{\mathrm{k}}\\ \text{\hspace{0.17em}}=4\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+6\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{BC}}=\left(11-5\right)\stackrel{^}{\mathrm{i}}+\left(3-0\right)\stackrel{^}{\mathrm{j}}+\left(7+2\right)\stackrel{^}{\mathrm{k}}\\ \text{\hspace{0.17em}}=6\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}+9\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{AC}}=\left(11-1\right)\stackrel{^}{\mathrm{i}}+\left(3+2\right)\stackrel{^}{\mathrm{j}}+\left(7+8\right)\stackrel{^}{\mathrm{k}}\\ \text{\hspace{0.17em}}=10\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+5\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}+15\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ |\stackrel{\to }{\mathrm{AB}}|=|4\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+6\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}|\\ \text{\hspace{0.17em}}=\sqrt{{4}^{2}+{2}^{2}+{6}^{2}}\\ =\sqrt{16+4+36}\\ =\sqrt{56}=2\sqrt{14}\\ |\stackrel{\to }{\mathrm{BC}}|=|6\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}+9\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}|\\ \text{\hspace{0.17em}}=\sqrt{{6}^{2}+{3}^{2}+{9}^{2}}\\ =\sqrt{36+9+81}\\ =\sqrt{126}=3\sqrt{14}\\ |\stackrel{\to }{\mathrm{CA}}|=|10\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+5\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}+15\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}|\\ \text{\hspace{0.17em}}=\sqrt{{10}^{2}+{5}^{2}+{15}^{2}}\\ =\sqrt{100+25+225}\\ =\sqrt{350}=5\sqrt{14}\\ \therefore |\stackrel{\to }{\mathrm{CA}}|=|\stackrel{\to }{\mathrm{AB}}|+|\stackrel{\to }{\mathrm{BC}}|\\ \text{Thus, the given points A, B, and C are collinear.}\\ \text{Now, let point B divide AC in the ratio}\mathrm{\lambda }\text{:1. Then, we have:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{OB}}=\frac{\mathrm{\lambda }\stackrel{\to }{\mathrm{OC}}+1\stackrel{\to }{\mathrm{OA}}}{\mathrm{\lambda }+1}\\ ⇒5\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}=\frac{\mathrm{\lambda }\left(11\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+7\stackrel{^}{\mathrm{k}}\right)+\left(\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}-8\stackrel{^}{\mathrm{k}}\right)}{\mathrm{\lambda }+1}\\ \left(\mathrm{\lambda }+1\right)\left(5\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\right)=\mathrm{\lambda }\left(11\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+7\stackrel{^}{\mathrm{k}}\right)+\left(\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}-8\stackrel{^}{\mathrm{k}}\right)\\ 5\left(\mathrm{\lambda }+1\right)\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-2\left(\mathrm{\lambda }+1\right)\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}=\left(11\mathrm{\lambda }+1\right)\stackrel{^}{\mathrm{i}}+\left(3\mathrm{\lambda }-2\right)\stackrel{^}{\mathrm{j}}+\left(7\mathrm{\lambda }-8\right)7\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}5\left(\mathrm{\lambda }+1\right)=\left(11\mathrm{\lambda }+1\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}5\mathrm{\lambda }+5=11\mathrm{\lambda }+1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4=6\mathrm{\lambda }\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\lambda }=\frac{4}{6}\\ =\frac{2}{3}\\ \text{Thus, point B divides AC in the ratio\hspace{0.17em}}2:3.\end{array}$

Q.62

$\begin{array}{l}\text{Show that}|\stackrel{\to }{a}|\stackrel{\to }{b}+|\stackrel{\to }{b}|\stackrel{\to }{a}\text{is perpendicular to}|\stackrel{\to }{a}|\stackrel{\to }{b}-|\stackrel{\to }{b}|\stackrel{\to }{a},\text{for any}\\ \text{two nonzero vectors}\stackrel{\to }{a}\text{and}\stackrel{\to }{b}.\end{array}$

Ans.

$\begin{array}{l}\left(|\stackrel{\to }{a}|\stackrel{\to }{b}+|\stackrel{\to }{b}|\stackrel{\to }{a}\right).\left(|\stackrel{\to }{a}|\stackrel{\to }{b}-|\stackrel{\to }{b}|\stackrel{\to }{a}\right)={|\stackrel{\to }{a}|}^{2}{|\stackrel{\to }{b}|}^{2}-|\stackrel{\to }{a}|\stackrel{\to }{b}|\stackrel{\to }{b}|\stackrel{\to }{a}+|\stackrel{\to }{b}|\stackrel{\to }{a}|\stackrel{\to }{a}|\stackrel{\to }{b}-{|\stackrel{\to }{b}|}^{2}{|\stackrel{\to }{a}|}^{2}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-|\stackrel{\to }{a}||\stackrel{\to }{b}|\stackrel{\to }{b}\stackrel{\to }{a}+|\stackrel{\to }{b}||\stackrel{\to }{a}|\stackrel{\to }{a}\stackrel{\to }{b}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0\text{}\text{}\text{}\text{}\text{}\left[âˆµ\stackrel{\to }{b}\stackrel{\to }{a}=\stackrel{\to }{a}\stackrel{\to }{b}\right]\\ \text{So,}\text{\hspace{0.17em}}\left(|\stackrel{\to }{a}|\stackrel{\to }{b}+|\stackrel{\to }{b}|\stackrel{\to }{a}\right)\text{and}\left(|\stackrel{\to }{a}|\stackrel{\to }{b}-|\stackrel{\to }{b}|\stackrel{\to }{a}\right)\text{are perpendicular to each other}\text{.}\end{array}$

Q.63

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{position}\mathrm{vector}\mathrm{of}\mathrm{a}\mathrm{point}\mathrm{R}\mathrm{which}\mathrm{divides}\mathrm{the}\mathrm{line}\\ \mathrm{joining}\mathrm{two}\mathrm{points}\mathrm{P}\mathrm{and}\mathrm{Q}\mathrm{whose}\mathrm{position}\mathrm{vectors}\mathrm{are}\\ \left(2\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}\right)\mathrm{and}\left(\stackrel{\to }{\mathrm{a}}-3\stackrel{\to }{\mathrm{b}}\right)\mathrm{externally}\mathrm{in}\mathrm{the}\mathrm{ratio}1:2.\mathrm{Also},\mathrm{show}\\ \mathrm{that}\mathrm{P}\mathrm{is}\mathrm{the}\mathrm{mid}\mathrm{point}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{segment}\mathrm{RQ}.\end{array}$

Ans.

$\begin{array}{l}\text{Given that}\stackrel{\to }{\mathrm{OP}}=\left(2\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}\right)\text{and}\stackrel{\to }{\mathrm{OQ}}=\left(\stackrel{\to }{\mathrm{a}}-3\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\right)\\ \text{Since, point R divides a line segment joining two points P and Q}\\ \text{externally in the ratio 1: 2. Then, by using the section formula,}\\ \text{we get:}\\ \text{\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{OR}}=\frac{2\left(2\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}\right)-\left(\stackrel{\to }{\mathrm{a}}-3\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\right)}{2-1}\\ =\frac{4\stackrel{\to }{\mathrm{a}}+2\stackrel{\to }{\mathrm{b}}-\stackrel{\to }{\mathrm{a}}+3\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}}{1}=3\stackrel{\to }{\mathrm{a}}+5\stackrel{\to }{\mathrm{b}}\text{}\\ \text{Therefore, the position vector of point R is\hspace{0.17em}}3\stackrel{\to }{\mathrm{a}}+5\stackrel{\to }{\mathrm{b}}.\\ \text{Position vector of the mid-point of RQ}=\frac{\stackrel{\to }{\mathrm{OQ}}+\stackrel{\to }{\mathrm{OR}}}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\stackrel{\to }{\mathrm{a}}-3\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}+3\stackrel{\to }{\mathrm{a}}+5\stackrel{\to }{\mathrm{b}}}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{4\stackrel{\to }{\mathrm{a}}+2\stackrel{\to }{\mathrm{b}}}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}=\stackrel{\to }{\mathrm{OP}}\\ \text{Hence, P is the mid-point of the line segment RQ.}\end{array}$

Q.64

$\begin{array}{l}\text{If}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}=0\text{and}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=\text{0, then what can be concluded about}\\ \text{the vector}\stackrel{\to }{\mathrm{b}}\text{?}\end{array}$

Ans.

$\begin{array}{l}\text{Since, it is given that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}=0\\ ⇒{|\stackrel{\to }{\mathrm{a}}|}^{2}=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}|\stackrel{\to }{\mathrm{a}}|=0\\ \therefore \text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}\text{is a zero vector.}\\ \text{Since, vector}\stackrel{\to }{\mathrm{b}}\text{satisfying}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=0\text{can be any vector.}\end{array}$

Q.65

$\begin{array}{l}\text{If}\stackrel{\to }{\mathrm{a}},\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{b}},\stackrel{\to }{\mathrm{c}}\text{are unit vectors such that}\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}=\stackrel{\to }{0},\text{find the}\\ \text{value of}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}+\stackrel{\to }{\mathrm{c}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}.\end{array}$

Ans.

$\begin{array}{l}\text{Given:}\stackrel{\to }{\mathrm{a}},\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{b}},\stackrel{\to }{\mathrm{c}}\text{are unit vectors such that}\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}=\stackrel{\to }{0}\\ \text{Then,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left[\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}\right]}^{2}=0\\ {|\stackrel{\to }{\mathrm{a}}|}^{2}+{|\stackrel{\to }{\mathrm{b}}|}^{2}+{|\stackrel{\to }{\mathrm{c}}|}^{2}+2\left[\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}+\stackrel{\to }{\mathrm{c}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}\right]=0\\ \text{\hspace{0.17em}\hspace{0.17em}}1+1+1+2\left[\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}+\stackrel{\to }{\mathrm{c}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}\right]=0\\ ⇒\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}+\stackrel{\to }{\mathrm{c}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}=-\frac{3}{2}\end{array}$

Q.66

$\begin{array}{l}\mathrm{The}\mathrm{two}\mathrm{adjacent}\mathrm{sides}\mathrm{of}\mathrm{a}\mathrm{parallelogram}\mathrm{are}2\stackrel{^}{\mathrm{i}}-4\stackrel{^}{\mathrm{j}}+5\stackrel{^}{\text{k}}\mathrm{and}\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}-3\stackrel{^}{\mathrm{k}}.\\ \mathrm{Find}\mathrm{the}\mathrm{unit}\mathrm{vector}\mathrm{parallel}\mathrm{to}\mathrm{its}\mathrm{diagonal}.\mathrm{Also},\mathrm{find}\mathrm{its}\mathrm{area}.\end{array}$

Ans.

$\begin{array}{l}\text{Adjacent sides of a parallelogram are:}\\ \stackrel{\to }{\mathrm{a}}=2\stackrel{^}{\mathrm{i}}-4\stackrel{^}{\mathrm{j}}+5\stackrel{^}{\mathrm{k}}\text{and}\stackrel{\to }{\mathrm{b}}=\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}-3\stackrel{^}{\mathrm{k}}\\ \text{Then, the diagonal of a parallelogram}=\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\stackrel{^}{\mathrm{i}}-4\stackrel{^}{\mathrm{j}}+5\stackrel{^}{\mathrm{k}}+\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}-3\stackrel{^}{\mathrm{k}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-6\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\\ \text{The unit vector parallel to the diagonal}=\frac{3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-6\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}}{|3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-6\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}|}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-6\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}}{\sqrt{{3}^{2}+{\left(-6\right)}^{2}+{2}^{2}}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-6\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}}{\sqrt{9+36+4}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-6\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}}{\sqrt{49}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3}{7}\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-\frac{6}{7}\stackrel{^}{\mathrm{j}}+\frac{2}{7}\stackrel{^}{\mathrm{k}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Now},\text{Area of parallelogram ABCD}=|\stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{b}}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{b}}=|\begin{array}{ccc}\stackrel{^}{\mathrm{i}}& \stackrel{^}{\mathrm{j}}& \stackrel{^}{\mathrm{k}}\\ 2& -4& 5\\ 1& -2& -3\end{array}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(12+10\right)\stackrel{^}{\mathrm{i}}-\left(-6-5\right)\stackrel{^}{\mathrm{j}}+\left(-4+4\right)\stackrel{^}{\mathrm{k}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=22\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+11\stackrel{^}{\mathrm{j}}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}|\stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{b}}|=11|2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=11\sqrt{{2}^{2}+{1}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=11\sqrt{5}\\ \text{Thus, the area of the parallelogram is}11\sqrt{5}\text{\hspace{0.17em}square units.}\end{array}$

Q.67

$\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{direction}\mathrm{cosines}\mathrm{of}\mathrm{a}\mathrm{vector}\mathrm{equally}\mathrm{inclined}\mathrm{to}\mathrm{the}\mathrm{axes}\phantom{\rule{0ex}{0ex}}\mathrm{OX},\mathrm{OY}\mathrm{and}\mathrm{OZ}\mathrm{are}\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}.$

Ans.

$\begin{array}{l}\text{Let a vector be equally inclined to axes OX, OY, and OZ at}\\ \text{angle}\mathrm{\alpha }\text{. Then, the direction cosines of the vector are cos}\mathrm{\alpha }\text{,}\\ \text{cos}\mathrm{\alpha }\text{, and cos}\mathrm{\alpha }\text{.}\\ {\text{Since, cos}}^{\text{2}}\mathrm{\alpha }+{\text{cos}}^{\text{2}}\mathrm{\beta }+{\text{cos}}^{\text{2}}\mathrm{\gamma }=1\\ \mathrm{So},{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}cos}}^{\text{2}}\mathrm{\alpha }+{\text{cos}}^{\text{2}}\mathrm{\alpha }+{\text{cos}}^{\text{2}}\mathrm{\alpha }=1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3{\text{\hspace{0.17em}cos}}^{\text{2}}\mathrm{\alpha }=1\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}cos}}^{\text{2}}\mathrm{\alpha }=\frac{1}{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}cos}\mathrm{\alpha }=\frac{1}{\sqrt{3}}\\ \text{Hence, the direction cosines of the vector which are equally}\\ \text{inclined to the axes are}\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}.\end{array}$

Q.68

$\begin{array}{l}\text{If either vector}\stackrel{\to }{\mathrm{a}}=0\text{or}\stackrel{\to }{\mathrm{b}}=0,\text{then}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=0.\text{But the converse}\\ \text{need not be true. Justify your answer with an example.}\end{array}$

Ans.

$\begin{array}{l}\text{Let}\stackrel{\to }{\mathrm{a}}=3\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\text{and}\stackrel{\to }{\mathrm{b}}=2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}-6\stackrel{^}{\mathrm{k}}\\ \text{Then,}\\ \text{\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=\left(3\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\right)\left(2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}-6\stackrel{^}{\mathrm{k}}\right)\end{array}$

$\begin{array}{l}=6+12-18\\ =0\\ \text{And}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}|\stackrel{\to }{\mathrm{a}}|=|3\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}|\\ =\sqrt{{3}^{2}+{4}^{2}+{3}^{2}}\\ =\sqrt{9+16+9}\\ =\sqrt{34}\ne 0\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}|\stackrel{\to }{\mathrm{b}}|=|2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}-6\stackrel{^}{\mathrm{k}}|\\ =\sqrt{{2}^{2}+{3}^{2}+{\left(-6\right)}^{2}}\\ =\sqrt{4+9+36}\\ =\sqrt{49}\\ =7\ne 0\\ \text{Hence, the converse of the given statement need not be true.}\end{array}$

Q.69

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\text{k}},\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=3\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+7\stackrel{^}{\mathrm{k}}\mathrm{and}\stackrel{\to }{\mathrm{c}}=2\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}.\mathrm{Find}\mathrm{a}\\ \mathrm{vector}\stackrel{\to }{\mathrm{d}}\text{\hspace{0.17em}}\mathrm{which}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{both}\stackrel{\to }{\mathrm{a}}\stackrel{}{\mathrm{a}}\mathrm{nd}\stackrel{\to }{\mathrm{b}},\stackrel{\to }{\mathrm{c}}.\stackrel{\to }{\mathrm{d}}=15.\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{vectors}\mathrm{are}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}},\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=3\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+7\stackrel{^}{\mathrm{k}}\mathrm{and}\\ \stackrel{\to }{\mathrm{c}}=2\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}.\text{Let}\stackrel{\to }{\mathrm{d}}={\mathrm{d}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{d}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{d}}_{3}\stackrel{^}{\mathrm{k}}\\ \mathrm{Since},\text{}\stackrel{\to }{\mathrm{d}}\text{\hspace{0.17em}\hspace{0.17em}is perpendicular to}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}\hspace{0.17em}and}\stackrel{\to }{\mathrm{b}},\text{so}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{d}}=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\left(\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right).\left({\mathrm{d}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{d}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{d}}_{3}\stackrel{^}{\mathrm{k}}\right)=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{d}}_{1}+4{\mathrm{d}}_{2}+2{\mathrm{d}}_{3}=0\text{\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{d}}=0\\ ⇒\left(3\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+7\stackrel{^}{\mathrm{k}}\right).\left({\mathrm{d}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{d}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{d}}_{3}\stackrel{^}{\mathrm{k}}\right)=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3{\mathrm{d}}_{1}-2{\mathrm{d}}_{2}+7{\mathrm{d}}_{3}=0\text{\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \mathrm{But},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{d}}=15\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(2\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\right).\left({\mathrm{d}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{d}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{d}}_{3}\stackrel{^}{\mathrm{k}}\right)=15\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2{\mathrm{d}}_{1}-{\mathrm{d}}_{2}+4{\mathrm{d}}_{3}=15\text{\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{iii}\right)\\ \text{On solving (i), (ii), and (iii), we get:}\\ {\mathrm{d}}_{1}=\frac{160}{3},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{d}}_{2}=-\frac{5}{3},{\mathrm{d}}_{3}=-\frac{70}{3}\\ \therefore \stackrel{\to }{\mathrm{d}}=\frac{160}{3}\stackrel{^}{\mathrm{i}}-\frac{5}{3}\stackrel{^}{\mathrm{j}}-\frac{70}{3}\stackrel{^}{\mathrm{k}}=\frac{1}{3}\left(160\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-5\stackrel{^}{\mathrm{j}}-70\stackrel{^}{\mathrm{k}}\right)\\ \text{Hence, the required vector is\hspace{0.17em}}\frac{1}{3}\left(160\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-5\stackrel{^}{\mathrm{j}}-70\stackrel{^}{\mathrm{k}}\right).\end{array}$

Q.70

$\begin{array}{l}\mathrm{The}\mathrm{scalar}\mathrm{product}\mathrm{of}\mathrm{the}\mathrm{vector}\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\text{k}}\mathrm{with}\mathrm{a}\mathrm{unit}\mathrm{vector}\\ \mathrm{along}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{vectors}2\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}-5\stackrel{^}{\mathrm{k}}\mathrm{and}\mathrm{\lambda }\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\mathrm{is}\mathrm{equal}\\ \mathrm{to}\mathrm{one}.\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{\lambda }.\end{array}$

Ans.

$\begin{array}{l}\mathrm{Sum}\mathrm{of}\mathrm{vectors}2\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}-5\stackrel{^}{\mathrm{k}}\mathrm{and}\mathrm{\lambda }\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}=2\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}-5\stackrel{^}{\mathrm{k}}+\mathrm{\lambda }\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(2+\mathrm{\lambda }\right)\stackrel{^}{\mathrm{i}}+6\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}-2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ \mathrm{Unit}\text{vector of}\left(2+\mathrm{\lambda }\right)\stackrel{^}{\mathrm{i}}+6\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}-2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}=\frac{\left(2+\mathrm{\lambda }\right)\stackrel{^}{\mathrm{i}}+6\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}-2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}}{|\left(2+\mathrm{\lambda }\right)\stackrel{^}{\mathrm{i}}+6\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}-2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}|}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(2+\mathrm{\lambda }\right)\stackrel{^}{\mathrm{i}}+6\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}-2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}}{\sqrt{{\left(2+\mathrm{\lambda }\right)}^{2}+{6}^{2}+{\left(-2\right)}^{2}}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(2+\mathrm{\lambda }\right)\stackrel{^}{\mathrm{i}}+6\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}-2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}}{\sqrt{4+4\mathrm{\lambda }+{\mathrm{\lambda }}^{2}+36+4}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(2+\mathrm{\lambda }\right)\stackrel{^}{\mathrm{i}}+6\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}-2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}}{\sqrt{44+4\mathrm{\lambda }+{\mathrm{\lambda }}^{2}}}\\ \mathrm{Scalar}\text{product of}\mathrm{unit}\text{vector and}\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)=1\left[\mathrm{Given}\right]\\ \left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right).\frac{\left(2+\mathrm{\lambda }\right)\stackrel{^}{\mathrm{i}}+6\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}–2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}}{\sqrt{44+4\mathrm{\lambda }+{\mathrm{\lambda }}^{2}}}=1\\ \left(2+\mathrm{\lambda }\right)+6–2=\sqrt{44+4\mathrm{\lambda }+{\mathrm{\lambda }}^{2}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(6+\mathrm{\lambda }\right)}^{2}=44+4\mathrm{\lambda }+{\mathrm{\lambda }}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}36+12\mathrm{\lambda }+{\mathrm{\lambda }}^{2}=44+4\mathrm{\lambda }+{\mathrm{\lambda }}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}8\mathrm{\lambda }=8\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\lambda }=1\\ \mathrm{Thus},\mathrm{the}\mathrm{value}\mathrm{of}\text{\hspace{0.17em}}\mathrm{\lambda }\mathrm{is}1.\end{array}$

Q.71

$\begin{array}{l}\mathrm{If}\stackrel{\to }{\mathrm{a}},\stackrel{\to }{\mathrm{b}},\stackrel{\to }{\mathrm{c}}\mathrm{are}\mathrm{mutually}\mathrm{perpendicular}\mathrm{vector}\mathrm{of}\mathrm{equal}\mathrm{magnitudes},\\ \mathrm{show}\mathrm{that}\mathrm{the}\mathrm{vector}\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}\mathrm{is}\mathrm{equally}\mathrm{inclined}\mathrm{to}\stackrel{\to }{\mathrm{a}},\stackrel{\to }{\mathrm{b}}\mathrm{and}\stackrel{\to }{\mathrm{c}}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{Since},\text{}\stackrel{\to }{\mathrm{a}},\stackrel{\to }{\mathrm{b}},\stackrel{\to }{\mathrm{c}}\mathrm{are}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{mutually}\mathrm{perpendicular}\mathrm{vectors},\text{\hspace{0.17em}}\mathrm{so}\\ \stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}=\stackrel{\to }{\mathrm{c}}.\stackrel{\to }{\mathrm{a}}=0\\ \mathrm{Magnitudes}\text{â€‹ of}\stackrel{\to }{\mathrm{a}},\stackrel{\to }{\mathrm{b}},\stackrel{\to }{\mathrm{c}}\mathrm{are}\text{equal, so}\\ |\stackrel{\to }{\mathrm{a}}|=|\stackrel{\to }{\mathrm{b}}|=|\stackrel{\to }{\mathrm{c}}|\\ \mathrm{Let}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}},\stackrel{\to }{\mathrm{b}},\stackrel{\to }{\mathrm{c}}\mathrm{are}\text{making angles}{\mathrm{\theta }}_{1}\text{,\hspace{0.17em}}{\mathrm{\theta }}_{2}\text{and}{\mathrm{\theta }}_{3}\text{with\hspace{0.17em}}\left(\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}\right)\\ \text{respectively.Then, we have}\\ \text{cos}{\mathrm{\theta }}_{1}=\frac{\left(\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}\right).\stackrel{\to }{\mathrm{a}}}{|\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}||\stackrel{\to }{\mathrm{a}}|}=\frac{{|\stackrel{\to }{\mathrm{a}}|}^{2}}{|\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}||\stackrel{\to }{\mathrm{a}}|}=\frac{|\stackrel{\to }{\mathrm{a}}|}{|\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}|}\\ \text{cos}{\mathrm{\theta }}_{2}=\frac{\left(\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}\right).\stackrel{\to }{\mathrm{b}}}{|\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}||\stackrel{\to }{\mathrm{b}}|}=\frac{{|\stackrel{\to }{\mathrm{b}}|}^{2}}{|\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}||\stackrel{\to }{\mathrm{b}}|}=\frac{|\stackrel{\to }{\mathrm{b}}|}{|\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}|}\\ \text{cos}{\mathrm{\theta }}_{3}=\frac{\left(\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}\right).\stackrel{\to }{\mathrm{c}}}{|\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}||\stackrel{\to }{\mathrm{c}}|}=\frac{{|\stackrel{\to }{\mathrm{c}}|}^{2}}{|\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}||\stackrel{\to }{\mathrm{c}}|}=\frac{|\stackrel{\to }{\mathrm{c}}|}{|\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}|}\\ \mathrm{Since},\text{\hspace{0.17em}}|\stackrel{\to }{\mathrm{a}}|=|\stackrel{\to }{\mathrm{b}}|=|\stackrel{\to }{\mathrm{c}}|,\text{so cos}{\mathrm{\theta }}_{1}=\text{cos}{\mathrm{\theta }}_{2}=\text{cos}{\mathrm{\theta }}_{3}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{\theta }}_{1}={\mathrm{\theta }}_{2}={\mathrm{\theta }}_{3}\\ \text{Hence, the vector}\left(\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}\right)\text{\hspace{0.17em}\hspace{0.17em}is equally inclined to\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{a}},\stackrel{\to }{\mathrm{b}}\text{and}\stackrel{\to }{\mathrm{c}}.\end{array}$

Q.72

$\mathrm{Prove}\mathrm{that}\left(\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}\right).\left(\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}\right)={|\stackrel{\to }{\mathrm{a}}|}^{2}+{|\stackrel{\to }{\mathrm{b}}|}^{2},\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{\mathrm{a}},\stackrel{\to }{\mathrm{b}}\mathrm{are}\mathrm{perpendicular},\mathrm{given}\stackrel{\to }{\mathrm{a}}\ne 0,\stackrel{\to }{\mathrm{b}}\ne 0.$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\left(\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}\right).\left(\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}\right)\\ \text{\hspace{0.17em}}=\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\\ \text{\hspace{0.17em}}={|\stackrel{\to }{\mathrm{a}}|}^{2}+2\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}+{|\stackrel{\to }{\mathrm{b}}|}^{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[âˆµ\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}\right]\\ \text{\hspace{0.17em}}={|\stackrel{\to }{\mathrm{a}}|}^{2}+2\left(0\right)+{|\stackrel{\to }{\mathrm{b}}|}^{2}\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=0\text{as}\stackrel{\to }{\mathrm{a}}\text{and}\stackrel{\to }{\mathrm{b}}\text{are}\\ \text{perpendicular to each other.}\end{array}\right]\\ \text{\hspace{0.17em}}={|\stackrel{\to }{\mathrm{a}}|}^{2}+{|\stackrel{\to }{\mathrm{b}}|}^{2}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.73

$\begin{array}{l}\mathrm{If}\mathrm{\theta }\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{two}\mathrm{vectors}\stackrel{\to }{\mathrm{a}}\mathrm{and}\stackrel{\to }{\mathrm{b}},\mathrm{then}\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}\ge 0\\ \mathrm{only}\mathrm{when}\\ \left(\mathrm{A}\right)0<\mathrm{\theta }<\frac{\mathrm{\pi }}{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \left(\mathrm{B}\right)0\le \mathrm{\theta }\le \frac{\mathrm{\pi }}{2}\\ \left(\mathrm{C}\right)0<\mathrm{\theta }<\mathrm{\pi }\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \left(\mathrm{D}\right)0\le \mathrm{\theta }\le \mathrm{\pi }\end{array}$

Ans.

$\begin{array}{l}\mathrm{If}\mathrm{\theta }\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{two}\mathrm{vectors}\stackrel{\to }{\mathrm{a}}\mathrm{and}\stackrel{\to }{\mathrm{b}},\\ \mathrm{then}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}\ge 0\\ |\stackrel{\to }{\mathrm{a}}|.|\stackrel{\to }{\mathrm{b}}|\mathrm{cos\theta }\ge 0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{cos\theta }\ge 0\left[âˆµ|\stackrel{\to }{\mathrm{a}}|\ne 0,\text{\hspace{0.17em}}|\stackrel{\to }{\mathrm{b}}|\ne 0\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}0\le \mathrm{\theta }\le \frac{\mathrm{\pi }}{2}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}\ge 0\text{if \hspace{0.17em}\hspace{0.17em}}0\le \mathrm{\theta }\le \frac{\mathrm{\pi }}{2}.\\ \mathrm{Thus},\text{option B is correct.}\end{array}$

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Ans.

$\begin{array}{l}\text{The vertices of ΔABC are given as A (1, 2, 3), B (–1, 0, 0),}\\ \text{and C (0, 1, 2)}\text{.}\end{array}$

$\begin{array}{l}\text{Also, it is given that}\angle \text{ABC is the angle between the vectors}\text{\hspace{0.17em}}\stackrel{\to }{\text{BA}}\\ \text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\stackrel{\to }{\text{BC}}\text{.}\\ \therefore \stackrel{\to }{\text{BA}}\text{=}\left(\text{1 + 1}\right)\stackrel{^}{\text{i}}\text{+}\left(\text{2 – 0}\right)\stackrel{^}{\text{j}}\text{+}\left(\text{3 – 0}\right)\stackrel{^}{\text{k}}\\ \text{}\text{\hspace{0.17em}}\text{= 2}\stackrel{^}{\text{i}}\text{+ 2}\stackrel{^}{\text{j}}\text{+ 3}\text{\hspace{0.17em}}\stackrel{^}{\text{k}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\stackrel{\to }{\text{BC}}\text{=}\left(\text{0 + 1}\right)\stackrel{^}{\text{i}}\text{+}\left(\text{1 – 0}\right)\stackrel{^}{\text{j}}\text{+}\left(\text{2 – 0}\right)\stackrel{^}{\text{k}}\\ \text{}\text{\hspace{0.17em}}\text{=}\stackrel{^}{\text{i}}\text{+}\stackrel{^}{\text{j}}\text{+ 2}\stackrel{^}{\text{k}}\\ \therefore \stackrel{\to }{\text{BA}}\text{\hspace{0.17em}}\cdot \text{\hspace{0.17em}}\stackrel{\to }{\text{BC}}\\ \text{}\text{\hspace{0.17em}}\text{=}\left(\text{2}\stackrel{^}{\text{i}}\text{+ 2}\stackrel{^}{\text{j}}\text{+ 3}\text{\hspace{0.17em}}\stackrel{^}{\text{k}}\right)\cdot \left(\stackrel{^}{\text{i}}\text{+}\stackrel{^}{\text{j}}\text{+ 2}\stackrel{^}{\text{k}}\right)\\ \text{}\text{\hspace{0.17em}}\text{= 2 + 2 + 6}\\ \text{}\text{\hspace{0.17em}}\text{= 10}\\ |\stackrel{\to }{\text{BA}}|\text{=}|\text{2}\stackrel{^}{\text{i}}\text{+ 2}\stackrel{^}{\text{j}}\text{+ 3}\text{\hspace{0.17em}}\stackrel{^}{\text{k}}|\\ \text{}\text{\hspace{0.17em}}\text{=}\sqrt{{\text{2}}^{\text{2}}{\text{+ 2}}^{\text{2}}{\text{+ 3}}^{\text{2}}}\\ \text{}\text{\hspace{0.17em}}\text{=}\sqrt{\text{4 + 4 + 9}}\\ \text{}\text{\hspace{0.17em}}\text{=}\sqrt{\text{17}}\\ |\stackrel{\to }{\text{BC}}|\text{=}|\stackrel{^}{\text{i}}\text{+}\stackrel{^}{\text{j}}\text{+ 2}\stackrel{^}{\text{k}}|\\ \text{}\text{\hspace{0.17em}}\text{=}\sqrt{{\text{1}}^{\text{2}}{\text{+ 1}}^{\text{2}}{\text{+ 2}}^{\text{2}}}\\ \text{}\text{\hspace{0.17em}}\text{=}\sqrt{\text{1 + 1 + 4}}\\ \text{}\text{\hspace{0.17em}}\text{=}\sqrt{\text{6}}\\ \text{Now, it is known that:}\text{\hspace{0.17em}}\\ \stackrel{\to }{\text{BA}}\text{\hspace{0.17em}}\cdot \text{\hspace{0.17em}}\stackrel{\to }{\text{BC}}\text{=}|\stackrel{\to }{\text{BA}}|\cdot |\stackrel{\to }{\text{BC}}|\text{cos}\left(\angle \text{ABC}\right)\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{10 =}\sqrt{\text{17}}\sqrt{\text{6}}\text{cos}\left(\angle \text{ABC}\right)\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}\text{cos}\left(\angle \text{ABC}\right)\text{=}\frac{\text{10}}{\sqrt{\text{102}}}\\ \text{}\angle {\text{ABC = cos}}^{\text{–1}}\left(\frac{\text{10}}{\sqrt{\text{102}}}\right)\end{array}$

Q.75

$\begin{array}{l}\text{If the vertices A, B, C of a triangle ABC are}\left(\text{1,2,3}\right)\text{,}\left(\text{-1,0,0}\right)\text{,}\\ \left(\text{0,1,2}\right)\text{respectively, then find}\angle \text{ABC.}\\ \left[\angle \text{ABC is the angle between the vectors}\stackrel{\to }{\text{BA}}\text{and}\stackrel{\to }{\text{BC}}\right].\end{array}$

Ans.

$\begin{array}{l}\text{The vertices of ΔABC are given as A (1, 2, 3), B (–1, 0, 0),}\\ \text{and C (0, 1, 2).}\end{array}$

$\begin{array}{l}\text{Also, it is given that}\angle \text{ABC is the angle between the vectors\hspace{0.17em}}\stackrel{\to }{\text{BA}}\\ \text{and\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\text{BC}}\text{.}\\ \therefore \stackrel{\to }{\text{BA}}\text{=}\left(\text{1 + 1}\right)\stackrel{^}{\text{i}}\text{+}\left(\text{2 – 0}\right)\stackrel{^}{\text{j}}\text{+}\left(\text{3 – 0}\right)\stackrel{^}{\text{k}}\\ \text{\hspace{0.17em}= 2}\stackrel{^}{\text{i}}\text{+ 2}\stackrel{^}{\text{j}}\text{+ 3\hspace{0.17em}}\stackrel{^}{\text{k}}\\ \text{\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\text{BC}}\text{=}\left(\text{0 + 1}\right)\stackrel{^}{\text{i}}\text{+}\left(\text{1 – 0}\right)\stackrel{^}{\text{j}}\text{+}\left(\text{2 – 0}\right)\stackrel{^}{\text{k}}\\ \text{\hspace{0.17em}=}\stackrel{^}{\text{i}}\text{+}\stackrel{^}{\text{j}}\text{+ 2}\stackrel{^}{\text{k}}\\ \therefore \stackrel{\to }{\text{BA}}\text{\hspace{0.17em}}\cdot \text{\hspace{0.17em}}\stackrel{\to }{\text{BC}}\\ \text{\hspace{0.17em}=}\left(\text{2}\stackrel{^}{\text{i}}\text{+ 2}\stackrel{^}{\text{j}}\text{+ 3\hspace{0.17em}}\stackrel{^}{\text{k}}\right)\cdot \left(\stackrel{^}{\text{i}}\text{+}\stackrel{^}{\text{j}}\text{+ 2}\stackrel{^}{\text{k}}\right)\\ \text{\hspace{0.17em}= 2 + 2 + 6}\\ \text{\hspace{0.17em}= 10}\\ |\stackrel{\to }{\text{BA}}|\text{=}|\text{2}\stackrel{^}{\text{i}}\text{+ 2}\stackrel{^}{\text{j}}\text{+ 3\hspace{0.17em}}\stackrel{^}{\text{k}}|\\ \text{\hspace{0.17em}=}\sqrt{{\text{2}}^{\text{2}}{\text{+ 2}}^{\text{2}}{\text{+ 3}}^{\text{2}}}\\ \text{\hspace{0.17em}=}\sqrt{\text{4 + 4 + 9}}\\ \text{\hspace{0.17em}=}\sqrt{\text{17}}\\ |\stackrel{\to }{\text{BC}}|\text{=}|\stackrel{^}{\text{i}}\text{+}\stackrel{^}{\text{j}}\text{+ 2}\stackrel{^}{\text{k}}|\\ \text{\hspace{0.17em}=}\sqrt{{\text{1}}^{\text{2}}{\text{+ 1}}^{\text{2}}{\text{+ 2}}^{\text{2}}}\\ \text{\hspace{0.17em}=}\sqrt{\text{1 + 1 + 4}}\\ \text{\hspace{0.17em}=}\sqrt{\text{6}}\\ \text{Now, it is known that:\hspace{0.17em}}\\ \stackrel{\to }{\text{BA}}\text{\hspace{0.17em}}\cdot \text{\hspace{0.17em}}\stackrel{\to }{\text{BC}}\text{=}|\stackrel{\to }{\text{BA}}|\cdot |\stackrel{\to }{\text{BC}}|\text{cos}\left(\angle \text{ABC}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}10 =}\sqrt{\text{17}}\sqrt{\text{6}}\text{cos}\left(\angle \text{ABC)}\end{array}$

$\begin{array}{l}\text{cos}\left(\angle \text{ABC}\right)\text{=}\frac{\text{10}}{\sqrt{\text{102}}}\\ \angle {\text{ABC = cos}}^{\text{–1}}\left(\frac{\text{10}}{\sqrt{\text{102}}}\right)\end{array}$

$\begin{array}{l}\text{\hspace{0.17em}}|\stackrel{\to }{\text{AC}}|\text{=}|\text{2}\stackrel{^}{\text{i}}\text{+ 8}\stackrel{^}{\text{j}}\text{– 8}\stackrel{^}{\text{k}}|\\ \text{=}\sqrt{{\text{2}}^{\text{2}}{\text{+ 8}}^{\text{2}}\text{+}{\left(\text{–8}\right)}^{\text{2}}}\\ \text{=}\sqrt{\text{4 + 64 + 64}}\\ \text{=}\sqrt{\text{132}}\text{= 2}\sqrt{\text{33}}\\ \therefore \text{\hspace{0.17em}}|\stackrel{\to }{\text{AC}}|\text{=}|\stackrel{\to }{\text{AB}}|\text{+}|\stackrel{\to }{\text{BC}}|\\ \text{Hence, the given points A, B, and C are collinear.}\end{array}$

Q.76

$\begin{array}{l}\text{Show that the vectors}2\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}},\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}-5\stackrel{^}{\mathrm{k}}\text{and}3\stackrel{^}{\mathrm{i}}-4\stackrel{^}{\mathrm{j}}-4\stackrel{^}{\mathrm{k}}\\ \text{form the vertices of a right angled triangle.}\end{array}$

Ans.

$\begin{array}{l}\text{Let vectors}2\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}},\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}-5\stackrel{^}{\mathrm{k}}\mathrm{a}\mathrm{nd}3\stackrel{^}{\mathrm{i}}-4\stackrel{^}{\mathrm{j}}-4\stackrel{^}{\mathrm{k}}\text{be position vectors}\\ \text{of points A, B and C respectively.}\\ \text{i.e.,}\stackrel{\to }{\mathrm{OA}}=2\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}},\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{OB}}=\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}-5\stackrel{^}{\mathrm{k}},\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{OC}}=3\stackrel{^}{\mathrm{i}}-4\stackrel{^}{\mathrm{j}}-4\stackrel{^}{\mathrm{k}}\\ \therefore \stackrel{\to }{\mathrm{AB}}=\left(1-2\right)\stackrel{^}{\mathrm{i}}+\left(-3+1\right)\stackrel{^}{\mathrm{j}}+\left(-5-1\right)\stackrel{^}{\mathrm{k}}=-\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}-6\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{BC}}=\left(3-1\right)\stackrel{^}{\mathrm{i}}+\left(-4+3\right)\stackrel{^}{\mathrm{j}}+\left(-4+5\right)\stackrel{^}{\mathrm{k}}=2\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{CA}}=\left(2-3\right)\stackrel{^}{\mathrm{i}}+\left(-1+4\right)\stackrel{^}{\mathrm{j}}+\left(1+4\right)\stackrel{^}{\mathrm{k}}=-\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+5\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ |\stackrel{\to }{\mathrm{AB}}|=|-\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}-6\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\left(-1\right)}^{2}+{\left(-2\right)}^{2}+{\left(-6\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{1+4+36}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{41}\\ |\stackrel{\to }{\mathrm{BC}}|=|2\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}|\end{array}$

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{2}^{2}+{\left(-1\right)}^{2}+{1}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{4+1+1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{6}\\ |\stackrel{\to }{\mathrm{CA}}|=|-\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+5\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\left(-1\right)}^{2}+{3}^{2}+{5}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{1+9+25}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{35}\\ \therefore {|\stackrel{\to }{\mathrm{BC}}|}^{2}+{|\stackrel{\to }{\mathrm{CA}}|}^{2}={\left(\sqrt{6}\right)}^{2}+{\left(\sqrt{35}\right)}^{2}\end{array}$

$\begin{array}{l}=6+35\\ =41\\ ={|\stackrel{\to }{\mathrm{AB}}|}^{2}\\ \text{Hence,}\mathrm{\Delta }\text{ABC is a right-angled triangle.}\end{array}$

Q.77

$\begin{array}{l}\text{If}\stackrel{\to }{\text{a}}\text{is a non zero vector of magnitude ‘a’ and λ a non zero}\\ \text{scalar, then λ\hspace{0.17em}}\stackrel{\to }{\text{a}}\text{is unit vector if}\\ \left(\text{A}\right)\text{\hspace{0.17em}\hspace{0.17em}λ = 1}\\ \left(\text{B}\right)\text{\hspace{0.17em}\hspace{0.17em}λ = –1}\\ \left(\text{C}\right)\text{\hspace{0.17em}a =}|\text{λ}|\text{}\\ \left(\text{D}\right)\text{a =}\frac{\text{1}}{|\text{λ}|}\end{array}$

Ans.

$\begin{array}{l}\text{Vector λ}\stackrel{\to }{\text{a}}\text{\hspace{0.17em}\hspace{0.17em}is a unit vector if}|\text{λ}\stackrel{\to }{\text{a}}|\text{= 1.}\\ \text{So,\hspace{0.17em}\hspace{0.17em}}|\text{λ}\stackrel{\to }{\text{a}}|\text{= 1}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}|\text{λ}||\stackrel{\to }{\text{a}}|\text{= 1}\end{array}$

$\begin{array}{l}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}|\stackrel{\to }{\text{a}}|\text{=}\frac{\text{1}}{|\text{λ}|}\left[\text{λ}\ne \text{0}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}a =}\frac{\text{1}}{|\text{λ}|}\left[|\stackrel{\to }{\text{a}}|\text{= a}\right]\\ \text{Hence, λ}\stackrel{\to }{\text{a}}\text{\hspace{0.17em}vector is a unit vector if\hspace{0.17em}a =}\frac{\text{1}}{|\text{λ}|}\text{.}\\ \text{The correct answer is D.}\end{array}$

Q.78

$\begin{array}{l}\text{Let â€‹â€„â€‹â€„and â€‹â€„â€‹â€‹â€„beâ€„â€‹two â€‹â€„unit â€‹â€„vectors â€‹â€„andâ€„â€‹}\mathrm{\theta }\text{â€‹â€„isâ€„â€‹the â€‹angle â€‹â€„between}\\ \text{them}.\text{â€‹â€„Then}\stackrel{\to }{\mathrm{a}}\text{â€‹â€„â€‹â€„}+\stackrel{\to }{\mathrm{b}}\text{â€„â€‹â€‹â€„is â€‹â€„a â€‹â€„unit â€‹â€„vectorâ€„â€‹if}\\ \left(\mathrm{A}\right)\text{â€„}\mathrm{\theta }=\frac{\mathrm{\pi }}{4}\\ \left(\mathrm{B}\right)\text{â€„}\mathrm{\theta }=\frac{\mathrm{\pi }}{3}\\ \left(\mathrm{C}\right)\text{â€„}\mathrm{\theta }=\frac{\mathrm{\pi }}{2}\\ \left(\mathrm{D}\right)\text{â€„}\mathrm{\theta }=\frac{2\mathrm{\pi }}{3}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Let}\stackrel{\to }{\mathrm{a}}\mathrm{and}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{be}\mathrm{two}\mathrm{unit}\mathrm{vectors}\text{i.e.,}|\stackrel{\to }{\mathrm{a}}|=|\stackrel{\to }{\mathrm{b}}|=1\\ \mathrm{and}\mathrm{\theta }\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{between}\text{}\mathrm{them}.\\ \mathrm{Then}\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}\mathrm{is}\mathrm{a}\mathrm{unit}\mathrm{vector}\text{if}|\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}|=1.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}|\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}|=1⇒{|\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}|}^{2}=1\\ ⇒\left(\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}\right).\left(\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}\right)=1\\ ⇒\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{|\stackrel{\to }{\mathrm{a}}|}^{2}\text{\hspace{0.17em}}+2\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}+\text{\hspace{0.17em}}{|\stackrel{\to }{\mathrm{b}}|}^{2}=1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1+2\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}+1=1\left[âˆµ|\stackrel{\to }{\mathrm{a}}|=|\stackrel{\to }{\mathrm{b}}|=1\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=-\frac{1}{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}|\stackrel{\to }{\mathrm{a}}|\text{\hspace{0.17em}}.\text{\hspace{0.17em}}|\stackrel{\to }{\mathrm{b}}|\mathrm{cos\theta }=-\frac{1}{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1\text{\hspace{0.17em}}.\text{\hspace{0.17em}}1\mathrm{cos\theta }=-\frac{1}{2}\left[âˆµ|\stackrel{\to }{\mathrm{a}}|=|\stackrel{\to }{\mathrm{b}}|=1\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{cos\theta }=\mathrm{cos}\frac{2\mathrm{\pi }}{3}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\theta }=\frac{2\mathrm{\pi }}{3}\\ \mathrm{Thus},\text{}\left(\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}\right)\text{is unit vector if}\mathrm{\theta }=\frac{2\mathrm{\pi }}{3}.\\ \mathrm{Then},\text{the correct option is D.}\end{array}$

Q.79

$\begin{array}{l}\text{Theâ€„â€‹valueâ€‹â€„ofâ€„}\stackrel{^}{\mathrm{i}}.\text{â€„â€‹}\left(\stackrel{^}{\mathrm{j}}\text{â€„}×\text{â€„â€‹}\stackrel{^}{\mathrm{k}}\right)\text{â€‹}+\text{â€‹}\stackrel{^}{\mathrm{j}}.\text{â€„â€‹}\left(\stackrel{^}{\mathrm{i}}\text{â€‹â€„}×\text{â€„}\stackrel{^}{\mathrm{k}}\right)+\text{â€‹}\stackrel{^}{\mathrm{k}}.\text{â€‹}\left(\stackrel{^}{\mathrm{i}}\text{â€‹â€„}×\text{â€„â€‹}\stackrel{^}{\mathrm{j}}\right)\text{â€‹â€„is}\\ \begin{array}{l}\left(\text{A}\right)\text{â€„}0\\ \left(\text{B}\right)\text{â€„}-1\\ \left(\text{C}\right)\text{â€„}1\\ \left(\text{D}\right)\text{â€„}3\end{array}\end{array}$

Ans.

$\begin{array}{l}\stackrel{^}{\mathrm{i}}.\left(\stackrel{^}{\mathrm{j}}×\stackrel{^}{\mathrm{k}}\right)+\stackrel{^}{\mathrm{j}}.\left(\stackrel{^}{\mathrm{i}}×\stackrel{^}{\mathrm{k}}\right)+\stackrel{^}{\mathrm{k}}.\left(\stackrel{^}{\mathrm{i}}×\stackrel{^}{\mathrm{j}}\right)=\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}.\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}.\left(-\stackrel{^}{\mathrm{j}}\right)+\stackrel{^}{\mathrm{k}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\left[\begin{array}{l}âˆµ\stackrel{^}{\mathrm{j}}×\stackrel{^}{\mathrm{k}}=\stackrel{^}{\mathrm{i}},\text{â€„}\stackrel{^}{\mathrm{i}}×\stackrel{^}{\mathrm{k}}=\stackrel{^}{\mathrm{j}}\\ \stackrel{^}{\mathrm{i}}×\stackrel{^}{\mathrm{j}}=\stackrel{^}{\mathrm{k}}\end{array}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1-1+1\left[\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}.\stackrel{^}{\mathrm{i}}=\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}.\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}}=\stackrel{^}{\mathrm{k}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}=1\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1\\ \text{Thus, the correct option is C.}\end{array}$

Q.80

$\begin{array}{l}\mathrm{If}\text{â€„}\mathrm{\theta }\text{â€„}\mathrm{is}\text{â€„}\mathrm{the}\text{â€„}\mathrm{angle}\text{â€„}\mathrm{between}\text{â€„}\mathrm{any}\text{â€„}\mathrm{two}\text{â€„}\mathrm{vectors}\text{â€„}\stackrel{\to }{\mathrm{a}}\text{â€„}\mathrm{and}\text{â€„}\stackrel{\to }{\mathrm{b}},\mathrm{then}\\ \left|\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}\right|=\left|\stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{b}}\right|\mathrm{when}\text{â€„}\mathrm{\theta }\text{â€„}\mathrm{is}\text{â€„}\mathrm{equal}\text{â€„}\mathrm{to}\\ \left(\mathrm{A}\right)\text{â€„}0\\ \left(\mathrm{B}\right)\text{â€„}\frac{\mathrm{\pi }}{4}\\ \left(\mathrm{C}\right)\text{â€„}\frac{\mathrm{\pi }}{2}\\ \left(\mathrm{D}\right)\text{â€„}\mathrm{\pi }\end{array}$

Ans.

$\begin{array}{l}\text{Let}\mathrm{\theta }\text{be the angle between two vectors}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}and\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\text{and}|\stackrel{\to }{\mathrm{a}}|\ne 0,\text{\hspace{0.17em}}|\stackrel{\to }{\mathrm{b}}|\ne 0.\\ \mathrm{Given}:\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}|\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}|=|\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}|\\ ⇒|\stackrel{\to }{\mathrm{a}}|.|\stackrel{\to }{\mathrm{b}}|\mathrm{cos\theta }=|\stackrel{\to }{\mathrm{a}}|.|\stackrel{\to }{\mathrm{b}}|\mathrm{sin\theta }\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{cos\theta }=\mathrm{sin\theta }\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[|\stackrel{\to }{\mathrm{a}}|\text{and}|\stackrel{\to }{\mathrm{b}}|\text{are positive.}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{tan\theta }=1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{tan}\frac{\mathrm{\pi }}{4}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\theta }=\frac{\mathrm{\pi }}{4}\\ \mathrm{So},\text{the correct option is B.}\end{array}$