# 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 –Â  Ex 10.1 Solutions – 5 QuestionsÂ

Class 12 Mathematics Chapter 10 – Ex 10.2 Solutions – 19 Questions

Class 12 Mathematics Chapter 10 –Â  Ex 10.3 Solutions – 18 Questions

Class 12 Mathematics Chapter 10 – Ex 10.4 Solutions – 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.Â

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 Î¸.
• If Î¸ = 0Â°, meaning both S and T are in the same direction, then T = |S| |T|.
• If Î¸ = 90Â°, meaning S and T are orthogonal then T = 0.
• If we have two vectors S = (S1, S2, S3 â€¦.. Sn) and T = (T1, T2, T3 â€¦.. Tn) then the dot product is given as S.T = (S1T1 + S2T2 + S3T3 â€¦.. SnTn)

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

## NCERT Mathematics Syllabus

### Term – 1

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

### Â Term – 2

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

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 MathematicsÂ

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

$\mathrm{Find}|\stackrel{â†’}{\mathrm{a}}Ã—\stackrel{â†’}{\mathrm{b}}|,\mathrm{if}\stackrel{â†’}{\mathrm{a}}=\stackrel{^}{\mathrm{i}}âˆ’7\stackrel{^}{\mathrm{j}}+7\stackrel{^}{\text{k}}\mathrm{and}\stackrel{â†’}{\mathrm{b}}=3\text{â€‰}\stackrel{^}{\mathrm{i}}âˆ’2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}.$

Ans.

$\begin{array}{l}\mathrm{Given}\text{vectors are:}\stackrel{â†’}{\mathrm{a}}=\stackrel{^}{\mathrm{i}}âˆ’7\stackrel{^}{\mathrm{j}}+7\stackrel{^}{\mathrm{k}}\mathrm{and}\stackrel{â†’}{\mathrm{b}}=3\text{â€‰}\stackrel{^}{\mathrm{i}}âˆ’2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\\ \mathrm{Then},\\ \text{â€‰â€‰â€‰}\stackrel{â†’}{\mathrm{a}}Ã—\stackrel{â†’}{\mathrm{b}}=|\begin{array}{ccc}\stackrel{^}{\mathrm{i}}& \stackrel{^}{\mathrm{j}}& \stackrel{^}{\mathrm{k}}\\ 1& âˆ’7& 7\\ 3& âˆ’2& 2\end{array}|\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}=\left(âˆ’14+14\right)\stackrel{^}{\mathrm{i}}+\left(2âˆ’21\right)\stackrel{^}{\mathrm{j}}+\left(âˆ’2+21\right)\stackrel{^}{\mathrm{k}}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}=âˆ’19\stackrel{^}{\mathrm{j}}+19\stackrel{^}{\mathrm{k}}\\ âˆ´|\stackrel{â†’}{\mathrm{a}}Ã—\stackrel{â†’}{\mathrm{b}}|=|âˆ’19\stackrel{^}{\mathrm{j}}+19\stackrel{^}{\mathrm{k}}|\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}=\sqrt{{\left(âˆ’19\right)}^{2}+{\left(19\right)}^{2}}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}=2\sqrt{19}\end{array}$

Q.2

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

Ans.

$\begin{array}{l}\mathrm{The}\text{given vectors are}\stackrel{â†’}{\mathrm{a}}=3\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\mathrm{and}\stackrel{â†’}{\mathrm{b}}=\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}âˆ’2\stackrel{^}{\mathrm{k}}.\\ \mathrm{Then},\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}=3\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}+\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}âˆ’2\stackrel{^}{\mathrm{k}}\\ \text{â€‰â€‰â€‰â€‰}=4\text{â€‰}\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}\\ \mathrm{and}\text{â€‰â€‰â€‰}\stackrel{â†’}{\mathrm{a}}âˆ’\stackrel{â†’}{\mathrm{b}}=3\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}âˆ’\stackrel{^}{\mathrm{i}}âˆ’2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\\ \text{â€‰â€‰â€‰â€‰}=2\text{â€‰}\stackrel{^}{\mathrm{i}}+4\text{â€‰}\stackrel{^}{\mathrm{k}}\\ \left(\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}\right)Ã—\left(\stackrel{â†’}{\mathrm{a}}âˆ’\stackrel{â†’}{\mathrm{b}}\right)\\ \text{â€‰â€‰â€‰â€‰}=|\begin{array}{ccc}\stackrel{^}{\mathrm{i}}& \stackrel{^}{\mathrm{j}}& \stackrel{^}{\mathrm{k}}\\ 4& 4& 0\\ 2& 0& 4\end{array}|\\ \text{â€‰â€‰â€‰â€‰}=\left(16âˆ’0\right)\stackrel{^}{\mathrm{i}}âˆ’\left(16âˆ’0\right)\stackrel{^}{\mathrm{j}}+\left(0âˆ’8\right)\stackrel{^}{\mathrm{k}}\\ \text{â€‰â€‰â€‰â€‰}=16\text{â€‰â€‰}\stackrel{^}{\mathrm{i}}âˆ’16\text{â€‰}\stackrel{^}{\mathrm{j}}âˆ’8\text{â€‰}\stackrel{^}{\mathrm{k}}\\ |\left(\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}\right)Ã—\left(\stackrel{â†’}{\mathrm{a}}âˆ’\stackrel{â†’}{\mathrm{b}}\right)|\\ \text{â€‰â€‰â€‰â€‰}=|16\text{â€‰â€‰}\stackrel{^}{\mathrm{i}}âˆ’16\text{â€‰}\stackrel{^}{\mathrm{j}}âˆ’8\text{â€‰}\stackrel{^}{\mathrm{k}}|\\ \text{â€‰â€‰â€‰â€‰}=\sqrt{{\left(16\right)}^{2}+{\left(âˆ’16\right)}^{2}+{\left(âˆ’8\right)}^{2}}\\ \text{â€‰â€‰â€‰â€‰}=24\\ \text{Hence, the unit vector perpendicular to each of the vectors}\\ \left(\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}\right)\text{â€‰â€‰and}\left(\stackrel{â†’}{\mathrm{a}}âˆ’\stackrel{â†’}{\mathrm{b}}\right)\text{is given by the relation,}\\ \text{â€‰â€‰â€‰â€‰}=Â±\frac{\left(\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}\right)Ã—\left(\stackrel{â†’}{\mathrm{a}}âˆ’\stackrel{â†’}{\mathrm{b}}\right)}{|\left(\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}\right)Ã—\left(\stackrel{â†’}{\mathrm{a}}âˆ’\stackrel{â†’}{\mathrm{b}}\right)|}\\ \text{â€‰â€‰â€‰â€‰}=Â±\frac{16\text{â€‰â€‰}\stackrel{^}{\mathrm{i}}âˆ’16\text{â€‰}\stackrel{^}{\mathrm{j}}âˆ’8\text{â€‰}\stackrel{^}{\mathrm{k}}}{24}\\ \text{â€‰â€‰â€‰â€‰}=Â±\frac{2\text{â€‰â€‰}\stackrel{^}{\mathrm{i}}âˆ’2\text{â€‰}\stackrel{^}{\mathrm{j}}âˆ’\text{â€‰}\stackrel{^}{\mathrm{k}}}{3}\end{array}$

Q.3

$\begin{array}{l}\mathrm{If}\mathrm{a}\mathrm{unit}\mathrm{vector}\stackrel{â†’}{\mathrm{a}}\mathrm{makes}\mathrm{angles}\frac{\mathrm{Ï€}}{3}\mathrm{with}\stackrel{^}{\mathrm{i}},\frac{\mathrm{Ï€}}{4}\mathrm{with}\text{â€‰}\stackrel{^}{\mathrm{j}}\text{}\mathrm{and}\mathrm{an}\\ \mathrm{acute}\mathrm{angle}\mathrm{Î¸}\mathrm{with}\stackrel{^}{\text{k}},\mathrm{then}\mathrm{find}\mathrm{Î¸}\mathrm{and}\mathrm{hence},\mathrm{the}\mathrm{components}\mathrm{of}\stackrel{â†’}{\mathrm{a}}.\end{array}$

Ans.

$\begin{array}{l}\text{Let unit vector}\stackrel{â†’}{\mathrm{a}}\text{â€‰}={\text{a}}_{\text{1}}\stackrel{^}{\mathrm{i}}+{\text{a}}_{\text{2}}\stackrel{^}{\mathrm{j}}+{\text{a}}_{\text{3}}\stackrel{^}{\mathrm{k}}\text{â€‰â€‰}\mathrm{and}\text{â€‰â€‰}|\stackrel{â†’}{\mathrm{a}}|=1.\\ \text{It is given thatâ€‰}\stackrel{â†’}{\mathrm{a}}\text{makes angles}\frac{\mathrm{Ï€}}{3}\text{â€‰â€‰with}\stackrel{^}{\mathrm{i}}\text{â€‰},\text{â€‰}\frac{\mathrm{Ï€}}{4}\text{â€‰â€‰with}\stackrel{^}{\mathrm{j}}\text{and an acute}\\ \text{angle}\mathrm{Î¸}\text{withâ€‰}\stackrel{^}{\mathrm{k}}.\\ \text{Then, we have:}\\ \text{cos}\frac{\mathrm{Ï€}}{3}=\frac{{\mathrm{a}}_{1}}{|\stackrel{â†’}{\mathrm{a}}|}â‡’\frac{1}{2}=\frac{{\mathrm{a}}_{1}}{1}\\ â‡’{\mathrm{a}}_{1}=\frac{1}{2}\\ \text{cos}\frac{\mathrm{Ï€}}{4}=\frac{{\mathrm{a}}_{2}}{|\stackrel{â†’}{\mathrm{a}}|}â‡’\frac{1}{\sqrt{2}}=\frac{{\mathrm{a}}_{2}}{1}\\ â‡’{\mathrm{a}}_{2}=\frac{1}{\sqrt{2}}\\ \text{cos}\mathrm{Î¸}=\frac{{\mathrm{a}}_{3}}{|\stackrel{â†’}{\mathrm{a}}|}â‡’\text{cos}\mathrm{Î¸}=\frac{{\mathrm{a}}_{3}}{1}\\ â‡’{\mathrm{a}}_{3}=\text{cos}\mathrm{Î¸}\\ \mathrm{Since},\text{â€‰}|\stackrel{â†’}{\mathrm{a}}|=1\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\sqrt{{{\mathrm{a}}_{1}}^{2}+{{\mathrm{a}}_{2}}^{2}+{{\mathrm{a}}_{3}}^{2}}=1\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{{\mathrm{a}}_{1}}^{2}+{{\mathrm{a}}_{2}}^{2}+{{\mathrm{a}}_{3}}^{2}=1\\ {\left(\frac{1}{2}\right)}^{2}+{\left(\frac{1}{\sqrt{2}}\right)}^{2}+{\text{cos}}^{\text{2}}\mathrm{Î¸}=1\\ {\text{â€‰â€‰cos}}^{\text{2}}\mathrm{Î¸}=1âˆ’\frac{1}{4}âˆ’\frac{1}{2}\\ {\text{â€‰â€‰cos}}^{\text{2}}\mathrm{Î¸}=\frac{1}{4}\\ â‡’\text{â€‰â€‰â€‰cos}\mathrm{Î¸}=\frac{1}{2}=\mathrm{cos}\frac{\mathrm{Ï€}}{3}\\ \mathrm{So},\text{â€‰â€‰â€‰}\mathrm{Î¸}=\frac{\mathrm{Ï€}}{3}\\ \mathrm{Thus},\text{}\mathrm{Î¸}=\frac{\mathrm{Ï€}}{3}\text{and and the components of}\stackrel{â†’}{\mathrm{a}}\text{â€‰â€‰areâ€‰}\left(\frac{1}{2},\frac{1}{\sqrt{2}},\frac{1}{2}\right).\end{array}$

Q.4

$\mathrm{Show}\mathrm{that}\left(\stackrel{â†’}{\mathrm{a}}âˆ’\stackrel{â†’}{\mathrm{b}}\right)Ã—\left(\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}\right)=2\left(\stackrel{â†’}{\mathrm{a}}Ã—\stackrel{â†’}{\mathrm{b}}\right)$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\left(\stackrel{â†’}{\mathrm{a}}âˆ’\stackrel{â†’}{\mathrm{b}}\right)Ã—\left(\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}\right)\\ \text{â€‰}=\stackrel{â†’}{\mathrm{a}}Ã—\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{a}}Ã—\stackrel{â†’}{\mathrm{b}}âˆ’\stackrel{â†’}{\mathrm{b}}Ã—\stackrel{â†’}{\mathrm{a}}âˆ’\stackrel{â†’}{\mathrm{b}}Ã—\stackrel{â†’}{\mathrm{b}}\\ \text{â€‰}=0+\stackrel{â†’}{\mathrm{a}}Ã—\stackrel{â†’}{\mathrm{b}}+\stackrel{â†’}{\mathrm{a}}Ã—\stackrel{â†’}{\mathrm{b}}âˆ’0\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\left[\stackrel{â†’}{\mathrm{b}}Ã—\stackrel{â†’}{\mathrm{a}}=\stackrel{â†’}{\mathrm{a}}Ã—\stackrel{â†’}{\mathrm{b}}\right]\\ \text{â€‰}=2\left(\stackrel{â†’}{\mathrm{a}}Ã—\stackrel{â†’}{\mathrm{b}}\right)=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

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

$\begin{array}{l}\text{Here},\text{â€‰}\mathrm{vector}\text{}\stackrel{â†’}{\mathrm{OP}}\text{represents the displacement of 40 km, 30Â°}\\ \text{East of North.}\end{array}$

Q.6 Classify the following measures as scalars and vectors. (i) 10 kg
(ii) 2 meters north-west
(iii) 40Â°
(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) 40Â° is a scalar quantity as it involves only magnitude.
(iv) 40 watts is a scalar quantity as it involves only magnitude.
(v) 10â€“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 In fig.(a square), identify the following vectors. i) Coinitial
(ii) Equal
(iii) Collinear but not equal

Ans.

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

Q.9

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

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

$\mathrm{Find}\mathrm{Î»}\mathrm{and}\mathrm{Î¼}\mathrm{if}\left(2\stackrel{^}{\mathrm{i}}+6\stackrel{^}{\mathrm{j}}+27\stackrel{^}{\text{k}}\right)Ã—\left(\stackrel{^}{\mathrm{i}}+\mathrm{Î»}\stackrel{^}{\mathrm{j}}+\mathrm{Î¼}\text{â€‰}\stackrel{^}{\mathrm{k}}\right)=0.$

Ans.

$\begin{array}{l}\left(2\stackrel{^}{\mathrm{i}}+6\stackrel{^}{\mathrm{j}}+27\stackrel{^}{\mathrm{k}}\right)Ã—\left(\stackrel{^}{\mathrm{i}}+\mathrm{Î»}\stackrel{^}{\mathrm{j}}+\mathrm{Î¼}\mathrm{â€‰}\stackrel{^}{\mathrm{k}}\right)=0\\ â‡’|\begin{array}{ccc}\stackrel{^}{\mathrm{i}}& \stackrel{^}{\mathrm{j}}& \stackrel{^}{\mathrm{k}}\\ 2& 6& 27\\ 1& \mathrm{Î»}& \mathrm{Î¼}\end{array}|=0\\ \left(6\mathrm{Î¼}âˆ’27\mathrm{Î»}\right)\stackrel{^}{\mathrm{i}}âˆ’\left(2\mathrm{Î¼}âˆ’27\right)\stackrel{^}{\mathrm{j}}+\left(2\mathrm{Î»}âˆ’6\right)\mathrm{â€‰}\stackrel{^}{\mathrm{k}}=0\stackrel{^}{\mathrm{i}}+0\stackrel{^}{\mathrm{j}}+0\stackrel{^}{\mathrm{k}}\\ \mathrm{On}\mathrm{comparing}\mathrm{the}\mathrm{corresponding}\mathrm{components},\mathrm{we}\mathrm{have}:\\ 6\mathrm{Î¼}âˆ’27\mathrm{Î»}=0,â€‰â€‰2\mathrm{Î¼}âˆ’27=0\mathrm{and}2\mathrm{Î»}âˆ’6=0\\ â‡’\mathrm{Î¼}=\frac{27}{2}\mathrm{and}\mathrm{Î»}=3\end{array}$

Q.11

$\begin{array}{l}\text{Compute the magnitude of the following vectors:}\\ \stackrel{â†’}{\mathrm{a}}=\stackrel{^}{\mathrm{i}}\text{â€‰}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}};\text{â€‰}\stackrel{â†’}{\mathrm{b}}=2\stackrel{^}{\mathrm{i}}\text{â€‰}âˆ’7\stackrel{^}{\mathrm{j}}âˆ’3\text{â€‰}\stackrel{^}{\mathrm{k}};\text{â€‰â€‰â€‰}\stackrel{â†’}{\mathrm{c}}=\frac{1}{\sqrt{3}}\stackrel{^}{\mathrm{i}}\text{â€‰}+\frac{1}{\sqrt{3}}\stackrel{^}{\mathrm{j}}âˆ’\frac{1}{\sqrt{3}}\stackrel{^}{\mathrm{k}};\end{array}$

Ans.

$\begin{array}{l}\text{The given vectors are:}\\ \stackrel{â†’}{\mathrm{a}}=\stackrel{^}{\mathrm{i}}\text{â€‰}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}};\text{â€‰}\stackrel{â†’}{\mathrm{b}}=2\stackrel{^}{\mathrm{i}}\text{â€‰}âˆ’7\stackrel{^}{\mathrm{j}}âˆ’3\text{â€‰}\stackrel{^}{\mathrm{k}};\text{â€‰â€‰â€‰}\stackrel{â†’}{\mathrm{c}}=\frac{1}{\sqrt{3}}\stackrel{^}{\mathrm{i}}\text{â€‰}+\frac{1}{\sqrt{3}}\stackrel{^}{\mathrm{j}}âˆ’\frac{1}{\sqrt{3}}\stackrel{^}{\mathrm{k}}\\ \text{Magnitude of}\stackrel{â†’}{\mathrm{a}}=|\stackrel{â†’}{\mathrm{a}}|\\ \text{â€‰â€‰â€‰â€‰}=\sqrt{{\left(1\right)}^{2}+{\left(1\right)}^{2}+{\left(1\right)}^{2}}\\ \text{â€‰â€‰â€‰â€‰}=\sqrt{3}\\ \text{Magnitude of}\stackrel{â†’}{\mathrm{b}}=|\stackrel{â†’}{\mathrm{b}}|\\ \text{â€‰â€‰â€‰â€‰}=\sqrt{{\left(2\right)}^{2}+{\left(âˆ’7\right)}^{2}+{\left(âˆ’3\right)}^{2}}\\ \text{â€‰â€‰â€‰â€‰}=\sqrt{62}\\ \text{Magnitude of}\stackrel{â†’}{\mathrm{c}}=|\stackrel{â†’}{\mathrm{c}}|\\ \text{â€‰â€‰â€‰â€‰}=\sqrt{{\left(\frac{1}{\sqrt{3}}\right)}^{2}+{\left(\frac{1}{\sqrt{3}}\right)}^{2}+{\left(\frac{1}{\sqrt{3}}\right)}^{2}}\\ \text{â€‰â€‰â€‰â€‰}=\sqrt{\frac{3}{3}}=1\end{array}$

Q.12 Write two different vectors having same magnitude.Â
Ans.

$\begin{array}{l}\begin{array}{l}\text{Two different vectors are:}\end{array}\\ \begin{array}{l}\stackrel{â†’}{\mathrm{a}}\text{}=\stackrel{^}{\mathrm{i}}\text{}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}};\text{}\stackrel{â†’}{\mathrm{b}}\text{}=\text{}2\stackrel{^}{\mathrm{i}}\text{}+3\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\end{array}\\ \begin{array}{l}\text{Then}\end{array}\\ \text{Magnitudeof}\stackrel{â†’}{\mathrm{a}}\text{=}\left|\stackrel{^}{\mathrm{i}}\text{}+\text{}2\stackrel{^}{\mathrm{j}}+\text{}3\stackrel{^}{\mathrm{k}}\text{}\right|\\ \begin{array}{l}\text{=}\sqrt{{\text{1}}^{\text{2}}{\text{+2}}^{\text{2}}{\text{+3}}^{\text{2}}}\end{array}\\ \begin{array}{l}\text{=}\sqrt{\text{1+4+9}}\end{array}\\ \begin{array}{l}\text{=}\sqrt{\text{14}}\end{array}\\ \begin{array}{l}\text{Magnitude of}\stackrel{â†’}{\mathrm{b}}=\left|2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right|\end{array}\\ \begin{array}{l}\text{=}\sqrt{{\text{2}}^{\text{2}}{\text{+3}}^{\text{2}}{\text{+1}}^{\text{2}}}\end{array}\\ \begin{array}{l}\text{=}\sqrt{\text{4+9+1}}\end{array}\\ \begin{array}{l}\text{=}\sqrt{\text{14}}\end{array}\\ \begin{array}{l}\text{Hence,}\stackrel{â†’}{\mathrm{a}}\text{and}\stackrel{â†’}{\mathrm{b}}\text{aretwodifferentvectorshavingthe same}\end{array}\\ \begin{array}{l}\text{magnitude.}\end{array}\\ \begin{array}{l}\text{The vectors are different because they have different directions.}\end{array}\end{array}$

Q.13 Write two different vectors having same direction.
Ans.

$\begin{array}{l}\begin{array}{l}\text{Two different vectors are:}\end{array}\\ \begin{array}{l}\stackrel{â†’}{\mathrm{a}}=2\stackrel{^}{\mathrm{i}}\text{}+2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}};\text{}\stackrel{â†’}{\mathrm{b}}=\stackrel{^}{\mathrm{i}}\text{}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\end{array}\\ \begin{array}{l}\text{Then,}\end{array}\\ \begin{array}{l}\text{Magnitudeof}\stackrel{â†’}{\text{a}}\end{array}\text{}=\text{}\begin{array}{l}\left|2\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right|\end{array}\\ \begin{array}{l}=\text{}\sqrt{{2}^{2}+{2}^{2}+{2}^{2}}\end{array}\\ \begin{array}{l}=\text{}\sqrt{4+4+4}\end{array}\\ \begin{array}{l}=\text{}\sqrt{12}\end{array}\\ =\text{}\begin{array}{l}2\sqrt{3}\end{array}\\ \text{Direction cosinesof}\stackrel{â†’}{\mathrm{a}}\text{}\mathrm{are}:\\ \begin{array}{l}\frac{2}{2\sqrt{3}},\frac{2}{2\sqrt{3}},\frac{2}{2\sqrt{3}}\text{}\mathrm{i}.\mathrm{e}.,\text{}\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\end{array}\\ \begin{array}{l}\text{Magnitudeof}\stackrel{â†’}{\mathrm{b}}=\left|\stackrel{^}{\mathrm{i}}\text{â€‰}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right|\end{array}\\ \begin{array}{l}=\sqrt{{1}^{2}+{1}^{2}+{1}^{2}}\end{array}\\ \begin{array}{l}=\sqrt{1+1+1}\end{array}\\ \begin{array}{l}=\sqrt{3}\end{array}\\ \begin{array}{l}\text{Direction cosinesof}\stackrel{â†’}{\mathrm{b}}\text{are:}\end{array}\\ \begin{array}{l}\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\end{array}\\ \begin{array}{l}\text{Direction cosinesof}\stackrel{â†’}{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{â†’}{\mathrm{b}}\text{aresame.Hence,these two}\end{array}\\ \begin{array}{l}\text{vectorshavesamedirection.}\end{array}\end{array}$

Q.14

$\begin{array}{l}\text{Find the values of}\mathrm{x}\text{and}\mathrm{y}\text{so that the vectorsâ€‰â€‰}2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}\text{and}\\ \mathrm{x}\stackrel{^}{\mathrm{i}}+\mathrm{y}\stackrel{^}{\mathrm{j}}\text{are equal.}\end{array}$

Ans.

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

Q.15 Find the scalar and vector components of the vector with initial point (2, 1) and erminal point (â€“ 5, 7).

Ans.

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

Q.16

$\begin{array}{l}\mathrm{Given}\mathrm{that}\stackrel{â†’}{\mathrm{a}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{b}}=0\mathrm{and}\stackrel{â†’}{\mathrm{a}}\text{â€‰}Ã—\text{â€‰}\stackrel{â†’}{\mathrm{b}}=0.\mathrm{What}\mathrm{can}\mathrm{you}\mathrm{conclude}\\ \mathrm{about}\mathrm{the}\mathrm{vectors}\stackrel{â†’}{\mathrm{a}}\mathrm{and}\text{â€¹}\stackrel{â†’}{\mathrm{b}}\text{â€‰}?\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\stackrel{â†’}{\mathrm{a}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{b}}=0\\ \mathrm{Either}\text{}|\stackrel{â†’}{\mathrm{a}}|=0\text{or}|\stackrel{â†’}{\mathrm{b}}|=0,\\ \text{or}\stackrel{â†’}{\mathrm{a}}âŠ¥\stackrel{â†’}{\mathrm{b}}\text{â€‰}\left(\mathrm{if}\text{}\stackrel{â†’}{\mathrm{a}}\text{â€‰â€‰}\mathrm{and}\text{â€‰â€‰}\stackrel{â†’}{\mathrm{b}}\text{are non-zero.}\right)\\ \left(\mathrm{ii}\right)\stackrel{â†’}{\mathrm{a}}\text{â€‰}Ã—\text{â€‰}\stackrel{â†’}{\mathrm{b}}=0\\ \mathrm{Either}\text{}|\stackrel{â†’}{\mathrm{a}}|=0\text{or}|\stackrel{â†’}{\mathrm{b}}|=0,\text{}\\ \text{or}\stackrel{â†’}{\mathrm{a}}\text{â€‰â€‰}\text{â€‰â€‰}\stackrel{â†’}{\mathrm{b}}\text{â€‰}\left(\mathrm{if}\text{}\stackrel{â†’}{\mathrm{a}}\text{â€‰â€‰}\mathrm{and}\text{â€‰â€‰}\stackrel{â†’}{\mathrm{b}}\text{are non-zero.}\right)\\ \text{But,}\stackrel{â†’}{\mathrm{a}}\text{â€‰â€‰and}\stackrel{â†’}{\mathrm{b}}\text{cannot be perpendicular and parallel simultaneously.}\\ \text{Hence,â€‰â€¹}|\stackrel{â†’}{\mathrm{a}}|=0\text{or}|\stackrel{â†’}{\mathrm{b}}|=0.\end{array}$

Q.17

$\begin{array}{l}\text{Find the sum of the vectors}\stackrel{â†’}{\mathrm{a}}=\stackrel{^}{\mathrm{i}}âˆ’2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}},\text{â€‰}\stackrel{â†’}{\mathrm{b}}=âˆ’2\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}+5\stackrel{^}{\mathrm{k}}\\ \text{and}\stackrel{â†’}{\mathrm{c}}=\stackrel{^}{\mathrm{i}}âˆ’6\stackrel{^}{\mathrm{j}}âˆ’7\text{â€‰}\stackrel{^}{\mathrm{k}}\text{â€‰}.\end{array}$

Ans.

$\begin{array}{l}\text{The given vectors are:}\\ \stackrel{â†’}{\mathrm{a}}=\stackrel{^}{\mathrm{i}}âˆ’2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}},\text{â€‰}\stackrel{â†’}{\mathrm{b}}=âˆ’2\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}+5\stackrel{^}{\mathrm{k}}\text{â€‰â€‰}\mathrm{and}\stackrel{â†’}{\mathrm{c}}=\stackrel{^}{\mathrm{i}}âˆ’6\stackrel{^}{\mathrm{j}}âˆ’7\text{â€‰}\stackrel{^}{\mathrm{k}}\text{â€‰}\\ \text{Then,}\\ \stackrel{â†’}{\mathrm{a}}\text{â€‰}+\stackrel{â†’}{\mathrm{b}}\text{â€‰â€‰}+\text{â€‰â€‰}\stackrel{â†’}{\mathrm{c}}=\stackrel{^}{\mathrm{i}}âˆ’2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}+\left(âˆ’2\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}+5\stackrel{^}{\mathrm{k}}\right)+\stackrel{^}{\mathrm{i}}âˆ’6\stackrel{^}{\mathrm{j}}âˆ’7\text{â€‰}\stackrel{^}{\mathrm{k}}\\ =âˆ’\text{â€‰}4\stackrel{^}{\mathrm{j}}âˆ’\stackrel{^}{\mathrm{k}}\end{array}$

Q.18

$\begin{array}{l}\text{Find the unit vector in the direction of the vectorâ€‰}\\ \stackrel{â†’}{\mathrm{a}}=\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}.\end{array}$

Ans.

$\begin{array}{l}\text{The unit vector}\stackrel{^}{\mathrm{a}}\text{â€‰â€‰in the direction of vectorâ€‰}\stackrel{â†’}{\mathrm{a}}=\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\text{â€‰â€‰is given}\\ \text{by}\stackrel{^}{\mathrm{a}}=\frac{\stackrel{â†’}{\mathrm{a}}}{|\stackrel{â†’}{\mathrm{a}}|}\\ =\frac{\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}}{|\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}|}\\ =\frac{\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}}{\sqrt{{1}^{2}+{1}^{2}+{2}^{2}}}\\ =\frac{\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}}{\sqrt{6}}\\ =\frac{\stackrel{^}{\mathrm{i}}}{\sqrt{6}}+\frac{\stackrel{^}{\mathrm{j}}}{\sqrt{6}}+\frac{2}{\sqrt{6}}\stackrel{^}{\mathrm{k}}\end{array}$

Q.19

$\begin{array}{l}\text{Find the unit vector in the direction of the vectorâ€‰}\stackrel{â†’}{\mathrm{PQ}},\text{}\\ \text{where P and Q are the points}\left(\text{1,2,3}\right)\text{and}\left(\text{4,5,6}\right)\\ \text{respectively.}\end{array}$

Ans.

$\begin{array}{l}\text{Here,}\stackrel{â†’}{\text{OP}}=\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\text{and}\stackrel{â†’}{\mathrm{OQ}}=4\stackrel{^}{\mathrm{i}}+5\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}\\ \text{So,â€‰}\stackrel{â†’}{\mathrm{PQ}}=\stackrel{â†’}{\mathrm{OQ}}âˆ’\stackrel{â†’}{\mathrm{OP}}\\ =\left(4\stackrel{^}{\mathrm{i}}+5\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}\right)âˆ’\left(\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\right)\\ \text{â€‰â€‰â€‰â€‰â€‰}=3\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\\ \text{Unitvectorofâ€‰}\stackrel{â†’}{\mathrm{PQ}}\\ \text{â€‰â€‰â€‰â€‰â€‰}=\frac{\text{â€‰}\stackrel{â†’}{\mathrm{PQ}}}{\left|\stackrel{â†’}{\mathrm{PQ}}\right|}\\ \text{â€‰â€‰â€‰â€‰â€‰}=\frac{3\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}}{\left|3\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\right|}\\ \text{â€‰â€‰â€‰â€‰â€‰}=\frac{3\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}}{\sqrt{{3}^{2}+{3}^{2}+{3}^{2}}}\\ \text{â€‰â€‰â€‰}=\frac{3\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}}{3\sqrt{3}}\\ \text{â€‰â€‰â€‰}=\frac{\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}}{\sqrt{3}}\\ \text{â€‰â€‰â€‰}=\frac{\stackrel{^}{\mathrm{i}}}{\sqrt{3}}+\frac{\stackrel{^}{\mathrm{j}}}{\sqrt{3}}+\frac{\stackrel{^}{\mathrm{k}}}{\sqrt{3}}\end{array}$

Q.20

$\begin{array}{l}\text{For given vectors,â€‰}\stackrel{â†’}{\mathrm{a}}=2\stackrel{^}{\mathrm{i}}âˆ’\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\text{and}\stackrel{â†’}{\mathrm{b}}=âˆ’\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}âˆ’\stackrel{^}{\mathrm{k}},\text{find the}\\ \text{unit vector in the direction of the vector}\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}.\end{array}$

Ans.

$\begin{array}{l}\text{Theâ€‰â€‰given vectors are:}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\stackrel{â†’}{\mathrm{a}}=2\stackrel{^}{\mathrm{i}}âˆ’\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\text{and}\stackrel{â†’}{\mathrm{b}}=âˆ’\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}âˆ’\stackrel{^}{\mathrm{k}}\\ \text{â€‰}\stackrel{â†’}{\mathrm{a}}\text{â€‰}+\stackrel{â†’}{\mathrm{b}}=2\stackrel{^}{\mathrm{i}}âˆ’\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}+\left(âˆ’\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}âˆ’\stackrel{^}{\mathrm{k}}\right)\\ \text{â€‰â€‰â€‰}=\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{k}}\\ |\text{â€‰}\stackrel{â†’}{\mathrm{a}}\text{â€‰}+\stackrel{â†’}{\mathrm{b}}|=|\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{k}}|\\ \text{â€‰â€‰â€‰}=\sqrt{{1}^{2}+{1}^{2}}\\ \text{â€‰â€‰â€‰}=\sqrt{2}\\ âˆ´\text{Unit vector of â€‰}\stackrel{â†’}{\mathrm{a}}\text{â€‰}+\stackrel{â†’}{\mathrm{b}}\\ \text{â€‰â€‰â€‰}=\frac{\text{â€‰}\stackrel{â†’}{\mathrm{a}}\text{â€‰}+\stackrel{â†’}{\mathrm{b}}}{|\text{â€‰}\stackrel{â†’}{\mathrm{a}}\text{â€‰}+\stackrel{â†’}{\mathrm{b}}|}=\frac{\stackrel{^}{\mathrm{i}}}{\sqrt{2}}+\frac{\stackrel{^}{\mathrm{k}}}{\sqrt{2}}\end{array}$

Q.21

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

Ans.

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

Q.22

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

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{vectors}\mathrm{are}\\ \stackrel{â†’}{\mathrm{a}}=2\stackrel{^}{\mathrm{i}}âˆ’3\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\\ \stackrel{â†’}{\mathrm{b}}=âˆ’4\stackrel{^}{\mathrm{i}}+6\stackrel{^}{\mathrm{j}}–8\stackrel{^}{\mathrm{k}}\\ â€‰â€‰â€‰â€‰=âˆ’2\left(2\stackrel{^}{\mathrm{i}}–3\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\right)\\ \stackrel{â†’}{\mathrm{b}}\mathrm{â€‰}=âˆ’2\stackrel{â†’}{\mathrm{a}},\mathrm{which}\mathrm{is}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\stackrel{â†’}{\mathrm{a}}=\mathrm{Î»}\stackrel{â†’}{\mathrm{b}},\\ \mathrm{where},\mathrm{Î»}=âˆ’2\\ \mathrm{Hence},\mathrm{the}\mathrm{given}\mathrm{vectors}\mathrm{are}\mathrm{collinear}.\end{array}$

Q.23

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

Ans.

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

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

Ans.

$\begin{array}{l}\text{The given points are}\mathrm{A}\left(1,\text{}2,â€“3\right)\text{and}\mathrm{B}\left(â€“1,â€“2,1\right).\\ âˆ´\text{â€‰â€‰â€‰â€‰â€‰}\stackrel{â†’}{\mathrm{AB}}=\left(âˆ’1âˆ’1\right)\stackrel{^}{\mathrm{i}}+\left(âˆ’2âˆ’2\right)\stackrel{^}{\mathrm{j}}+\left(1+3\right)\stackrel{^}{\mathrm{k}}\\ \text{â€‰â€‰â€‰â€‰}=âˆ’2\stackrel{^}{\mathrm{i}}âˆ’4\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\\ âˆ´|\stackrel{â†’}{\mathrm{AB}}|=\sqrt{{\left(âˆ’2\right)}^{2}+{\left(âˆ’4\right)}^{2}+{4}^{2}}\\ \text{â€‰â€‰â€‰â€‰}=\sqrt{4+16+16}\\ \text{â€‰â€‰â€‰â€‰}=\sqrt{36}\\ \text{â€‰â€‰â€‰â€‰}=6\\ \text{Hence, the direction cosines of â€‰}\stackrel{â†’}{\mathrm{AB}}\text{â€‰â€‰areâ€‰â€‰}\left(\frac{âˆ’2}{6},\frac{âˆ’4}{6},\frac{4}{6}\right)=\left(\frac{âˆ’1}{3},\frac{âˆ’2}{3},\frac{2}{3}\right).\end{array}$

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.

$\begin{array}{l}\text{The given vector is}\stackrel{â†’}{\mathrm{a}}=\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}},\text{then}\\ \text{â€‰â€‰â€‰â€‰â€‰}|\stackrel{â†’}{\mathrm{a}}|=|\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}|\\ =\sqrt{{1}^{2}+{1}^{2}+{1}^{2}}\\ =\sqrt{3}\\ \text{The direction cosines of}\stackrel{â†’}{\mathrm{a}}\text{â€‰are}\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right).\\ \text{Let}\mathrm{Î±},\text{â€‰}\mathrm{Î²}\text{and}\mathrm{Î³}\text{be the angles formed by}\stackrel{â†’}{\mathrm{a}}\text{â€‰with the positive}\\ \text{directions of x, y, and z axes.}\\ \text{Then, we have cos}\mathrm{Î±}=\frac{1}{\sqrt{3}},\text{cos}\mathrm{Î²}=\frac{1}{\sqrt{3}}\text{,â€‰cos}\mathrm{Î³}=\frac{1}{\sqrt{3}}.\\ \text{Hence, the given vector is equally inclined to axes OX,}\\ \text{OY, and OZ.}\end{array}$

Q.26

$\begin{array}{l}\mathrm{Let}\mathrm{the}\mathrm{vectors}\stackrel{â†’}{\mathrm{a}},\text{â€‰}\stackrel{â†’}{\text{â€‰}\mathrm{b}},\text{â€‰}\stackrel{â†’}{\mathrm{c}}\text{}\mathrm{be}\mathrm{given}\mathrm{as}{\mathrm{a}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{a}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{a}}_{3}\stackrel{^}{\text{k}},\text{â€‰}\\ {\mathrm{b}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{b}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{b}}_{3}\stackrel{^}{\mathrm{k}},\text{â€‰â€‰}{\mathrm{c}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{c}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{c}}_{3}\stackrel{^}{\mathrm{k}}.\mathrm{Then}\mathrm{show}\mathrm{that}\\ \stackrel{â†’}{\mathrm{a}}Ã—\left(\stackrel{â†’}{\mathrm{b}}+\stackrel{â†’}{\mathrm{c}}\right)=\stackrel{â†’}{\mathrm{a}}Ã—\stackrel{â†’}{\mathrm{b}}+\stackrel{â†’}{\mathrm{a}}Ã—\stackrel{â†’}{\mathrm{c}}.\end{array}$

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{â€‰}{\mathrm{b}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{b}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{b}}_{3}\stackrel{^}{\mathrm{k}},\text{â€‰â€‰}{\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{â†’}{\mathrm{a}}Ã—\left(\stackrel{â†’}{\mathrm{b}}+\stackrel{â†’}{\mathrm{c}}\right)\\ =\left({\mathrm{a}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{a}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{a}}_{3}\stackrel{^}{\mathrm{k}}\right)Ã—\left({\mathrm{b}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{b}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{b}}_{3}\stackrel{^}{\mathrm{k}}+\text{â€‰}{\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)Ã—\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)âˆ’{\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)âˆ’{\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)âˆ’{\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}âˆ’{\mathrm{a}}_{3}{\mathrm{b}}_{2}âˆ’{\mathrm{a}}_{3}{\mathrm{c}}_{2}\right\}\stackrel{^}{\mathrm{i}}+\left\{{\mathrm{a}}_{1}{\mathrm{b}}_{3}+{\mathrm{a}}_{1}{\mathrm{c}}_{3}âˆ’{\mathrm{a}}_{3}{\mathrm{b}}_{1}âˆ’{\mathrm{a}}_{3}{\mathrm{c}}_{1}\right\}\stackrel{^}{\mathrm{j}}\\ +\left\{{\mathrm{a}}_{1}{\mathrm{b}}_{2}+{\mathrm{a}}_{1}{\mathrm{c}}_{2}âˆ’{\mathrm{a}}_{2}{\mathrm{b}}_{3}âˆ’{\mathrm{a}}_{2}{\mathrm{c}}_{3}\right\}\stackrel{^}{\mathrm{k}}\text{â€‰}...\left(\mathrm{i}\right)\\ \stackrel{â†’}{\mathrm{a}}Ã—\stackrel{â†’}{\mathrm{b}}=\left({\mathrm{a}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{a}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{a}}_{3}\stackrel{^}{\mathrm{k}}\right)Ã—\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}âˆ’{\mathrm{a}}_{3}{\mathrm{b}}_{2}\right)\stackrel{^}{\mathrm{i}}+\left({\mathrm{a}}_{1}{\mathrm{b}}_{3}âˆ’{\mathrm{a}}_{3}{\mathrm{b}}_{1}\right)\stackrel{^}{\mathrm{j}}+\left({\mathrm{a}}_{1}{\mathrm{b}}_{2}âˆ’{\mathrm{a}}_{2}{\mathrm{b}}_{1}\right)\stackrel{^}{\mathrm{k}}\\ \stackrel{â†’}{\mathrm{a}}Ã—\stackrel{â†’}{\mathrm{c}}=\left({\mathrm{a}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{a}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{a}}_{3}\stackrel{^}{\mathrm{k}}\right)Ã—\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}âˆ’{\mathrm{a}}_{3}{\mathrm{b}}_{2}\right)\stackrel{^}{\mathrm{i}}+\left({\mathrm{a}}_{1}{\mathrm{b}}_{3}âˆ’{\mathrm{a}}_{3}{\mathrm{b}}_{1}\right)\stackrel{^}{\mathrm{j}}+\left({\mathrm{a}}_{1}{\mathrm{b}}_{2}âˆ’{\mathrm{a}}_{2}{\mathrm{b}}_{1}\right)\stackrel{^}{\mathrm{k}}\\ \stackrel{â†’}{\mathrm{a}}Ã—\stackrel{â†’}{\mathrm{b}}+\stackrel{â†’}{\mathrm{a}}Ã—\stackrel{â†’}{\mathrm{c}}\\ =\left({\mathrm{a}}_{2}{\mathrm{b}}_{3}âˆ’{\mathrm{a}}_{3}{\mathrm{b}}_{2}\right)\stackrel{^}{\mathrm{i}}+\left({\mathrm{a}}_{1}{\mathrm{b}}_{3}âˆ’{\mathrm{a}}_{3}{\mathrm{b}}_{1}\right)\stackrel{^}{\mathrm{j}}+\left({\mathrm{a}}_{1}{\mathrm{b}}_{2}âˆ’{\mathrm{a}}_{2}{\mathrm{b}}_{1}\right)\stackrel{^}{\mathrm{k}}+\left({\mathrm{a}}_{2}{\mathrm{b}}_{3}âˆ’{\mathrm{a}}_{3}{\mathrm{b}}_{2}\right)\stackrel{^}{\mathrm{i}}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}+\left({\mathrm{a}}_{1}{\mathrm{b}}_{3}âˆ’{\mathrm{a}}_{3}{\mathrm{b}}_{1}\right)\stackrel{^}{\mathrm{j}}+\left({\mathrm{a}}_{1}{\mathrm{b}}_{2}âˆ’{\mathrm{a}}_{2}{\mathrm{b}}_{1}\right)\stackrel{^}{\mathrm{k}}\\ =\left({\mathrm{a}}_{2}{\mathrm{b}}_{3}âˆ’{\mathrm{a}}_{3}{\mathrm{b}}_{2}+{\mathrm{a}}_{2}{\mathrm{b}}_{3}âˆ’{\mathrm{a}}_{3}{\mathrm{b}}_{2}\right)\stackrel{^}{\mathrm{i}}+\left({\mathrm{a}}_{1}{\mathrm{b}}_{3}âˆ’{\mathrm{a}}_{3}{\mathrm{b}}_{1}+{\mathrm{a}}_{1}{\mathrm{b}}_{3}âˆ’{\mathrm{a}}_{3}{\mathrm{b}}_{1}\right)\stackrel{^}{\mathrm{j}}\\ +\left({\mathrm{a}}_{1}{\mathrm{b}}_{2}âˆ’{\mathrm{a}}_{2}{\mathrm{b}}_{1}+{\mathrm{a}}_{1}{\mathrm{b}}_{2}âˆ’{\mathrm{a}}_{2}{\mathrm{b}}_{1}\right)\stackrel{^}{\mathrm{k}}\\ =\left({\mathrm{a}}_{2}{\mathrm{b}}_{3}+{\mathrm{a}}_{2}{\mathrm{b}}_{3}âˆ’{\mathrm{a}}_{3}{\mathrm{b}}_{2}âˆ’{\mathrm{a}}_{3}{\mathrm{b}}_{2}\right)\stackrel{^}{\mathrm{i}}+\left({\mathrm{a}}_{1}{\mathrm{b}}_{3}+{\mathrm{a}}_{1}{\mathrm{b}}_{3}âˆ’{\mathrm{a}}_{3}{\mathrm{b}}_{1}âˆ’{\mathrm{a}}_{3}{\mathrm{b}}_{1}\right)\stackrel{^}{\mathrm{j}}\\ +\left({\mathrm{a}}_{1}{\mathrm{b}}_{2}+{\mathrm{a}}_{1}{\mathrm{b}}_{2}âˆ’{\mathrm{a}}_{2}{\mathrm{b}}_{1}âˆ’{\mathrm{a}}_{2}{\mathrm{b}}_{1}\right)\stackrel{^}{\mathrm{k}}\text{â€‰â€‰}...\left(\mathrm{ii}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we have}\\ \text{â€‰â€‰}\stackrel{â†’}{\mathrm{a}}Ã—\left(\stackrel{â†’}{\mathrm{b}}Ã—\stackrel{â†’}{\mathrm{c}}\right)=\stackrel{â†’}{\mathrm{a}}Ã—\stackrel{â†’}{\mathrm{b}}+\stackrel{â†’}{\mathrm{a}}Ã—\stackrel{â†’}{\mathrm{c}}.\text{â€‰â€‰}\\ \text{â€‰Hence, the given result is proved.}\end{array}$

Q.27

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

Ans.

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

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

Ans.

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

Q.29

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

Ans.

$\begin{array}{l}\text{Position vectors of points A, B, and C are respectively given as:}\\ \stackrel{â†’}{\mathrm{a}}=3\stackrel{^}{\mathrm{i}}âˆ’4\stackrel{^}{\mathrm{j}}âˆ’4\stackrel{^}{\mathrm{k}},\text{â€‰}\stackrel{â†’}{\mathrm{b}}=2\stackrel{^}{\mathrm{i}}âˆ’\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\text{â€‰and}\stackrel{â†’}{\mathrm{c}}=\stackrel{^}{\mathrm{i}}âˆ’3\stackrel{^}{\mathrm{j}}âˆ’5\stackrel{^}{\mathrm{k}}\\ âˆ´\text{â€‰â€‰â€‰â€‰â€‰}\stackrel{â†’}{\mathrm{AB}}=\stackrel{â†’}{\mathrm{b}}âˆ’\stackrel{â†’}{\mathrm{a}}\\ \text{â€‰â€‰â€‰â€‰â€‰}=\left(2\stackrel{^}{\mathrm{i}}âˆ’\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)âˆ’\left(3\stackrel{^}{\mathrm{i}}âˆ’4\stackrel{^}{\mathrm{j}}âˆ’4\stackrel{^}{\mathrm{k}}\right)\\ \text{â€‰â€‰â€‰â€‰â€‰}=âˆ’\text{â€‰}\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+5\text{â€‰}\stackrel{^}{\mathrm{k}}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\stackrel{â†’}{\mathrm{BC}}=\stackrel{â†’}{\mathrm{c}}âˆ’\stackrel{â†’}{\mathrm{b}}\\ \text{â€‰â€‰â€‰â€‰â€‰}=\left(\stackrel{^}{\mathrm{i}}âˆ’3\stackrel{^}{\mathrm{j}}âˆ’5\stackrel{^}{\mathrm{k}}\right)âˆ’\left(2\stackrel{^}{\mathrm{i}}âˆ’\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)\\ \text{â€‰â€‰â€‰â€‰â€‰}=âˆ’\text{â€‰}\stackrel{^}{\mathrm{i}}âˆ’2\stackrel{^}{\mathrm{j}}âˆ’6\text{â€‰}\stackrel{^}{\mathrm{k}}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\stackrel{â†’}{\mathrm{CA}}=\stackrel{â†’}{\mathrm{a}}âˆ’\stackrel{â†’}{\mathrm{c}}\\ \text{â€‰â€‰â€‰â€‰â€‰}=\left(3\stackrel{^}{\mathrm{i}}âˆ’4\stackrel{^}{\mathrm{j}}âˆ’4\stackrel{^}{\mathrm{k}}\right)âˆ’\left(\stackrel{^}{\mathrm{i}}âˆ’3\stackrel{^}{\mathrm{j}}âˆ’5\stackrel{^}{\mathrm{k}}\right)\end{array}$ $\begin{array}{c}\text{â€‰â€‰â€‰â€‰â€‰}=2\text{â€‰}\stackrel{^}{\mathrm{i}}âˆ’\stackrel{^}{\mathrm{j}}+\text{â€‰}\stackrel{^}{\mathrm{k}}\\ \text{So,â€‰}{|\stackrel{â†’}{\mathrm{AB}}|}^{2}={|âˆ’\text{â€‰}\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+5\text{â€‰}\stackrel{^}{\mathrm{k}}|}^{2}\\ \text{â€‰â€‰â€‰â€‰â€‰}=35\\ \text{â€‰â€‰â€‰â€‰}{|\stackrel{â†’}{\mathrm{BC}}|}^{2}={|âˆ’\text{â€‰}\stackrel{^}{\mathrm{i}}âˆ’2\stackrel{^}{\mathrm{j}}âˆ’6\text{â€‰}\stackrel{^}{\mathrm{k}}|}^{2}\\ \text{â€‰â€‰â€‰â€‰â€‰}=41\\ \text{â€‰â€‰â€‰â€‰}{|\stackrel{â†’}{\mathrm{CA}}|}^{2}={|2\text{â€‰}\stackrel{^}{\mathrm{i}}âˆ’\stackrel{^}{\mathrm{j}}+\text{â€‰}\stackrel{^}{\mathrm{k}}|}^{2}\\ \text{â€‰â€‰â€‰â€‰â€‰}=6\\ {|\stackrel{â†’}{\mathrm{AB}}|}^{2}=\text{â€‰â€‰â€‰}{|\stackrel{â†’}{\mathrm{BC}}|}^{2}+{|\stackrel{â†’}{\mathrm{CA}}|}^{2}\\ =36+5=41\end{array}$ $\text{Hence, ABC is a right-angled triangle}$

Q.30n

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

Ans.

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

Ans.

$\begin{array}{l}\text{The vertices of triangle ABC are given as A (1, 1, 2), B (2, 3, 5)}\\ \text{and C (1, 5, 5).}\\ \text{The adjacent sidesâ€‰â€‰}\stackrel{â†’}{\mathrm{AB}}\text{and}\stackrel{â†’}{\mathrm{BC}}\text{â€‰â€‰of}\mathrm{Î”}\text{ABC are given as:}\\ \text{â€‰â€‰â€‰}\stackrel{â†’}{\mathrm{AB}}=\left(2âˆ’1\right)\stackrel{^}{\mathrm{i}}+\left(3âˆ’1\right)\stackrel{^}{\mathrm{j}}+\left(5âˆ’2\right)\stackrel{^}{\mathrm{k}}\\ =\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\text{â€‰}\stackrel{^}{\mathrm{k}}\\ \text{â€‰â€‰â€‰}\stackrel{â†’}{\mathrm{BC}}=\left(1âˆ’2\right)\stackrel{^}{\mathrm{i}}+\left(5âˆ’3\right)\stackrel{^}{\mathrm{j}}+\left(5âˆ’5\right)\stackrel{^}{\mathrm{k}}\\ =âˆ’\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}\\ âˆ´\stackrel{â†’}{\mathrm{AB}}Ã—\stackrel{â†’}{\mathrm{BC}}\\ =\frac{1}{2}|\begin{array}{ccc}\stackrel{^}{\mathrm{i}}& \stackrel{^}{\mathrm{j}}& \stackrel{^}{\mathrm{k}}\\ 1& 2& 3\\ âˆ’1& 2& 0\end{array}|\\ =\frac{1}{2}\left\{\left(0âˆ’6\right)\stackrel{^}{\mathrm{i}}âˆ’\left(0+3\right)\stackrel{^}{\mathrm{j}}+\left(2+2\right)\stackrel{^}{\mathrm{k}}\right\}\\ =\frac{1}{2}\left(âˆ’6\stackrel{^}{\mathrm{i}}âˆ’3\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\right)\\ \mathrm{The}\text{area}\mathrm{Î”ABC}\\ =\frac{1}{2}|\stackrel{â†’}{\mathrm{AB}}Ã—\stackrel{â†’}{\mathrm{BC}}|\\ =\frac{1}{2}|âˆ’\text{â€‰}6\stackrel{^}{\mathrm{i}}âˆ’3\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}|\\ =\frac{1}{2}\sqrt{{\left(âˆ’\text{â€‰}6\right)}^{2}+{\left(âˆ’3\right)}^{2}+{4}^{2}}\\ =\frac{1}{2}\sqrt{36+9+16}\\ =\frac{1}{2}\sqrt{61}\text{square units}\\ \text{Thus, the area of}\mathrm{Î”}\text{ABC is}\frac{1}{2}\sqrt{61}\text{square units}.\end{array}$

Q.32

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

Ans.

$\begin{array}{l}\text{The area of the parallelogram whose adjacent sides areâ€‰}\stackrel{â†’}{\mathrm{a}}\text{and}\stackrel{â†’}{\mathrm{b}}\\ \mathrm{is}\text{}|\stackrel{â†’}{\mathrm{a}}Ã—\stackrel{â†’}{\mathrm{b}}|.\\ \text{Adjacent sides are in vector form, which are as follows:}\\ \stackrel{â†’}{\mathrm{a}}=\stackrel{^}{\mathrm{i}}âˆ’\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\text{and}\\ \stackrel{â†’}{\mathrm{b}}=2\stackrel{^}{\mathrm{i}}âˆ’7\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\\ âˆ´\stackrel{â†’}{\mathrm{a}}Ã—\stackrel{â†’}{\mathrm{b}}=|\begin{array}{ccc}\stackrel{^}{\mathrm{i}}& \stackrel{^}{\mathrm{j}}& \stackrel{^}{\mathrm{k}}\\ 1& âˆ’1& 3\\ 2& âˆ’7& 1\end{array}|\\ \text{â€‰â€‰â€‰â€‰}=\left(âˆ’1+21\right)\stackrel{^}{\mathrm{i}}+\left(1âˆ’6\right)\stackrel{^}{\mathrm{j}}+\left(âˆ’7+2\right)\stackrel{^}{\mathrm{k}}\\ \text{â€‰â€‰â€‰â€‰}=20\text{â€‰}\stackrel{^}{\mathrm{i}}âˆ’5\text{â€‰}\stackrel{^}{\mathrm{j}}âˆ’5\text{â€‰}\stackrel{^}{\mathrm{k}}\\ \text{â€‰}|\stackrel{â†’}{\mathrm{a}}Ã—\stackrel{â†’}{\mathrm{b}}|=|20\text{â€‰}\stackrel{^}{\mathrm{i}}âˆ’5\text{â€‰}\stackrel{^}{\mathrm{j}}âˆ’5\text{â€‰}\stackrel{^}{\mathrm{k}}|\\ \text{â€‰â€‰â€‰â€‰}=\sqrt{{20}^{2}+{\left(âˆ’5\right)}^{2}+{\left(âˆ’5\right)}^{2}}\\ \text{â€‰â€‰â€‰â€‰}=\sqrt{400+25+25}\\ \text{â€‰â€‰â€‰â€‰}=\sqrt{450}\\ \text{â€‰â€‰â€‰â€‰}=15\sqrt{2}\\ \text{Thus, the area of the given parallelogram is 15}\sqrt{2}\text{â€‰square units.}\end{array}$

Q.33

$\begin{array}{l}\mathrm{Let}\mathrm{the}\mathrm{vectors}\stackrel{â†’}{\mathrm{a}}\mathrm{and}\stackrel{â†’}{\mathrm{b}}\mathrm{be}\mathrm{such}\mathrm{that}|\stackrel{â†’}{\mathrm{a}}|=3\mathrm{and}|\stackrel{â†’}{\mathrm{b}}|=\frac{\sqrt{3}}{2},\mathrm{then}\\ \stackrel{â†’}{\mathrm{a}}Ã—\stackrel{â†’}{\mathrm{b}}\mathrm{is}\mathrm{a}\mathrm{unit}\mathrm{vector},\mathrm{if}\mathrm{the}\mathrm{angle}\mathrm{between}\stackrel{â†’}{\mathrm{a}}\mathrm{and}\stackrel{â†’}{\mathrm{b}}\mathrm{is}\\ \left(\mathrm{A}\right)\frac{\mathrm{Ï€}}{6}\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\\ \left(\mathrm{B}\right)\frac{\mathrm{Ï€}}{4}\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\\ \left(\mathrm{C}\right)\frac{\mathrm{Ï€}}{3}\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\\ \left(\mathrm{D}\right)\frac{\mathrm{Ï€}}{2}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Given}:\text{â€‰â€‰}|\stackrel{â†’}{\mathrm{a}}|=3\text{and}|\stackrel{â†’}{\mathrm{b}}|=\frac{\sqrt{2}}{3}\\ \mathrm{Since},\text{}\stackrel{â†’}{\mathrm{a}}Ã—\stackrel{â†’}{\mathrm{b}}=|\stackrel{â†’}{\mathrm{a}}||\stackrel{â†’}{\mathrm{b}}|\mathrm{sinÎ¸}\text{â€‰}\stackrel{^}{\mathrm{n}},\text{where â€‰}\stackrel{^}{\mathrm{n}}\text{is a unit vector perpendicular}\\ \text{to both}\stackrel{â†’}{\mathrm{a}}\text{â€‰and}\stackrel{â†’}{\mathrm{b}}\text{â€¹ and}\mathrm{Î¸}\text{â€¹ is the angle between}\stackrel{â†’}{\mathrm{a}}\text{â€‰and}\stackrel{â†’}{\mathrm{b}}\text{â€¹}.\\ \mathrm{Since},\text{}\stackrel{â†’}{\mathrm{a}}Ã—\stackrel{â†’}{\mathrm{b}}\text{is a unit vector, so}|\stackrel{â†’}{\mathrm{a}}Ã—\stackrel{â†’}{\mathrm{b}}|=1.\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}|\stackrel{â†’}{\mathrm{a}}Ã—\stackrel{â†’}{\mathrm{b}}|=1\\ â‡’||\stackrel{â†’}{\mathrm{a}}||\stackrel{â†’}{\mathrm{b}}|\mathrm{sinÎ¸}\text{â€‰}\stackrel{^}{\mathrm{n}}|=1\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰}|\stackrel{â†’}{\mathrm{a}}||\stackrel{â†’}{\mathrm{b}}||\mathrm{sinÎ¸}|\text{â€‰}=1\\ \text{â€‰â€‰}3Ã—\frac{\sqrt{2}}{3}Ã—\mathrm{sinÎ¸}=1\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{sinÎ¸}=\frac{1}{\sqrt{2}}=\mathrm{sin}\frac{\mathrm{Ï€}}{4}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰}\mathrm{Î¸}=\frac{\mathrm{Ï€}}{4}\\ \text{Hence,}\stackrel{â†’}{\mathrm{a}}Ã—\stackrel{â†’}{\mathrm{b}}\text{â€‰is a unit vector if the angle between}\stackrel{â†’}{\mathrm{a}}\text{â€‰and}\stackrel{â†’}{\mathrm{b}}\text{isâ€‰}\frac{\mathrm{Ï€}}{4}.\\ \text{The correct answer is B.}\end{array}$

Q.34

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

Ans.

$\begin{array}{l}\text{The position vectors of vertices A, B, C, and D of rectangle}\\ \text{ABCD are given as:}\\ \stackrel{â†’}{\mathrm{OA}}=âˆ’\stackrel{^}{\mathrm{i}}+\frac{1}{2}\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}},\text{}\stackrel{â†’}{\mathrm{OB}}=\text{â€‰}\stackrel{^}{\mathrm{i}}+\frac{1}{2}\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}},\text{â€‰}\stackrel{â†’}{\mathrm{OC}}=\text{â€‰}\stackrel{^}{\mathrm{i}}âˆ’\frac{1}{2}\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\text{and}\\ \stackrel{â†’}{\mathrm{OD}}=âˆ’\text{â€‰}\stackrel{^}{\mathrm{i}}âˆ’\frac{1}{2}\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\\ \text{The adjacent sides}\stackrel{â†’}{\mathrm{AB}}\text{â€‰and}\stackrel{â†’}{\mathrm{BC}}\text{â€‰of the given rectangle are given as:}\\ \stackrel{â†’}{\mathrm{AB}}=\left(1+1\right)\stackrel{^}{\mathrm{i}}+\left(\frac{1}{2}âˆ’\frac{1}{2}\right)\stackrel{^}{\mathrm{j}}+\left(4âˆ’4\right)\stackrel{^}{\mathrm{k}}=2\stackrel{^}{\mathrm{i}}\\ \stackrel{â†’}{\mathrm{BC}}=\left(1âˆ’1\right)\stackrel{^}{\mathrm{i}}+\left(âˆ’\frac{1}{2}âˆ’\frac{1}{2}\right)\stackrel{^}{\mathrm{j}}+\left(4âˆ’4\right)\stackrel{^}{\mathrm{k}}=âˆ’\stackrel{^}{\mathrm{j}}\\ âˆ´\stackrel{â†’}{\mathrm{AB}}\text{â€‰}Ã—\text{â€‰}\stackrel{â†’}{\mathrm{BC}}\\ =|\begin{array}{ccc}\text{â€‰}\stackrel{^}{\mathrm{i}}& \stackrel{^}{\mathrm{j}}& \stackrel{^}{\mathrm{k}}\\ 2& 0& 0\\ 0& âˆ’1& 0\end{array}|\\ =\left(âˆ’2âˆ’0\right)\stackrel{^}{\mathrm{k}}\\ =âˆ’2\stackrel{^}{\mathrm{k}}\\ |\stackrel{â†’}{\mathrm{AB}}\text{â€‰}Ã—\text{â€‰}\stackrel{â†’}{\mathrm{BC}}|\\ =|âˆ’2\stackrel{^}{\mathrm{k}}|\\ =2\\ \text{Now, it is known that the area of a parallelogram whose adjacent}\\ \text{sides areâ€‰}\stackrel{â†’}{\mathrm{a}}\text{and}\stackrel{â†’}{\mathrm{b}}\text{â€‰â€‰}\mathrm{is}\text{}|\stackrel{â†’}{\mathrm{a}}Ã—\stackrel{â†’}{\mathrm{b}}|.\\ \text{Hence, the area of the given rectangle isâ€‰}|\stackrel{â†’}{\mathrm{AB}}\text{â€‰}Ã—\text{â€‰}\stackrel{â†’}{\mathrm{BC}}|=2\text{square units.}\\ \text{The correct answer is C.}\end{array}$

Q.35

$\mathrm{Find}\left[\stackrel{â†’}{\mathrm{a}}\text{â€‰}\stackrel{â†’}{\mathrm{b}}\text{â€‰}\stackrel{â†’}{\mathrm{c}}\right]\mathrm{if}\stackrel{â†’}{\mathrm{a}}=\stackrel{^}{\mathrm{i}}\text{â€‰}âˆ’2\text{â€‰}\stackrel{^}{\mathrm{j}}\text{â€‰}+3\text{â€‰}\stackrel{^}{\text{k}},\stackrel{â†’}{\mathrm{b}}=2\stackrel{^}{\mathrm{i}}\text{â€‰}âˆ’3\text{â€‰}\stackrel{^}{\mathrm{j}}\text{â€‰}+\text{â€‰}\stackrel{^}{\mathrm{k}},\text{â€‰}\stackrel{â†’}{\mathrm{c}}=3\text{â€‰}\stackrel{^}{\mathrm{i}}\text{â€‰}+\text{â€‰}\stackrel{^}{\mathrm{j}}\text{â€‰}âˆ’2\text{â€‰}\stackrel{^}{\mathrm{k}}.$

Ans.

$\begin{array}{l}\left[\stackrel{â†’}{\mathrm{a}}\text{â€‰}\stackrel{â†’}{\mathrm{b}}\text{â€‰}\stackrel{â†’}{\mathrm{c}}\right]=|\begin{array}{ccc}1& âˆ’2& 3\\ 2& âˆ’3& 1\\ 3& 1& âˆ’2\end{array}|\\ \text{â€‰â€‰}=1|\begin{array}{cc}âˆ’3& 1\\ 1& âˆ’2\end{array}|+2|\begin{array}{cc}2& 1\\ 3& âˆ’2\end{array}|+3|\begin{array}{cc}2& âˆ’3\\ 3& 1\end{array}|\\ \text{â€‰â€‰}=1\left(6âˆ’1\right)+2\left(âˆ’4âˆ’3\right)+3\left(2+9\right)\\ \text{â€‰â€‰}=1\left(5\right)+2\left(âˆ’7\right)+3\left(11\right)\\ \text{â€‰â€‰}=5âˆ’14+33\\ \text{â€‰â€‰}=24\\ âˆ´\left[\stackrel{â†’}{\mathrm{a}}\text{â€‰}\stackrel{â†’}{\mathrm{b}}\text{â€‰}\stackrel{â†’}{\mathrm{c}}\right]\text{â€‰}=24\end{array}$

Q.36

$\mathrm{In}\mathrm{triangle},\mathrm{which}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{is}\mathrm{not}\mathrm{true}:\phantom{\rule{0ex}{0ex}}\begin{array}{l}\left(\mathrm{A}\right)\text{â€‰}\stackrel{â†’}{\mathrm{AB}}\text{â€‰}+\text{â€‰}\stackrel{â†’}{\mathrm{BC}}+\text{â€‰}\stackrel{â†’}{\mathrm{CA}}=0\\ \left(\mathrm{B}\right)\text{â€‰}\stackrel{â†’}{\mathrm{AB}}\text{â€‰}+\text{â€‰}\stackrel{â†’}{\mathrm{BC}}â€“\mathrm{AC}=0\\ \left(\mathrm{C}\right)\text{â€‰}\stackrel{â†’}{\mathrm{AB}}+\text{â€‰}\stackrel{â†’}{\mathrm{BC}}â€“\stackrel{â†’}{\mathrm{CA}}=0\\ \left(\mathrm{D}\right)\text{â€‰}\stackrel{â†’}{\mathrm{AB}}\text{â€‰}â€“\text{â€‰}\stackrel{â†’}{\mathrm{CB}}+\stackrel{â†’}{\mathrm{CA}}=0\end{array}$

Ans.

$\begin{array}{l}\text{On applying the triangle law of addition in the given triangle,}\\ \text{we have:}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}\stackrel{â†’}{\mathrm{AB}}+\stackrel{â†’}{\mathrm{BC}}=\stackrel{â†’}{\mathrm{AC}}...\left(\text{i}\right)\\ â‡’\stackrel{â†’}{\mathrm{AB}}+\stackrel{â†’}{\mathrm{BC}}=âˆ’\text{â€‰}\stackrel{â†’}{\mathrm{CA}}\\ â‡’\stackrel{â†’}{\mathrm{AB}}+\stackrel{â†’}{\mathrm{BC}}+\text{â€‰}\stackrel{â†’}{\mathrm{CA}}=\stackrel{â†’}{0}...\left(\text{ii}\right)\\ âˆ´\text{The equation given in alternative A is true.}\\ \text{â€‰â€‰â€‰}\stackrel{â†’}{\mathrm{AB}}+\stackrel{â†’}{\mathrm{BC}}=\text{â€‰}\stackrel{â†’}{\mathrm{AC}}\\ â‡’\text{â€‰}\stackrel{â†’}{\mathrm{AB}}+\stackrel{â†’}{\mathrm{BC}}âˆ’\stackrel{â†’}{\mathrm{AC}}=\stackrel{â†’}{0}\\ âˆ´\text{The equation given in alternative B is true.}\\ \text{From equation}\left(\text{ii}\right)\text{, we have:}\\ \stackrel{â†’}{\mathrm{AB}}âˆ’\stackrel{â†’}{\mathrm{CB}}+\text{â€‰}\stackrel{â†’}{\mathrm{AC}}=\stackrel{â†’}{0}\\ âˆ´\text{The equation given in alternative D is true.}\\ \text{Now, the equation given in alternative C,}\\ \stackrel{â†’}{\mathrm{AB}}+\stackrel{â†’}{\mathrm{BC}}âˆ’\stackrel{â†’}{\mathrm{CA}}=\stackrel{â†’}{0}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰}\stackrel{â†’}{\mathrm{AB}}+\stackrel{â†’}{\mathrm{BC}}=\stackrel{â†’}{\mathrm{CA}}\text{â€‰â€‰}...\left(\mathrm{iii}\right)\\ \text{From equation}\left(\text{i}\right)\text{and}\left(\text{iii}\right)\text{,â€¹â€‰we have}\end{array}$ $\begin{array}{l}\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\stackrel{â†’}{\mathrm{CA}}=\stackrel{â†’}{\mathrm{AC}}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}âˆ’\stackrel{â†’}{\mathrm{AC}}=\stackrel{â†’}{\mathrm{AC}}\\ \text{â€‰}â‡’\stackrel{â†’}{\mathrm{AC}}+\stackrel{â†’}{\mathrm{AC}}=\stackrel{â†’}{0}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}2\stackrel{â†’}{\mathrm{AC}}=\stackrel{â†’}{0}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\stackrel{â†’}{\mathrm{AC}}=\stackrel{â†’}{0},\text{which is not true.}\\ \text{Hence, the equation given in alternative C is incorrect.}\\ \text{The correct answer is C.}\end{array}$

Q.37

$\begin{array}{l}\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{vectors}\stackrel{â†’}{\mathrm{a}}=\stackrel{^}{\mathrm{i}}\text{â€‰}âˆ’2\text{â€‰}\stackrel{^}{\mathrm{j}}\text{â€‰}+3\text{â€‰}\stackrel{^}{\text{k}},\stackrel{â†’}{\mathrm{b}}=âˆ’2\stackrel{^}{\mathrm{i}}\text{â€‰}+3\text{â€‰}\stackrel{^}{\mathrm{j}}\text{â€‰}âˆ’4\text{â€‰}\stackrel{^}{\mathrm{k}},\text{â€‰}\\ \stackrel{â†’}{\mathrm{c}}=\text{â€‰}\stackrel{^}{\mathrm{i}}\text{â€‰}âˆ’\mathbf{3}\text{â€‰}\stackrel{^}{\mathrm{j}}\text{â€‰}+5\text{â€‰}\stackrel{^}{\mathrm{k}}\text{}\mathrm{are}\mathrm{coplanar}.\end{array}$

Ans.

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

Q.38

$\begin{array}{l}\mathrm{Find}\text{â€‰}\mathrm{Î»}\mathrm{if}\mathrm{the}\mathrm{vectors}\stackrel{^}{\mathrm{i}}\text{â€‰}âˆ’\text{â€‰}\stackrel{^}{\mathrm{j}}\text{â€‰}+\text{â€‰}\stackrel{^}{\text{k}},3\stackrel{^}{\mathrm{i}}\text{â€‰}+\text{â€‰}\stackrel{^}{\mathrm{j}}\text{â€‰}+2\text{â€‰}\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\text{â€‰}\stackrel{^}{\mathrm{i}}\text{â€‰}+\mathrm{Î»}\text{â€‰}\stackrel{^}{\mathrm{j}}\text{â€‰}âˆ’3\text{â€‰}\stackrel{^}{\mathrm{k}}\text{}\mathrm{are}\\ \mathrm{coplanar}.\end{array}$

Ans.

$\begin{array}{l}\left[\stackrel{â†’}{\mathrm{a}}\text{â€‰}\stackrel{â†’}{\mathrm{b}}\text{â€‰}\stackrel{â†’}{\mathrm{c}}\right]=|\begin{array}{ccc}1& âˆ’1& 1\\ 3& 1& 2\\ 1& \mathrm{Î»}& âˆ’3\end{array}|\\ \text{â€‰â€‰â€‰â€‰}\left[\stackrel{â†’}{\mathrm{a}}\text{â€‰}\stackrel{â†’}{\mathrm{b}}\text{â€‰}\stackrel{â†’}{\mathrm{c}}\right]\text{â€‰â€‰}=1|\begin{array}{cc}1& 2\\ \mathrm{Î»}& âˆ’3\end{array}|âˆ’\left(âˆ’1\right)|\begin{array}{cc}3& 2\\ 1& âˆ’3\end{array}|+1|\begin{array}{cc}3& 1\\ 1& \mathrm{Î»}\end{array}|\\ \text{â€‰â€‰}=1\left(âˆ’3âˆ’2\mathrm{Î»}\right)+1\left(âˆ’9âˆ’2\right)+1\left(3\mathrm{Î»}âˆ’1\right)\\ \text{â€‰â€‰}=âˆ’3âˆ’2\mathrm{Î»}âˆ’11+3\mathrm{Î»}âˆ’1\\ \text{â€‰â€‰}=\mathrm{Î»}âˆ’15\\ \mathrm{Since},\text{given vectors are coplanar. So,}\\ \text{â€‰â€‰â€‰â€‰}\left[\stackrel{â†’}{\mathrm{a}}\text{â€‰}\stackrel{â†’}{\mathrm{b}}\text{â€‰}\stackrel{â†’}{\mathrm{c}}\right]=0\\ â‡’\mathrm{Î»}âˆ’15=0\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰}\mathrm{Î»}=15\end{array}$

Q.39

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

Ans.

Q.40

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

Ans.

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

$\begin{array}{l}\text{If}\stackrel{â†’}{\mathrm{a}}\text{}\stackrel{â†’}{\mathrm{b}}\text{are two collinear vectors, then which of the following}\\ \text{are incorrect:}\\ \left(\text{A}\right)\text{â€‰}\stackrel{â†’}{\mathrm{b}}=\mathrm{Î»}\stackrel{â†’}{\mathrm{a}},\text{for some scalar}\mathrm{Î»}\\ \left(\text{B}\right)\text{â€‰}\stackrel{â†’}{\mathrm{a}}=Â±\text{â€‰}\stackrel{â†’}{\mathrm{b}}\\ \left(\text{C}\right)\text{the respective components of}\stackrel{â†’}{\mathrm{a}}\text{and}\stackrel{â†’}{\mathrm{b}}\text{are proportional}\\ \left(\text{D}\right)\text{both the vectors}\stackrel{â†’}{\mathrm{a}}\text{and}\stackrel{â†’}{\mathrm{b}}\text{have same direction, but}\\ \text{â€‰â€‰â€‰â€‰â€‰different magnitudes.}\end{array}$

Ans.

$\begin{array}{l}\text{If}\stackrel{â†’}{\mathrm{a}}\text{â€‰â€‰and}\stackrel{â†’}{\mathrm{b}}\text{are two collinear vectors, then they are parallel.}\\ \text{Therefore, we have:}\\ \text{â€‰â€‰}\stackrel{â†’}{\mathrm{b}}=\mathrm{Î»}\stackrel{â†’}{\mathrm{a}}\left(\mathrm{Î»}\text{is a scalar quantities.}\right)\\ \text{Ifâ€‰}\mathrm{Î»}=Â±1,\text{then â€‰}\stackrel{â†’}{\mathrm{b}}=Â±1\stackrel{â†’}{\mathrm{a}}.\\ \text{Ifâ€‰}\stackrel{â†’}{\mathrm{a}}={\mathrm{a}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{a}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{a}}_{3}\stackrel{^}{\mathrm{k}}\text{and}\stackrel{â†’}{\mathrm{b}}={\mathrm{b}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{b}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{b}}_{3}\stackrel{^}{\mathrm{k}},\text{â€‰then}\\ \text{â€‰â€‰}\stackrel{â†’}{\mathrm{b}}=\mathrm{Î»}\stackrel{â†’}{\mathrm{a}}â‡’{\mathrm{b}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{b}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{b}}_{3}\stackrel{^}{\mathrm{k}}=\mathrm{Î»}\left({\mathrm{a}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{a}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{a}}_{3}\stackrel{^}{\mathrm{k}}\right)\end{array}$ $\begin{array}{l}â‡’{\mathrm{b}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{b}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{b}}_{3}\stackrel{^}{\mathrm{k}}=\left({\mathrm{Î»a}}_{1}\right)\stackrel{^}{\mathrm{i}}+\left({\mathrm{Î»a}}_{2}\right)\stackrel{^}{\mathrm{j}}+\left({\mathrm{Î»a}}_{3}\right)\stackrel{^}{\mathrm{k}}\\ â‡’{\mathrm{b}}_{1}={\mathrm{Î»a}}_{1},{\mathrm{b}}_{2}={\mathrm{Î»a}}_{2},{\mathrm{b}}_{3}={\mathrm{Î»a}}_{3}\\ â‡’\frac{{\mathrm{b}}_{1}}{{\mathrm{a}}_{1}}=\frac{{\mathrm{b}}_{2}}{{\mathrm{a}}_{2}}=\frac{{\mathrm{b}}_{3}}{{\mathrm{a}}_{3}}=\mathrm{Î»}\\ \text{Thus, the respective components of}\stackrel{â†’}{\mathrm{a}}\text{â€‰â€‰and}\stackrel{â†’}{\mathrm{b}}\text{â€‰â€‰are proportional.}\\ \text{However, vectors}\stackrel{â†’}{\mathrm{a}}\text{â€‰â€‰and}\stackrel{â†’}{\mathrm{b}}\text{â€‰â€‰can have different directions.}\\ \text{Hence, the statement given in D is incorrect.}\\ \text{The correct answer is D.}\end{array}$

Q.42

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

Ans.

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

Q.43

$\begin{array}{l}\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{vectors}\text{â€‰}\stackrel{â†’}{\mathrm{a}}\text{â€‰},\stackrel{â†’}{\mathrm{b}},\text{â€‰}\stackrel{â†’}{\mathrm{c}}\mathrm{are}\mathrm{coplanar}\mathrm{if}\stackrel{â†’}{\mathrm{a}\text{â€‰}}+\stackrel{â†’}{\mathrm{b}\text{â€‰}},\stackrel{â†’}{\mathrm{b}\text{â€‰}}+\stackrel{â†’}{\mathrm{c}\text{â€‰}}\text{â€‰}\\ \mathrm{and}\stackrel{â†’}{\mathrm{c}\text{â€‰}}+\stackrel{â†’}{\mathrm{a}\text{â€‰}}\text{â€‰}\mathrm{are}\mathrm{coplanar}.\end{array}$

Ans.

$\begin{array}{l}\left[\stackrel{â†’}{\mathrm{a}\text{â€‰}}+\stackrel{â†’}{\mathrm{b}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{b}\text{â€‰}}+\stackrel{â†’}{\mathrm{c}\text{â€‰}},\stackrel{â†’}{\mathrm{c}\text{â€‰}}+\stackrel{â†’}{\mathrm{a}\text{â€‰}}\right]=\left(\stackrel{â†’}{\mathrm{a}\text{â€‰}}+\stackrel{â†’}{\mathrm{b}\text{â€‰}}\right).\left(\left(\stackrel{â†’}{\mathrm{b}\text{â€‰}}+\stackrel{â†’}{\mathrm{c}\text{â€‰}}\right)Ã—\left(\stackrel{â†’}{\mathrm{c}\text{â€‰}}+\stackrel{â†’}{\mathrm{a}\text{â€‰}}\right)\right)\\ \text{â€‰â€‰}=\left(\stackrel{â†’}{\mathrm{a}\text{â€‰}}+\stackrel{â†’}{\mathrm{b}\text{â€‰}}\right).\left(\stackrel{â†’}{\mathrm{b}\text{â€‰}}Ã—\stackrel{â†’}{\mathrm{c}\text{â€‰}}+\stackrel{â†’}{\mathrm{b}\text{â€‰}}Ã—\stackrel{â†’}{\mathrm{a}\text{â€‰}}+\stackrel{â†’}{\mathrm{c}\text{â€‰}}Ã—\stackrel{â†’}{\mathrm{c}\text{â€‰}}+\stackrel{â†’}{\mathrm{c}\text{â€‰}}Ã—\stackrel{â†’}{\mathrm{a}\text{â€‰}}\right)\\ \text{â€‰â€‰}=\left(\stackrel{â†’}{\mathrm{a}\text{â€‰}}+\stackrel{â†’}{\mathrm{b}\text{â€‰}}\right).\left(\stackrel{â†’}{\mathrm{b}\text{â€‰}}Ã—\stackrel{â†’}{\mathrm{c}\text{â€‰}}+\stackrel{â†’}{\mathrm{b}\text{â€‰}}Ã—\stackrel{â†’}{\mathrm{a}\text{â€‰}}+0+\stackrel{â†’}{\mathrm{c}\text{â€‰}}Ã—\stackrel{â†’}{\mathrm{a}\text{â€‰}}\right)\\ \text{â€‰â€‰}=\stackrel{â†’}{\mathrm{a}\text{â€‰}.}\left(\stackrel{â†’}{\mathrm{b}\text{â€‰}}Ã—\stackrel{â†’}{\mathrm{c}\text{â€‰}}\right)\text{â€‰}+\stackrel{â†’}{\mathrm{a}\text{â€‰}}.\left(\stackrel{â†’}{\mathrm{b}\text{â€‰}}Ã—\stackrel{â†’}{\mathrm{a}\text{â€‰}}\right)\text{â€‰}+\stackrel{â†’}{\mathrm{a}\text{â€‰}}\left(\stackrel{â†’}{\mathrm{c}\text{â€‰}}Ã—\stackrel{â†’}{\mathrm{a}\text{â€‰}}\right)+\stackrel{â†’}{\mathrm{b}\text{â€‰}.}\left(\stackrel{â†’}{\mathrm{b}\text{â€‰}}Ã—\stackrel{â†’}{\mathrm{c}\text{â€‰}}\right)\text{â€‰}\\ +\text{â€‰}\stackrel{â†’}{\mathrm{b}\text{â€‰}}.\left(\stackrel{â†’}{\mathrm{b}\text{â€‰}}Ã—\stackrel{â†’}{\mathrm{a}\text{â€‰}}\right)\text{â€‰}+\stackrel{â†’}{\mathrm{b}\text{â€‰}}.\left(\stackrel{â†’}{\mathrm{c}\text{â€‰}}Ã—\stackrel{â†’}{\mathrm{a}\text{â€‰}}\right)\\ \text{â€‰â€‰}=\left[\stackrel{â†’}{\mathrm{a}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{b}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{c}\text{â€‰}}\right]+\left[\stackrel{â†’}{\mathrm{a}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{b}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{a}\text{â€‰}}\right]+\left[\stackrel{â†’}{\mathrm{a}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{c}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{a}\text{â€‰}}\right]+\left[\stackrel{â†’}{\mathrm{b}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{b}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{c}\text{â€‰}}\right]\\ \text{â€‰â€‰â€‰â€‰â€‰}+\left[\stackrel{â†’}{\mathrm{b}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{b}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{a}\text{â€‰}}\right]+\left[\stackrel{â†’}{\mathrm{b}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{c}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{a}\text{â€‰}}\right]\\ \text{â€‰â€‰}=\left[\stackrel{â†’}{\mathrm{a}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{b}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{c}\text{â€‰}}\right]+0+0+0\text{â€‰}+0+\left[\stackrel{â†’}{\mathrm{a}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{b}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{c}\text{â€‰}}\right]\\ \left[\left[\stackrel{â†’}{\mathrm{a}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{b}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{a}\text{â€‰}}\right]=\left[\stackrel{â†’}{\mathrm{a}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{c}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{a}\text{â€‰}}\right]=\left[\stackrel{â†’}{\mathrm{b}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{b}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{c}\text{â€‰}}\right]=\left[\stackrel{â†’}{\mathrm{b}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{b}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{a}\text{â€‰}}\right]=0\right]\\ \text{â€‰â€‰}=2\left[\stackrel{â†’}{\mathrm{a}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{b}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{c}\text{â€‰}}\right]\\ \mathrm{Since},\stackrel{â†’}{\mathrm{a}\text{â€‰}}+\stackrel{â†’}{\mathrm{b}\text{â€‰}},\stackrel{â†’}{\mathrm{b}\text{â€‰}}+\stackrel{â†’}{\mathrm{c}\text{â€‰}}\text{â€‰}\mathrm{and}\stackrel{â†’}{\mathrm{c}\text{â€‰}}+\stackrel{â†’}{\mathrm{a}\text{â€‰}}\text{â€‰}\mathrm{are}\mathrm{coplanar}.\text{}\mathrm{So},\\ \left[\stackrel{â†’}{\mathrm{a}\text{â€‰}}+\stackrel{â†’}{\mathrm{b}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{b}\text{â€‰}}+\stackrel{â†’}{\mathrm{c}\text{â€‰}},\stackrel{â†’}{\mathrm{c}\text{â€‰}}+\stackrel{â†’}{\mathrm{a}\text{â€‰}}\right]=0\\ â‡’\text{â€‰â€‰}2\left[\stackrel{â†’}{\mathrm{a}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{b}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{c}\text{â€‰}}\right]=0\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰}\left[\stackrel{â†’}{\mathrm{a}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{b}\text{â€‰}},\text{â€‰}\stackrel{â†’}{\mathrm{c}\text{â€‰}}\right]=0\\ \mathrm{So},\text{}\mathrm{the}\mathrm{vectors}\text{â€‰}\stackrel{â†’}{\mathrm{a}}\text{â€‰},\stackrel{â†’}{\mathrm{b}},\text{â€‰}\stackrel{â†’}{\mathrm{c}}\mathrm{are}\mathrm{coplanar}.\end{array}$

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

Ans.

$\begin{array}{l}\text{If is}\stackrel{â†’}{\mathrm{r}}\text{â€‰a unit vector in the XY-plane, thenâ€‰}\stackrel{â†’}{\mathrm{r}}\text{â€‰}=\mathrm{cosÎ¸}\text{â€‰}\stackrel{^}{\mathrm{i}}+\mathrm{sinÎ¸}\stackrel{^}{\mathrm{j}}.\\ \text{Here,}\mathrm{Î¸}\text{is the angle made by the unit vector with the}\\ \text{positive direction of the x-axis.}\\ \text{Therefore, for}\mathrm{Î¸}\text{= 30Â°:}\\ \text{â€‰}\stackrel{â†’}{\mathrm{r}}\text{â€‰}=\mathrm{cos}\text{30Â°â€‰â€‰}\stackrel{^}{\mathrm{i}}+\mathrm{sin}\text{30Â°â€‰}\stackrel{^}{\mathrm{j}}=\frac{\sqrt{3}}{2}\stackrel{^}{\mathrm{i}}+\frac{1}{2}\stackrel{^}{\mathrm{j}}\\ \text{Thus, the required unit vector isâ€‰}\frac{\sqrt{3}}{2}\stackrel{^}{\mathrm{i}}+\frac{1}{2}\stackrel{^}{\mathrm{j}}.\end{array}$

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

Ans.

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

Q.46 A girl walks 4 km towards west, then she walks 3 km in a direction 30Â° east of orth and stops. Determine the girlâ€™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 girlâ€™s position can be shown as:

$\begin{array}{l}\text{Now, we have:}\\ \text{â€‰â€‰}\stackrel{â†’}{\mathrm{OA}}=âˆ’\text{â€‰}4\text{â€‰}\stackrel{^}{\mathrm{i}}\\ \text{â€‰â€‰â€‰}\stackrel{â†’}{\mathrm{AB}}=\text{â€‰}\stackrel{^}{\mathrm{i}}|\stackrel{â†’}{\mathrm{AB}}|\mathrm{cos}60\mathrm{Â°}+\text{â€‰}\stackrel{^}{\mathrm{j}}\text{â€‰}|\stackrel{â†’}{\mathrm{AB}}|\text{sin60Â°}\\ =\text{â€‰}\stackrel{^}{\mathrm{i}}\left(3\right)\frac{1}{2}+\text{â€‰}\stackrel{^}{\mathrm{j}}\left(3\right)\text{}\frac{\sqrt{3}}{2}\\ =\frac{3}{2}\stackrel{^}{\mathrm{i}}+\frac{3\sqrt{3}}{2}\stackrel{^}{\mathrm{j}}\\ \text{By the triangle law of vector addition, we have:}\\ \text{â€‰â€‰â€‰}\stackrel{â†’}{\mathrm{OB}}=\stackrel{â†’}{\mathrm{OA}}+\stackrel{â†’}{\mathrm{AB}}\\ =âˆ’\text{â€‰}4\text{â€‰}\stackrel{^}{\mathrm{i}}+\frac{3}{2}\stackrel{^}{\mathrm{i}}+\frac{3\sqrt{3}}{2}\stackrel{^}{\mathrm{j}}\\ =\left(âˆ’\text{â€‰}4+\frac{3}{2}\right)\stackrel{^}{\mathrm{i}}+\frac{3\sqrt{3}}{2}\stackrel{^}{\mathrm{j}}\\ =âˆ’\frac{5}{2}\stackrel{^}{\mathrm{i}}+\frac{3\sqrt{3}}{2}\stackrel{^}{\mathrm{j}}\\ \text{Hence, the girl’s displacement from her initial point of departure}\\ \text{isâ€‰â€‰}âˆ’\frac{5}{2}\stackrel{^}{\mathrm{i}}+\frac{3\sqrt{3}}{2}\stackrel{^}{\mathrm{j}}.\end{array}$

Q.47

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

Ans.

$\begin{array}{l}\mathrm{In}\text{}\mathrm{Î”ABC},\text{â€‰}\mathrm{let}\text{â€‰}\stackrel{â†’}{\mathrm{CB}}=\stackrel{â†’}{\mathrm{a}},\text{â€‰}\stackrel{â†’}{\mathrm{CA}}=\stackrel{â†’}{\mathrm{b}}\text{and}\stackrel{â†’}{\mathrm{AB}}=\stackrel{â†’}{\mathrm{c}}.\\ \text{Now, by the triangle law of vector addition, we haveâ€‰â€‰â€‰}\stackrel{â†’}{\mathrm{a}}=\stackrel{â†’}{\mathrm{b}}+\stackrel{â†’}{\mathrm{c}}.\\ \mathrm{Since},\text{sum of any two sides of a triangle is always greater than third side.}\\ âˆ´|\stackrel{â†’}{\mathrm{a}}|<|\stackrel{â†’}{\mathrm{b}}|+|\stackrel{â†’}{\mathrm{c}}|\\ \text{Hence, it is not true thatâ€‰}|\stackrel{â†’}{\mathrm{a}}|=|\stackrel{â†’}{\mathrm{b}}|+|\stackrel{â†’}{\mathrm{c}}|.\end{array}$

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.

$\begin{array}{l}\mathrm{x}\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)\mathrm{is}\mathrm{a}\mathrm{unit}\mathrm{vector},\text{}\\ \text{soâ€‰â€‰}|\mathrm{x}\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)|=1\\ â‡’\mathrm{x}|\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)|=1\\ \text{â€‰}\mathrm{x}\sqrt{{1}^{2}+{1}^{2}+{1}^{2}}=1\\ â‡’\mathrm{x}\sqrt{3}=1\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰}\mathrm{x}=Â±\frac{1}{\sqrt{3}}\\ \text{Thus, the required value of x isâ€‰}Â±\frac{1}{\sqrt{3}}.\end{array}$

Q.49

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

Ans.

$\begin{array}{l}\mathrm{The}\text{given}\mathrm{vectorsare}\\ \stackrel{â†’}{\mathrm{a}}=2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}âˆ’\stackrel{^}{\mathrm{k}}\mathrm{and}\stackrel{â†’}{\mathrm{b}}=\stackrel{^}{\mathrm{i}}âˆ’2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\\ \mathrm{Let}\text{}\stackrel{â†’}{\mathrm{c}}\text{be the resultant vectors of}\stackrel{â†’}{\mathrm{a}}\text{and}\stackrel{â†’}{\mathrm{b}}.\\ \mathrm{Then},\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}\stackrel{â†’}{\mathrm{c}}=\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}âˆ’\stackrel{^}{\mathrm{k}}+\stackrel{^}{\mathrm{i}}âˆ’2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\\ =3\text{â€‰}\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}\\ âˆ´|\stackrel{â†’}{\mathrm{c}}|=|3\text{â€‰}\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}|\\ =\sqrt{{3}^{2}+{1}^{2}}\\ =\sqrt{10}\\ \mathrm{So},\text{â€‰}\stackrel{^}{\mathrm{c}}=\frac{3\text{â€‰}\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}}{\sqrt{10}}\\ \text{Thus, the vector of magnitude 5 units and parallel to the}\\ \text{resultant of vectorsâ€‰}\stackrel{â†’}{\mathrm{a}}\text{and}\stackrel{â†’}{\mathrm{b}}\text{â€‰}\mathrm{is}\\ Â±5.\stackrel{^}{\mathrm{c}}=Â±5.\text{â€‰}\frac{3\text{â€‰}\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}}{\sqrt{10}}=Â±\text{â€‰}\frac{3\sqrt{10}}{2}\stackrel{^}{\mathrm{i}}Â±\frac{\sqrt{10}}{2}\stackrel{^}{\mathrm{j}}\end{array}$

Q.50

$\begin{array}{l}\text{Find the angle between two vectors}\stackrel{â†’}{\mathrm{a}}\text{and}\stackrel{â†’}{\mathrm{b}}\text{with}\\ \text{magnitude}\sqrt{\text{3}}\text{andâ€‰â€‰2, respectively having}\stackrel{â†’}{\mathrm{a}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{b}}=\sqrt{\text{6}}.\end{array}$

Ans.

$\begin{array}{l}\text{It is given that,}\\ |\stackrel{â†’}{\mathrm{a}}|=\sqrt{3},|\stackrel{â†’}{\mathrm{b}}|=2\text{and}\stackrel{â†’}{\mathrm{a}}.\stackrel{â†’}{\mathrm{b}}=\sqrt{6}\\ \text{Now, we know that}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\stackrel{â†’}{\mathrm{a}}.\stackrel{â†’}{\mathrm{b}}=|\stackrel{â†’}{\mathrm{a}}||\stackrel{â†’}{\mathrm{b}}|\mathrm{cosÎ¸}\\ â‡’\sqrt{6}=\sqrt{3}.2\mathrm{cosÎ¸}\\ â‡’\mathrm{cosÎ¸}=\frac{\sqrt{6}}{\sqrt{3}.2}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\frac{1}{\sqrt{2}}\end{array}$ $\begin{array}{l}â‡’\mathrm{cosÎ¸}=\mathrm{cos}\frac{\mathrm{Ï€}}{4}\\ â‡’\text{â€‰â€‰â€‰}\mathrm{Î¸}=\frac{\mathrm{Ï€}}{4}\\ \text{Hence, the angle between the given vectorsâ€‰}\stackrel{â†’}{\mathrm{a}}\text{and}\stackrel{â†’}{\mathrm{b}}\text{is}\frac{\mathrm{Ï€}}{4}.\end{array}$

Q.51

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

Ans.

$\begin{array}{l}\text{The given vectors areâ€‰}\stackrel{â†’}{\mathrm{a}}=\stackrel{^}{\mathrm{i}}âˆ’2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\text{and}\stackrel{â†’}{\mathrm{b}}=3\stackrel{^}{\mathrm{i}}âˆ’2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}.\\ \text{â€‰â€‰â€‰â€‰â€‰}|\stackrel{â†’}{\mathrm{a}}|=|\stackrel{^}{\mathrm{i}}âˆ’2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}|\\ =\sqrt{{1}^{2}+{\left(âˆ’2\right)}^{2}+{3}^{2}}\\ =\sqrt{1+4+9}\\ =\sqrt{14}\\ \text{â€‰â€‰â€‰â€‰â€‰}|\stackrel{â†’}{\mathrm{b}}|=|3\stackrel{^}{\mathrm{i}}âˆ’2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}|\\ =\sqrt{{3}^{2}+{\left(âˆ’2\right)}^{2}+{1}^{2}}\\ =\sqrt{9+4+1}\\ =\sqrt{14}\\ \text{Now,}\\ \text{â€‰â€‰â€‰â€‰}\stackrel{â†’}{\mathrm{a}}.\stackrel{â†’}{\mathrm{b}}=\left(\stackrel{^}{\mathrm{i}}âˆ’2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\right).\left(3\stackrel{^}{\mathrm{i}}âˆ’2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)\\ =3+4+3=10\end{array}$ $\begin{array}{l}\text{Also, we know that}\\ \text{â€‰}\stackrel{â†’}{\mathrm{a}}.\stackrel{â†’}{\mathrm{b}}=|\stackrel{â†’}{\mathrm{a}}|.|\stackrel{â†’}{\mathrm{b}}|\mathrm{cosÎ¸}\\ 10=\sqrt{14}.\sqrt{14}\mathrm{cosÎ¸}\\ \mathrm{cosÎ¸}=\frac{10}{14}\\ =\frac{5}{7}\\ \text{â€‰â€‰â€‰â€‰â€‰}\mathrm{Î¸}={\mathrm{cos}}^{âˆ’1}\left(\frac{5}{7}\right)\end{array}$

Q.52

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

Ans.

$\begin{array}{l}\text{Let}\stackrel{â†’}{\mathrm{a}}=\stackrel{^}{\mathrm{i}}âˆ’\stackrel{^}{\mathrm{j}}\text{and}\stackrel{â†’}{\mathrm{b}}=\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}\\ \text{Now, projection of vector}\stackrel{â†’}{\mathrm{a}}\text{â€‰â€‰on}\stackrel{â†’}{\mathrm{b}}\text{â€‰â€‰is given by,}\\ \frac{1}{|\stackrel{â†’}{\mathrm{b}}|}\left(\stackrel{â†’}{\mathrm{a}}.\stackrel{â†’}{\mathrm{b}}\right)=\frac{1}{|\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}|}\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}\right).\left(\stackrel{^}{\mathrm{i}}âˆ’\stackrel{^}{\mathrm{j}}\right)\\ \text{â€‰â€‰â€‰â€‰â€‰}=\frac{1}{\sqrt{{1}^{2}+{1}^{2}}}\left(1âˆ’1\right)=0\\ \text{Hence, the projection of vector}\stackrel{â†’}{\mathrm{a}}\text{â€‰onâ€‰â€‰}\stackrel{â†’}{\mathrm{b}}\text{is 0.}\end{array}$

Q.53

$\begin{array}{l}\text{Find the projection of the vector}\stackrel{^}{\mathrm{i}}+\mathbf{3}\stackrel{^}{\mathrm{j}}+7\stackrel{^}{k}\text{on the vector}\\ 7\text{â€‰}\stackrel{^}{\mathrm{i}}âˆ’\stackrel{^}{\mathrm{j}}+8\text{â€‰}\stackrel{^}{\mathrm{k}}.\end{array}$

Ans.

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

Q.54

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

Ans.

$\begin{array}{l}\text{Letâ€‰â€‰}\stackrel{â†’}{\mathrm{a}}=\frac{1}{7}\left(2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}\right)\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\stackrel{â†’}{\mathrm{b}}=\frac{1}{7}\left(3\text{â€‰}\stackrel{^}{\mathrm{i}}âˆ’6\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)\text{â€‰}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\stackrel{â†’}{\mathrm{c}}=\frac{1}{7}\left(6\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}âˆ’3\stackrel{^}{\mathrm{k}}\right)\\ â‡’\text{â€‰}|\stackrel{â†’}{\mathrm{a}}|=|\frac{1}{7}\left(2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}\right)|\end{array}$ $\begin{array}{l}\text{â€‰â€‰}=\frac{1}{7}\sqrt{{2}^{2}+{3}^{2}+{6}^{2}}\\ \text{â€‰â€‰}=\frac{1}{7}\sqrt{4+9+36}\\ \text{â€‰â€‰}=\frac{7}{7}=1\\ \text{â€‰â€‰â€‰â€‰}|\stackrel{â†’}{\mathrm{b}}|=|\frac{1}{7}\left(3\text{â€‰}\stackrel{^}{\mathrm{i}}âˆ’6\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)\text{â€‰}|\\ \text{â€‰â€‰}=\frac{1}{7}\sqrt{{3}^{2}+{\left(âˆ’6\right)}^{2}+{2}^{2}}\\ \text{â€‰â€‰}=\frac{1}{7}\sqrt{9+36+4}\\ \text{â€‰â€‰}=\frac{7}{7}=1\\ \text{and}\\ \text{â€‰â€‰â€‰â€‰}|\stackrel{â†’}{\mathrm{c}}|=|\frac{1}{7}\left(6\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}âˆ’3\stackrel{^}{\mathrm{k}}\right)\text{â€‰}|\\ \text{â€‰â€‰}=\frac{1}{7}\sqrt{{6}^{2}+{2}^{2}+{\left(âˆ’3\right)}^{2}}\\ \text{â€‰â€‰}=\frac{1}{7}\sqrt{36+4+9}\\ \text{â€‰â€‰}=\frac{7}{7}=1\\ \text{Thus, each of the given three vectors is a unit vector.}\\ \text{Now,}\\ \text{â€‰â€‰}\stackrel{â†’}{\mathrm{a}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{b}}=\frac{1}{7}\left(2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}\right).\frac{1}{7}\left(3\text{â€‰}\stackrel{^}{\mathrm{i}}âˆ’6\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)\\ =\frac{1}{49}\left(6âˆ’18+12\right)=0\end{array}$ $\begin{array}{l}\stackrel{â†’}{\mathrm{b}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{c}}=\frac{1}{7}\left(3\text{â€‰}\stackrel{^}{\mathrm{i}}âˆ’6\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right).\frac{1}{7}\left(6\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}âˆ’3\stackrel{^}{\mathrm{k}}\right)\\ =\frac{1}{49}\left(18âˆ’12âˆ’6\right)=0\\ \stackrel{â†’}{\mathrm{c}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{a}}=\frac{1}{7}\left(6\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}âˆ’3\stackrel{^}{\mathrm{k}}\right).\frac{1}{7}\left(2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}\right)\\ =\frac{1}{49}\left(12+6âˆ’18\right)=0\\ \text{Hence, the given three vectors are mutually perpendicular}\\ \text{to each other.}\end{array}$

Q.55

$\text{Find}\left|\stackrel{â†’}{\text{a}}\right|\text{and}\left|\stackrel{â†’}{\text{b}}\right|\text{, if}\left(\stackrel{â†’}{\text{a}}+\stackrel{â†’}{\text{b}}\right)\text{.}\left(\stackrel{â†’}{\text{a}}â€“\stackrel{â†’}{\text{b}}\right)=\text{8 and}\left|\stackrel{â†’}{\text{a}}\right|=\text{8}\left|\stackrel{â†’}{\text{b}}\right|\text{.}$

Ans.

$\begin{array}{l}\text{We have,}\\ \left(\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}\right).\left(\stackrel{â†’}{\mathrm{a}}âˆ’\stackrel{â†’}{\mathrm{b}}\right)=8\text{and}|\stackrel{â†’}{\mathrm{a}}|=8|\stackrel{â†’}{\mathrm{b}}|\\ \text{So,}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\left(\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}\right).\left(\stackrel{â†’}{\mathrm{a}}âˆ’\stackrel{â†’}{\mathrm{b}}\right)=8\\ â‡’\stackrel{â†’}{\mathrm{a}}.\stackrel{â†’}{\mathrm{a}}âˆ’\stackrel{â†’}{\mathrm{a}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{b}}+\stackrel{â†’}{\mathrm{b}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{a}}âˆ’\stackrel{â†’}{\mathrm{b}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{b}}=8\\ â‡’{|\stackrel{â†’}{\mathrm{a}}|}^{2}âˆ’{|\stackrel{â†’}{\mathrm{b}}|}^{2}=8\left[\stackrel{â†’}{\mathrm{a}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{b}}=\stackrel{â†’}{\mathrm{b}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{a}}\right]\\ â‡’\text{â€‰â€‰â€‰â€‰}{\left(8|\stackrel{â†’}{\mathrm{b}}|\right)}^{2}âˆ’{|\stackrel{â†’}{\mathrm{b}}|}^{2}=8\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰}64{|\stackrel{â†’}{\mathrm{b}}|}^{2}âˆ’{|\stackrel{â†’}{\mathrm{b}}|}^{2}=8\\ \end{array}$ $\begin{array}{l}â‡’\text{}\text{}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}|\stackrel{â†’}{b}|=\sqrt{\frac{8}{63}}=\frac{2\sqrt{2}}{3\sqrt{7}}\\ âˆ´\text{}\text{}\text{}\text{}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}|\stackrel{â†’}{a}|=8|\stackrel{â†’}{b}|\\ \text{}\text{}\text{}\text{}\text{}\text{â€‰}=8Ã—\frac{2\sqrt{2}}{3\sqrt{7}}\\ \text{}\text{}\text{}\text{}\text{}\text{â€‰}=\frac{16\sqrt{2}}{3\sqrt{7}}\end{array}$

Q.56

$\text{Evaluate the product}\left(3\stackrel{â†’}{\mathrm{a}}âˆ’5\stackrel{â†’}{\mathrm{b}}\right).\left(2\stackrel{â†’}{\mathrm{a}}+7\stackrel{â†’}{\mathrm{b}}\right).$

Ans.

$\begin{array}{l}\left(3\stackrel{â†’}{\mathrm{a}}âˆ’5\stackrel{â†’}{\mathrm{b}}\right).\left(2\stackrel{â†’}{\mathrm{a}}+7\stackrel{â†’}{\mathrm{b}}\right)=6\stackrel{â†’}{\mathrm{a}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{a}}+21\stackrel{â†’}{\mathrm{a}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{b}}âˆ’10\stackrel{â†’}{\mathrm{b}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{a}}âˆ’35\text{â€‰}\stackrel{â†’}{\mathrm{b}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{b}}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=6{|\stackrel{â†’}{\mathrm{a}}|}^{2}+21\stackrel{â†’}{\mathrm{a}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{b}}âˆ’10\stackrel{â†’}{\mathrm{b}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{a}}âˆ’35\text{â€‰}{|\stackrel{â†’}{\mathrm{b}}|}^{2}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=6{|\stackrel{â†’}{\mathrm{a}}|}^{2}+11\stackrel{â†’}{\mathrm{a}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{b}}âˆ’35\text{â€‰}{|\stackrel{â†’}{\mathrm{b}}|}^{2}\text{â€‰â€‰â€‰}\left[\stackrel{â†’}{\mathrm{a}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{b}}=\stackrel{â†’}{\mathrm{b}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{a}}\right]\end{array}$

Q.57

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

Ans.

$\begin{array}{l}\text{Let}\mathrm{Î¸}\text{be the angle between the vectors}\stackrel{â†’}{\text{a}}\text{and}\stackrel{â†’}{\text{b}}\text{.â€‰Then,}\\ \stackrel{â†’}{\mathrm{a}}\text{â€‰.â€‰}\stackrel{â†’}{\mathrm{b}}=|\stackrel{â†’}{\mathrm{a}}||\stackrel{â†’}{\mathrm{b}}|\mathrm{cosÎ¸}\\ \text{â€‰â€‰â€‰}\frac{1}{2}=|\stackrel{â†’}{\mathrm{a}}||\stackrel{â†’}{\mathrm{a}}|\mathrm{cos}60Â°\left[|\stackrel{â†’}{\mathrm{a}}|=|\stackrel{â†’}{\mathrm{b}}|\right]\end{array}$ $\begin{array}{l}\text{â€‰â€‰â€‰}\frac{1}{2}={|\stackrel{â†’}{\mathrm{a}}|}^{2}Ã—\frac{1}{2}\\ \text{â€‰â€‰â€‰â€‰}1={|\stackrel{â†’}{\mathrm{a}}|}^{2}\\ â‡’\text{â€‰â€‰}|\stackrel{â†’}{\mathrm{a}}|=1\\ \text{So,â€‰â€‰â€‰â€‰â€‰}|\stackrel{â†’}{\mathrm{b}}|=|\stackrel{â†’}{\mathrm{a}}|=1\end{array}$

Q.58

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

Ans.

$\begin{array}{l}\text{Since,}\\ \left(\stackrel{â†’}{\mathrm{x}}âˆ’\stackrel{â†’}{\mathrm{a}}\right).\left(\stackrel{â†’}{\mathrm{x}}+\stackrel{â†’}{\mathrm{a}}\right)=12\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\stackrel{â†’}{\mathrm{x}}.\stackrel{â†’}{\mathrm{x}}+\stackrel{â†’}{\mathrm{x}}.\stackrel{â†’}{\mathrm{a}}âˆ’\stackrel{â†’}{\mathrm{a}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{x}}âˆ’\stackrel{â†’}{\mathrm{a}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{a}}=12\\ \text{â€‰â€‰}{|\stackrel{â†’}{\mathrm{x}}|}^{2}âˆ’{|\stackrel{â†’}{\mathrm{a}}|}^{2}=12\left[\stackrel{â†’}{\mathrm{x}}.\stackrel{â†’}{\mathrm{a}}=\stackrel{â†’}{\mathrm{a}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{x}}\right]\\ â‡’\text{â€‰â€‰}{|\stackrel{â†’}{\mathrm{x}}|}^{2}âˆ’{1}^{2}=12\left[\stackrel{â†’}{\mathrm{a}}\text{is a unit vector so,}|\stackrel{â†’}{\mathrm{a}}|=1\right]\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{|\stackrel{â†’}{\mathrm{x}}|}^{2}=12+1\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}|\stackrel{â†’}{\mathrm{x}}|=\sqrt{13}\end{array}$

Q.59

$\begin{array}{l}\mathrm{If}\stackrel{â†’}{\mathrm{a}}=\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\text{k}},\text{â€‰â€‰}\stackrel{â†’}{\mathrm{b}}=2\stackrel{^}{\mathrm{i}}âˆ’\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\mathrm{and}\stackrel{â†’}{\mathrm{c}}=\stackrel{^}{\mathrm{i}}âˆ’2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}},\mathrm{find}\mathrm{a}\mathrm{unit}\\ \mathrm{vector}\mathrm{parallel}\mathrm{to}\mathrm{the}\mathrm{vector}2\stackrel{â†’}{\mathrm{a}}âˆ’\stackrel{â†’}{\mathrm{b}}+3\stackrel{â†’}{\mathrm{c}}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\text{given vectors are:}\\ \stackrel{â†’}{\mathrm{a}}=\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}},\text{â€‰â€‰}\stackrel{â†’}{\mathrm{b}}=2\stackrel{^}{\mathrm{i}}âˆ’\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\mathrm{and}\stackrel{â†’}{\mathrm{c}}=\stackrel{^}{\mathrm{i}}âˆ’2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\\ \mathrm{So},\text{â€‰}\\ \text{â€‰â€‰}2\stackrel{â†’}{\mathrm{a}}âˆ’\stackrel{â†’}{\mathrm{b}}+3\stackrel{â†’}{\mathrm{c}}=2\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)âˆ’\left(2\stackrel{^}{\mathrm{i}}âˆ’\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\right)+3\left(\stackrel{^}{\mathrm{i}}âˆ’2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)\\ \text{â€‰â€‰â€‰â€‰}=2\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}âˆ’2\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}âˆ’3\stackrel{^}{\mathrm{k}}+3\stackrel{^}{\mathrm{i}}âˆ’6\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\\ \text{â€‰â€‰â€‰â€‰}=3\text{â€‰}\stackrel{^}{\mathrm{i}}âˆ’3\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\\ |2\stackrel{â†’}{\mathrm{a}}âˆ’\stackrel{â†’}{\mathrm{b}}+3\stackrel{â†’}{\mathrm{c}}|=|3\text{â€‰}\stackrel{^}{\mathrm{i}}âˆ’3\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}|\\ \text{â€‰â€‰â€‰â€‰}=\sqrt{{3}^{2}+{\left(âˆ’3\right)}^{2}+{2}^{2}}\\ \text{â€‰â€‰â€‰â€‰}=\sqrt{9+9+4}\\ \text{â€‰â€‰â€‰â€‰}=\sqrt{22}\\ \text{Thus, the unit vector alongâ€‰}\left(2\stackrel{â†’}{\mathrm{a}}âˆ’\stackrel{â†’}{\mathrm{b}}+3\stackrel{â†’}{\mathrm{c}}\right)\text{â€‰}\mathrm{is}\\ \frac{2\stackrel{â†’}{\mathrm{a}}âˆ’\stackrel{â†’}{\mathrm{b}}+3\stackrel{â†’}{\mathrm{c}}}{|2\stackrel{â†’}{\mathrm{a}}âˆ’\stackrel{â†’}{\mathrm{b}}+3\stackrel{â†’}{\mathrm{c}}|}=\frac{3\text{â€‰}\stackrel{^}{\mathrm{i}}âˆ’3\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}}{\sqrt{22}}\\ \text{â€‰â€‰â€‰â€‰}=\frac{3}{\sqrt{22}}\text{â€‰}\stackrel{^}{\mathrm{i}}âˆ’\frac{3}{\sqrt{22}}\stackrel{^}{\mathrm{j}}+\frac{2}{\sqrt{22}}\stackrel{^}{\mathrm{k}}\end{array}$

Q.60

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

Ans.

$\begin{array}{l}\text{The given vectors are:}\\ \stackrel{â†’}{\mathrm{a}}=2\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}},\stackrel{â†’}{\mathrm{b}}=âˆ’\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\text{and}\stackrel{â†’}{\mathrm{c}}=3\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}\end{array}$ $\begin{array}{l}âˆ´\text{â€‰â€‰â€‰â€‰â€‰}\stackrel{â†’}{\mathrm{a}}+\mathrm{Î»}\stackrel{â†’}{\mathrm{b}}=\left(2\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\right)+\mathrm{Î»}\left(âˆ’\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\left(2âˆ’\mathrm{Î»}\right)\stackrel{^}{\mathrm{i}}+\left(2+2\mathrm{Î»}\right)\stackrel{^}{\mathrm{j}}+\left(3+\mathrm{Î»}\right)\stackrel{^}{\mathrm{k}}\\ \text{Since,â€‰}\left(\stackrel{â†’}{\mathrm{a}}+\mathrm{Î»}\stackrel{â†’}{\mathrm{b}}\right)\text{is perpendicular to}\stackrel{â†’}{\mathrm{c}}\text{, so}\\ \left(\stackrel{â†’}{\mathrm{a}}+\mathrm{Î»}\stackrel{â†’}{\mathrm{b}}\right).\stackrel{â†’}{\mathrm{c}}=0\\ â‡’\left\{\left(2âˆ’\mathrm{Î»}\right)\stackrel{^}{\mathrm{i}}+\left(2+2\mathrm{Î»}\right)\stackrel{^}{\mathrm{j}}+\left(3+\mathrm{Î»}\right)\stackrel{^}{\mathrm{k}}\right\}.\left(3\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}\right)=0\\ â‡’3\left(2âˆ’\mathrm{Î»}\right)+\left(2+2\mathrm{Î»}\right)=0\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}6âˆ’3\mathrm{Î»}+2+2\mathrm{Î»}=0\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}8âˆ’\mathrm{Î»}=0\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{Î»}=8\\ \text{Therefore, the required value of}\mathrm{Î»}\text{is 8.}\end{array}$

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

Ans.

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

Q.62

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

Ans.

$\begin{array}{l}\left(|\stackrel{â†’}{a}|\stackrel{â†’}{b}+|\stackrel{â†’}{b}|\stackrel{â†’}{a}\right).\left(|\stackrel{â†’}{a}|\stackrel{â†’}{b}âˆ’|\stackrel{â†’}{b}|\stackrel{â†’}{a}\right)={|\stackrel{â†’}{a}|}^{2}{|\stackrel{â†’}{b}|}^{2}âˆ’|\stackrel{â†’}{a}|\stackrel{â†’}{b}|\stackrel{â†’}{b}|\stackrel{â†’}{a}+|\stackrel{â†’}{b}|\stackrel{â†’}{a}|\stackrel{â†’}{a}|\stackrel{â†’}{b}âˆ’{|\stackrel{â†’}{b}|}^{2}{|\stackrel{â†’}{a}|}^{2}\\ \text{}\text{}\text{}\text{}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}=âˆ’|\stackrel{â†’}{a}||\stackrel{â†’}{b}|\stackrel{â†’}{b}\stackrel{â†’}{a}+|\stackrel{â†’}{b}||\stackrel{â†’}{a}|\stackrel{â†’}{a}\stackrel{â†’}{b}\\ \text{}\text{}\text{}\text{}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}=0\text{}\text{}\text{}\text{}\text{}\left[\stackrel{â†’}{b}\stackrel{â†’}{a}=\stackrel{â†’}{a}\stackrel{â†’}{b}\right]\\ \text{So,}\text{â€‰}\left(|\stackrel{â†’}{a}|\stackrel{â†’}{b}+|\stackrel{â†’}{b}|\stackrel{â†’}{a}\right)\text{and}\left(|\stackrel{â†’}{a}|\stackrel{â†’}{b}âˆ’|\stackrel{â†’}{b}|\stackrel{â†’}{a}\right)\text{are perpendicular to each other}\text{.}\end{array}$

Q.63

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

Ans.

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

Q.64

$\begin{array}{l}\text{If}\stackrel{â†’}{\mathrm{a}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{a}}=0\text{and}\stackrel{â†’}{\mathrm{a}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{b}}=\text{0, then what can be concluded about}\\ \text{the vector}\stackrel{â†’}{\mathrm{b}}\text{?}\end{array}$

Ans.

$\begin{array}{l}\text{Since, it is given that}\\ \text{â€‰â€‰â€‰â€‰â€‰}\stackrel{â†’}{\mathrm{a}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{a}}=0\\ â‡’{|\stackrel{â†’}{\mathrm{a}}|}^{2}=0\\ â‡’\text{â€‰â€‰â€‰}|\stackrel{â†’}{\mathrm{a}}|=0\\ âˆ´\text{â€‰}\stackrel{â†’}{\mathrm{a}}\text{is a zero vector.}\\ \text{Since, vector}\stackrel{â†’}{\mathrm{b}}\text{satisfying}\stackrel{â†’}{\mathrm{a}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{b}}=0\text{can be any vector.}\end{array}$

Q.65

$\begin{array}{l}\text{If}\stackrel{â†’}{\mathrm{a}},\text{â€‰â€‰}\stackrel{â†’}{\mathrm{b}},\stackrel{â†’}{\mathrm{c}}\text{are unit vectors such that}\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}+\stackrel{â†’}{\mathrm{c}}=\stackrel{â†’}{0},\text{find the}\\ \text{value of}\stackrel{â†’}{\mathrm{a}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{b}}+\stackrel{â†’}{\mathrm{b}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{c}}+\stackrel{â†’}{\mathrm{c}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{a}}.\end{array}$

Ans.

$\begin{array}{l}\text{Given:}\stackrel{â†’}{\mathrm{a}},\text{â€‰â€‰}\stackrel{â†’}{\mathrm{b}},\stackrel{â†’}{\mathrm{c}}\text{are unit vectors such that}\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}+\stackrel{â†’}{\mathrm{c}}=\stackrel{â†’}{0}\\ \text{Then,â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{\left[\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}+\stackrel{â†’}{\mathrm{c}}\right]}^{2}=0\\ {|\stackrel{â†’}{\mathrm{a}}|}^{2}+{|\stackrel{â†’}{\mathrm{b}}|}^{2}+{|\stackrel{â†’}{\mathrm{c}}|}^{2}+2\left[\stackrel{â†’}{\mathrm{a}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{b}}+\stackrel{â†’}{\mathrm{b}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{c}}+\stackrel{â†’}{\mathrm{c}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{a}}\right]=0\\ \text{â€‰â€‰}1+1+1+2\left[\stackrel{â†’}{\mathrm{a}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{b}}+\stackrel{â†’}{\mathrm{b}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{c}}+\stackrel{â†’}{\mathrm{c}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{a}}\right]=0\\ â‡’\stackrel{â†’}{\mathrm{a}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{b}}+\stackrel{â†’}{\mathrm{b}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{c}}+\stackrel{â†’}{\mathrm{c}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{a}}=âˆ’\frac{3}{2}\end{array}$

Q.66

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

Ans.

$\begin{array}{l}\text{Adjacent sides of a parallelogram are:}\\ \stackrel{â†’}{\mathrm{a}}=2\stackrel{^}{\mathrm{i}}âˆ’4\stackrel{^}{\mathrm{j}}+5\stackrel{^}{\mathrm{k}}\text{and}\stackrel{â†’}{\mathrm{b}}=\stackrel{^}{\mathrm{i}}âˆ’2\stackrel{^}{\mathrm{j}}âˆ’3\stackrel{^}{\mathrm{k}}\\ \text{Then, the diagonal of a parallelogram}=\text{â€‰}\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}\\ \text{â€‰â€‰â€‰â€‰â€‰}=2\stackrel{^}{\mathrm{i}}âˆ’4\stackrel{^}{\mathrm{j}}+5\stackrel{^}{\mathrm{k}}+\stackrel{^}{\mathrm{i}}âˆ’2\stackrel{^}{\mathrm{j}}âˆ’3\stackrel{^}{\mathrm{k}}\\ \text{â€‰â€‰â€‰â€‰â€‰}=3\text{â€‰}\stackrel{^}{\mathrm{i}}âˆ’6\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\\ \text{The unit vector parallel to the diagonal}=\frac{3\text{â€‰}\stackrel{^}{\mathrm{i}}âˆ’6\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}}{|3\text{â€‰}\stackrel{^}{\mathrm{i}}âˆ’6\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}|}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\frac{3\text{â€‰}\stackrel{^}{\mathrm{i}}âˆ’6\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}}{\sqrt{{3}^{2}+{\left(âˆ’6\right)}^{2}+{2}^{2}}}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\frac{3\text{â€‰}\stackrel{^}{\mathrm{i}}âˆ’6\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}}{\sqrt{9+36+4}}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\frac{3\text{â€‰}\stackrel{^}{\mathrm{i}}âˆ’6\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}}{\sqrt{49}}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\frac{3}{7}\text{â€‰}\stackrel{^}{\mathrm{i}}âˆ’\frac{6}{7}\stackrel{^}{\mathrm{j}}+\frac{2}{7}\stackrel{^}{\mathrm{k}}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{Now},\text{Area of parallelogram ABCD}=|\stackrel{â†’}{\mathrm{a}}Ã—\stackrel{â†’}{\mathrm{b}}|\\ \text{â€‰â€‰â€‰â€‰â€‰}\stackrel{â†’}{\mathrm{a}}Ã—\stackrel{â†’}{\mathrm{b}}=|\begin{array}{ccc}\stackrel{^}{\mathrm{i}}& \stackrel{^}{\mathrm{j}}& \stackrel{^}{\mathrm{k}}\\ 2& âˆ’4& 5\\ 1& âˆ’2& âˆ’3\end{array}|\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\left(12+10\right)\stackrel{^}{\mathrm{i}}âˆ’\left(âˆ’6âˆ’5\right)\stackrel{^}{\mathrm{j}}+\left(âˆ’4+4\right)\stackrel{^}{\mathrm{k}}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=22\text{â€‰}\stackrel{^}{\mathrm{i}}+11\stackrel{^}{\mathrm{j}}\\ âˆ´\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}|\stackrel{â†’}{\mathrm{a}}Ã—\stackrel{â†’}{\mathrm{b}}|=11|2\text{â€‰}\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}|\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=11\sqrt{{2}^{2}+{1}^{2}}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=11\sqrt{5}\\ \text{Thus, the area of the parallelogram is}11\sqrt{5}\text{â€‰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.

$\begin{array}{l}\text{Let a vector be equally inclined to axes OX, OY, and OZ at}\\ \text{angle}\mathrm{Î±}\text{. Then, the direction cosines of the vector are cos}\mathrm{Î±}\text{,}\\ \text{cos}\mathrm{Î±}\text{, and cos}\mathrm{Î±}\text{.}\\ {\text{Since, cos}}^{\text{2}}\mathrm{Î±}+{\text{cos}}^{\text{2}}\mathrm{Î²}+{\text{cos}}^{\text{2}}\mathrm{Î³}=1\\ \mathrm{So},{\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰cos}}^{\text{2}}\mathrm{Î±}+{\text{cos}}^{\text{2}}\mathrm{Î±}+{\text{cos}}^{\text{2}}\mathrm{Î±}=1\\ \text{â€‰â€‰â€‰â€‰}3{\text{â€‰cos}}^{\text{2}}\mathrm{Î±}=1\\ {\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰cos}}^{\text{2}}\mathrm{Î±}=\frac{1}{3}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰cos}\mathrm{Î±}=\frac{1}{\sqrt{3}}\\ \text{Hence, the direction cosines of the vector which are equally}\\ \text{inclined to the axes are}\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}.\end{array}$

Q.68

$\begin{array}{l}\text{If either vector}\stackrel{â†’}{\mathrm{a}}=0\text{or}\stackrel{â†’}{\mathrm{b}}=0,\text{then}\stackrel{â†’}{\mathrm{a}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{b}}=0.\text{But the converse}\\ \text{need not be true. Justify your answer with an example.}\end{array}$

Ans.

$\begin{array}{l}\text{Let}\stackrel{â†’}{\mathrm{a}}=3\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\text{and}\stackrel{â†’}{\mathrm{b}}=2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}âˆ’6\stackrel{^}{\mathrm{k}}\\ \text{Then,}\\ \text{â€‰â€‰}\stackrel{â†’}{\mathrm{a}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{b}}=\left(3\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\right)\left(2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}âˆ’6\stackrel{^}{\mathrm{k}}\right)\end{array}$

Q.69

$\begin{array}{l}\mathrm{Let}\text{â€‰}\stackrel{â†’}{\mathrm{a}}=\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\text{k}},\text{â€‰â€‰}\stackrel{â†’}{\mathrm{b}}=3\stackrel{^}{\mathrm{i}}âˆ’2\stackrel{^}{\mathrm{j}}+7\stackrel{^}{\mathrm{k}}\mathrm{and}\stackrel{â†’}{\mathrm{c}}=2\stackrel{^}{\mathrm{i}}âˆ’\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}.\mathrm{Find}\mathrm{a}\\ \mathrm{vector}\stackrel{â†’}{\mathrm{d}}\text{â€‰}\mathrm{which}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{both}\stackrel{â†’}{\mathrm{a}}\stackrel{}{\mathrm{a}}\mathrm{nd}\stackrel{â†’}{\mathrm{b}},\stackrel{â†’}{\mathrm{c}}.\stackrel{â†’}{\mathrm{d}}=15.\end{array}$

Ans.

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

Q.70

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

Ans.

$\begin{array}{l}\mathrm{Sum}\mathrm{of}\mathrm{vectors}2\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}âˆ’5\stackrel{^}{\mathrm{k}}\mathrm{and}\mathrm{Î»}\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}=2\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}âˆ’5\stackrel{^}{\mathrm{k}}+\mathrm{Î»}\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\left(2+\mathrm{Î»}\right)\stackrel{^}{\mathrm{i}}+6\text{â€‰}\stackrel{^}{\mathrm{j}}âˆ’2\text{â€‰}\stackrel{^}{\mathrm{k}}\\ \mathrm{Unit}\text{vector of}\left(2+\mathrm{Î»}\right)\stackrel{^}{\mathrm{i}}+6\text{â€‰}\stackrel{^}{\mathrm{j}}âˆ’2\text{â€‰}\stackrel{^}{\mathrm{k}}=\frac{\left(2+\mathrm{Î»}\right)\stackrel{^}{\mathrm{i}}+6\text{â€‰}\stackrel{^}{\mathrm{j}}âˆ’2\text{â€‰}\stackrel{^}{\mathrm{k}}}{|\left(2+\mathrm{Î»}\right)\stackrel{^}{\mathrm{i}}+6\text{â€‰}\stackrel{^}{\mathrm{j}}âˆ’2\text{â€‰}\stackrel{^}{\mathrm{k}}|}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\frac{\left(2+\mathrm{Î»}\right)\stackrel{^}{\mathrm{i}}+6\text{â€‰}\stackrel{^}{\mathrm{j}}âˆ’2\text{â€‰}\stackrel{^}{\mathrm{k}}}{\sqrt{{\left(2+\mathrm{Î»}\right)}^{2}+{6}^{2}+{\left(âˆ’2\right)}^{2}}}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\frac{\left(2+\mathrm{Î»}\right)\stackrel{^}{\mathrm{i}}+6\text{â€‰}\stackrel{^}{\mathrm{j}}âˆ’2\text{â€‰}\stackrel{^}{\mathrm{k}}}{\sqrt{4+4\mathrm{Î»}+{\mathrm{Î»}}^{2}+36+4}}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\frac{\left(2+\mathrm{Î»}\right)\stackrel{^}{\mathrm{i}}+6\text{â€‰}\stackrel{^}{\mathrm{j}}âˆ’2\text{â€‰}\stackrel{^}{\mathrm{k}}}{\sqrt{44+4\mathrm{Î»}+{\mathrm{Î»}}^{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{Î»}\right)\stackrel{^}{\mathrm{i}}+6\text{â€‰}\stackrel{^}{\mathrm{j}}–2\text{â€‰}\stackrel{^}{\mathrm{k}}}{\sqrt{44+4\mathrm{Î»}+{\mathrm{Î»}}^{2}}}=1\\ \left(2+\mathrm{Î»}\right)+6–2=\sqrt{44+4\mathrm{Î»}+{\mathrm{Î»}}^{2}}\\ â‡’\text{â€‰â€‰â€‰}{\left(6+\mathrm{Î»}\right)}^{2}=44+4\mathrm{Î»}+{\mathrm{Î»}}^{2}\\ â‡’\text{â€‰â€‰}36+12\mathrm{Î»}+{\mathrm{Î»}}^{2}=44+4\mathrm{Î»}+{\mathrm{Î»}}^{2}\\ â‡’\text{â€‰â€‰â€‰â€‰}8\mathrm{Î»}=8\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{Î»}=1\\ \mathrm{Thus},\mathrm{the}\mathrm{value}\mathrm{of}\text{â€‰}\mathrm{Î»}\mathrm{is}1.\end{array}$

Q.71

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

Ans.

$\begin{array}{l}\mathrm{Since},\text{}\stackrel{â†’}{\mathrm{a}},\stackrel{â†’}{\mathrm{b}},\stackrel{â†’}{\mathrm{c}}\mathrm{are}\text{â€‰â€‰}\mathrm{mutually}\mathrm{perpendicular}\mathrm{vectors},\text{â€‰}\mathrm{so}\\ \stackrel{â†’}{\mathrm{a}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{b}}=\stackrel{â†’}{\mathrm{b}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{c}}=\stackrel{â†’}{\mathrm{c}}.\stackrel{â†’}{\mathrm{a}}=0\\ \mathrm{Magnitudes}\text{â€¹ of}\stackrel{â†’}{\mathrm{a}},\stackrel{â†’}{\mathrm{b}},\stackrel{â†’}{\mathrm{c}}\mathrm{are}\text{equal, so}\\ |\stackrel{â†’}{\mathrm{a}}|=|\stackrel{â†’}{\mathrm{b}}|=|\stackrel{â†’}{\mathrm{c}}|\\ \mathrm{Let}\text{â€‰}\stackrel{â†’}{\mathrm{a}},\stackrel{â†’}{\mathrm{b}},\stackrel{â†’}{\mathrm{c}}\mathrm{are}\text{making angles}{\mathrm{Î¸}}_{1}\text{,â€‰}{\mathrm{Î¸}}_{2}\text{and}{\mathrm{Î¸}}_{3}\text{withâ€‰}\left(\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}+\stackrel{â†’}{\mathrm{c}}\right)\\ \text{respectively.Then, we have}\\ \text{cos}{\mathrm{Î¸}}_{1}=\frac{\left(\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}+\stackrel{â†’}{\mathrm{c}}\right).\stackrel{â†’}{\mathrm{a}}}{|\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}+\stackrel{â†’}{\mathrm{c}}||\stackrel{â†’}{\mathrm{a}}|}=\frac{{|\stackrel{â†’}{\mathrm{a}}|}^{2}}{|\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}+\stackrel{â†’}{\mathrm{c}}||\stackrel{â†’}{\mathrm{a}}|}=\frac{|\stackrel{â†’}{\mathrm{a}}|}{|\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}+\stackrel{â†’}{\mathrm{c}}|}\\ \text{cos}{\mathrm{Î¸}}_{2}=\frac{\left(\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}+\stackrel{â†’}{\mathrm{c}}\right).\stackrel{â†’}{\mathrm{b}}}{|\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}+\stackrel{â†’}{\mathrm{c}}||\stackrel{â†’}{\mathrm{b}}|}=\frac{{|\stackrel{â†’}{\mathrm{b}}|}^{2}}{|\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}+\stackrel{â†’}{\mathrm{c}}||\stackrel{â†’}{\mathrm{b}}|}=\frac{|\stackrel{â†’}{\mathrm{b}}|}{|\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}+\stackrel{â†’}{\mathrm{c}}|}\\ \text{cos}{\mathrm{Î¸}}_{3}=\frac{\left(\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}+\stackrel{â†’}{\mathrm{c}}\right).\stackrel{â†’}{\mathrm{c}}}{|\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}+\stackrel{â†’}{\mathrm{c}}||\stackrel{â†’}{\mathrm{c}}|}=\frac{{|\stackrel{â†’}{\mathrm{c}}|}^{2}}{|\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}+\stackrel{â†’}{\mathrm{c}}||\stackrel{â†’}{\mathrm{c}}|}=\frac{|\stackrel{â†’}{\mathrm{c}}|}{|\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}+\stackrel{â†’}{\mathrm{c}}|}\\ \mathrm{Since},\text{â€‰}|\stackrel{â†’}{\mathrm{a}}|=|\stackrel{â†’}{\mathrm{b}}|=|\stackrel{â†’}{\mathrm{c}}|,\text{so cos}{\mathrm{Î¸}}_{1}=\text{cos}{\mathrm{Î¸}}_{2}=\text{cos}{\mathrm{Î¸}}_{3}\\ âˆ´\text{â€‰â€‰â€‰â€‰}{\mathrm{Î¸}}_{1}={\mathrm{Î¸}}_{2}={\mathrm{Î¸}}_{3}\\ \text{Hence, the vector}\left(\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}+\stackrel{â†’}{\mathrm{c}}\right)\text{â€‰â€‰is equally inclined toâ€‰â€‰}\stackrel{â†’}{\mathrm{a}},\stackrel{â†’}{\mathrm{b}}\text{and}\stackrel{â†’}{\mathrm{c}}.\end{array}$

Q.72

Ans.

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

Q.73

$\begin{array}{l}\mathrm{If}\mathrm{Î¸}\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{two}\mathrm{vectors}\stackrel{â†’}{\mathrm{a}}\mathrm{and}\stackrel{â†’}{\mathrm{b}},\mathrm{then}\stackrel{â†’}{\mathrm{a}}.\stackrel{â†’}{\mathrm{b}}â‰¥0\\ \mathrm{only}\mathrm{when}\\ \left(\mathrm{A}\right)0<\mathrm{Î¸}<\frac{\mathrm{Ï€}}{2}\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\\ \left(\mathrm{B}\right)0â‰¤\mathrm{Î¸}â‰¤\frac{\mathrm{Ï€}}{2}\\ \left(\mathrm{C}\right)0<\mathrm{Î¸}<\mathrm{Ï€}\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\\ \left(\mathrm{D}\right)0â‰¤\mathrm{Î¸}â‰¤\mathrm{Ï€}\end{array}$

Ans.

Q.74

Ans.

$\begin{array}{l}\text{The vertices of Î”ABC are given as A (1, 2, 3), B (â€“1, 0, 0),}\\ \text{and C (0, 1, 2)}\text{.}\end{array}$ $\begin{array}{l}\end{array}$

Q.75

Ans.

$\begin{array}{l}\text{The vertices of Î”ABC are given as A (1, 2, 3), B (â€“1, 0, 0),}\\ \text{and C (0, 1, 2).}\end{array}$ $\begin{array}{l}\text{â€‰}|\stackrel{â†’}{\text{AC}}|\text{=}|\text{2}\stackrel{^}{\text{i}}\text{+ 8}\stackrel{^}{\text{j}}\text{â€“ 8}\stackrel{^}{\text{k}}|\\ \text{=}\sqrt{{\text{2}}^{\text{2}}{\text{+ 8}}^{\text{2}}\text{+}{\left(\text{â€“8}\right)}^{\text{2}}}\\ \text{=}\sqrt{\text{4 + 64 + 64}}\\ \text{=}\sqrt{\text{132}}\text{= 2}\sqrt{\text{33}}\\ âˆ´\text{â€‰}|\stackrel{â†’}{\text{AC}}|\text{=}|\stackrel{â†’}{\text{AB}}|\text{+}|\stackrel{â†’}{\text{BC}}|\\ \text{Hence, the given points A, B, and C are collinear.}\end{array}$

Q.76

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

Ans.

$\begin{array}{l}\text{Let vectors}2\stackrel{^}{\mathrm{i}}âˆ’\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}},\stackrel{^}{\mathrm{i}}âˆ’3\stackrel{^}{\mathrm{j}}âˆ’5\stackrel{^}{\mathrm{k}}\mathrm{a}\mathrm{nd}3\stackrel{^}{\mathrm{i}}âˆ’4\stackrel{^}{\mathrm{j}}âˆ’4\stackrel{^}{\mathrm{k}}\text{be position vectors}\\ \text{of points A, B and C respectively.}\\ \text{i.e.,}\stackrel{â†’}{\mathrm{OA}}=2\stackrel{^}{\mathrm{i}}âˆ’\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}},\text{â€‰â€‰}\stackrel{â†’}{\mathrm{OB}}=\stackrel{^}{\mathrm{i}}âˆ’3\stackrel{^}{\mathrm{j}}âˆ’5\stackrel{^}{\mathrm{k}},\text{â€‰â€‰}\stackrel{â†’}{\mathrm{OC}}=3\stackrel{^}{\mathrm{i}}âˆ’4\stackrel{^}{\mathrm{j}}âˆ’4\stackrel{^}{\mathrm{k}}\\ âˆ´\stackrel{â†’}{\mathrm{AB}}=\left(1âˆ’2\right)\stackrel{^}{\mathrm{i}}+\left(âˆ’3+1\right)\stackrel{^}{\mathrm{j}}+\left(âˆ’5âˆ’1\right)\stackrel{^}{\mathrm{k}}=âˆ’\stackrel{^}{\mathrm{i}}âˆ’2\stackrel{^}{\mathrm{j}}âˆ’6\text{â€‰}\stackrel{^}{\mathrm{k}}\\ \text{â€‰â€‰â€‰}\stackrel{â†’}{\mathrm{BC}}=\left(3âˆ’1\right)\stackrel{^}{\mathrm{i}}+\left(âˆ’4+3\right)\stackrel{^}{\mathrm{j}}+\left(âˆ’4+5\right)\stackrel{^}{\mathrm{k}}=2\stackrel{^}{\mathrm{i}}âˆ’\stackrel{^}{\mathrm{j}}+\text{â€‰}\stackrel{^}{\mathrm{k}}\\ \text{â€‰â€‰â€‰}\stackrel{â†’}{\mathrm{CA}}=\left(2âˆ’3\right)\stackrel{^}{\mathrm{i}}+\left(âˆ’1+4\right)\stackrel{^}{\mathrm{j}}+\left(1+4\right)\stackrel{^}{\mathrm{k}}=âˆ’\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+5\text{â€‰}\stackrel{^}{\mathrm{k}}\\ |\stackrel{â†’}{\mathrm{AB}}|=|âˆ’\stackrel{^}{\mathrm{i}}âˆ’2\stackrel{^}{\mathrm{j}}âˆ’6\text{â€‰}\stackrel{^}{\mathrm{k}}|\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\sqrt{{\left(âˆ’1\right)}^{2}+{\left(âˆ’2\right)}^{2}+{\left(âˆ’6\right)}^{2}}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\sqrt{1+4+36}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\sqrt{41}\\ |\stackrel{â†’}{\mathrm{BC}}|=|2\stackrel{^}{\mathrm{i}}âˆ’\stackrel{^}{\mathrm{j}}+\text{â€‰}\stackrel{^}{\mathrm{k}}|\end{array}$ $\begin{array}{l}\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\sqrt{{2}^{2}+{\left(âˆ’1\right)}^{2}+{1}^{2}}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\sqrt{4+1+1}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\sqrt{6}\\ |\stackrel{â†’}{\mathrm{CA}}|=|âˆ’\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+5\text{â€‰}\stackrel{^}{\mathrm{k}}|\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\sqrt{{\left(âˆ’1\right)}^{2}+{3}^{2}+{5}^{2}}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\sqrt{1+9+25}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\sqrt{35}\\ âˆ´{|\stackrel{â†’}{\mathrm{BC}}|}^{2}+{|\stackrel{â†’}{\mathrm{CA}}|}^{2}={\left(\sqrt{6}\right)}^{2}+{\left(\sqrt{35}\right)}^{2}\end{array}$ $\begin{array}{l}=6+35\\ =41\\ ={|\stackrel{â†’}{\mathrm{AB}}|}^{2}\\ \text{Hence,}\mathrm{Î”}\text{ABC is a right-angled triangle.}\end{array}$

Q.77

$\begin{array}{l}\text{If}\stackrel{â†’}{\text{a}}\text{is a non zero vector of magnitude ‘a’ and Î» a non zero}\\ \text{scalar, then Î»â€‰}\stackrel{â†’}{\text{a}}\text{is unit vector if}\\ \left(\text{A}\right)\text{â€‰â€‰Î» = 1}\\ \left(\text{B}\right)\text{â€‰â€‰Î» = â€“1}\\ \left(\text{C}\right)\text{â€‰a =}|\text{Î»}|\text{}\\ \left(\text{D}\right)\text{a =}\frac{\text{1}}{|\text{Î»}|}\end{array}$

Ans.

$\begin{array}{l}\text{Vector Î»}\stackrel{â†’}{\text{a}}\text{â€‰â€‰is a unit vector if}|\text{Î»}\stackrel{â†’}{\text{a}}|\text{= 1.}\\ \text{So,â€‰â€‰}|\text{Î»}\stackrel{â†’}{\text{a}}|\text{= 1}\\ â‡’\text{â€‰â€‰â€‰â€‰}|\text{Î»}||\stackrel{â†’}{\text{a}}|\text{= 1}\end{array}$

Q.78

$\begin{array}{l}\text{Let â€¹â€¹and â€¹â€¹â€¹beâ€¹two â€¹unit â€¹vectors â€¹andâ€¹}\mathrm{Î¸}\text{â€¹isâ€¹the â€¹angle â€¹between}\\ \text{them}.\text{â€¹Then}\stackrel{â†’}{\mathrm{a}}\text{â€¹â€¹}+\stackrel{â†’}{\mathrm{b}}\text{â€¹â€¹is â€¹a â€¹unit â€¹vectorâ€¹if}\\ \left(\mathrm{A}\right)\text{}\mathrm{Î¸}=\frac{\mathrm{Ï€}}{4}\\ \left(\mathrm{B}\right)\text{}\mathrm{Î¸}=\frac{\mathrm{Ï€}}{3}\\ \left(\mathrm{C}\right)\text{}\mathrm{Î¸}=\frac{\mathrm{Ï€}}{2}\\ \left(\mathrm{D}\right)\text{}\mathrm{Î¸}=\frac{2\mathrm{Ï€}}{3}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Let}\stackrel{â†’}{\mathrm{a}}\mathrm{and}\stackrel{â†’}{\mathrm{b}}\text{â€‰â€‰}\mathrm{be}\mathrm{two}\mathrm{unit}\mathrm{vectors}\text{i.e.,}|\stackrel{â†’}{\mathrm{a}}|=|\stackrel{â†’}{\mathrm{b}}|=1\\ \mathrm{and}\mathrm{Î¸}\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{between}\text{}\mathrm{them}.\\ \mathrm{Then}\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}\mathrm{is}\mathrm{a}\mathrm{unit}\mathrm{vector}\text{if}|\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}|=1.\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}|\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}|=1â‡’{|\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}|}^{2}=1\\ â‡’\left(\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}\right).\left(\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}\right)=1\\ â‡’\stackrel{â†’}{\mathrm{a}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{a}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{b}}+\stackrel{â†’}{\mathrm{b}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{b}}=1\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{|\stackrel{â†’}{\mathrm{a}}|}^{2}\text{â€‰}+2\stackrel{â†’}{\mathrm{a}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{b}}+\text{â€‰}{|\stackrel{â†’}{\mathrm{b}}|}^{2}=1\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}1+2\stackrel{â†’}{\mathrm{a}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{b}}+1=1\left[|\stackrel{â†’}{\mathrm{a}}|=|\stackrel{â†’}{\mathrm{b}}|=1\right]\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\stackrel{â†’}{\mathrm{a}}\text{â€‰}.\text{â€‰}\stackrel{â†’}{\mathrm{b}}=âˆ’\frac{1}{2}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}|\stackrel{â†’}{\mathrm{a}}|\text{â€‰}.\text{â€‰}|\stackrel{â†’}{\mathrm{b}}|\mathrm{cosÎ¸}=âˆ’\frac{1}{2}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}1\text{â€‰}.\text{â€‰}1\mathrm{cosÎ¸}=âˆ’\frac{1}{2}\left[|\stackrel{â†’}{\mathrm{a}}|=|\stackrel{â†’}{\mathrm{b}}|=1\right]\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{cosÎ¸}=\mathrm{cos}\frac{2\mathrm{Ï€}}{3}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{Î¸}=\frac{2\mathrm{Ï€}}{3}\\ \mathrm{Thus},\text{}\left(\stackrel{â†’}{\mathrm{a}}+\stackrel{â†’}{\mathrm{b}}\right)\text{is unit vector if}\mathrm{Î¸}=\frac{2\mathrm{Ï€}}{3}.\\ \mathrm{Then},\text{the correct option is D.}\end{array}$

Q.79

$\begin{array}{l}\text{Theâ€¹valueâ€¹of}\stackrel{^}{\mathrm{i}}.\text{â€¹}\left(\stackrel{^}{\mathrm{j}}\text{}Ã—\text{â€¹}\stackrel{^}{\mathrm{k}}\right)\text{â€¹}+\text{â€¹}\stackrel{^}{\mathrm{j}}.\text{â€¹}\left(\stackrel{^}{\mathrm{i}}\text{â€¹}Ã—\text{}\stackrel{^}{\mathrm{k}}\right)+\text{â€¹}\stackrel{^}{\mathrm{k}}.\text{â€¹}\left(\stackrel{^}{\mathrm{i}}\text{â€¹}Ã—\text{â€¹}\stackrel{^}{\mathrm{j}}\right)\text{â€¹is}\\ \begin{array}{l}\left(\text{A}\right)\text{}0\\ \left(\text{B}\right)\text{}âˆ’1\\ \left(\text{C}\right)\text{}1\\ \left(\text{D}\right)\text{}3\end{array}\end{array}$

Ans.

$\begin{array}{l}\stackrel{^}{\mathrm{i}}.\left(\stackrel{^}{\mathrm{j}}Ã—\stackrel{^}{\mathrm{k}}\right)+\stackrel{^}{\mathrm{j}}.\left(\stackrel{^}{\mathrm{i}}Ã—\stackrel{^}{\mathrm{k}}\right)+\stackrel{^}{\mathrm{k}}.\left(\stackrel{^}{\mathrm{i}}Ã—\stackrel{^}{\mathrm{j}}\right)=\stackrel{^}{\mathrm{i}}\text{â€‰}.\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}\text{â€‰}.\left(âˆ’\stackrel{^}{\mathrm{j}}\right)+\stackrel{^}{\mathrm{k}}\text{â€‰}.\text{â€‰}\stackrel{^}{\mathrm{k}}\left[\begin{array}{l}\stackrel{^}{\mathrm{j}}Ã—\stackrel{^}{\mathrm{k}}=\stackrel{^}{\mathrm{i}},\text{}\stackrel{^}{\mathrm{i}}Ã—\stackrel{^}{\mathrm{k}}=\stackrel{^}{\mathrm{j}}\\ \stackrel{^}{\mathrm{i}}Ã—\stackrel{^}{\mathrm{j}}=\stackrel{^}{\mathrm{k}}\end{array}\right]\\ \text{â€‰â€‰â€‰}=1âˆ’1+1\left[\stackrel{^}{\mathrm{i}}\text{â€‰}.\stackrel{^}{\mathrm{i}}=\stackrel{^}{\mathrm{j}}\text{â€‰}.\stackrel{^}{\mathrm{j}}\text{â€‰}=\stackrel{^}{\mathrm{k}}\text{â€‰}.\text{â€‰}\stackrel{^}{\mathrm{k}}=1\right]\\ \text{â€‰â€‰â€‰}=1\\ \text{Thus, the correct option is C.}\end{array}$

Q.80

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

Ans.