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.

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

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

Key Topics Covered in NCERT Solutions Class 12 Mathematics Chapter 10

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

List of NCERT Solutions Class 12 Mathematics Chapter 10 Exercises

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

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

Class 12 Mathematics Chapter 10 – Ex 10.1 Solutions – 5 Questions

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

Class 12 Mathematics Chapter 10 – 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.

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

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

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

NCERT Mathematics Syllabus

Class 12 NCERT Mathematics Syllabus

Term – 1

Unit Name

Chapter Name

Relations and Function

Relations and Functions

Inverse Trigonometric Functions

Algebra

Matrices

Determinants

Calculus

Continuity and Differentiability

Application of Derivatives

Linear Programming

Term – 2

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

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

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

NCERT Mathematics Exam Pattern

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

NCERT Exemplar Class 12 Mathematics

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

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

Key Features of NCERT Solutions Class 12 Mathematics Chapter 10

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

The following are the important aspects to consider:

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

Q.1

$Find | \vec{a} \times \vec{b} |, if \vec{a} = \hat{i} - 7 \hat{j} + 7 \hat{k} and \vec{b} = 3 \hat{i} - 2 \hat{j} + 2 \hat{k} .$

Ans.

$\begin{array}{l} Given vectors are: \vec{a} = \hat{i} - 7 \hat{j} + 7 \hat{k} and \vec{b} = 3 \hat{i} - 2 \hat{j} + 2 \hat{k} \\ Then, \\ \vec{a} \times \vec{b} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 7 & 7 \\ 3 & - 2 & 2 \end{matrix} | \\ = (- 14 + 14) \hat{i} + (2 - 21) \hat{j} + (- 2 + 21) \hat{k} \\ = - 19 \hat{j} + 19 \hat{k} \\ ∴ | \vec{a} \times \vec{b} | = | - 19 \hat{j} + 19 \hat{k} | \\ = \sqrt{{(- 19)}^{2} + {(19)}^{2}} \\ = 2 \sqrt{19} \end{array}$

Q.2

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

Ans.

$\begin{array}{l} The given vectors are \vec{a} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k} and \vec{b} = \hat{i} + 2 \hat{j} - 2 \hat{k} . \\ Then, \vec{a} + \vec{b} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k} + \hat{i} + 2 \hat{j} - 2 \hat{k} \\ = 4 \hat{i} + 4 \hat{j} \\ and \vec{a} - \vec{b} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k} - \hat{i} - 2 \hat{j} + 2 \hat{k} \\ = 2 \hat{i} + 4 \hat{k} \\ (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) \\ = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{matrix} | \\ = (16 - 0) \hat{i} - (16 - 0) \hat{j} + (0 - 8) \hat{k} \\ = 16 \hat{i} - 16 \hat{j} - 8 \hat{k} \\ | (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) | \\ = | 16 \hat{i} - 16 \hat{j} - 8 \hat{k} | \\ = \sqrt{{(16)}^{2} + {(- 16)}^{2} + {(- 8)}^{2}} \\ = 24 \\ Hence, the unit vector perpendicular to each of the vectors \\ (\vec{a} + \vec{b}) and (\vec{a} - \vec{b}) is given by the relation, \\ = \pm \frac{(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})}{| (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) |} \\ = \pm \frac{16 \hat{i} - 16 \hat{j} - 8 \hat{k}}{24} \\ = \pm \frac{2 \hat{i} - 2 \hat{j} - \hat{k}}{3} \end{array}$

Q.3

$\begin{array}{l} If a unit vector \vec{a} makes angles \frac{π}{3} with \hat{i}, \frac{π}{4} with \hat{j} and an \\ acute angle θ with \hat{k}, then find θ and hence, the components of \vec{a} . \end{array}$

Ans.

$\begin{array}{l} Let unit vector \vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k} and | \vec{a} | = 1 . \\ It is given that \vec{a} makes angles \frac{π}{3} with \hat{i}, \frac{π}{4} with \hat{j} and an acute \\ angle θ with \hat{k} . \\ Then, we have: \\ cos \frac{π}{3} = \frac{a_{1}}{| \vec{a} |} \Rightarrow \frac{1}{2} = \frac{a_{1}}{1} \\ \Rightarrow a_{1} = \frac{1}{2} \\ cos \frac{π}{4} = \frac{a_{2}}{| \vec{a} |} \Rightarrow \frac{1}{\sqrt{2}} = \frac{a_{2}}{1} \\ \Rightarrow a_{2} = \frac{1}{\sqrt{2}} \\ cos θ = \frac{a_{3}}{| \vec{a} |} \Rightarrow cos θ = \frac{a_{3}}{1} \\ \Rightarrow a_{3} = cos θ \\ Since, | \vec{a} | = 1 \\ \Rightarrow \sqrt{{a_{1}}^{2} + {a_{2}}^{2} + {a_{3}}^{2}} = 1 \\ \Rightarrow {a_{1}}^{2} + {a_{2}}^{2} + {a_{3}}^{2} = 1 \\ {(\frac{1}{2})}^{2} + {(\frac{1}{\sqrt{2}})}^{2} + {cos}^{2} θ = 1 \\ {cos}^{2} θ = 1 - \frac{1}{4} - \frac{1}{2} \\ {cos}^{2} θ = \frac{1}{4} \\ \Rightarrow cos θ = \frac{1}{2} = \cos \frac{π}{3} \\ So, θ = \frac{π}{3} \\ Thus, θ = \frac{π}{3} and and the components of \vec{a} are (\frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2}) . \end{array}$

Q.4

$Show that (\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = 2 (\vec{a} \times \vec{b})$

Ans.

$\begin{array}{l} L . H . S . = (\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) \\ = \vec{a} \times \vec{a} + \vec{a} \times \vec{b} - \vec{b} \times \vec{a} - \vec{b} \times \vec{b} \\ = 0 + \vec{a} \times \vec{b} + \vec{a} \times \vec{b} - 0 [\vec{b} \times \vec{a} = \vec{a} \times \vec{b}] \\ = 2 (\vec{a} \times \vec{b}) = R . H . S . \end{array}$

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

$\begin{array}{l} Here, vector \vec{OP} represents the displacement of 40 km, 30° \\ East of North. \end{array}$

Q.6 Classify the following measures as scalars and vectors. (i) 10 kg
(ii) 2 meters north-west
(iii) 40°
(iv) 40 watt
(v) 10^-19 coulomb
(vi) 20 m/s²

Ans.

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

Q.7 Classify the following as scalar and vector quantities. (i) time period
(ii) distance
(iii) force
(iv) velocity
(v) work done

Ans.

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

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

Ans.

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

Q.9

$Answer the following as true or false. \begin{array}{l} (i) \vec{a} and - \vec{a} are collinear . \\ (ii) Two collinear vectors are always equal inmagnitude . \\ (iii) Two vectors having same magnitude are collinear . \\ (iv) Two collinear vectors having the same magnitude \\ are equal . \end{array}$

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

$Find λ and μ if (2 \hat{i} + 6 \hat{j} + 27 \hat{k}) \times (\hat{i} + λ \hat{j} + μ \hat{k}) = 0 .$

Ans.

$\begin{array}{l} (2 \hat{i} + 6 \hat{j} + 27 \hat{k}) \times (\hat{i} + λ \hat{j} + μ \hat{k}) = 0 \\ \Rightarrow | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 27 \\ 1 & λ & μ \end{matrix} | = 0 \\ (6 μ - 27 λ) \hat{i} - (2 μ - 27) \hat{j} + (2 λ - 6) \hat{k} = 0 \hat{i} + 0 \hat{j} + 0 \hat{k} \\ On comparing the corresponding components, we have : \\ 6 μ - 27 λ = 0, 2 μ - 27 = 0 and 2 λ - 6 = 0 \\ \Rightarrow μ = \frac{27}{2} and λ = 3 \end{array}$

Q.11

$\begin{array}{l} Compute the magnitude of the following vectors: \\ \vec{a} = \hat{i} + \hat{j} + \hat{k}; \vec{b} = 2 \hat{i} - 7 \hat{j} - 3 \hat{k}; \vec{c} = \frac{1}{\sqrt{3}} \hat{i} + \frac{1}{\sqrt{3}} \hat{j} - \frac{1}{\sqrt{3}} \hat{k}; \end{array}$

Ans.

$\begin{array}{l} The given vectors are: \\ \vec{a} = \hat{i} + \hat{j} + \hat{k}; \vec{b} = 2 \hat{i} - 7 \hat{j} - 3 \hat{k}; \vec{c} = \frac{1}{\sqrt{3}} \hat{i} + \frac{1}{\sqrt{3}} \hat{j} - \frac{1}{\sqrt{3}} \hat{k} \\ Magnitude of \vec{a} = | \vec{a} | \\ = \sqrt{{(1)}^{2} + {(1)}^{2} + {(1)}^{2}} \\ = \sqrt{3} \\ Magnitude of \vec{b} = | \vec{b} | \\ = \sqrt{{(2)}^{2} + {(- 7)}^{2} + {(- 3)}^{2}} \\ = \sqrt{62} \\ Magnitude of \vec{c} = | \vec{c} | \\ = \sqrt{{(\frac{1}{\sqrt{3}})}^{2} + {(\frac{1}{\sqrt{3}})}^{2} + {(\frac{1}{\sqrt{3}})}^{2}} \\ = \sqrt{\frac{3}{3}} = 1 \end{array}$

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

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

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

$\begin{array}{l} \begin{array}{l} Two different vectors are: \end{array} \\ \begin{array}{l} \vec{a} = 2 \hat{i} + 2 \hat{j} + 2 \hat{k}; \vec{b} = \hat{i} + \hat{j} + \hat{k} \end{array} \\ \begin{array}{l} Then, \end{array} \\ \begin{array}{l} Magnitudeof \vec{a} \end{array} = \begin{array}{l} |2 \hat{i} + 2 \hat{j} + 2 \hat{k}| \end{array} \\ \begin{array}{l} = \sqrt{2^{2} + 2^{2} + 2^{2}} \end{array} \\ \begin{array}{l} = \sqrt{4 + 4 + 4} \end{array} \\ \begin{array}{l} = \sqrt{12} \end{array} \\ = \begin{array}{l} 2 \sqrt{3} \end{array} \\ Direction cosinesof \vec{a} are : \\ \begin{array}{l} \frac{2}{2 \sqrt{3}}, \frac{2}{2 \sqrt{3}}, \frac{2}{2 \sqrt{3}} i . e ., \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \end{array} \\ \begin{array}{l} Magnitudeof \vec{b} = |\hat{i} + \hat{j} + \hat{k}| \end{array} \\ \begin{array}{l} = \sqrt{1^{2} + 1^{2} + 1^{2}} \end{array} \\ \begin{array}{l} = \sqrt{1 + 1 + 1} \end{array} \\ \begin{array}{l} = \sqrt{3} \end{array} \\ \begin{array}{l} Direction cosinesof \vec{b} are: \end{array} \\ \begin{array}{l} \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \end{array} \\ \begin{array}{l} Direction cosinesof \vec{a} and \vec{b} aresame.Hence,these two \end{array} \\ \begin{array}{l} vectorshavesamedirection. \end{array} \end{array}$

Q.14

$\begin{array}{l} Find the values of x and y so that the vectors 2 \hat{i} + 3 \hat{j} and \\ x \hat{i} + y \hat{j} are equal. \end{array}$

Ans.

$\begin{array}{l} Since, \\ 2 \hat{i} + 3 \hat{j} = x \hat{i} + y \hat{j} \\ So, on comparing coefficients of \hat{i} and \hat{j}, we get \\ x = 2 and y = 3 . \end{array}$

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

Ans.

$\begin{array}{l} The vector with the initial point P (2, 1) and terminal point \\ Q (-5, 7) can be given by, \\ \vec{PQ} = (- 5 - 2) \hat{i} + (7 - 1) \hat{j} \\ = - 7 \hat{i} + 6 \hat{j} \\ Hence, the required scalar components are -7 and 6 while the \\ vector components are - 7 \hat{i} and 6 \hat{j} . \end{array}$

Q.16

$\begin{array}{l} Given that \vec{a} . \vec{b} = 0 and \vec{a} \times \vec{b} = 0 . What can you conclude \\ about the vectors \vec{a} and ‹ \vec{b} ? \end{array}$

Ans.

$\begin{array}{l} (i) \vec{a} . \vec{b} = 0 \\ Either | \vec{a} | = 0 or | \vec{b} | = 0, \\ or \vec{a} ⊥ \vec{b} (if \vec{a} and \vec{b} are non-zero.) \\ (ii) \vec{a} \times \vec{b} = 0 \\ Either | \vec{a} | = 0 or | \vec{b} | = 0, \\ or \vec{a} \vec{b} (if \vec{a} and \vec{b} are non-zero.) \\ But, \vec{a} and \vec{b} cannot be perpendicular and parallel simultaneously. \\ Hence, ‹ | \vec{a} | = 0 or | \vec{b} | = 0 . \end{array}$

Q.17

$\begin{array}{l} Find the sum of the vectors \vec{a} = \hat{i} - 2 \hat{j} + \hat{k}, \vec{b} = - 2 \hat{i} + 4 \hat{j} + 5 \hat{k} \\ and \vec{c} = \hat{i} - 6 \hat{j} - 7 \hat{k} . \end{array}$

Ans.

$\begin{array}{l} The given vectors are: \\ \vec{a} = \hat{i} - 2 \hat{j} + \hat{k}, \vec{b} = - 2 \hat{i} + 4 \hat{j} + 5 \hat{k} and \vec{c} = \hat{i} - 6 \hat{j} - 7 \hat{k} \\ Then, \\ \vec{a} + \vec{b} + \vec{c} = \hat{i} - 2 \hat{j} + \hat{k} + (- 2 \hat{i} + 4 \hat{j} + 5 \hat{k}) + \hat{i} - 6 \hat{j} - 7 \hat{k} \\ = - 4 \hat{j} - \hat{k} \end{array}$

Q.18

$\begin{array}{l} Find the unit vector in the direction of the vector \\ \vec{a} = \hat{i} + \hat{j} + 2 \hat{k} . \end{array}$

Ans.

$\begin{array}{l} The unit vector \hat{a} in the direction of vector \vec{a} = \hat{i} + \hat{j} + 2 \hat{k} is given \\ by \hat{a} = \frac{\vec{a}}{| \vec{a} |} \\ = \frac{\hat{i} + \hat{j} + 2 \hat{k}}{| \hat{i} + \hat{j} + 2 \hat{k} |} \\ = \frac{\hat{i} + \hat{j} + 2 \hat{k}}{\sqrt{1^{2} + 1^{2} + 2^{2}}} \\ = \frac{\hat{i} + \hat{j} + 2 \hat{k}}{\sqrt{6}} \\ = \frac{\hat{i}}{\sqrt{6}} + \frac{\hat{j}}{\sqrt{6}} + \frac{2}{\sqrt{6}} \hat{k} \end{array}$

Q.19

$\begin{array}{l} Find the unit vector in the direction of the vector \vec{PQ}, \\ where P and Q are the points (1,2,3) and (4,5,6) \\ respectively. \end{array}$

Ans.

$\begin{array}{l} Here, \vec{OP} = \hat{i} + 2 \hat{j} + 3 \hat{k} and \vec{OQ} = 4 \hat{i} + 5 \hat{j} + 6 \hat{k} \\ So, \vec{PQ} = \vec{OQ} - \vec{OP} \\ = (4 \hat{i} + 5 \hat{j} + 6 \hat{k}) - (\hat{i} + 2 \hat{j} + 3 \hat{k}) \\ = 3 \hat{i} + 3 \hat{j} + 3 \hat{k} \\ Unitvectorof \vec{PQ} \\ = \frac{\vec{PQ}}{|\vec{PQ}|} \\ = \frac{3 \hat{i} + 3 \hat{j} + 3 \hat{k}}{|3 \hat{i} + 3 \hat{j} + 3 \hat{k}|} \\ = \frac{3 \hat{i} + 3 \hat{j} + 3 \hat{k}}{\sqrt{3^{2} + 3^{2} + 3^{2}}} \\ = \frac{3 \hat{i} + 3 \hat{j} + 3 \hat{k}}{3 \sqrt{3}} \\ = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}} \\ = \frac{\hat{i}}{\sqrt{3}} + \frac{\hat{j}}{\sqrt{3}} + \frac{\hat{k}}{\sqrt{3}} \end{array}$

Q.20

$\begin{array}{l} For given vectors, \vec{a} = 2 \hat{i} - \hat{j} + 2 \hat{k} and \vec{b} = - \hat{i} + \hat{j} - \hat{k}, find the \\ unit vector in the direction of the vector \vec{a} + \vec{b} . \end{array}$

Ans.

$\begin{array}{l} The given vectors are: \\ \vec{a} = 2 \hat{i} - \hat{j} + 2 \hat{k} and \vec{b} = - \hat{i} + \hat{j} - \hat{k} \\ \vec{a} + \vec{b} = 2 \hat{i} - \hat{j} + 2 \hat{k} + (- \hat{i} + \hat{j} - \hat{k}) \\ = \hat{i} + \hat{k} \\ | \vec{a} + \vec{b} | = | \hat{i} + \hat{k} | \\ = \sqrt{1^{2} + 1^{2}} \\ = \sqrt{2} \\ ∴ Unit vector of \vec{a} + \vec{b} \\ = \frac{\vec{a} + \vec{b}}{| \vec{a} + \vec{b} |} = \frac{\hat{i}}{\sqrt{2}} + \frac{\hat{k}}{\sqrt{2}} \end{array}$

Q.21

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

Ans.

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

Q.22

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

Ans.

$\begin{array}{l} The given vectors are \\ \vec{a} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k} \\ \vec{b} = - 4 \hat{i} + 6 \hat{j} - 8 \hat{k} \\ = - 2 (2 \hat{i} - 3 \hat{j} + 4 \hat{k}) \\ \vec{b} = - 2 \vec{a}, which is in the form of \vec{a} = λ \vec{b}, \\ where, λ = - 2 \\ Hence, the given vectors are collinear . \end{array}$

Q.23

$Find the direction cosines of the vector \hat{i} + 2 \hat{j} + 3 \hat{k} .$

Ans.

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

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

Ans.

$\begin{array}{l} The given points are A (1, 2, - 3) and B (- 1, - 2, 1) . \\ ∴ \vec{AB} = (- 1 - 1) \hat{i} + (- 2 - 2) \hat{j} + (1 + 3) \hat{k} \\ = - 2 \hat{i} - 4 \hat{j} + 4 \hat{k} \\ ∴ | \vec{AB} | = \sqrt{{(- 2)}^{2} + {(- 4)}^{2} + 4^{2}} \\ = \sqrt{4 + 16 + 16} \\ = \sqrt{36} \\ = 6 \\ Hence, the direction cosines of \vec{AB} are (\frac{- 2}{6}, \frac{- 4}{6}, \frac{4}{6}) = (\frac{- 1}{3}, \frac{- 2}{3}, \frac{2}{3}) . \end{array}$

Q.25

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

Ans.

$\begin{array}{l} The given vector is \vec{a} = \hat{i} + \hat{j} + \hat{k}, then \\ | \vec{a} | = | \hat{i} + \hat{j} + \hat{k} | \\ = \sqrt{1^{2} + 1^{2} + 1^{2}} \\ = \sqrt{3} \\ The direction cosines of \vec{a} are (\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}) . \\ Let α, β and γ be the angles formed by \vec{a} with the positive \\ directions of x, y, and z axes. \\ Then, we have cos α = \frac{1}{\sqrt{3}}, cos β = \frac{1}{\sqrt{3}}, cos γ = \frac{1}{\sqrt{3}} . \\ Hence, the given vector is equally inclined to axes OX, \\ OY, and OZ. \end{array}$

Q.26

$\begin{array}{l} Let the vectors \vec{a}, \vec{b}, \vec{c} be given as a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}, \\ b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}, c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k} . Then show that \\ \vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c} . \end{array}$

Ans.

$\begin{array}{l} The given vectors are: a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}, b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}, c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k} . \\ L . H . S . = \vec{a} \times (\vec{b} + \vec{c}) \\ = (a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}) \times (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} + c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}) \\ = (a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}) \times {(b_{1} + c_{1}) \hat{i} + (b_{2} + c_{2}) \hat{j} + (b_{3} + c_{3}) \hat{k}} \\ = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ (b_{1} + c_{1}) & (b_{2} + c_{2}) & (b_{3} + c_{3}) \end{matrix} | \\ = {a_{2} (b_{3} + c_{3}) - a_{3} (b_{2} + c_{2})} \hat{i} + {a_{1} (b_{3} + c_{3}) - a_{3} (b_{1} + c_{1})} \hat{j} \\ + {a_{1} (b_{2} + c_{2}) - a_{2} (b_{3} + c_{3})} \hat{k} \\ = {a_{2} b_{3} + a_{2} c_{3} - a_{3} b_{2} - a_{3} c_{2}} \hat{i} + {a_{1} b_{3} + a_{1} c_{3} - a_{3} b_{1} - a_{3} c_{1}} \hat{j} \\ + {a_{1} b_{2} + a_{1} c_{2} - a_{2} b_{3} - a_{2} c_{3}} \hat{k} . .. (i) \\ \vec{a} \times \vec{b} = (a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}) \times (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) \\ = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{matrix} | \\ = (a_{2} b_{3} - a_{3} b_{2}) \hat{i} + (a_{1} b_{3} - a_{3} b_{1}) \hat{j} + (a_{1} b_{2} - a_{2} b_{1}) \hat{k} \\ \vec{a} \times \vec{c} = (a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}) \times (c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}) \\ = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix} | \\ = (a_{2} b_{3} - a_{3} b_{2}) \hat{i} + (a_{1} b_{3} - a_{3} b_{1}) \hat{j} + (a_{1} b_{2} - a_{2} b_{1}) \hat{k} \\ \vec{a} \times \vec{b} + \vec{a} \times \vec{c} \\ = (a_{2} b_{3} - a_{3} b_{2}) \hat{i} + (a_{1} b_{3} - a_{3} b_{1}) \hat{j} + (a_{1} b_{2} - a_{2} b_{1}) \hat{k} + (a_{2} b_{3} - a_{3} b_{2}) \hat{i} \\ + (a_{1} b_{3} - a_{3} b_{1}) \hat{j} + (a_{1} b_{2} - a_{2} b_{1}) \hat{k} \\ = (a_{2} b_{3} - a_{3} b_{2} + a_{2} b_{3} - a_{3} b_{2}) \hat{i} + (a_{1} b_{3} - a_{3} b_{1} + a_{1} b_{3} - a_{3} b_{1}) \hat{j} \\ + (a_{1} b_{2} - a_{2} b_{1} + a_{1} b_{2} - a_{2} b_{1}) \hat{k} \\ = (a_{2} b_{3} + a_{2} b_{3} - a_{3} b_{2} - a_{3} b_{2}) \hat{i} + (a_{1} b_{3} + a_{1} b_{3} - a_{3} b_{1} - a_{3} b_{1}) \hat{j} \\ + (a_{1} b_{2} + a_{1} b_{2} - a_{2} b_{1} - a_{2} b_{1}) \hat{k} . .. (ii) \\ From equation (i) and equation (ii), we have \\ \vec{a} \times (\vec{b} \times \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c} . \\ Hence, the given result is proved. \end{array}$

Q.27

$\begin{array}{l} Find the position vector of a point R which divides the \\ line joining two points P and Q whose position vectors \\ are \hat{i} + 2 \hat{j} - \hat{k} and - \hat{i} + \hat{j} + \hat{k} respectively, in the ration 2:1 \\ (i) internally \\ (ii) externally \end{array}$

Ans.

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

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

Ans.

$\begin{array}{l} The position vector of mid-point R of the vector joining points \\ P (2, 3, 4) and Q (4, 1,-2) is given by, \\ \vec{OR} = \frac{(2 \hat{i} + 3 \hat{j} + 4 \hat{k}) + (4 \hat{i} + \hat{j} - 2 \hat{k})}{2} \\ = \frac{6 \hat{i} + 4 \hat{j} + 2 \hat{k}}{2} \\ = 3 \hat{i} + 2 \hat{j} + \hat{k} \end{array}$

Q.29

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

Ans.

$\begin{array}{l} Position vectors of points A, B, and C are respectively given as: \\ \vec{a} = 3 \hat{i} - 4 \hat{j} - 4 \hat{k}, \vec{b} = 2 \hat{i} - \hat{j} + \hat{k} and \vec{c} = \hat{i} - 3 \hat{j} - 5 \hat{k} \\ ∴ \vec{AB} = \vec{b} - \vec{a} \\ = (2 \hat{i} - \hat{j} + \hat{k}) - (3 \hat{i} - 4 \hat{j} - 4 \hat{k}) \\ = - \hat{i} + 3 \hat{j} + 5 \hat{k} \\ \vec{BC} = \vec{c} - \vec{b} \\ = (\hat{i} - 3 \hat{j} - 5 \hat{k}) - (2 \hat{i} - \hat{j} + \hat{k}) \\ = - \hat{i} - 2 \hat{j} - 6 \hat{k} \\ \vec{CA} = \vec{a} - \vec{c} \\ = (3 \hat{i} - 4 \hat{j} - 4 \hat{k}) - (\hat{i} - 3 \hat{j} - 5 \hat{k}) \end{array}$ $\begin{matrix} = 2 \hat{i} - \hat{j} + \hat{k} \\ So, {| \vec{AB} |}^{2} = {| - \hat{i} + 3 \hat{j} + 5 \hat{k} |}^{2} \\ = 35 \\ {| \vec{BC} |}^{2} = {| - \hat{i} - 2 \hat{j} - 6 \hat{k} |}^{2} \\ = 41 \\ {| \vec{CA} |}^{2} = {| 2 \hat{i} - \hat{j} + \hat{k} |}^{2} \\ = 6 \\ {| \vec{AB} |}^{2} = {| \vec{BC} |}^{2} + {| \vec{CA} |}^{2} \\ = 36 + 5 = 41 \end{matrix}$ $Hence, ABC is a right-angled triangle$

Q.30n

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

Ans.

$\begin{array}{l} Let us consider any parallel non-zero vectors so that \vec{a} \times \vec{b} = 0 \\ Let \vec{a} = 4 \hat{i} + 6 \hat{j} + 8 \hat{k} and \vec{b} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \\ Then, \\ \vec{a} \times \vec{b} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 6 & 8 \\ 2 & 3 & 4 \end{matrix} | \\ = (24 - 24) \hat{i} + (16 - 16) \hat{j} + (12 - 12) \hat{k} \\ = 0 \hat{i} + 0 \hat{j} + 0 \hat{k} \\ = \vec{0} \\ And \\ | \vec{a} | = | 4 \hat{i} + 6 \hat{j} + 8 \hat{k} | \\ = \sqrt{4^{2} + 6^{2} + 8^{2}} \\ = \sqrt{16 + 36 + 64} \\ = \sqrt{116} \\ ∴ | \vec{a} | \neq 0 \\ | \vec{b} | = | 2 \hat{i} + 3 \hat{j} + 4 \hat{k} | \\ = \sqrt{2^{2} + 3^{2} + 4^{2}} \\ = \sqrt{4 + 9 + 16} \\ = \sqrt{29} \\ ∴ | \vec{b} | \neq 0 \\ Hence, the converse of the given statement need not be true. \end{array}$

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

Ans.

$\begin{array}{l} The vertices of triangle ABC are given as A (1, 1, 2), B (2, 3, 5) \\ and C (1, 5, 5). \\ The adjacent sides \vec{AB} and \vec{BC} of Δ ABC are given as: \\ \vec{AB} = (2 - 1) \hat{i} + (3 - 1) \hat{j} + (5 - 2) \hat{k} \\ = \hat{i} + 2 \hat{j} + 3 \hat{k} \\ \vec{BC} = (1 - 2) \hat{i} + (5 - 3) \hat{j} + (5 - 5) \hat{k} \\ = - \hat{i} + 2 \hat{j} \\ ∴ \vec{AB} \times \vec{BC} \\ = \frac{1}{2} | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ - 1 & 2 & 0 \end{matrix} | \\ = \frac{1}{2} {(0 - 6) \hat{i} - (0 + 3) \hat{j} + (2 + 2) \hat{k}} \\ = \frac{1}{2} (- 6 \hat{i} - 3 \hat{j} + 4 \hat{k}) \\ The area ΔABC \\ = \frac{1}{2} | \vec{AB} \times \vec{BC} | \\ = \frac{1}{2} | - 6 \hat{i} - 3 \hat{j} + 4 \hat{k} | \\ = \frac{1}{2} \sqrt{{(- 6)}^{2} + {(- 3)}^{2} + 4^{2}} \\ = \frac{1}{2} \sqrt{36 + 9 + 16} \\ = \frac{1}{2} \sqrt{61} square units \\ Thus, the area of Δ ABC is \frac{1}{2} \sqrt{61} square units . \end{array}$

Q.32

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

Ans.

$\begin{array}{l} The area of the parallelogram whose adjacent sides are \vec{a} and \vec{b} \\ is | \vec{a} \times \vec{b} | . \\ Adjacent sides are in vector form, which are as follows: \\ \vec{a} = \hat{i} - \hat{j} + 3 \hat{k} and \\ \vec{b} = 2 \hat{i} - 7 \hat{j} + \hat{k} \\ ∴ \vec{a} \times \vec{b} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 1 & 3 \\ 2 & - 7 & 1 \end{matrix} | \\ = (- 1 + 21) \hat{i} + (1 - 6) \hat{j} + (- 7 + 2) \hat{k} \\ = 20 \hat{i} - 5 \hat{j} - 5 \hat{k} \\ | \vec{a} \times \vec{b} | = | 20 \hat{i} - 5 \hat{j} - 5 \hat{k} | \\ = \sqrt{20^{2} + {(- 5)}^{2} + {(- 5)}^{2}} \\ = \sqrt{400 + 25 + 25} \\ = \sqrt{450} \\ = 15 \sqrt{2} \\ Thus, the area of the given parallelogram is 15 \sqrt{2} square units. \end{array}$

Q.33

$\begin{array}{l} Let the vectors \vec{a} and \vec{b} be such that | \vec{a} | = 3 and | \vec{b} | = \frac{\sqrt{3}}{2}, then \\ \vec{a} \times \vec{b} is a unit vector, if the angle between \vec{a} and \vec{b} is \\ (A) \frac{π}{6} \\ (B) \frac{π}{4} \\ (C) \frac{π}{3} \\ (D) \frac{π}{2} \end{array}$

Ans.

$\begin{array}{l} Given : | \vec{a} | = 3 and | \vec{b} | = \frac{\sqrt{2}}{3} \\ Since, \vec{a} \times \vec{b} = | \vec{a} | | \vec{b} | sinθ \hat{n}, where \hat{n} is a unit vector perpendicular \\ to both \vec{a} and \vec{b} ‹ and θ ‹ is the angle between \vec{a} and \vec{b} ‹ . \\ Since, \vec{a} \times \vec{b} is a unit vector, so | \vec{a} \times \vec{b} | = 1 . \\ | \vec{a} \times \vec{b} | = 1 \\ \Rightarrow | | \vec{a} | | \vec{b} | sinθ \hat{n} | = 1 \\ \Rightarrow | \vec{a} | | \vec{b} | | sinθ | = 1 \\ 3 \times \frac{\sqrt{2}}{3} \times sinθ = 1 \\ sinθ = \frac{1}{\sqrt{2}} = \sin \frac{π}{4} \\ \Rightarrow θ = \frac{π}{4} \\ Hence, \vec{a} \times \vec{b} is a unit vector if the angle between \vec{a} and \vec{b} is \frac{π}{4} . \\ The correct answer is B. \end{array}$

Q.34

$\begin{array}{l} Area of a rectangle having vertices A, B, C and D with \\ position vector - \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k}, \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k}, \hat{i} - \frac{1}{2} \hat{j} + 4 \hat{k} \\ and - \hat{i} - \frac{1}{2} \hat{j} + 4 \hat{k} respectively is \\ (A) \frac{1}{2} \\ (B) 1 \\ (C) 2 \\ (D) 4 \end{array}$

Ans.

$\begin{array}{l} The position vectors of vertices A, B, C, and D of rectangle \\ ABCD are given as: \\ \vec{OA} = - \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k}, \vec{OB} = \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k}, \vec{OC} = \hat{i} - \frac{1}{2} \hat{j} + 4 \hat{k} and \\ \vec{OD} = - \hat{i} - \frac{1}{2} \hat{j} + 4 \hat{k} \\ The adjacent sides \vec{AB} and \vec{BC} of the given rectangle are given as: \\ \vec{AB} = (1 + 1) \hat{i} + (\frac{1}{2} - \frac{1}{2}) \hat{j} + (4 - 4) \hat{k} = 2 \hat{i} \\ \vec{BC} = (1 - 1) \hat{i} + (- \frac{1}{2} - \frac{1}{2}) \hat{j} + (4 - 4) \hat{k} = - \hat{j} \\ ∴ \vec{AB} \times \vec{BC} \\ = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 0 \\ 0 & - 1 & 0 \end{matrix} | \\ = (- 2 - 0) \hat{k} \\ = - 2 \hat{k} \\ | \vec{AB} \times \vec{BC} | \\ = | - 2 \hat{k} | \\ = 2 \\ Now, it is known that the area of a parallelogram whose adjacent \\ sides are \vec{a} and \vec{b} is | \vec{a} \times \vec{b} | . \\ Hence, the area of the given rectangle is | \vec{AB} \times \vec{BC} | = 2 square units. \\ The correct answer is C. \end{array}$

Q.35

$Find [\vec{a} \vec{b} \vec{c}] if \vec{a} = \hat{i} - 2 \hat{j} + 3 \hat{k}, \vec{b} = 2 \hat{i} - 3 \hat{j} + \hat{k}, \vec{c} = 3 \hat{i} + \hat{j} - 2 \hat{k} .$

Ans.

$\begin{array}{l} [\vec{a} \vec{b} \vec{c}] = | \begin{matrix} 1 & - 2 & 3 \\ 2 & - 3 & 1 \\ 3 & 1 & - 2 \end{matrix} | \\ = 1 | \begin{matrix} - 3 & 1 \\ 1 & - 2 \end{matrix} | + 2 | \begin{matrix} 2 & 1 \\ 3 & - 2 \end{matrix} | + 3 | \begin{matrix} 2 & - 3 \\ 3 & 1 \end{matrix} | \\ = 1 (6 - 1) + 2 (- 4 - 3) + 3 (2 + 9) \\ = 1 (5) + 2 (- 7) + 3 (11) \\ = 5 - 14 + 33 \\ = 24 \\ ∴ [\vec{a} \vec{b} \vec{c}] = 24 \end{array}$

Q.36

$In triangle, which of the following is not true : \begin{array}{l} (A) \vec{AB} + \vec{BC} + \vec{CA} = 0 \\ (B) \vec{AB} + \vec{BC} - AC = 0 \\ (C) \vec{AB} + \vec{BC} - \vec{CA} = 0 \\ (D) \vec{AB} - \vec{CB} + \vec{CA} = 0 \end{array}$

Ans.

$\begin{array}{l} On applying the triangle law of addition in the given triangle, \\ we have: \\ \vec{AB} + \vec{BC} = \vec{AC} . .. (i) \\ \Rightarrow \vec{AB} + \vec{BC} = - \vec{CA} \\ \Rightarrow \vec{AB} + \vec{BC} + \vec{CA} = \vec{0} . .. (ii) \\ ∴ The equation given in alternative A is true. \\ \vec{AB} + \vec{BC} = \vec{AC} \\ \Rightarrow \vec{AB} + \vec{BC} - \vec{AC} = \vec{0} \\ ∴ The equation given in alternative B is true. \\ From equation (ii), we have: \\ \vec{AB} - \vec{CB} + \vec{AC} = \vec{0} \\ ∴ The equation given in alternative D is true. \\ Now, the equation given in alternative C, \\ \vec{AB} + \vec{BC} - \vec{CA} = \vec{0} \\ \Rightarrow \vec{AB} + \vec{BC} = \vec{CA} . .. (iii) \\ From equation (i) and (iii),‹ we have \end{array}$ $\begin{array}{l} \vec{CA} = \vec{AC} \\ \Rightarrow - \vec{AC} = \vec{AC} \\ \Rightarrow \vec{AC} + \vec{AC} = \vec{0} \\ \Rightarrow 2 \vec{AC} = \vec{0} \\ \Rightarrow \vec{AC} = \vec{0}, which is not true. \\ Hence, the equation given in alternative C is incorrect. \\ The correct answer is C. \end{array}$

Q.37

$\begin{array}{l} Show that the vectors \vec{a} = \hat{i} - 2 \hat{j} + 3 \hat{k}, \vec{b} = - 2 \hat{i} + 3 \hat{j} - 4 \hat{k}, \\ \vec{c} = \hat{i} - 3 \hat{j} + 5 \hat{k} are coplanar . \end{array}$

Ans.

$\begin{array}{l} [\vec{a} \vec{b} \vec{c}] = | \begin{matrix} 1 & - 2 & 3 \\ - 2 & 3 & - 4 \\ 1 & - 3 & 5 \end{matrix} | \\ = 1 | \begin{matrix} 3 & - 4 \\ - 3 & 5 \end{matrix} | + 2 | \begin{matrix} - 2 & - 4 \\ 1 & 5 \end{matrix} | + 3 | \begin{matrix} - 2 & 3 \\ 1 & - 3 \end{matrix} | \\ = 1 (15 - 12) + 2 (- 10 + 4) + 3 (6 - 3) \\ = 1 (3) + 2 (- 6) + 3 (3) \\ = 3 - 12 + 9 \\ = 0 \\ Since, [\vec{a} \vec{b} \vec{c}] = 0 . So, given vectors are coplanar. \end{array}$

Q.38

$\begin{array}{l} Find λ if the vectors \hat{i} - \hat{j} + \hat{k}, 3 \hat{i} + \hat{j} + 2 \hat{k} and \hat{i} + λ \hat{j} - 3 \hat{k} are \\ coplanar . \end{array}$

Ans.

$\begin{array}{l} [\vec{a} \vec{b} \vec{c}] = | \begin{matrix} 1 & - 1 & 1 \\ 3 & 1 & 2 \\ 1 & λ & - 3 \end{matrix} | \\ [\vec{a} \vec{b} \vec{c}] = 1 | \begin{matrix} 1 & 2 \\ λ & - 3 \end{matrix} | - (- 1) | \begin{matrix} 3 & 2 \\ 1 & - 3 \end{matrix} | + 1 | \begin{matrix} 3 & 1 \\ 1 & λ \end{matrix} | \\ = 1 (- 3 - 2 λ) + 1 (- 9 - 2) + 1 (3 λ - 1) \\ = - 3 - 2 λ - 11 + 3 λ - 1 \\ = λ - 15 \\ Since, given vectors are coplanar. So, \\ [\vec{a} \vec{b} \vec{c}] = 0 \\ \Rightarrow λ - 15 = 0 \\ \Rightarrow λ = 15 \end{array}$

Q.39

$\begin{array}{l} Let \vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = \hat{i}, \vec{c} = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k} . Then \\ (a) If c_{1} = 1 and c_{2} = 2, find c_{3} which makes \vec{a}, \vec{b} and \vec{c} coplanar . \\ (b) If c_{1} = - 1 and c_{3} = 1, show that no value of c_{1} can \\ make \vec{a}, \vec{b} and \vec{c} coplanar . \end{array}$

Ans.

$\begin{array}{l} (i) Given : \vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = \hat{i} and \\ \vec{c} = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k} \\ Putting c_{1} = 1 {and c}_{2} = 2, we get \\ \vec{c} = 1 \hat{i} + 2 \hat{j} + c_{3} \hat{k} \\ Since, \vec{a}, \vec{b} and \vec{c} are coplanar . Then, \\ [\vec{a} \vec{b} \vec{c}] = | \begin{matrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 2 & c_{3} \end{matrix} | \\ [\vec{a} \vec{b} \vec{c}] = 1 | \begin{matrix} 0 & 0 \\ 2 & c_{3} \end{matrix} | - 1 | \begin{matrix} 1 & 0 \\ 1 & c_{3} \end{matrix} | + 1 | \begin{matrix} 1 & 0 \\ 1 & 2 \end{matrix} | \\ = 1 (0) - 1 (c_{3} - 0) + 1 (2 - 0) \\ = 0 - c_{3} + 2 \\ Since, given vectors are coplanar. So, \\ [\vec{a} \vec{b} \vec{c}] = 0 \\ \Rightarrow - c_{3} + 2 = 0 \\ \Rightarrow c_{3} = 2 \\ (ii) Given : \vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = \hat{i} and \\ \vec{c} = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k} \\ Putting c_{2} = - 1 {and c}_{3} = 1, we get \\ \vec{c} = c_{1} \hat{i} - 1 \hat{j} + 1 \hat{k} \\ Since, \vec{a}, \vec{b} and \vec{c} are coplanar . Then, \\ [\vec{a} \vec{b} \vec{c}] = | \begin{matrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ c_{1} & - 1 & 1 \end{matrix} | \\ [\vec{a} \vec{b} \vec{c}] = 1 | \begin{matrix} 0 & 0 \\ - 1 & 1 \end{matrix} | - 1 | \begin{matrix} 1 & 0 \\ c_{1} & 1 \end{matrix} | + 1 | \begin{matrix} 1 & 0 \\ c_{1} & - 1 \end{matrix} | \\ = 1 (0) - 1 (1 - 0) + 1 (- 1 - 0) \\ = 0 - 1 - 1 \\ = - 2 \\ Since, \\ [\vec{a} \vec{b} \vec{c}] \neq 0 \\ So, there will be no effect on the value of [\vec{a} \vec{b} \vec{c}] by any \\ {value of c}_{1} {.Thus, no value of c}_{1} can make \vec{a}, \vec{b} and \vec{c} coplanar. \end{array}$

Q.40

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

Ans.

$\begin{array}{l} Let the four points with position vectors be \\ OA = 4 \hat{i} + 8 \hat{j} + 12 \hat{k}, OB = 2 \hat{i} + 4 \hat{j} + 6 \hat{k}, OC = 3 \hat{i} + 5 \hat{j} + 4 \hat{k} \\ and OD = 5 \hat{i} + 8 \hat{j} + 5 \hat{k}, then \\ \vec{AB} = \vec{OB} - \vec{OA} \\ = (2 \hat{i} + 4 \hat{j} + 6 \hat{k}) - (4 \hat{i} + 8 \hat{j} + 12 \hat{k}) \\ = - 2 \hat{i} - 4 \hat{j} - 6 \hat{k} \\ \vec{AC} = \vec{OC} - \vec{OA} \\ = (3 \hat{i} + 5 \hat{j} + 4 \hat{k}) - (4 \hat{i} + 8 \hat{j} + 12 \hat{k}) \\ = - \hat{i} - 3 \hat{j} - 8 \hat{k} \\ \vec{AD} = \vec{OD} - \vec{OA} \\ = (5 \hat{i} + 8 \hat{j} + 5 \hat{k}) - (4 \hat{i} + 8 \hat{j} + 12 \hat{k}) \\ = \hat{i} + 0 \hat{j} - 7 \hat{k} \\ Thus, \\ [\vec{AB} \vec{AC} \vec{AD}] = | \begin{matrix} - 2 & - 4 & - 6 \\ - 1 & - 3 & - 8 \\ 1 & 0 & - 7 \end{matrix} | \\ = - 2 | \begin{matrix} - 3 & - 8 \\ 0 & - 7 \end{matrix} | - (- 4) | \begin{matrix} - 1 & - 8 \\ 1 & - 7 \end{matrix} | + (- 6) | \begin{matrix} - 1 & - 3 \\ 1 & 0 \end{matrix} | \\ = - 2 (21 - 0) + 4 (7 + 8) - 6 (0 + 3) \\ = - 42 + 60 - 18 \\ = 0 \\ Thus, the given four points with position vectors are coplanar. \end{array}$

Q.41

$\begin{array}{l} If \vec{a} \vec{b} are two collinear vectors, then which of the following \\ are incorrect: \\ (A) \vec{b} = λ \vec{a}, for some scalar λ \\ (B) \vec{a} = \pm \vec{b} \\ (C) the respective components of \vec{a} and \vec{b} are proportional \\ (D) both the vectors \vec{a} and \vec{b} have same direction, but \\ different magnitudes. \end{array}$

Ans.

$\begin{array}{l} If \vec{a} and \vec{b} are two collinear vectors, then they are parallel. \\ Therefore, we have: \\ \vec{b} = λ \vec{a} (λ is a scalar quantities.) \\ If λ = \pm 1, then \vec{b} = \pm 1 \vec{a} . \\ If \vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k} and \vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}, then \\ \vec{b} = λ \vec{a} \Rightarrow b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} = λ (a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}) \end{array}$ $\begin{array}{l} \Rightarrow b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} = ({λa}_{1}) \hat{i} + ({λa}_{2}) \hat{j} + ({λa}_{3}) \hat{k} \\ \Rightarrow b_{1} = {λa}_{1}, b_{2} = {λa}_{2}, b_{3} = {λa}_{3} \\ \Rightarrow \frac{b_{1}}{a_{1}} = \frac{b_{2}}{a_{2}} = \frac{b_{3}}{a_{3}} = λ \\ Thus, the respective components of \vec{a} and \vec{b} are proportional. \\ However, vectors \vec{a} and \vec{b} can have different directions. \\ Hence, the statement given in D is incorrect. \\ The correct answer is D. \end{array}$

Q.42

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

Ans.

$\begin{array}{l} Let the four points with position vectors be \\ OA = 3 \hat{i} + 2 \hat{j} + \hat{k}, OB = 4 \hat{i} + x \hat{j} + 5 \hat{k}, OC = 4 \hat{i} + 2 \hat{j} - 2 \hat{k} \\ andOD = 6 \hat{i} + 5 \hat{j} - 1 \hat{k}, then \\ \vec{AB} = \vec{OB} - \vec{OA} \\ = (4 \hat{i} + x \hat{j} + 5 \hat{k}) - (3 \hat{i} + 2 \hat{j} + \hat{k}) \\ = \hat{i} + (x - 2) \hat{j} + 4 \hat{k} \\ \vec{AC} = \vec{OC} - \vec{OA} \\ = (4 \hat{i} + 2 \hat{j} - 2 \hat{k}) - (4 \hat{i} + 8 \hat{j} + 12 \hat{k}) \\ = 0 \hat{i} - 6 \hat{j} - 14 \hat{k} \\ \vec{AD} = \vec{OD} - \vec{OA} \\ = (6 \hat{i} + 5 \hat{j} - 1 \hat{k}) - (4 \hat{i} + 8 \hat{j} + 12 \hat{k}) \\ = 2 \hat{i} - 3 \hat{j} - 13 \hat{k} \\ Since, given vectors are coplanar. So, \\ [\vec{AB} \vec{AC} \vec{AD}] = 0 \\ ∴ [\vec{AB} \vec{AC} \vec{AD}] = | \begin{matrix} 1 & x - 2 & 4 \\ 0 & - 6 & - 14 \\ 2 & - 3 & - 13 \end{matrix} | \\ = 1 | \begin{matrix} - 6 & - 14 \\ - 3 & - 13 \end{matrix} | - (x - 2) | \begin{matrix} 0 & - 14 \\ 2 & - 13 \end{matrix} | + 4 | \begin{matrix} 0 & - 6 \\ 2 & - 3 \end{matrix} | \\ = 1 (78 - 42) - (x - 2) (0 + 28) + 4 (0 + 12) \\ = 36 - 28 x + 56 + 48 \\ = 140 - 28 x \\ \Rightarrow 140 - 28 x = 0 \\ \Rightarrow x = \frac{140}{28} = 5 \\ Thus, the value of x is 5. \end{array}$

Q.43

$\begin{array}{l} Show that the vectors \vec{a}, \vec{b}, \vec{c} are coplanar if \vec{a} + \vec{b}, \vec{b} + \vec{c} \\ and \vec{c} + \vec{a} are coplanar . \end{array}$

Ans.

$\begin{array}{l} [\vec{a} + \vec{b}, \vec{b} + \vec{c}, \vec{c} + \vec{a}] = (\vec{a} + \vec{b}) . ((\vec{b} + \vec{c}) \times (\vec{c} + \vec{a})) \\ = (\vec{a} + \vec{b}) . (\vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{c} + \vec{c} \times \vec{a}) \\ = (\vec{a} + \vec{b}) . (\vec{b} \times \vec{c} + \vec{b} \times \vec{a} + 0 + \vec{c} \times \vec{a}) \\ = \vec{a .} (\vec{b} \times \vec{c}) + \vec{a} . (\vec{b} \times \vec{a}) + \vec{a} (\vec{c} \times \vec{a}) + \vec{b .} (\vec{b} \times \vec{c}) \\ + \vec{b} . (\vec{b} \times \vec{a}) + \vec{b} . (\vec{c} \times \vec{a}) \\ = [\vec{a}, \vec{b}, \vec{c}] + [\vec{a}, \vec{b}, \vec{a}] + [\vec{a}, \vec{c}, \vec{a}] + [\vec{b}, \vec{b}, \vec{c}] \\ + [\vec{b}, \vec{b}, \vec{a}] + [\vec{b}, \vec{c}, \vec{a}] \\ = [\vec{a}, \vec{b}, \vec{c}] + 0 + 0 + 0 + 0 + [\vec{a}, \vec{b}, \vec{c}] \\ [[\vec{a}, \vec{b}, \vec{a}] = [\vec{a}, \vec{c}, \vec{a}] = [\vec{b}, \vec{b}, \vec{c}] = [\vec{b}, \vec{b}, \vec{a}] = 0] \\ = 2 [\vec{a}, \vec{b}, \vec{c}] \\ Since, \vec{a} + \vec{b}, \vec{b} + \vec{c} and \vec{c} + \vec{a} are coplanar . So, \\ [\vec{a} + \vec{b}, \vec{b} + \vec{c}, \vec{c} + \vec{a}] = 0 \\ \Rightarrow 2 [\vec{a}, \vec{b}, \vec{c}] = 0 \\ \Rightarrow [\vec{a}, \vec{b}, \vec{c}] = 0 \\ So, the vectors \vec{a}, \vec{b}, \vec{c} are coplanar . \end{array}$

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

Ans.

$\begin{array}{l} If is \vec{r} a unit vector in the XY-plane, then \vec{r} = cosθ \hat{i} + sinθ \hat{j} . \\ Here, θ is the angle made by the unit vector with the \\ positive direction of the x-axis. \\ Therefore, for θ = 30°: \\ \vec{r} = \cos 30° \hat{i} + \sin 30° \hat{j} = \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j} \\ Thus, the required unit vector is \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j} . \end{array}$

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

Ans.

$\begin{array}{l} The vector joining the points P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) is: \\ \vec{PQ} = (x_{2} - x_{1}) \hat{i} + (y_{2} - y_{1}) \hat{j} + (z_{2} - z_{1}) \hat{k} \\ | \vec{PQ} | = | (x_{2} - x_{1}) \hat{i} + (y_{2} - y_{1}) \hat{j} + (z_{2} - z_{1}) \hat{k} | \\ = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} + {(z_{2} - z_{1})}^{2}} \\ Hence, the scalar components and the magnitude of the vector \\ joining the given points are respectively \\ {(x_{2} - x_{1}), (y_{2} - y_{1}), (z_{2} - z_{1})} and \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} + {(z_{2} - z_{1})}^{2}} . \end{array}$

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

Ans.

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

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

Q.47

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

Ans.

$\begin{array}{l} In ΔABC, let \vec{CB} = \vec{a}, \vec{CA} = \vec{b} and \vec{AB} = \vec{c} . \\ Now, by the triangle law of vector addition, we have \vec{a} = \vec{b} + \vec{c} . \\ Since, sum of any two sides of a triangle is always greater than third side. \\ ∴ | \vec{a} | < | \vec{b} | + | \vec{c} | \\ Hence, it is not true that | \vec{a} | = | \vec{b} | + | \vec{c} | . \end{array}$

Q.48

$Find the value of x for which x (\hat{i} + \hat{j} + \hat{k}) is a unit vector .$

Ans.

$\begin{array}{l} x (\hat{i} + \hat{j} + \hat{k}) is a unit vector, \\ so | x (\hat{i} + \hat{j} + \hat{k}) | = 1 \\ \Rightarrow x | (\hat{i} + \hat{j} + \hat{k}) | = 1 \\ x \sqrt{1^{2} + 1^{2} + 1^{2}} = 1 \\ \Rightarrow x \sqrt{3} = 1 \\ \Rightarrow x = \pm \frac{1}{\sqrt{3}} \\ Thus, the required value of x is \pm \frac{1}{\sqrt{3}} . \end{array}$

Q.49

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

Ans.

$\begin{array}{l} The given vectorsare \\ \vec{a} = 2 \hat{i} + 3 \hat{j} - \hat{k} and \vec{b} = \hat{i} - 2 \hat{j} + \hat{k} \\ Let \vec{c} be the resultant vectors of \vec{a} and \vec{b} . \\ Then, \\ \vec{c} = \vec{a} + \vec{b} \\ = 2 \hat{i} + 3 \hat{j} - \hat{k} + \hat{i} - 2 \hat{j} + \hat{k} \\ = 3 \hat{i} + \hat{j} \\ ∴ | \vec{c} | = | 3 \hat{i} + \hat{j} | \\ = \sqrt{3^{2} + 1^{2}} \\ = \sqrt{10} \\ So, \hat{c} = \frac{3 \hat{i} + \hat{j}}{\sqrt{10}} \\ Thus, the vector of magnitude 5 units and parallel to the \\ resultant of vectors \vec{a} and \vec{b} is \\ \pm 5 . \hat{c} = \pm 5 . \frac{3 \hat{i} + \hat{j}}{\sqrt{10}} = \pm \frac{3 \sqrt{10}}{2} \hat{i} \pm \frac{\sqrt{10}}{2} \hat{j} \end{array}$

Q.50

$\begin{array}{l} Find the angle between two vectors \vec{a} and \vec{b} with \\ magnitude \sqrt{3} and 2, respectively having \vec{a} . \vec{b} = \sqrt{6} . \end{array}$

Ans.

$\begin{array}{l} It is given that, \\ | \vec{a} | = \sqrt{3}, | \vec{b} | = 2 and \vec{a} . \vec{b} = \sqrt{6} \\ Now, we know that \\ \vec{a} . \vec{b} = | \vec{a} | | \vec{b} | cosθ \\ \Rightarrow \sqrt{6} = \sqrt{3} . 2 cosθ \\ \Rightarrow cosθ = \frac{\sqrt{6}}{\sqrt{3} . 2} \\ = \frac{1}{\sqrt{2}} \end{array}$ $\begin{array}{l} \Rightarrow cosθ = \cos \frac{π}{4} \\ \Rightarrow θ = \frac{π}{4} \\ Hence, the angle between the given vectors \vec{a} and \vec{b} is \frac{π}{4} . \end{array}$

Q.51

$Find the angle between the vectors \hat{i} - 2 \hat{j} + 3 \hat{k} and 3 \hat{i} - 2 \hat{j} + \hat{k} .$

Ans.

$\begin{array}{l} The given vectors are \vec{a} = \hat{i} - 2 \hat{j} + 3 \hat{k} and \vec{b} = 3 \hat{i} - 2 \hat{j} + \hat{k} . \\ | \vec{a} | = | \hat{i} - 2 \hat{j} + 3 \hat{k} | \\ = \sqrt{1^{2} + {(- 2)}^{2} + 3^{2}} \\ = \sqrt{1 + 4 + 9} \\ = \sqrt{14} \\ | \vec{b} | = | 3 \hat{i} - 2 \hat{j} + \hat{k} | \\ = \sqrt{3^{2} + {(- 2)}^{2} + 1^{2}} \\ = \sqrt{9 + 4 + 1} \\ = \sqrt{14} \\ Now, \\ \vec{a} . \vec{b} = (\hat{i} - 2 \hat{j} + 3 \hat{k}) . (3 \hat{i} - 2 \hat{j} + \hat{k}) \\ = 3 + 4 + 3 = 10 \end{array}$ $\begin{array}{l} Also, we know that \\ \vec{a} . \vec{b} = | \vec{a} | . | \vec{b} | cosθ \\ 10 = \sqrt{14} . \sqrt{14} cosθ \\ cosθ = \frac{10}{14} \\ = \frac{5}{7} \\ θ = \cos^{- 1} (\frac{5}{7}) \end{array}$

Q.52

$Find the projection of the vector \hat{i} - \hat{j} on the vector \hat{i} + \hat{j} .$

Ans.

$\begin{array}{l} Let \vec{a} = \hat{i} - \hat{j} and \vec{b} = \hat{i} + \hat{j} \\ Now, projection of vector \vec{a} on \vec{b} is given by, \\ \frac{1}{| \vec{b} |} (\vec{a} . \vec{b}) = \frac{1}{| \hat{i} + \hat{j} |} (\hat{i} + \hat{j}) . (\hat{i} - \hat{j}) \\ = \frac{1}{\sqrt{1^{2} + 1^{2}}} (1 - 1) = 0 \\ Hence, the projection of vector \vec{a} on \vec{b} is 0. \end{array}$

Q.53

$\begin{array}{l} Find the projection of the vector \hat{i} + 3 \hat{j} + 7 \hat{k} on the vector \\ 7 \hat{i} - \hat{j} + 8 \hat{k} . \end{array}$

Ans.

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

Q.54

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

Ans.

$\begin{array}{l} Let \vec{a} = \frac{1}{7} (2 \hat{i} + 3 \hat{j} + 6 \hat{k}) \\ \vec{b} = \frac{1}{7} (3 \hat{i} - 6 \hat{j} + 2 \hat{k}) \\ \vec{c} = \frac{1}{7} (6 \hat{i} + 2 \hat{j} - 3 \hat{k}) \\ \Rightarrow | \vec{a} | = | \frac{1}{7} (2 \hat{i} + 3 \hat{j} + 6 \hat{k}) | \end{array}$ $\begin{array}{l} = \frac{1}{7} \sqrt{2^{2} + 3^{2} + 6^{2}} \\ = \frac{1}{7} \sqrt{4 + 9 + 36} \\ = \frac{7}{7} = 1 \\ | \vec{b} | = | \frac{1}{7} (3 \hat{i} - 6 \hat{j} + 2 \hat{k}) | \\ = \frac{1}{7} \sqrt{3^{2} + {(- 6)}^{2} + 2^{2}} \\ = \frac{1}{7} \sqrt{9 + 36 + 4} \\ = \frac{7}{7} = 1 \\ and \\ | \vec{c} | = | \frac{1}{7} (6 \hat{i} + 2 \hat{j} - 3 \hat{k}) | \\ = \frac{1}{7} \sqrt{6^{2} + 2^{2} + {(- 3)}^{2}} \\ = \frac{1}{7} \sqrt{36 + 4 + 9} \\ = \frac{7}{7} = 1 \\ Thus, each of the given three vectors is a unit vector. \\ Now, \\ \vec{a} . \vec{b} = \frac{1}{7} (2 \hat{i} + 3 \hat{j} + 6 \hat{k}) . \frac{1}{7} (3 \hat{i} - 6 \hat{j} + 2 \hat{k}) \\ = \frac{1}{49} (6 - 18 + 12) = 0 \end{array}$ $\begin{array}{l} \vec{b} . \vec{c} = \frac{1}{7} (3 \hat{i} - 6 \hat{j} + 2 \hat{k}) . \frac{1}{7} (6 \hat{i} + 2 \hat{j} - 3 \hat{k}) \\ = \frac{1}{49} (18 - 12 - 6) = 0 \\ \vec{c} . \vec{a} = \frac{1}{7} (6 \hat{i} + 2 \hat{j} - 3 \hat{k}) . \frac{1}{7} (2 \hat{i} + 3 \hat{j} + 6 \hat{k}) \\ = \frac{1}{49} (12 + 6 - 18) = 0 \\ Hence, the given three vectors are mutually perpendicular \\ to each other. \end{array}$

Q.55

$Find |\vec{a}| and |\vec{b}|, if (\vec{a} + \vec{b}) . (\vec{a} - \vec{b}) = 8 and |\vec{a}| = 8 |\vec{b}| .$

Ans.

$\begin{array}{l} We have, \\ (\vec{a} + \vec{b}) . (\vec{a} - \vec{b}) = 8 and | \vec{a} | = 8 | \vec{b} | \\ So, \\ (\vec{a} + \vec{b}) . (\vec{a} - \vec{b}) = 8 \\ \Rightarrow \vec{a} . \vec{a} - \vec{a} . \vec{b} + \vec{b} . \vec{a} - \vec{b} . \vec{b} = 8 \\ \Rightarrow {| \vec{a} |}^{2} - {| \vec{b} |}^{2} = 8 [\vec{a} . \vec{b} = \vec{b} . \vec{a}] \\ \Rightarrow {(8 | \vec{b} |)}^{2} - {| \vec{b} |}^{2} = 8 \\ \Rightarrow 64 {| \vec{b} |}^{2} - {| \vec{b} |}^{2} = 8 \end{array}$ $\begin{array}{l} \Rightarrow | \vec{b} | = \sqrt{\frac{8}{63}} = \frac{2 \sqrt{2}}{3 \sqrt{7}} \\ ∴ | \vec{a} | = 8 | \vec{b} | \\ = 8 \times \frac{2 \sqrt{2}}{3 \sqrt{7}} \\ = \frac{16 \sqrt{2}}{3 \sqrt{7}} \end{array}$

Q.56

$Evaluate the product (3 \vec{a} - 5 \vec{b}) . (2 \vec{a} + 7 \vec{b}) .$

Ans.

$\begin{array}{l} (3 \vec{a} - 5 \vec{b}) . (2 \vec{a} + 7 \vec{b}) = 6 \vec{a} . \vec{a} + 21 \vec{a} . \vec{b} - 10 \vec{b} . \vec{a} - 35 \vec{b} . \vec{b} \\ = 6 {| \vec{a} |}^{2} + 21 \vec{a} . \vec{b} - 10 \vec{b} . \vec{a} - 35 {| \vec{b} |}^{2} \\ = 6 {| \vec{a} |}^{2} + 11 \vec{a} . \vec{b} - 35 {| \vec{b} |}^{2} [\vec{a} . \vec{b} = \vec{b} . \vec{a}] \end{array}$

Q.57

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

Ans.

$\begin{array}{l} Let θ be the angle between the vectors \vec{a} and \vec{b} . Then, \\ \vec{a} . \vec{b} = | \vec{a} | | \vec{b} | cosθ \\ \frac{1}{2} = | \vec{a} | | \vec{a} | \cos 60 ° [| \vec{a} | = | \vec{b} |] \end{array}$ $\begin{array}{l} \frac{1}{2} = {| \vec{a} |}^{2} \times \frac{1}{2} \\ 1 = {| \vec{a} |}^{2} \\ \Rightarrow | \vec{a} | = 1 \\ So, | \vec{b} | = | \vec{a} | = 1 \end{array}$

Q.58

$Find | \vec{x} |, if for a unit vector \vec{a}, (\vec{x} - \vec{a}) . (\vec{x} + \vec{a}) = 12 .$

Ans.

$\begin{array}{l} Since, \\ (\vec{x} - \vec{a}) . (\vec{x} + \vec{a}) = 12 \\ \vec{x} . \vec{x} + \vec{x} . \vec{a} - \vec{a} . \vec{x} - \vec{a} . \vec{a} = 12 \\ {| \vec{x} |}^{2} - {| \vec{a} |}^{2} = 12 [\vec{x} . \vec{a} = \vec{a} . \vec{x}] \\ \Rightarrow {| \vec{x} |}^{2} - 1^{2} = 12 [\vec{a} is a unit vector so, | \vec{a} | = 1] \\ \Rightarrow {| \vec{x} |}^{2} = 12 + 1 \\ \Rightarrow | \vec{x} | = \sqrt{13} \end{array}$

Q.59

$\begin{array}{l} If \vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = 2 \hat{i} - \hat{j} + 3 \hat{k} and \vec{c} = \hat{i} - 2 \hat{j} + \hat{k}, find a unit \\ vector parallel to the vector 2 \vec{a} - \vec{b} + 3 \vec{c} . \end{array}$

Ans.

$\begin{array}{l} The given vectors are: \\ \vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = 2 \hat{i} - \hat{j} + 3 \hat{k} and \vec{c} = \hat{i} - 2 \hat{j} + \hat{k} \\ So, \\ 2 \vec{a} - \vec{b} + 3 \vec{c} = 2 (\hat{i} + \hat{j} + \hat{k}) - (2 \hat{i} - \hat{j} + 3 \hat{k}) + 3 (\hat{i} - 2 \hat{j} + \hat{k}) \\ = 2 \hat{i} + 2 \hat{j} + 2 \hat{k} - 2 \hat{i} + \hat{j} - 3 \hat{k} + 3 \hat{i} - 6 \hat{j} + 3 \hat{k} \\ = 3 \hat{i} - 3 \hat{j} + 2 \hat{k} \\ | 2 \vec{a} - \vec{b} + 3 \vec{c} | = | 3 \hat{i} - 3 \hat{j} + 2 \hat{k} | \\ = \sqrt{3^{2} + {(- 3)}^{2} + 2^{2}} \\ = \sqrt{9 + 9 + 4} \\ = \sqrt{22} \\ Thus, the unit vector along (2 \vec{a} - \vec{b} + 3 \vec{c}) is \\ \frac{2 \vec{a} - \vec{b} + 3 \vec{c}}{| 2 \vec{a} - \vec{b} + 3 \vec{c} |} = \frac{3 \hat{i} - 3 \hat{j} + 2 \hat{k}}{\sqrt{22}} \\ = \frac{3}{\sqrt{22}} \hat{i} - \frac{3}{\sqrt{22}} \hat{j} + \frac{2}{\sqrt{22}} \hat{k} \end{array}$

Q.60

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

Ans.

$\begin{array}{l} The given vectors are: \\ \vec{a} = 2 \hat{i} + 2 \hat{j} + 3 \hat{k}, \vec{b} = - \hat{i} + 2 \hat{j} + \hat{k} and \vec{c} = 3 \hat{i} + \hat{j} \end{array}$ $\begin{array}{l} ∴ \vec{a} + λ \vec{b} = (2 \hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (- \hat{i} + 2 \hat{j} + \hat{k}) \\ = (2 - λ) \hat{i} + (2 + 2 λ) \hat{j} + (3 + λ) \hat{k} \\ Since, (\vec{a} + λ \vec{b}) is perpendicular to \vec{c}, so \\ (\vec{a} + λ \vec{b}) . \vec{c} = 0 \\ \Rightarrow {(2 - λ) \hat{i} + (2 + 2 λ) \hat{j} + (3 + λ) \hat{k}} . (3 \hat{i} + \hat{j}) = 0 \\ \Rightarrow 3 (2 - λ) + (2 + 2 λ) = 0 \\ \Rightarrow 6 - 3 λ + 2 + 2 λ = 0 \\ \Rightarrow 8 - λ = 0 \\ \Rightarrow λ = 8 \\ Therefore, the required value of λ is 8. \end{array}$

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

Ans.

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

Q.62

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

Ans.

$\begin{array}{l} (| \vec{a} | \vec{b} + | \vec{b} | \vec{a}) . (| \vec{a} | \vec{b} - | \vec{b} | \vec{a}) = {| \vec{a} |}^{2} {| \vec{b} |}^{2} - | \vec{a} | \vec{b} | \vec{b} | \vec{a} + | \vec{b} | \vec{a} | \vec{a} | \vec{b} - {| \vec{b} |}^{2} {| \vec{a} |}^{2} \\ = - | \vec{a} | | \vec{b} | \vec{b} \vec{a} + | \vec{b} | | \vec{a} | \vec{a} \vec{b} \\ = 0 [\vec{b} \vec{a} = \vec{a} \vec{b}] \\ So, (| \vec{a} | \vec{b} + | \vec{b} | \vec{a}) and (| \vec{a} | \vec{b} - | \vec{b} | \vec{a}) are perpendicular to each other . \end{array}$

Q.63

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

Ans.

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

Q.64

$\begin{array}{l} If \vec{a} . \vec{a} = 0 and \vec{a} . \vec{b} = 0, then what can be concluded about \\ the vector \vec{b} ? \end{array}$

Ans.

$\begin{array}{l} Since, it is given that \\ \vec{a} . \vec{a} = 0 \\ \Rightarrow {| \vec{a} |}^{2} = 0 \\ \Rightarrow | \vec{a} | = 0 \\ ∴ \vec{a} is a zero vector. \\ Since, vector \vec{b} satisfying \vec{a} . \vec{b} = 0 can be any vector. \end{array}$

Q.65

$\begin{array}{l} If \vec{a}, \vec{b}, \vec{c} are unit vectors such that \vec{a} + \vec{b} + \vec{c} = \vec{0}, find the \\ value of \vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a} . \end{array}$

Ans.

$\begin{array}{l} Given: \vec{a}, \vec{b}, \vec{c} are unit vectors such that \vec{a} + \vec{b} + \vec{c} = \vec{0} \\ Then, {[\vec{a} + \vec{b} + \vec{c}]}^{2} = 0 \\ {| \vec{a} |}^{2} + {| \vec{b} |}^{2} + {| \vec{c} |}^{2} + 2 [\vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}] = 0 \\ 1 + 1 + 1 + 2 [\vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}] = 0 \\ \Rightarrow \vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a} = - \frac{3}{2} \end{array}$

Q.66

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

Ans.

$\begin{array}{l} Adjacent sides of a parallelogram are: \\ \vec{a} = 2 \hat{i} - 4 \hat{j} + 5 \hat{k} and \vec{b} = \hat{i} - 2 \hat{j} - 3 \hat{k} \\ Then, the diagonal of a parallelogram = \vec{a} + \vec{b} \\ = 2 \hat{i} - 4 \hat{j} + 5 \hat{k} + \hat{i} - 2 \hat{j} - 3 \hat{k} \\ = 3 \hat{i} - 6 \hat{j} + 2 \hat{k} \\ The unit vector parallel to the diagonal = \frac{3 \hat{i} - 6 \hat{j} + 2 \hat{k}}{| 3 \hat{i} - 6 \hat{j} + 2 \hat{k} |} \\ = \frac{3 \hat{i} - 6 \hat{j} + 2 \hat{k}}{\sqrt{3^{2} + {(- 6)}^{2} + 2^{2}}} \\ = \frac{3 \hat{i} - 6 \hat{j} + 2 \hat{k}}{\sqrt{9 + 36 + 4}} \\ = \frac{3 \hat{i} - 6 \hat{j} + 2 \hat{k}}{\sqrt{49}} \\ = \frac{3}{7} \hat{i} - \frac{6}{7} \hat{j} + \frac{2}{7} \hat{k} \\ Now, Area of parallelogram ABCD = | \vec{a} \times \vec{b} | \\ \vec{a} \times \vec{b} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & - 4 & 5 \\ 1 & - 2 & - 3 \end{matrix} | \\ = (12 + 10) \hat{i} - (- 6 - 5) \hat{j} + (- 4 + 4) \hat{k} \\ = 22 \hat{i} + 11 \hat{j} \\ ∴ | \vec{a} \times \vec{b} | = 11 | 2 \hat{i} + \hat{j} | \\ = 11 \sqrt{2^{2} + 1^{2}} \\ = 11 \sqrt{5} \\ Thus, the area of the parallelogram is 11 \sqrt{5} square units. \end{array}$

Q.67

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

Ans.

$\begin{array}{l} Let a vector be equally inclined to axes OX, OY, and OZ at \\ angle α . Then, the direction cosines of the vector are cos α, \\ cos α, and cos α . \\ {Since, cos}^{2} α + {cos}^{2} β + {cos}^{2} γ = 1 \\ So, {cos}^{2} α + {cos}^{2} α + {cos}^{2} α = 1 \\ 3 {cos}^{2} α = 1 \\ {cos}^{2} α = \frac{1}{3} \\ cos α = \frac{1}{\sqrt{3}} \\ Hence, the direction cosines of the vector which are equally \\ inclined to the axes are \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} . \end{array}$

Q.68

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

Ans.

$\begin{array}{l} Let \vec{a} = 3 \hat{i} + 4 \hat{j} + 3 \hat{k} and \vec{b} = 2 \hat{i} + 3 \hat{j} - 6 \hat{k} \\ Then, \\ \vec{a} . \vec{b} = (3 \hat{i} + 4 \hat{j} + 3 \hat{k}) (2 \hat{i} + 3 \hat{j} - 6 \hat{k}) \end{array}$ $\begin{array}{l} = 6 + 12 - 18 \\ = 0 \\ And \\ | \vec{a} | = | 3 \hat{i} + 4 \hat{j} + 3 \hat{k} | \\ = \sqrt{3^{2} + 4^{2} + 3^{2}} \\ = \sqrt{9 + 16 + 9} \\ = \sqrt{34} \neq 0 \\ | \vec{b} | = | 2 \hat{i} + 3 \hat{j} - 6 \hat{k} | \\ = \sqrt{2^{2} + 3^{2} + {(- 6)}^{2}} \\ = \sqrt{4 + 9 + 36} \\ = \sqrt{49} \\ = 7 \neq 0 \\ Hence, the converse of the given statement need not be true. \end{array}$

Q.69

$\begin{array}{l} Let \vec{a} = \hat{i} + 4 \hat{j} + 2 \hat{k}, \vec{b} = 3 \hat{i} - 2 \hat{j} + 7 \hat{k} and \vec{c} = 2 \hat{i} - \hat{j} + 4 \hat{k} . Find a \\ vector \vec{d} which is perpendicular to both \vec{a} \overset{}{a} nd \vec{b}, \vec{c} . \vec{d} = 15 . \end{array}$

Ans.

$\begin{array}{l} The given vectors are \vec{a} = \hat{i} + 4 \hat{j} + 2 \hat{k}, \vec{b} = 3 \hat{i} - 2 \hat{j} + 7 \hat{k} and \\ \vec{c} = 2 \hat{i} - \hat{j} + 4 \hat{k} . Let \vec{d} = d_{1} \hat{i} + d_{2} \hat{j} + d_{3} \hat{k} \\ Since, \vec{d} is perpendicular to \vec{a} and \vec{b}, so \\ \vec{a} . \vec{d} = 0 \\ \Rightarrow (\hat{i} + 4 \hat{j} + 2 \hat{k}) . (d_{1} \hat{i} + d_{2} \hat{j} + d_{3} \hat{k}) = 0 \\ \Rightarrow d_{1} + 4 d_{2} + 2 d_{3} = 0 . .. (i) \\ \vec{b} . \vec{d} = 0 \\ \Rightarrow (3 \hat{i} - 2 \hat{j} + 7 \hat{k}) . (d_{1} \hat{i} + d_{2} \hat{j} + d_{3} \hat{k}) = 0 \\ \Rightarrow 3 d_{1} - 2 d_{2} + 7 d_{3} = 0 . .. (ii) \\ But, \vec{c} . \vec{d} = 15 \\ \Rightarrow (2 \hat{i} - \hat{j} + 4 \hat{k}) . (d_{1} \hat{i} + d_{2} \hat{j} + d_{3} \hat{k}) = 15 \\ \Rightarrow 2 d_{1} - d_{2} + 4 d_{3} = 15 . .. (iii) \\ On solving (i), (ii), and (iii), we get: \\ d_{1} = \frac{160}{3}, d_{2} = - \frac{5}{3}, d_{3} = - \frac{70}{3} \\ ∴ \vec{d} = \frac{160}{3} \hat{i} - \frac{5}{3} \hat{j} - \frac{70}{3} \hat{k} = \frac{1}{3} (160 \hat{i} - 5 \hat{j} - 70 \hat{k}) \\ Hence, the required vector is \frac{1}{3} (160 \hat{i} - 5 \hat{j} - 70 \hat{k}) . \end{array}$

Q.70

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

Ans.

$\begin{array}{l} Sum of vectors 2 \hat{i} + 4 \hat{j} - 5 \hat{k} and λ \hat{i} + 2 \hat{j} + 3 \hat{k} = 2 \hat{i} + 4 \hat{j} - 5 \hat{k} + λ \hat{i} + 2 \hat{j} + 3 \hat{k} \\ = (2 + λ) \hat{i} + 6 \hat{j} - 2 \hat{k} \\ Unit vector of (2 + λ) \hat{i} + 6 \hat{j} - 2 \hat{k} = \frac{(2 + λ) \hat{i} + 6 \hat{j} - 2 \hat{k}}{| (2 + λ) \hat{i} + 6 \hat{j} - 2 \hat{k} |} \\ = \frac{(2 + λ) \hat{i} + 6 \hat{j} - 2 \hat{k}}{\sqrt{{(2 + λ)}^{2} + 6^{2} + {(- 2)}^{2}}} \\ = \frac{(2 + λ) \hat{i} + 6 \hat{j} - 2 \hat{k}}{\sqrt{4 + 4 λ + λ^{2} + 36 + 4}} \\ = \frac{(2 + λ) \hat{i} + 6 \hat{j} - 2 \hat{k}}{\sqrt{44 + 4 λ + λ^{2}}} \\ Scalar product of unit vector and (\hat{i} + \hat{j} + \hat{k}) = 1 [Given] \\ (\hat{i} + \hat{j} + \hat{k}) . \frac{(2 + λ) \hat{i} + 6 \hat{j} - 2 \hat{k}}{\sqrt{44 + 4 λ + λ^{2}}} = 1 \\ (2 + λ) + 6 - 2 = \sqrt{44 + 4 λ + λ^{2}} \\ \Rightarrow {(6 + λ)}^{2} = 44 + 4 λ + λ^{2} \\ \Rightarrow 36 + 12 λ + λ^{2} = 44 + 4 λ + λ^{2} \\ \Rightarrow 8 λ = 8 \\ \Rightarrow λ = 1 \\ Thus, the value of λ is 1 . \end{array}$

Q.71

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

Ans.

$\begin{array}{l} Since, \vec{a}, \vec{b}, \vec{c} are mutually perpendicular vectors, so \\ \vec{a} . \vec{b} = \vec{b} . \vec{c} = \vec{c} . \vec{a} = 0 \\ Magnitudes ‹ of \vec{a}, \vec{b}, \vec{c} are equal, so \\ | \vec{a} | = | \vec{b} | = | \vec{c} | \\ Let \vec{a}, \vec{b}, \vec{c} are making angles θ_{1}, θ_{2} and θ_{3} with (\vec{a} + \vec{b} + \vec{c}) \\ respectively.Then, we have \\ cos θ_{1} = \frac{(\vec{a} + \vec{b} + \vec{c}) . \vec{a}}{| \vec{a} + \vec{b} + \vec{c} | | \vec{a} |} = \frac{{| \vec{a} |}^{2}}{| \vec{a} + \vec{b} + \vec{c} | | \vec{a} |} = \frac{| \vec{a} |}{| \vec{a} + \vec{b} + \vec{c} |} \\ cos θ_{2} = \frac{(\vec{a} + \vec{b} + \vec{c}) . \vec{b}}{| \vec{a} + \vec{b} + \vec{c} | | \vec{b} |} = \frac{{| \vec{b} |}^{2}}{| \vec{a} + \vec{b} + \vec{c} | | \vec{b} |} = \frac{| \vec{b} |}{| \vec{a} + \vec{b} + \vec{c} |} \\ cos θ_{3} = \frac{(\vec{a} + \vec{b} + \vec{c}) . \vec{c}}{| \vec{a} + \vec{b} + \vec{c} | | \vec{c} |} = \frac{{| \vec{c} |}^{2}}{| \vec{a} + \vec{b} + \vec{c} | | \vec{c} |} = \frac{| \vec{c} |}{| \vec{a} + \vec{b} + \vec{c} |} \\ Since, | \vec{a} | = | \vec{b} | = | \vec{c} |, so cos θ_{1} = cos θ_{2} = cos θ_{3} \\ ∴ θ_{1} = θ_{2} = θ_{3} \\ Hence, the vector (\vec{a} + \vec{b} + \vec{c}) is equally inclined to \vec{a}, \vec{b} and \vec{c} . \end{array}$

Q.72

$Prove that (\vec{a} + \vec{b}) . (\vec{a} + \vec{b}) = {| \vec{a} |}^{2} + {| \vec{b} |}^{2}, if and only if \vec{a}, \vec{b} are perpendicular, given \vec{a} \neq 0, \vec{b} \neq 0 .$

Ans.

$\begin{array}{l} L . H . S . = (\vec{a} + \vec{b}) . (\vec{a} + \vec{b}) \\ = \vec{a} . \vec{a} + \vec{a} . \vec{b} + \vec{b} . \vec{a} + \vec{b} . \vec{b} \\ = {| \vec{a} |}^{2} + 2 \vec{a} . \vec{b} + {| \vec{b} |}^{2} [\vec{a} . \vec{b} = \vec{b} . \vec{a}] \\ = {| \vec{a} |}^{2} + 2 (0) + {| \vec{b} |}^{2} [\begin{array}{l} \vec{a} . \vec{b} = 0 as \vec{a} and \vec{b} are \\ perpendicular to each other. \end{array}] \\ = {| \vec{a} |}^{2} + {| \vec{b} |}^{2} = R . H . S . \end{array}$

Q.73

$\begin{array}{l} If θ is the angle between two vectors \vec{a} and \vec{b}, then \vec{a} . \vec{b} \geq 0 \\ only when \\ (A) 0 < θ < \frac{π}{2} \\ (B) 0 \leq θ \leq \frac{π}{2} \\ (C) 0 < θ < π \\ (D) 0 \leq θ \leq π \end{array}$

Ans.

$\begin{array}{l} If θ is the angle between two vectors \vec{a} and \vec{b}, \\ then \vec{a} . \vec{b} \geq 0 \\ | \vec{a} | . | \vec{b} | cosθ \geq 0 \\ \Rightarrow cosθ \geq 0 [| \vec{a} | \neq 0, | \vec{b} | \neq 0] \\ \Rightarrow 0 \leq θ \leq \frac{π}{2} \\ So, \vec{a} . \vec{b} \geq 0 if 0 \leq θ \leq \frac{π}{2} . \\ Thus, option B is correct. \end{array}$

Q.74

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

Ans.

$\begin{array}{l} The vertices of ΔABC are given as A (1, 2, 3), B (–1, 0, 0), \\ and C (0, 1, 2) . \end{array}$ $\begin{array}{l} Also, it is given that ∠ ABC is the angle between the vectors \vec{BA} \\ and \vec{BC} . \\ ∴ \vec{BA} = (1 + 1) \hat{i} + (2 – 0) \hat{j} + (3 – 0) \hat{k} \\ = 2 \hat{i} + 2 \hat{j} + 3 \hat{k} \\ \vec{BC} = (0 + 1) \hat{i} + (1 – 0) \hat{j} + (2 – 0) \hat{k} \\ = \hat{i} + \hat{j} + 2 \hat{k} \\ ∴ \vec{BA} \cdot \vec{BC} \\ = (2 \hat{i} + 2 \hat{j} + 3 \hat{k}) \cdot (\hat{i} + \hat{j} + 2 \hat{k}) \\ = 2 + 2 + 6 \\ = 10 \\ | \vec{BA} | = | 2 \hat{i} + 2 \hat{j} + 3 \hat{k} | \\ = \sqrt{2^{2} {+ 2}^{2} {+ 3}^{2}} \\ = \sqrt{4 + 4 + 9} \\ = \sqrt{17} \\ | \vec{BC} | = | \hat{i} + \hat{j} + 2 \hat{k} | \\ = \sqrt{1^{2} {+ 1}^{2} {+ 2}^{2}} \\ = \sqrt{1 + 1 + 4} \\ = \sqrt{6} \\ Now, it is known that: \\ \vec{BA} \cdot \vec{BC} = | \vec{BA} | \cdot | \vec{BC} | cos (∠ ABC) \\ \Rightarrow 10 = \sqrt{17} \sqrt{6} cos (∠ ABC) \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} cos (∠ ABC) = \frac{10}{\sqrt{102}} \\ ∠ {ABC = cos}^{–1} (\frac{10}{\sqrt{102}}) \end{array}$

Q.75

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

Ans.

$\begin{array}{l} The vertices of ΔABC are given as A (1, 2, 3), B (–1, 0, 0), \\ and C (0, 1, 2). \end{array}$ $\begin{array}{l} Also, it is given that ∠ ABC is the angle between the vectors \vec{BA} \\ and \vec{BC} . \\ ∴ \vec{BA} = (1 + 1) \hat{i} + (2 – 0) \hat{j} + (3 – 0) \hat{k} \\ = 2 \hat{i} + 2 \hat{j} + 3 \hat{k} \\ \vec{BC} = (0 + 1) \hat{i} + (1 – 0) \hat{j} + (2 – 0) \hat{k} \\ = \hat{i} + \hat{j} + 2 \hat{k} \\ ∴ \vec{BA} \cdot \vec{BC} \\ = (2 \hat{i} + 2 \hat{j} + 3 \hat{k}) \cdot (\hat{i} + \hat{j} + 2 \hat{k}) \\ = 2 + 2 + 6 \\ = 10 \\ | \vec{BA} | = | 2 \hat{i} + 2 \hat{j} + 3 \hat{k} | \\ = \sqrt{2^{2} {+ 2}^{2} {+ 3}^{2}} \\ = \sqrt{4 + 4 + 9} \\ = \sqrt{17} \\ | \vec{BC} | = | \hat{i} + \hat{j} + 2 \hat{k} | \\ = \sqrt{1^{2} {+ 1}^{2} {+ 2}^{2}} \\ = \sqrt{1 + 1 + 4} \\ = \sqrt{6} \\ Now, it is known that: \\ \vec{BA} \cdot \vec{BC} = | \vec{BA} | \cdot | \vec{BC} | cos (∠ ABC) \\ \Rightarrow 10 = \sqrt{17} \sqrt{6} cos (∠ ABC) \end{array}$ $\begin{array}{l} cos (∠ ABC) = \frac{10}{\sqrt{102}} \\ ∠ {ABC = cos}^{–1} (\frac{10}{\sqrt{102}}) \end{array}$ $\begin{array}{l} | \vec{AC} | = | 2 \hat{i} + 8 \hat{j} – 8 \hat{k} | \\ = \sqrt{2^{2} {+ 8}^{2} + {(–8)}^{2}} \\ = \sqrt{4 + 64 + 64} \\ = \sqrt{132} = 2 \sqrt{33} \\ ∴ | \vec{AC} | = | \vec{AB} | + | \vec{BC} | \\ Hence, the given points A, B, and C are collinear. \end{array}$

Q.76

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

Ans.

$\begin{array}{l} Let vectors 2 \hat{i} - \hat{j} + \hat{k}, \hat{i} - 3 \hat{j} - 5 \hat{k} a nd 3 \hat{i} - 4 \hat{j} - 4 \hat{k} be position vectors \\ of points A, B and C respectively. \\ i.e., \vec{OA} = 2 \hat{i} - \hat{j} + \hat{k}, \vec{OB} = \hat{i} - 3 \hat{j} - 5 \hat{k}, \vec{OC} = 3 \hat{i} - 4 \hat{j} - 4 \hat{k} \\ ∴ \vec{AB} = (1 - 2) \hat{i} + (- 3 + 1) \hat{j} + (- 5 - 1) \hat{k} = - \hat{i} - 2 \hat{j} - 6 \hat{k} \\ \vec{BC} = (3 - 1) \hat{i} + (- 4 + 3) \hat{j} + (- 4 + 5) \hat{k} = 2 \hat{i} - \hat{j} + \hat{k} \\ \vec{CA} = (2 - 3) \hat{i} + (- 1 + 4) \hat{j} + (1 + 4) \hat{k} = - \hat{i} + 3 \hat{j} + 5 \hat{k} \\ | \vec{AB} | = | - \hat{i} - 2 \hat{j} - 6 \hat{k} | \\ = \sqrt{{(- 1)}^{2} + {(- 2)}^{2} + {(- 6)}^{2}} \\ = \sqrt{1 + 4 + 36} \\ = \sqrt{41} \\ | \vec{BC} | = | 2 \hat{i} - \hat{j} + \hat{k} | \end{array}$ $\begin{array}{l} = \sqrt{2^{2} + {(- 1)}^{2} + 1^{2}} \\ = \sqrt{4 + 1 + 1} \\ = \sqrt{6} \\ | \vec{CA} | = | - \hat{i} + 3 \hat{j} + 5 \hat{k} | \\ = \sqrt{{(- 1)}^{2} + 3^{2} + 5^{2}} \\ = \sqrt{1 + 9 + 25} \\ = \sqrt{35} \\ ∴ {| \vec{BC} |}^{2} + {| \vec{CA} |}^{2} = {(\sqrt{6})}^{2} + {(\sqrt{35})}^{2} \end{array}$ $\begin{array}{l} = 6 + 35 \\ = 41 \\ = {| \vec{AB} |}^{2} \\ Hence, Δ ABC is a right-angled triangle. \end{array}$

Q.77

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

Ans.

$\begin{array}{l} Vector λ \vec{a} is a unit vector if | λ \vec{a} | = 1. \\ So, | λ \vec{a} | = 1 \\ \Rightarrow | λ | | \vec{a} | = 1 \end{array}$ $\begin{array}{l} \Rightarrow | \vec{a} | = \frac{1}{| λ |} [λ \neq 0] \\ \Rightarrow a = \frac{1}{| λ |} [| \vec{a} | = a] \\ Hence, λ \vec{a} vector is a unit vector if a = \frac{1}{| λ |} . \\ The correct answer is D. \end{array}$

Q.78

$\begin{array}{l} Let ‹‹and ‹‹‹be‹two ‹unit ‹vectors ‹and‹ θ ‹is‹the ‹angle ‹between \\ them . ‹Then \vec{a} ‹‹ + \vec{b} ‹‹is ‹a ‹unit ‹vector‹if \\ (A) θ = \frac{π}{4} \\ (B) θ = \frac{π}{3} \\ (C) θ = \frac{π}{2} \\ (D) θ = \frac{2 π}{3} \end{array}$

Ans.

$\begin{array}{l} Let \vec{a} and \vec{b} be two unit vectors i.e., | \vec{a} | = | \vec{b} | = 1 \\ and θ is the angle between them . \\ Then \vec{a} + \vec{b} is a unit vector if | \vec{a} + \vec{b} | = 1 . \\ | \vec{a} + \vec{b} | = 1 \Rightarrow {| \vec{a} + \vec{b} |}^{2} = 1 \\ \Rightarrow (\vec{a} + \vec{b}) . (\vec{a} + \vec{b}) = 1 \\ \Rightarrow \vec{a} . \vec{a} + \vec{a} . \vec{b} + \vec{b} . \vec{a} + \vec{b} . \vec{b} = 1 \\ \Rightarrow {| \vec{a} |}^{2} + 2 \vec{a} . \vec{b} + {| \vec{b} |}^{2} = 1 \\ \Rightarrow 1 + 2 \vec{a} . \vec{b} + 1 = 1 [| \vec{a} | = | \vec{b} | = 1] \\ \Rightarrow \vec{a} . \vec{b} = - \frac{1}{2} \\ \Rightarrow | \vec{a} | . | \vec{b} | cosθ = - \frac{1}{2} \\ \Rightarrow 1 . 1 cosθ = - \frac{1}{2} [| \vec{a} | = | \vec{b} | = 1] \\ \Rightarrow cosθ = \cos \frac{2 π}{3} \\ \Rightarrow θ = \frac{2 π}{3} \\ Thus, (\vec{a} + \vec{b}) is unit vector if θ = \frac{2 π}{3} . \\ Then, the correct option is D. \end{array}$

Q.79

$\begin{array}{l} The‹value‹of \hat{i} . ‹ (\hat{j} \times ‹ \hat{k}) ‹ + ‹ \hat{j} . ‹ (\hat{i} ‹ \times \hat{k}) + ‹ \hat{k} . ‹ (\hat{i} ‹ \times ‹ \hat{j}) ‹is \\ \begin{array}{l} (A) 0 \\ (B) - 1 \\ (C) 1 \\ (D) 3 \end{array} \end{array}$

Ans.

$\begin{array}{l} \hat{i} . (\hat{j} \times \hat{k}) + \hat{j} . (\hat{i} \times \hat{k}) + \hat{k} . (\hat{i} \times \hat{j}) = \hat{i} . \hat{i} + \hat{j} . (- \hat{j}) + \hat{k} . \hat{k} [\begin{array}{l} \hat{j} \times \hat{k} = \hat{i}, \hat{i} \times \hat{k} = \hat{j} \\ \hat{i} \times \hat{j} = \hat{k} \end{array}] \\ = 1 - 1 + 1 [\hat{i} . \hat{i} = \hat{j} . \hat{j} = \hat{k} . \hat{k} = 1] \\ = 1 \\ Thus, the correct option is C. \end{array}$

Q.80

$\begin{array}{l} If θ is the angle between any two vectors \vec{a} and \vec{b}, then \\ |\vec{a} . \vec{b}| = |\vec{a} \times \vec{b}| when θ is equal to \\ (A) 0 \\ (B) \frac{π}{4} \\ (C) \frac{π}{2} \\ (D) π \end{array}$

Ans.

$\begin{array}{l} Let θ be the angle between two vectors \vec{a} and \vec{b} and | \vec{a} | \neq 0, | \vec{b} | \neq 0 . \\ Given : \\ | \vec{a} . \vec{b} | = | \vec{a} \times \vec{b} | \\ \Rightarrow | \vec{a} | . | \vec{b} | cosθ = | \vec{a} | . | \vec{b} | sinθ \\ \Rightarrow cosθ = sinθ [| \vec{a} | and | \vec{b} | are positive.] \\ \Rightarrow tanθ = 1 \\ = \tan \frac{π}{4} \\ \Rightarrow θ = \frac{π}{4} \\ So, the correct option is B. \end{array}$

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NCERT Solutions Class 12 Mathematics Chapter 10

Key Topics Covered in NCERT Solutions Class 12 Mathematics Chapter 10

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Related Chapters

Chapter 1 - Relations and Functions

Chapter 2 - Inverse Trigonometric Functions

Chapter 3 - Matrices

Chapter 4 - Determinants

Chapter 5 - Continuity and Differentiability

Chapter 6 - Application of Derivatives

Chapter 7 - Integrals

Chapter 8 - Application of Integrals

Chapter 9 - Differential Equations

Chapter 11 - Three Dimensional Geometry

Chapter 12 - Linear Programming

Chapter 13 - Probability

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