49. If either a→ = 0 and b→ = 0 then a→×b→=0 Is the converse true? Justify your answer with an example.

We take any parallel non- zero vectors so that a→×b→=0 .

13. For given vectors a=2i^−j^+2k^ and b=−i^+j^−k^ , find the unit vector in the direction of a→+b→

Given, a→=2i^−j^+2k^&b→=−i^+j^−k^ The sum of given vectors is given by

19. Find the position vector of a point R which divides the line joining two points P and Q. whose position vectors are i^+2j^−k^ and - i^+j^+k^ respectively, in the ratio 2 : 1 (i) Internally (ii) Externally

(i) The position vector of point R dividing the join of P and Q. internally in the ratio 2:1 is, =(−2i^+i^)+(2j^+2j^)+(2k^+k^)3=−i^+4j^+k^3=−13i^+43j^+13k^ (ii) The position vector of the point k dividing the join of P and Q. externally in the ratio 2:1 A15. (ii) OR→=2(−i^+j^+k^)−1(i^+2j^−k^)2−1=−2i^+2j^+2k^−i^−2j^+k^=−3i^+k^

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra – PDF Download

Q: 25. Kindly Consider the following

Now, a→.b→= (i^−2j^+3k^) (3i^−2j^+k^)=1.3+ (−2). (−2)+3.1=3+4+3=10 Also, we know,

Updated on Aug 6, 2025 15:51 IST

By Pallavi Pathak, Assistant Manager Content

Vector Algebra Class 12 covers the basic concepts of vectors, their algebraic and geometric properties, and various operations on vectors. Vector quantities include velocity, displacement, force, acceleration, momentum, weight, and electric field intensity. The chapter introduces the types of vectors, multiplication of a vector by a scalar, addition, and the product of vectors.
The subject matter experts at Shiksha created the Vectors Class 12 NCERT Solutions for students to deepen their concept clarity and score high in the CBSE Board and competitive exams like JEE Mains. This page also provides the Vector Algebra Class 12 PDF for students to download and prepare for the examination.
To get the topic-wise PDF and notes PDF for Class 12 Maths, check - Class 12 Maths Notes PDF for CBSE Exams.

Table of content

Glance at Vector Algebra Class 12
Class 12 Math Chapter 10 Vector Algebra : Key Topics, Weightage
Important Formulas of Class 12 Vector Algebra
Class 12 Maths Vector Algebra NCERT Solutions PDF – Download for Free
Class 12 Vector Algebra Exercise-wise NCERT Solutions
Class 12 Vector Algebra Exercise 10.1 Solutions

Glance at Vector Algebra Class 12

Here is a quick review of Class 12 Maths Vector Algebra:

The position vector of a point P(x, y, z) is given by $\vec{OP} (\vec{r}) = x \hat{i} + y \hat{j} + z \hat{k}$ and its magnitude by $\sqrt{x^{2} + y^{2} + z^{2}}$
The scalar components of a vector represent its projections along the respective axes, and they are in direction ratios.
The direction ratios (a, b, c), magnitude (r), and direction cosines (l,m,n) of any vector are related by - $l = \frac{a}{r}, m = \frac{b}{r}, n = \frac{c}{r}$
The chapter covers the vector sum of three sides of a triangle, two coinitial vectors, and the multiplication of vector.

To get access to the short revision notes of Chemistry, Physics, and Maths, check - NCERT Class 12 Notes.

Class 12 Math Chapter 10 Vector Algebra : Key Topics, Weightage

Before starting the preparation of any chapter, it is good to know the topics covered in it. See below the topics covered in the Vector Algebra Class 12:

Exercise	Topics Covered
10.1	Introduction
10.2	Some Basic Concepts
10.3	Types of Vectors
10.4	Addition of Vectors
10.5	Multiplication of a Vector by a Scalar
10.6	Product of Two Vectors

Vector Algebra Class 12 Weightage in JEE Main

Exam	Number of Questions	Weightage
JEE Main	2 questions	8%

Class 12 Maths Vector Algebra NCERT Solutions PDF – Download for Free

Students should download the Vector Algebra Class 12 PDF from the link given below. The solutions are given in a step-by-step format. It is easy to understand and improves the problem-solving skills of students.

Class 12 Math Chapter 10 Vector Solution PDF: Free PDF Download

Class 12 Vector Algebra Exercise-wise NCERT Solutions

This chapter introduces concepts like vector operations, dot and cross products, scalar triple products, and their real-world significance. Mastering these topics is crucial for excelling in board exams and competitive entrance tests.

Class 12 Vector Algebra Exercise 10.1 Solutions

Class 12 Vector Algebra Exercise 10.1 deals with key concepts such as vectors, their representation, operations like addition, subtraction, and scalar multiplication, as well as their geometric interpretations. Shiksha has provided detailed solutions to all the problems in Vector Algebra Exercise 10.1, ensuring that students grasp the concepts thoroughly. Vector Algebra Class 12 Exercise 10.1 Solutions consists of 5 Questions. Students can check the complete solution for VEctor Algebra Exercise 10.1 below;

Class 12 Vector Algebra Exercise 10.1 Solutions

Q1. Represent graphically a displacement of 40 km, 30° east of north.

A.1. $40 k m, 3 0^{0}$ east of north.

(i) 10 kg

(ii) 2 meters north-west

(iii) 40°

(iv) 40 Watt

(v) 10^–19 coulomb

(vi) 20 m/sec²

A.2. (i) 10kg involves only magnitude. So, it is scalar quantity.

(ii) 2 meters north-west involves both magnitude and direction. So, it is vector quantity.

(iii) 40⁰ involves only magnitude. So, it is scalar quantity.

(iv) 40⁰ watts involves only magnitude. So, it is scalar quantity.

(v) 10^-19 coulomb involves only magnitude. So, it is scalar quantity.

(vi) 20m/s^-2 involves magnitude and direction. So, it is vector quantity.

Q3. Classify the following as scalar and vector quantities:

(i) time period

(ii) distance

(iii) force

(iv) velocity

(v) work done

A.3. (i) Time period involves only magnitude. So, it is scalar quantity.

(ii) Distance involves only magnitude. So, it is scalar quantity.

(iii) Force involves both magnitude and direction. So, it is vector quantity.

(iv) Velocity involves both magnitude and direction. So, it is vector quantity.

(v) Work done involves only magnitude. So, it is scalar quantity.

Q..4. In the adjoining figure, (a square) identify the following vectors:

(ii) Equal

(iii) Collinear but not equal.

A.4.

(a) Vector $\vec{a}$ and $\vec{d}$ are co initial same initial point.

(b) $\vec{b}$ and $\vec{d}$ same magnitude & direction.

Commonly asked questions

25. Kindly Consider the following

Now,

$\begin{array}{l} \vec{a} . \vec{b} = (\hat{i} - 2 \hat{j} + 3 \hat{k}) (3 \hat{i} - 2 \hat{j} + \hat{k}) \\ = 1.3 + (- 2) . (- 2) + 3.1 \\ = 3 + 4 + 3 = 10 \end{array}$

Also, we know,

49. If either $\vec{a}$ = 0 and $\vec{b}$ = 0 then $\vec{a} \times \vec{b} = 0$ Is the converse true? Justify your answer with an example.

We take any parallel non- zero vectors so that $\vec{a} \times \vec{b} = 0$ .

13. For given vectors $a = 2 \hat{i} - \hat{j} + 2 \hat{k} a n d b = - \hat{i} + \hat{j} - \hat{k}$ , find the unit vector in the direction of $\vec{a} + \vec{b}$

Given, $\vec{a} = 2 \hat{i} - \hat{j} + 2 \hat{k} & \vec{b} = - \hat{i} + \hat{j} - \hat{k}$

The sum of given vectors is given by

14. Kindly Consider the following

Kindly go through the solution

19. 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 ratio 2 : 1

(i) Internally

(ii) Externally

(i) The position vector of point R dividing the join of P and Q. internally in

the ratio 2:1 is,

$\begin{array}{l} = \frac{(- 2 \hat{i} + \hat{i}) + (\hat{2 j} + \hat{2 j}) + (2 \hat{k} + \hat{k})}{3} = \frac{- \hat{i} + 4 \hat{j} + \hat{k}}{3} \\ = - \frac{1}{3} \hat{i} + \frac{4}{3} \hat{j} + \frac{1}{3} \hat{k} \end{array}$

(ii) The position vector of the point k dividing the join of P and Q. externally in the ratio 2:1

A15. (ii)

$\begin{array}{l} \vec{O R} = \frac{2 (- \hat{i} + \hat{j} + \hat{k}) - 1 (\hat{i} + \hat{2 j} - \hat{k})}{2 - 1} \\ = - 2 \hat{i} + \hat{2 j} + 2 \hat{k} - \hat{i} - \hat{2 j} + \hat{k} \\ = - 3 \hat{i} + \hat{k} \end{array}$

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

Given,

$\begin{array}{l} \vec{a} = \hat{i} + 4 \hat{j} + 2 \hat{k} \\ \vec{b} = 3 \hat{i} - 2 \hat{j} + 7 \hat{k} \\ \vec{c} = 2 \hat{i} - \hat{j} + 4 \hat{k} \end{array}$

Let, $\vec{d} = d_{1} \hat{i} + d_{2} \hat{j} + d_{3} \hat{k}$

Since, $\vec{d}$ is perpendicular to both $\vec{a} & \vec{b}$

$\begin{array}{l} \vec{d} . \vec{a} = 0 \\ \Rightarrow d_{1} + d_{2} 4 + d_{3} 2 = 0 - - - - - (1) \\ \vec{d} . \vec{b} = 0 \\ \Rightarrow d_{1} 3 + d_{2} (- 2) + d_{3} (7) = 0 \\ \Rightarrow d_{1} 3 - 2 d_{2} + 7 d_{3} = 0 - - - - - (2) \end{array}$

We know,

$\begin{array}{l} \vec{c} . \vec{d} = 15 \\ \Rightarrow 2 d_{1} - d_{2} + 4 d_{3} = 15 - - - - - (3) \\ F r o m, (1) \\ d_{1} + 4 d_{2} + 2 d_{3} = 0 \\ d_{1} = - 4 d_{2} - 2 d_{3} \end{array}$

Putting this value in (3) we get

$\begin{array}{l} \Rightarrow 2 (- 4 d_{2} - 2 d_{3}) - d_{2} + 4 d_{3} = 15 \\ \Rightarrow - 8 d_{2} - 4 d_{3} - d_{2} + 4 d_{3} = 15 \\ \Rightarrow - 9 d_{2} = 15 \\ \Rightarrow d_{2} = \frac{15}{9} = - \frac{5}{3} \end{array}$

Putting $d_{1} & d_{2}$ value in (2), we get

$\begin{array}{l} \Rightarrow 3 d_{1} - 2 d_{2} + 7 d_{3} = 0 \\ \Rightarrow 3 (- 4 d_{2} - 2 d_{3}) - 2 (- \frac{5}{3}) + 7 d_{3} = 0 \\ \Rightarrow - 12 \times (- \frac{5}{3}) - 6 d_{3} + \frac{10}{3} + 7 d_{3} = 0 \\ \Rightarrow 20 + d_{3} + \frac{10}{3} = 0 \\ \Rightarrow d_{3} = - 20 - \frac{10}{3} = \frac{- 60 - 10}{3} = \frac{- 70}{3} \\ N o w, d_{1} = - 4 \times - \frac{5}{3} - 2 \times - \frac{70}{3} \\ = \frac{20}{3} + \frac{140}{3} = \frac{160}{3} \\ ∴ d_{1} = \frac{160}{3}, d_{2} = - \frac{5}{3}, d_{3} = \frac{- 70}{3} \\ \vec{d} = \frac{160}{3} \hat{i} - \frac{5}{3} \hat{j} \frac{- 70}{3} \hat{k} = \frac{1}{3} (160 \hat{i} - 5 \hat{j} - 70 \hat{k}) \end{array}$

$∴$ The reQ.uired vector is $\frac{1}{3} (160 \hat{i} - 5 \hat{j} - 70 \hat{k})$

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

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

$\begin{array}{l} (\vec{a} + \vec{b}) . (\vec{a} - \vec{b}) = 8 \\ a n d | \vec{a} | = 8 | \vec{b} | \\ (\vec{a} + \vec{b}) . (\vec{a} - \vec{b}) = 8 \\ \vec{a} . \vec{a} - \vec{a} . \vec{b} + \vec{b} . \vec{a} - \vec{b} . \vec{b} = 8 \\ {| \vec{a} |}^{2} - {| \vec{b} |}^{2} = 8 \\ {(8 | \vec{b} |)}^{2} - {| \vec{b} |}^{2} = 8 \\ 64 {| \vec{b} |}^{2} - {| \vec{b} |}^{2} = 8 \\ 63 {| \vec{b} |}^{2} = 8 \\ | \vec{b} | =\sqrt8/\sqrt63 (∴ m a g n i t u d e o f a v e c t o r i s n o n - n e g a t i v e) \\ | \vec{b} | = \frac{2√2}{3√7} \\ A n d \\ | \vec{a} | = 8 | \vec{b} | = \frac{2 \times 2√2}{3√7} = \frac{1 6√2}{3√7} \end{array}$

23. If $\vec{a}$ and $\vec{b}$ are two collinear vectors, then which of the following are incorrect:

(A) $\vec{b}$ = λ $\vec{a}$ for some scalar λ

(B) $\vec{a}$ = ± $\vec{b}$

(D) Both the vectors $\vec{a}$ and $\vec{b}$ have same direction, but different magnitudes.

We know,

If $\vec{a}$ and $\vec{b}$ are two collinear vector, they are parallel.

So,

$\begin{array}{l} \vec{b} = λ \vec{a} \\ I f, λ = \pm 1, t h e n, \vec{a} = \pm \vec{b} \\ I f, \vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k} \\ \vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}, t h e n \\ \vec{b} = λ \vec{a} \\ b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} = λ (a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}) \\ = (λ a_{1}) \hat{i} + (λ a_{2}) \hat{j} + (λ a_{3}) \hat{k} \\ b_{1} = λ a_{1}, b_{2} = λ a_{2}, b_{3} = λ a_{3} \\ \frac{b_{1}}{a_{1}} = \frac{b_{2}}{a_{2}} = \frac{b_{3}}{a_{3}} = λ \end{array}$

Hence, the respective component are proportional but, vector $\vec{a}$ and $\vec{b}$ can have different direction.

Thus, the statement given in D is incorrect.

The correct answer is D.

3. Classify the following as scalar and vector quantities:

(i) Time period

(ii) Distance

(iii) Force

(iv) Velocity

(v) Work done

(i) Time period involves only magnitude. So, it is scalar quantity.

(ii) Distance involves only magnitude. So, it is scalar quantity.

(iii) Force involves both magnitude and direction. So, it is vector quantity.

(iv) Velocity involves both magnitude and direction. So, it is vector quantity.

(v) Work done involves only magnitude. So, it is scalar quantity.

47. Given that $\vec{a} . \vec{b} = 0$ and $\vec{a} \times \vec{b} = 0$ What can you conclude about the vectors $\vec{a}$ and $\vec{b}$ ?

Given,

$\vec{a} . \vec{b} = 0$ and $\vec{a} \times \vec{b} = 0$

For,

$\vec{a} . \vec{b} = 0$ , then either $| \vec{a} | = 0$ or $| \vec{b} | = 0$ or $\vec{a} ⊥ \vec{b}$

For,

$\vec{a} \times \vec{b} = 0$ , then either $| \vec{a} | = 0$ or $| \vec{b} | = 0$ or $\vec{a} ? \vec{b}$

$∴$ In case $\vec{a}$ and $\vec{b}$ are non- zero on both side.

But $\vec{a}$ and $\vec{b}$ cannot be both perpendicular and parallel simultaneously.

So, we can conclude that

$| \vec{a} | = 0$ or $| \vec{b} | = 0$

48. 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 $\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$

Given,

$\begin{array}{l} \vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k} \\ \vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} \\ \vec{c} = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k} \\ ∴ (\vec{b} + \vec{c}) = (b_{1} + c_{1}) \hat{i} + (b_{2} + c_{2}) \hat{j} + (b_{3} + c_{3}) \hat{k} \\ N o w, \end{array}$

$\begin{array}{l} = \hat{i} {a_{2} (b_{2} + c_{3}) - a_{3} (b_{2} + c_{2})} - \hat{j} {a_{1} (b_{3} + c_{3}) - a_{3} (b_{1} + c_{1})} \\ + \hat{k} {a_{1} (b_{2} + c_{2}) - a_{2} (b_{1} + c_{1})} \\ = \hat{i} {a_{2} b_{2} + a_{2} c_{3} - a_{3} b_{2} - a_{3} c_{2}} - \hat{j} {a_{1} b_{3} + a_{1} c_{3} - a_{3} b_{1} - a_{3} c_{1}} \\ + \hat{k} {a_{1} b_{2} + a_{1} c_{2} - a_{2} b_{1} - a_{2} c_{2}} - - - - (1) \end{array}$

$\begin{array}{l} = \hat{i} (a_{2} b_{3} - a_{3} b_{2}) - \hat{j} (a_{1} b_{3} - a_{3} b_{1}) + \hat{k} (a_{1} b_{2} - a_{2} b_{1}) - - - - (2) \\ A n d, \end{array}$

$\hat{i} (a_{2} c_{3} - a_{3} c_{2}) - \hat{j} (a_{1} c_{3} - a_{3} c_{1}) + \hat{k} (a_{1} c_{2} - a_{2} c_{1}) - - - - (3)$

Adding (2) and (3), we get

$\begin{array}{l} (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c}) = \hat{i} (a_{2} b_{3} - a_{3} b_{2}) - \hat{j} (a_{1} b_{3} - a_{3} b_{1}) + \hat{k} (a_{1} b_{2} - a_{2} b_{1}) \\ + \hat{i} (a_{2} c_{3} - a_{3} c_{2}) - \hat{j} (a_{1} c_{3} - a_{3} c_{1}) + \hat{k} (a_{1} c_{2} - a_{2} c_{1}) \\ (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c}) = \hat{i} (a_{2} b_{3} - a_{3} b_{2} + a_{2} c_{3} - a_{3} c_{2}) + \hat{j} (- a_{1} b_{3} + a_{3} b_{1} - a_{1} c_{3} + a_{3} c_{1}) \\ + \hat{k} (a_{1} b_{2} - a_{2} b_{1} + a_{1} c_{2} - a_{2} c_{1}) \\ = \hat{i} (a_{2} b_{3} + a_{2} c_{3} - a_{3} c_{2} - a_{3} b_{2}) - \hat{j} (a_{1} b_{3} + a_{1} c_{3} - a_{3} b_{1} - a_{3} c_{1}) \\ + \hat{k} (a_{1} b_{2} + a_{1} c_{2} - a_{2} b_{1} - a_{2} c_{1}) - - - - - (4) \end{array}$

From (1) and (4), we have

$\vec{a} (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$

Hence, proved.

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

Given, $P (1, 2, 3) & Q (4, 5, 6)$

So,

$\begin{array}{l} \vec{P Q} = (4 - 1) \hat{i} + (5 - 2) \hat{j} + (6 - 3) \hat{k} \\ = 3 \hat{i} + 3 \hat{j} + 3 \hat{k} \end{array}$

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

Given, A(1,2,-3) and (-1,-2,1)

Now,

$\begin{array}{l} | \vec{A B} | = (- 1 - 1) \hat{i} + (- 2 - 2) \hat{j} + (1 - (- 3)) \hat{k} \\ = - 2 \hat{i} - 4 \hat{j} + 4 \hat{k} \end{array}$

Then, $\begin{array}{l} \end{array}$

Let, l, m, n be direction cosine,

$l = \frac{x}{| \vec{A B} |} = \frac{- 2}{6} = \frac{- 1}{3}; m = \frac{y}{| \vec{A B} |} = \frac{- 4}{6} = \frac{- 2}{3}; n = \frac{z}{| \vec{A B} |} = \frac{4}{6} = \frac{2}{3}$

Therefore, direction cosine of $\vec{A B}$ are $(\frac{- 1}{3}, \frac{- 2}{3}, \frac{2}{3})$

7. Write two different vectors having same magnitude.

Two different vectors having same magnitude: -

(i) $2 \hat{i} + \hat{j} + 3 \hat{k}$

(ii) $\hat{i} + 3 \hat{j} + 2 \hat{k}$

15. Show that the vectors $2 \hat{i} - 3 \hat{j} + 4 \hat{k} a n d \vec{b} = - 4 \hat{i} + \hat{6 j} - 8 \hat{k}$ are collinear.

Let, $\vec{a} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k} & \vec{b} = - 4 \hat{i} + \hat{6 j} - 8 \hat{k}$

It is seen that

$\begin{array}{l} \vec{b} = - 4 \hat{i} + \hat{6 j} - 8 \hat{k} \\ = - 2 (2 \hat{i} - 3 \hat{j} + 4 \hat{k}) \\ = - 2 \vec{a} \\ ∴ \vec{b} = λ \vec{a} \end{array}$

Where, $λ = - 2$

Therefore, we can say that the given vector are collinear.

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

Let $\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k}$

53. Area of a rectangle having vertices A, B, C and D with position vectors $- \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k}, \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k}, a n d - \hat{i} \frac{1}{2} \hat{j} + 4 \hat{k}$ respectively is:

(A) $\frac{1}{2}$

(B) 1

(C) 2

(D) 4

(c) Given,

$\begin{array}{l} A = - \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k} \\ B = \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k} \\ C = \hat{i} - \frac{1}{2} \hat{j} + 4 \hat{k} \\ D = - \hat{i} - \frac{1}{2} \hat{j} + 4 \hat{k} \\ A \vec{B} = (1 + 1) \hat{i} + (\frac{1}{2} - \frac{1}{2}) \hat{j} + (4 - 4) \hat{k} \\ = 2 \hat{i} \\ B \vec{C} = (1 - 1) \hat{i} + (- \frac{1}{2} - \frac{1}{2}) \hat{j} + (4 - 4) \hat{k} \\ = - \hat{j} \end{array}$

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

$(2 \hat{i} + 4 \hat{j} - 5 \hat{k}) + (λ \hat{i} + 2 \hat{j} + 3 \hat{k}) = (2 + λ) \hat{i} + 6 \hat{j} - 2 \hat{k}$

The unit vector along $(2 \hat{i} + 4 \hat{j} - 5 \hat{k}) + (λ \hat{i} + 2 \hat{j} + 3 \hat{k})$ is given as;

By Q.uestion, scalar product of $(\hat{i} + \hat{j} + \hat{k})$ with this unit vector is 1.

67. If are mutually perpendicular vectors of equal magnitudes, show that the vector $\vec{a} + \vec{b} + \vec{c}$ -is equally inclined to $a, b,$ and $\vec{c}$ .

Given that $\vec{a}, \vec{b} & \vec{c}$ are mutually perpendicular vectors, we have

$\begin{array}{l} \vec{a} . \vec{b} = \vec{b} . \vec{c} = \vec{c} . \vec{a} = 0 \\ | \vec{a} | = | \vec{b} | = | \vec{c} | \end{array}$

Let, vector $\vec{a} + \vec{b} + \vec{c}$ be inclined to $\vec{a}, \vec{b} & \vec{c}$ at angles, $θ_{1}, θ_{2} & θ_{3}$ respectively.

$\begin{array}{l} ∴ c o s θ_{1} = \frac{(\vec{a} + \vec{b} + \vec{c}) . \vec{a}}{| \vec{a} + \vec{b} + \vec{c} | | \vec{a} |} = \frac{\vec{a} . \vec{a} + \vec{b} . \vec{a} + \vec{c} . \vec{a}}{| \vec{a} + \vec{b} + \vec{c} | | \vec{a} |} \\ = \frac{{| \vec{a} |}^{2}}{| \vec{a} + \vec{b} + \vec{c} | | \vec{a} |} [\vec{b} . \vec{a} = \vec{c} . \vec{a} = 0] \\ = \frac{| \vec{a} |}{| \vec{a} + \vec{b} + \vec{c} |} \end{array}$

$\begin{array}{l} ∴ c o s θ_{2} = \frac{(\vec{a} + \vec{b} + \vec{c}) . \vec{b}}{| \vec{a} + \vec{b} + \vec{c} | | \vec{b} |} \\ = \frac{\vec{a} . \vec{b} + \vec{b} . \vec{b} + \vec{b} . \vec{c}}{| \vec{a} + \vec{b} + \vec{c} | | \vec{b} |} [\vec{a} . \vec{b} = \vec{b} . \vec{c} = 0] \\ = \frac{{| \vec{b} |}^{2}}{| \vec{a} + \vec{b} + \vec{c} | | \vec{b} |} = \frac{| \vec{b} |}{| \vec{a} + \vec{b} + \vec{c} |} \\ ∴ c o s θ_{3} = \frac{(\vec{a} + \vec{b} + \vec{c}) . \vec{c}}{| \vec{a} + \vec{b} + \vec{c} | | \vec{c} |} \\ = \frac{\vec{a} . \vec{c} + \vec{b} . \vec{c} + \vec{c} . \vec{c}}{| \vec{a} + \vec{b} + \vec{c} | | \vec{c} |} [\vec{a} . \vec{c} = \vec{b} . \vec{c} = 0] \\ = \frac{{| \vec{c} |}^{2}}{| \vec{a} + \vec{b} + \vec{c} | | \vec{c} |} = \frac{| \vec{c} |}{| \vec{a} + \vec{b} + \vec{c} |} \\ n o w, a s, | \vec{a} | = | \vec{b} | = | \vec{c} |, \\ c o s θ_{1} = c o s θ_{2} = c o s θ_{3} \\ ∴ θ_{1} = θ_{2} = θ_{3} \end{array}$

Therefore, the vector $(\vec{a} + \vec{b} + \vec{c})$ are equally inclined to $\vec{a}, \vec{b} & \vec{c.}$

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

Let the vector with initial point P (2,1) and terminal point Q. (-5,7) can be shown as,

$\begin{array}{l} \vec{P Q} = (- 5, - 2) \hat{i} + (7, - 1) \hat{j} \\ \vec{P Q} = - 7 \hat{i} + 6 \hat{j} \end{array}$

The scalar components are -7 and 6.

The vector components are -7i and 6j.

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

Here,

Let, $\vec{a} = \hat{i} + \hat{j} + \hat{k}$

Then,

8. Find the values of x and y so that the vectors $2 \hat{i} + 3 \hat{j} a n d x \hat{i} + y \hat{j}$ are equal.

Note that two vector are equal only if their corresponding components are equal.

Thus, the given vectors $\vec{a}$ and $\vec{b}$ will be equal if and only if $x = 2 & y = 3$

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

The Position vector of mid-point R of the vector joining point P (2,3,4) and Q (4,1, -2) is given by;

$\begin{array}{l} \vec{O R} = \frac{(2 \hat{i} + 3 \hat{j} + 4 \hat{k}) + (4 \hat{i} + \hat{j} - \hat{2 k})}{2} \\ = \frac{(2 + 4) \hat{i} + (3 + 1) \hat{j} + (4 - 2) \hat{k}}{2} \\ = \frac{6 \hat{i} + 4 \hat{j} + 2 \hat{k}}{2} = \frac{6 \hat{i}}{2} + \frac{4 \hat{j}}{2} + \frac{2 \hat{k}}{2} \\ = 3 \hat{i} + 2 \hat{j} + k \end{array}$

24. Find the angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitude √3 and 2 respectively having $\vec{a}$ . $\vec{b}$ = √6

Given,

=√3

$, | \vec{b}$
$= 2 a n d \vec{a} . \vec{b} =\sqrt6$

We have,

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

$\begin{array}{l} (\vec{x} - \vec{a}) . (\vec{x} - \vec{a}) = 12 \\ \vec{x} . \vec{x} + \vec{x} . \vec{a} - \vec{a} . \vec{x} - \vec{a} . \vec{a} = 12 \\ {| \vec{x} |}^{2} - {| \vec{a} |}^{2} = 12 \\ {| \vec{x} |}^{2} - 1 = 12 [| \vec{a} | = 1 a s a \vec{a} i s u n i t v e c t o r] \\ {| \vec{x} |}^{2} = 12 + 1 = 13 \\ ∴ | \vec{x} | =\sqrt13 \end{array}$

36. If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors such that $\vec{a} + \vec{b} + \vec{c}$ = 0 find the value of $\vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}$

$| \vec{a} + \vec{b} + \vec{c} | = (\vec{a} + \vec{b} + \vec{c}) . (\vec{a} + \vec{b} + \vec{c})$

$\begin{array}{l} = \vec{a} . \vec{a} + \vec{a} . \vec{b} + \vec{a} . \vec{c} + \vec{b} . \vec{a} + \vec{b} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a} + \vec{c} . \vec{b} + \vec{c} . \vec{c} \\ = {| \vec{a} |}^{2} + {| \vec{b} |}^{2} + {| \vec{c} |}^{2} + 2 (\vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}) \\ = 1 + 1 + 1 + 2 (\vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}) \\ = 3 + 2 (\vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}) \\ \Rightarrow \vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a} = \frac{- 3}{2} \end{array}$

40. Show that the vectors $2 \hat{i} - \hat{j} + \hat{k} a n d 3 \hat{i} - 4 \hat{j} - 4 \hat{k}$ form the vertices of a right angled triangle.

Let vector $2 \hat{i} - \hat{j} + \hat{k}, \hat{i} - 3 \hat{j} - 5 \hat{k}$ and $3 \hat{i} - 4 \hat{j} - 4 \hat{k}$ be position vector of point A, B, C respectively.

So,

$\begin{array}{l} O \vec{A} = 2 \hat{i} - \hat{j} + \hat{k} \\ O \vec{B} = \hat{i} - 3 \hat{j} - 5 \hat{k} \\ O \vec{C} = 3 \hat{i} - 4 \hat{j} - 4 \hat{k} \end{array}$

Now, vectors $A \vec{B}, B \vec{C}$ and $A \vec{C}$ represents the sides of $? A B C$ .

Hence,

70. Choose the correct answer:

Let $\vec{a}$ and $\vec{b}$ be two unit vectors andθ 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 = $\frac{π}{3}$

Let, $\vec{a} & \vec{b}$ be two unit vectors and $θ$ be the angle between them.

Then, $| \vec{a} | = | \vec{b} | = 1$

Now, $\vec{a} + \vec{b}$ is a unit vector if $| \vec{a} + \vec{b} | = 1$

$\begin{array}{l} | \vec{a} + \vec{b} | = 1 \\ {(\vec{a} + \vec{b})}^{2} = 1 \\ (\vec{a} + \vec{b}) . (\vec{a} + \vec{b}) = 1 \\ \vec{a} . \vec{a} + \vec{a} . \vec{b} + \vec{b} . \vec{a} + \vec{b} . \vec{b} = 1 \\ {| \vec{a} |}^{2} + 2 \vec{a} . \vec{b} + {| \vec{b} |}^{2} = 1 \\ 1^{2} + 2 | \vec{a} | . | \vec{b} | c o s θ + 1^{2} = 1 \end{array}$

$1 + 2.1.1 c o s θ + 1 = 1$ [ $∴$ $\vec{a} & \vec{b}$ is unit vector.]

$\begin{array}{l} 2 c o s θ = 1 - 2 \\ c o s θ = \frac{- 1}{2} = \frac{2 π}{3} \\ ∴ θ = \frac{2 π}{3} \end{array}$

Therefore, the correct answer is (D)

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

Given,

$A (1, 1, 2), B (2, 3, 5) C (1, 5, 5)$

We have,

$\begin{array}{l} A \vec{B} = \hat{i} + 2 \hat{j} + 3 \hat{k} \\ A \vec{C} = 4 \hat{j} + 3 \hat{k} \end{array}$

The area of given triangle is $\frac{1}{2} | A \vec{B} \times A \vec{C} |$

51. Find the area of the parallelogram whose adjacent sides are determined by the vectors

$a = \vec{i} - \vec{j} + 3 \vec{k} a n d b = 2 \vec{i} - 7 \vec{j} + \vec{k}$

Given,

$\begin{array}{l} \vec{a} = \hat{i} - \hat{j} + 3 \hat{k} \\ \vec{b} = 2 \hat{i} - 7 \hat{j} + \hat{k} \end{array}$

The area of a parallelogram with $\vec{a}$ and $\vec{b}$ as its adjacent sides is given by $| \vec{a} \times \vec{b} |$

52. Let the vectors $\vec{a}$ and $\vec{b}$ such that | $\vec{a}$ | = 3 and | $\vec{b}$ | = √2/3 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) $\frac{π}{2}$

(B) Given,

$∴$ Hence, $\vec{a} \times \vec{b}$ is a unit vector if angle between $\vec{a}$ and $\vec{b}$ is $\frac{π}{4}$

54. Write down a unit vector in XY-plane making an angle of 30° with the positive direction of x-axis.

Let $\vec{r}$ be unit vector in the XY-plane then, $\vec{r} = c o s θ \hat{i} + s i n θ \hat{j}$

$θ$ is the angle made by the unit vector with the positive direction of the X-axis.

Then, $θ = 3 0^{0}$

$∴$ ReQ.uired unit vector $= \frac{}{2} \hat{i} + \frac{1}{2} \hat{j}$

55. Find the scalar components and magnitude of the vector joining the points $P (x_{1}, y_{1}, z_{1}) a n d Q (x_{2}, y_{2}, z_{2})$

Given,

Point $P (x_{1}, y_{1}, z_{1}) & Q (x_{2}, y_{2}, z_{2})$

$\vec{P Q}$ = Position vector of Q.- Position vector of P

$= (x_{2} - x_{1}) \hat{i} + (y_{2} - y_{1}) \hat{j} + (z_{2} - z_{1}) \hat{k}$

56. A girl 4 Km towards west, then she walks 3 Km in a direction 30° east of north and stops. Determine the girls displacement from her initial point of departure.

$∴$ The girl’s displacement from her initial point of departure is $= \frac{- 5}{3} \hat{i} + \frac{3√3}{2} \hat{j}$

57. If $\vec{a} = \vec{b} + \vec{c},$ then is it true that $| \vec{a} | = | \vec{b} | + | \vec{c} |$ Justify your answer.

Let us take a $? A B C$ , which $\vec{C B} = \vec{a}, \vec{C A} = \vec{b} & \vec{A B} = \vec{c}$

So, by triangle law of vector addition, we have $\vec{a} = \vec{b} + \vec{c}$

And, we know that $| \vec{a} | | \vec{b} | & | \vec{c} |$ represent, the sides of $? A B C$

Also, it is known that the sum of the length of any slides of a triangle is greater than the third side. $| \vec{a} | < | \vec{b} | + | \vec{c} |$

Hence, it is not true that $| \vec{a} | = | \vec{b} | + | \vec{c} |$

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

Given,

$x (\hat{i} + \hat{j} + \hat{k})$ is a unit vector.

So, $| x (\hat{i} + \hat{j} + \hat{k}) | = 1$

Now, $| x (\hat{i} + \hat{j} + \hat{k}) | = 1$ $\begin{array}{l} \end{array}$

59. Find a vector of magnitude 5 units and parallel to the resultant of the vectors $a = 2 \hat{i} + 3 \hat{j} - \hat{k} a n d b = \hat{i} - 2 \hat{j} + \hat{k}$

We know,

$\begin{array}{l} \vec{a} = 2 \hat{i} + 3 \hat{j} - \hat{k} \\ \vec{b} = \hat{i} - 2 \hat{j} + \hat{k} \end{array}$

Let, $\vec{c}$ be the resultant of $\vec{a}$ and $\vec{b}$

Then,

60. If $\vec{a} = \vec{i} + \vec{j} + \vec{k}, \vec{b} = 2 \vec{i} - \vec{j} + 3 \vec{k} a n d c = \vec{i} - 2 \vec{j} + \vec{k}$ find a unit vector parallel to the vector $2 \vec{a} - \vec{b} + 3 \vec{c}$

Given,

$\begin{array}{l} \vec{a} = \hat{i} + \hat{j} + \hat{k} \\ \vec{b} = \hat{2 i} - \hat{j} + 3 \hat{k} \\ \vec{c} = \hat{i} - 2 \hat{j} + \hat{k} \end{array}$

Then,

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

Given,

$\begin{array}{l} A (1, - 2, - 8) \\ B (5, 0, - 2) \\ C (11, 3, 7) \end{array}$

Now,

Thus, A,B and C are collinear.

Let, $λ : 1$ be the ratio that point B divides AC.

We have,

$\begin{array}{l} \vec{O B} = \frac{λ \vec{O C} + \vec{O A}}{λ + 1} \\ 5 \hat{i} - 2 \hat{k} = \frac{λ (11 \hat{i} + 3 \hat{j} + 7 \hat{k}) + (\hat{i} - 2 \hat{j} - 8 \hat{k)}}{λ + 1} \\ (5 \hat{i} - 2 \hat{k}) (λ + 1) = 11 λ \hat{i} + 3 λ \hat{j} + 7 λ \hat{k} + \hat{i} - 2 \hat{j} - 8 \hat{k} \\ 5 (λ + 1) \hat{i} - 2 (λ + 1) \hat{k} = (11 λ + 1) \hat{i} + (3 λ - 2) \hat{j} + (7 λ - 8) \hat{k} \end{array}$

On eQ.uating the corresponding component , we get

$\begin{array}{l} 5 (λ + 1) = 11 λ + 1 \\ 5 λ + 5 = 11 λ + 1 \\ 5 - 1 = 11 λ - 5 λ \\ 4 = 6 λ \\ ∴ λ = \frac{4}{6} = \frac{2}{3} \end{array}$

Hence, point B divides AC in the ratio $2 : 3$

62. Find the position vector of a point R which divides the line joining the 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 middle point of line segment RQ.

Given,

$\begin{array}{l} P (2 \vec{a} + \vec{b}) i . e, O P = 2 \vec{a} + \vec{b} \\ Q (\vec{a} - 3 \vec{b}) i . e, O Q = \vec{a} - 3 \vec{b} \end{array}$

It is given that point R divides a line segment joining two points P and Q.

externally in the ratio 1:2 Then,

$\begin{array}{l} \vec{O R} = \frac{2 (2 \vec{a} + \vec{b}) - (\vec{a} - 3 \vec{b})}{2 - 1} \\ = \frac{4 \vec{a} + \vec{2 b} - \vec{a} + 3 \vec{b}}{1} \\ \vec{O R} = 3 \vec{a} + 5 \vec{b} \end{array}$

$∴$ Position vector of the mid-point of RQ.

$\begin{array}{l} = \frac{\vec{O Q} + \vec{O R}}{2} \\ = \frac{(\vec{a} - 3 \vec{b}) + (3 \vec{a} + 5 \vec{b})}{2} \\ = \frac{\vec{a} - 3 \vec{b} + 3 \vec{a} + 5 \vec{b}}{2} \\ = \frac{4 \vec{a} - 2 \vec{b}}{2} = 2 \vec{a} - \vec{b} = \vec{O R} H e n c e p r o v e d \end{array}$

63. Two adjacent sides of a parallelogram are $2 \hat{i} - 4 \hat{j} + 5 \hat{k} a n d \hat{i} - 2 \hat{j} - 3 \hat{k}$ . Find the unit vector parallel to its diagonal. Also, find its area.

Given,

Adjacent sides of parallelogram are

$\begin{array}{l} \vec{a} = 2 \hat{i} - 4 \hat{j} + 5 \hat{k} \\ \vec{b} = \hat{i} - 2 \hat{j} - 3 \hat{k} \end{array}$

$∴$ Diagonal of parallelogram = $\vec{a} + \vec{b}$

$\begin{array}{l} \Rightarrow \vec{a} + \vec{b} = (2 + 1) \hat{i} + (- 4 + (- 2)) \hat{j} + (5 + (- 3)) \hat{k} \\ = 3 \hat{i} - 6 \hat{j} + 2 \hat{k} \end{array}$

Thus, the unit vector parallel to diagonal

64. Show that the direction cosines of a vector equally inclined to the axes OX,Oy and OZ

Kindly go through the solution

68. 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 $\vec{a} \neq \vec{0}, \vec{b} \neq \vec{0}$ .

$(\vec{a} + \vec{b}) . (\vec{a} + \vec{b}) = {| \vec{a} |}^{2} + {| \vec{b} |}^{2}$

$\vec{a} . \vec{a} + \vec{a} . \vec{b} + \vec{a} . \vec{b} + \vec{b} . \vec{b} = {| \vec{a} |}^{2} + {| \vec{b} |}^{2}$

( Distributive of scalar product over addition )

$\Rightarrow {| \vec{a} |}^{2} + 2 \vec{a} . \vec{b} + {| \vec{b} |}^{2} = {| \vec{a} |}^{2} + {| \vec{b} |}^{2}$

( Scalar product is commutative , $\vec{a} . \vec{b} = \vec{b} . \vec{a}$ )

$\begin{array}{l} \Rightarrow 2 \vec{a} . \vec{b} = {| \vec{a} |}^{2} - {| \vec{a} |}^{2} + {| \vec{b} |}^{2} - {| \vec{b} |}^{2} \\ \Rightarrow 2 \vec{a} . \vec{b} = 0 \\ \Rightarrow \vec{a} . \vec{b} = 0 \end{array}$

$∴$ Therefore, $\vec{a} & \vec{b}$ are perpendicular.

69. Choose the correct answer:

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) $0 \leq θ \leq π$

Let, $θ$ in triangle between two vector $\vec{a} & \vec{b}$

Then, without loss of generality, $\vec{a} & \vec{b}$ are non-zero vector so that $| \vec{a} | & | \vec{b} |$

are positive.

We know, $\vec{a} . \vec{b} = | \vec{a} | | \vec{b} | c o s θ$

So, $\vec{a} . \vec{b} \geq 0$

$\begin{array}{l} \Rightarrow | \vec{a} | | \vec{b} | c o s θ \geq 0 \\ c o s θ \geq 0 [∴ | \vec{a} | & | \vec{b} | a r e p o s i t i v e] \\ 0 \leq θ \leq \frac{π}{2} \end{array}$

Therefore, $\vec{a} . \vec{b} \geq 0$ , when $0 \leq θ \leq \frac{π}{2}$

Hence, the correct answer is B.

10. Find the sum of the vectors: $\vec{a} = \hat{i} - 2 \hat{j} + \hat{k}, \vec{b} = - 2 \hat{i} - 4 \hat{j} + 5 \hat{k} a n d \vec{c} = \hat{i} - 6 \hat{j} - 7 \hat{k}$

The given vectors are

$\vec{a} = \hat{i} - 2 \hat{j} + \hat{k}$

$\vec{b} = \hat{- 2 i} + 4 \hat{j} + 5 \hat{k}$

$\vec{c} = \hat{i} - 6 \hat{j} - 7 \hat{k}$

The sum of the vector is

$\vec{a} + \vec{b} + \vec{c} = (a_{1} + a_{2} + a_{3}) \hat{i} + (b_{1} + b_{2} + b_{3}) \hat{\hat{j} + (c_{1} + c_{2} + c_{3}) \hat{k}}$

$\begin{array}{l} = (1 - 2 + 1) \hat{i} + (- 2 + 4 - 6) \hat{j} + (1 + 5 - 7) \hat{k} \\ = 0 . \hat{i} + (- 4) \hat{j} + (- 1) \hat{k} \\ = - 4 \hat{j} - k \end{array}$

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

We have,

$\begin{array}{l} \vec{a} = 3 \hat{i} - 4 \hat{j} - 4 \hat{k} \\ \vec{b} = 2 \hat{i} - \hat{j} + \hat{k} \\ \vec{c} = \hat{i} - 3 \hat{j} - 5 \hat{k} \end{array}$

$\begin{array}{l} \vec{A B} = (2 - 3) \hat{i} + (- 1 - (- 4)) \hat{j} + (1 - (- 4)) \hat{k} \\ = - \hat{i} + 3 \hat{j} + 5 \hat{k} \\ \vec{B C} = (1 - 2) \hat{i} + (- 3 - (- 1)) \hat{j} + (- 5 - 1) \hat{k} \\ = - \hat{i} - 2 \hat{j} - 6 \hat{k} \\ \vec{C A} = (1 - 3) \hat{i} + (- 3 - (- 4)) \hat{j} + (- 5 - (- 4)) \hat{k} \\ = - 2 \hat{i} + \hat{j} - \hat{k} \end{array}$

Now, $\begin{array}{l} \end{array}$

Hence,

${| \vec{A B} |}^{2} + {| \vec{C A} |}^{2} = 35 + 6 = 41 = {| \vec{B C} |}^{2}$

Hence, given points from the vertices of a right angled triangle.

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

Kindly go through the solution

22. In triangle ABC (Fig. below), which of the following is not true:

(A) $\vec{A B} + \vec{B C} + \vec{C A} = \vec{0}$

By triangle law of addition in given triangle, we get:

$\begin{array}{l} \vec{A B} + \vec{B C} = \vec{A C} - - - - - (1) \\ \vec{A B} + \vec{B C} = - \vec{C A} \end{array}$

$\vec{A B} + \vec{B C} + \vec{C A} = \vec{0} - - - - - (2)$

So, (A) is true.

(B) $\vec{A B} + \vec{B C} - \vec{A C} = 0$

$\begin{array}{l} \vec{A B} + \vec{B C} = \vec{A C} \\ \vec{A B} + \vec{B C} - \vec{A C} = 0 \end{array}$

So, (B) is true.

(C) $\vec{A B} + \vec{B C} - \vec{C A} = 0$

$\begin{array}{l} \vec{A B} + \vec{B C} = \vec{C A} - - - - - (3) \\ F r o m, (1) & (3), \\ \vec{A C} = \vec{C A} \\ \vec{A C} = - \vec{A C} \\ \vec{A C} + \vec{A C} = 0 \\ 2 \vec{A C} = 0 \end{array}$

$∴$ The eQ.uation in alternative C $\vec{A C} = \vec{0}$ , which is not true, is incorrect.

(D) $\vec{A B} - \vec{C B} + \vec{C A} = \vec{0}$

$\begin{array}{l} F r o m, e q n (2) w e h a v e \\ \vec{A B} - \vec{C B} + \vec{C A} = 0 \end{array}$

The, equation given is alternative is D is true.

$∴$ The correct answer is C.

26. Find the projection of the vector

$\hat{i} - \hat{j}$ on the vector $\hat{i} + \hat{j}$

Let,

$\begin{array}{l} \vec{a} = \hat{i} - \hat{j} \\ \vec{b} = \hat{i} + \hat{j} \end{array}$

The projection of vector $\vec{a}$ on $\vec{b}$ is given by,

$∴$ The projection of vector $\vec{a}$ on $\vec{b}$ is 0.

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

Let,

$\begin{array}{l} \vec{a} = \hat{i} + 3 \hat{j} + 7 \hat{k} \\ \vec{b} = 7 \hat{i} - \hat{j} + 8 \hat{k} \end{array}$

The project of vector $\vec{a}$ on $\vec{b}$ is.

28. Kindly Consider the following

Here, each of the given three vector is a unit vector.

$\begin{array}{l} \vec{a} . \vec{b} = \frac{2}{7} \times \frac{3}{7} + \frac{3}{7} \times (\frac{- 6}{7}) + \frac{6}{7} \times \frac{2}{7} \\ = \frac{6}{49} + (\frac{- 18}{49}) + \frac{12}{49} = \frac{6 - 18 + 12}{49} = 0 \\ \vec{b} . \vec{c} = \frac{3}{7} \times \frac{6}{7} + (\frac{- 6}{7}) \times \frac{2}{7} + \frac{2}{7} \times (- \frac{3}{7}) \\ = \frac{18}{49} - \frac{12}{49} + (\frac{- 6}{49}) = \frac{18 - 12 - 6}{49} = 0 \\ \vec{c} . \vec{a} = \frac{6}{7} \times \frac{2}{7} + \frac{2}{7} \times \frac{3}{7} + (- \frac{3}{7}) \times \frac{6}{7} \\ = \frac{12}{49} + \frac{6}{49} - \frac{18}{49} = 0 \end{array}$

Therefore, the given three vectors are mutually perpendicular to each other.

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

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

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

31. Find the magnitude of two vectors $\vec{a}$ and $\vec{b}$ having the same magnitude such that the angle between them is 60° and their scalar product is 1/2.

Let $θ$ be the angle between the vectors $| \vec{a} |$ and $| \vec{b} |$ .

It is given that $| \vec{a} | = | \vec{b} |, \vec{a} . \vec{b} = \frac{1}{2} a n d θ = 6 0^{?} - - - - (1)$

We know, $\vec{a} . \vec{b} = | \vec{a} | | \vec{b} | c o s θ$

$\begin{array}{l} ∴ \frac{1}{2} = | \vec{a} | | \vec{a} | c o s 6 0^{?} (u s i n g (1)) \\ \frac{1}{2} = {| \vec{a} |}^{2} \times \frac{1}{2} \\ {| \vec{a} |}^{2} = 1 \\ | \vec{a} | = | \vec{b} | = 1 \end{array}$

$∴$ Magnitude of two vector=1

33. If $\vec{a} = 2 \hat{i} + 2 \hat{j} + 3 \hat{k}, \vec{b} = - \hat{i} + 1 \vec{j} + \vec{k} a n d c = 3 \hat{i} + \hat{j}$ are such that $\vec{a} + λ \vec{b}$ is perpendicular to $\vec{c}$ then find the value of λ

Given,

$\begin{array}{l} \vec{a} = 2 \hat{i} + 2 \hat{j} + 3 \hat{k} \\ \vec{b} = - \hat{i} + 2 \hat{j} + \hat{k} \\ \vec{c} = 3 \hat{i} + \hat{j} \end{array}$

Now,

$\begin{array}{l} \vec{a} + λ \vec{b} = (2 \hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (- \hat{i} + 2 \hat{j} + \hat{k}) \\ = (2 \hat{i} + 2 \hat{j} + 3 \hat{k}) + (- λ \hat{i} + 2 λ \hat{j} + λ \hat{k}) \\ = (2 - λ) \hat{i} + (2 + 2 λ) \hat{j} + (3 + λ) \hat{k} \end{array}$

If $(\vec{a} + λ \vec{b})$ is perpendicular to $\vec{c}$ , then $(\vec{a} + λ \vec{b}) . \vec{c} = 0$

$\begin{array}{l} = [(2 - λ) \hat{i} + (2 + 2 λ) \hat{j} + (3 + λ) \hat{k}] . (3 \hat{i} + \hat{j}) \\ = 3 (2 - λ) + 1 (2 + 2 λ) + 0 (3 + λ) \\ = 6 - 3 λ + 2 + 2 λ + 0 \\ = 8 - λ \\ \Rightarrow λ = 8 \end{array}$

Therefore, the required value of $λ$ is 8.

34. Show that $| \vec{a} | \vec{b} + | \vec{b} |$ is perpendicular to $| \vec{a} | \vec{b} - | \vec{b} | \vec{a}$ for any two non-zero vectors $\vec{a}$ and $\vec{b}$

$(| \vec{a} | \vec{b} + | \vec{b} | \vec{a}) . (| \vec{a} | \vec{b} - | \vec{b} | \vec{a})$

$\begin{array}{l} = | \vec{a} | \vec{b} . | \vec{a} | \vec{b} - | \vec{a} | \vec{b} . | \vec{b} | \vec{a} + | \vec{b} | \vec{a} . | \vec{a} | \vec{b} - | \vec{b} | \vec{a} . | \vec{b} | \vec{a} \\ = {| \vec{a} |}^{2} \vec{b} . \vec{b} - {| \vec{b} |}^{2} \vec{a} . \vec{a} \\ = {| \vec{a} |}^{2} {| \vec{b} |}^{2} - {| \vec{b} |}^{2} {| \vec{a} |}^{2} \\ = 0 \end{array}$

$∴$ Therefore, $| \vec{a} | \vec{b} + | \vec{b} | \vec{a}$ and $| \vec{a} | \vec{b} - | \vec{b} | \vec{a}$ are perpendicular.

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

We know,

$\vec{a} . \vec{a} = 0$ and $\vec{a} . \vec{b} = 0$

Now,

$\vec{a} . \vec{a} = 0 \Rightarrow {| \vec{a} |}^{2} \Rightarrow | \vec{a} | = 0$

$∴$ $\vec{a}$ is a zero vector.

Thus, vector $\vec{b}$ satisfying $\vec{a} . \vec{b} = 0$ can be any vector.

37. If either vector $\vec{a} = 0 o r \vec{b} = 0, t h e n \vec{a} . \vec{b} = 0$ . But the converse need not be true. Justify your answer with an example.

Consider

$\begin{array}{l} \vec{a} = 2 \hat{i} + 4 \hat{j} + 3 \hat{k} \\ \vec{b} = 3 \hat{i} + 3 \hat{j} - 6 \hat{k} \end{array}$ and

Then,

$\begin{array}{l} \vec{a} . \vec{b} = 2.3 + 4.3 + 3 . (- 6) \\ = 6 + 12 - 18 = 0 \end{array}$

Therefore, the converse of the given statement need not be true.

38. 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 $\bar{B A}$ and $\bar{B C}$ ]

Vertices of $? A B C$ are given as

$\begin{array}{l} A (1, 2, 3), B (- 1, 0, 0), C (0, 1, 2) \end{array}$

$∴ ? A B C$ is the angle between the vectors $B \vec{A}$ and $B \vec{C}$

39. Show that the points A (1, 2, 7), B (2, 6, 3) and C (3, 10, –1) are collinear.

Given, point are

41. If $\vec{a}$ is a nonzero vector of magnitude ‘a’ and λ a nonzero scalar, then λ $\vec{a}$ is unit vector if

(A) λ = 1

(B) λ = –1

(C) a = | λ

|(D) a = 1/|λ|

Vector $λ \vec{a}$ is a unit vector if $| λ \vec{a} | = 1$

Now,

$\begin{array}{l} | λ \vec{a} | = 1 \\ | λ | | \vec{a} | = 1 \\ | \vec{a} | = \frac{1}{| λ |} [λ \neq 0] \\ a = \frac{1}{| λ |} [| \vec{a} | = a] \end{array}$

$∴$ Therefore, vectar $λ \vec{a}$ is a unit vector if a= $\frac{1}{| λ |}$ .

Option (D)is correct.

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

Kindly go through the solution

43. Find a unit vector perpendicular to each of the vectors $\vec{a} + \vec{b} a n d \vec{a} - \vec{b}, w h e r e \vec{a} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k} a n d \vec{b} = \hat{i} - 2 \hat{j} - 2 \hat{k}$

Given,

$\begin{array}{l} \vec{a} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k} \\ \vec{b} = \hat{i} + 2 \hat{j} - 2 \hat{k} \\ \vec{a} + \vec{b} = 4 \hat{i} + 4 \hat{j}, \vec{a} - \vec{b} = 2 \hat{i} + 4 \hat{j} \end{array}$

A vector which is perpendicular to both $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ is given by

Say

Therefore, the unit vector is

$\begin{array}{l} \frac{\vec{c}}{| \vec{c} |} = \pm \frac{16 \hat{i} - 16 \hat{j} - 8 \hat{k}}{24} \\ = \pm \frac{16}{24} \hat{i} \pm \frac{16}{24} \hat{j} \pm \frac{8}{24} \hat{k} \\ = \pm \frac{2}{3} \hat{i} \pm \frac{2}{3} \hat{j} \pm \frac{1}{3} k \end{array}$

71. Choose the correct answer:

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:

(A) 0

(B) -1

(C) 1

(D) 3

$\hat{i} (\hat{j} \times \hat{k}) + \hat{j} (\hat{i} \times \hat{k}) + \hat{k} (\hat{i} \times \hat{j})$

$\begin{array}{l} = \hat{i} . \hat{i} + \hat{j} (- \hat{j}) + \hat{k} . \hat{k} \\ = 1 - \hat{j} . \hat{j} + 1 \\ = 1 - 1 + 1 \\ = 1 \end{array}$

Therefore, the correct answer is (C)

72. If θ be 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

Let, $θ$ be angle between two vector $\vec{a} & \vec{b}$ .

Then, without loss of generality, $\vec{a} & \vec{b}$ are non-zero vectors, so that $\vec{a} & \vec{b}$

are positive.

$\begin{array}{l} | \vec{a} . \vec{b} | = | \vec{a} \times \vec{b} | \\ | \vec{a} | | \vec{b} | c o s θ = | \vec{a} | | \vec{b} | s i n θ \\ c o s θ = s i n θ [| \vec{a} | & | \vec{b} | a r e p o s i t i v e] \\ t a n θ = 1 = \frac{π}{4} ∴ θ = \frac{π}{4} . \end{array}$

$∴$ Therefore, the correct answer is B.

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

$40 k m, 3 0^{0}$ east of north.

2. Check 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/sec²

(i) 10kg involves only magnitude. So, it is scalar quantity.

(ii) 2 meters north-west involves both magnitude and direction. So, it is vector quantity.

(iii) 40⁰ involves only magnitude. So, it is scalar quantity.

(iv) 40⁰ watts involves only magnitude. So, it is scalar quantity.

(v) 10^-19 coulomb involves only magnitude. So, it is scalar quantity.

(vi) 20m/s^-2 involves magnitude and direction. So, it is vector quantity.

4. In the adjoining figure, (a square) identify the following vectors:

(i) Coinitial

(ii) Equal

(iii) Collinear but not equal.

(a) Vector $\vec{a}$ and $\vec{d}$ are co initial same initial point.

(b) $\vec{b}$ and $\overset{?}{d}$ same magnitude & direction.

(c) $\vec{a}$ and $\vec{c}$ are collinear but not equal they are parallels their direction are not same.

5. Answer the following as true or false:

(i) $\vec{a}$ and - $\vec{a}$ are collinear.

(ii) Two collinear vectors are always equal in magnitude.

(iii) Two vectors having same magnitude are collinear.

(iv) Two collinear vectors having the same magnitude are equal.

(i) True, as vector quantity $\vec{a}$ and - $\vec{a}$ are parallel to same line.

(ii) False, as collinear vector are those vectors that are parallel to same line, but it is not necessary that they are equal also.

(iii) False, as two vectors having same magnitude may have different directions, so they are not collinear.

(iv) False, as two collinear vectors having same magnitude are not equal whey they are opposite in direction.

7. Compute the magnitude of the following vectors:

Kindly go through the solution

44. If a unit vector $\vec{a}$ makes an angle π/3 with $\hat{i}, \frac{π}{4} w i t h \hat{j}$ and an acute angle θ with $\hat{k}$ then find θ and hence, the components of $\vec{a}$ .

Let $\vec{a} = (a_{1}, a_{2}, a_{3})$ as component

We know,

$\vec{a}$ is a unit vector, $| \vec{a} | = 1$

Given that,

$\vec{a}$ marks angles $\frac{π}{3}$ with $\hat{i}$ , $\frac{π}{4}$ with $\hat{j}$ and $θ$ with $\hat{k}$ acute angle.

Now,

$c o s \frac{π}{3} = \frac{a_{1}}{| \vec{a} |} \Rightarrow \frac{1}{2} = a_{1} [| \vec{a} | = 1] c o s \frac{π}{4} = \frac{a_{2}}{| \vec{a} |} \Rightarrow1/\sqrt2$
$\begin{array}{l} \frac{}{} = a_{2} \\ c o s θ = \frac{a_{3}}{| \vec{a} |} \Rightarrow a_{3} = c o s θ \end{array}$

We know,

$| \vec{a} | = 1$

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

Show that

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

46. Find λ and μ if $(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = 2 (\vec{a} \times \vec{b})$

$(2 \hat{i} + 6 \hat{j} + 27 \hat{k}) \times (\hat{i} + λ \hat{j} + μ \hat{k}) = \vec{0}$

$\begin{array}{l} \Rightarrow \hat{i} (6 μ - 27 λ) - \hat{j} (2 μ - 27) + \hat{k} (2 λ - 6) \\ = 0 \hat{i} + 0 \hat{j} + 0 \hat{k} \end{array}$

On comparing both side components,

$\begin{array}{l} 6 μ - 27 λ = 0, \\ 2 μ - 27 = 0 \\ 2 μ = 27 \\ μ = \frac{27}{2}, \\ 2 λ - 6 = 0 \\ 2 λ = 6 \\ λ = \frac{6}{2} = 3 \end{array}$

$∴$ Therefore, the value of $μ = \frac{27}{2}$ and $λ = 3$

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