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NCERT Solutions for Class 12 Maths Chapter 10 - Vector Algebra

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NCERT Solutions for Class 12 Maths Chapter 10 - Vector Algebra

NCERT Solutions Class 12 Maths Chapter Vector Algebra can now be easily acquired by just visiting the main website of Vedantu. Here you can get complete solutions that are comprehensive and well-made to ease students’ doubts. Our Vector Algebra Class 12 Solutions can be referred for understanding, practice and revision of the concept. Vector Algebra Class 12 PDF has been created and designed by highly qualified and experienced teachers who are involved in research work in a particular subject for many years. Solutions are available especially for the convenience of students who want to explore more about a topic and to procure in-depth knowledge. 


Class:

NCERT Solutions for Class 12

Subject:

Class 12 Maths

Chapter Name:

Chapter 10 - Vector Algebra

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2023-24

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes



Important Topics Covered in NCERT Solutions Class 12 Maths Chapter 10

Before solving an exercise it is important to have complete knowledge of concepts. The idea of a syllabus helps students to make their strategy for examination. The topic along with the subtopic of NCERT Class 12 Maths Chapter 10 are given below.

Sections

Topics

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 Chapter at a Glance - Class 12 NCERT Solutions

  • A linear programming problem is one that is concerned with finding the optimal value (maximum or minimum) of a linear function of several variables (called objective function) subject to the conditions that the variables are nonnegative and satisfy a set of linear inequalities (called linear constraints). Variables are sometimes called decision variables and are non-negative.

  • Position vector of a point $P(x, y, z)$ is given as $\overrightarrow{O P}(-r)-x \hat{i}+\hat{y}+z \hat{k}$, and its magnitude by $\sqrt{x^2+y^2+z^2}$.

  • The scalar components of a vector are its direction ratios, and represent its projections along the respective axes.

  • The magnitude $(r)$, direction ratios $(a, b, c)$ and direction cosines $(l, m, n)$ of any vector are related as:

$l=\frac{a}{r}, m=\frac{b}{r}, n=\frac{c}{r}$

  • The vector sum of the three sides of a triangle taken in order is 0 .

  • The vector sum of two coinitial vectors is given by the diagonal of the parallelogram whose adjacent sides are the given vectors.

  • The multiplication of a given vector by a scalar $\lambda$, changes the magnitude of the vector by the multiple $|2|$, and keeps the direction same (or makes it opposite) according as the value of A is positive (or negative).

  • For a given vector $a$, the vector $\bar{a}=\frac{\vec{a}}{|\vec{a}|}$ gives the unit vector in the direction of $\dot{a}$.

  • The position vector of a point $\mathrm{R}$ dividing a line segment joining the points $\mathrm{P}$ and $\mathrm{Q}$ whose position vectors are $d$ and $b$ respectively, in the ratio $m: n$

(i) internally, is given by $\frac{m+m b}{m+n}$

(ii) externally, is given by $\frac{m b \vec{b}-n \vec{a}}{m-n}$

 The scalar product of two given vectors $a$ and $b$ haying angle $\theta$ between them is defined as $a b=|a||b| \cos \theta$.

  • Also, when $d \dot{b}$ is given the angle ' $\theta$ ' between the vectors $\dot{a}$ and $b$ may be determined by $\cos \theta=\frac{a b}{|\vec{a}||\vec{b}|}$

  • If $\theta$ is the angle between two vectors $a$ and $b$, then their cross product tis given as $a \times b=|a||b| \sin \theta \hat{b}$ where $\hat{n}$ is a unit vector perpendicular to the plane containing $d$ and $b$. Such that $d, b, \hat{n}$ form a right handed system of coordinate axes.

  • If we have two vectors $a$ and $b$, given in component form as $d=a_1 \hat{i}+a_1 \hat{j}+a_1 \hat{k}$ and $b=b_1 \hat{i}+b_2 \hat{j}+b_1 \hat{k}$ and $\lambda$ any scalar,

then $\vec{a}+\vec{b}=\left(a_1+b_1\right) \hat{i}+\left(a_2+b_2\right) \hat{j}+\left(a_5+b_5\right) \hat{k} ; \hat{\lambda} \vec{a}-\left(\hat{\lambda} a_1\right) \hat{i}+\left(\hat{\lambda} a_2\right) \hat{j}+\left(\hat{\lambda} a_3\right) \hat{k} ; a \vec{b}=a_1 b_1+a_2 b_1+a_3 b_3 ;$ and

$a \times b=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ a_1 & b_1 & c_1 \\a_2 & b_2 & c_2\end{array}\right|$

Table of Contents
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1. NCERT Solutions for Class 12 Maths Chapter 10 - Vector Algebra
  • NCERT Solutions for Class 12 Maths – Free PDF Download
  • Chapter 10 – Vector Algebra
2. 10.3 Types of Vectors
3. 10.6 Product of Two Vectors
  • Key Features of NCERT Solutions for Class 12 Maths Chapter 10
FAQs
Competitive Exams after 12th Science
More Free Study Material for Vectors Algebra
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Exercises under NCERT Class 12 Maths Chapter 10 – Vector Algebra

Chapter 10 of NCERT Solutions for Class 12 Maths deals with Vector Algebra. This chapter introduces students to the concept of vectors and their algebraic operations. It covers topics such as scalar and vector products, triple products, and their applications in geometry and physics.


The chapter is divided into four exercises and one miscellaneous exercise. Here is a brief summary of each exercise:


Exercise 10.1: This exercise has ten questions that ask students to find the direction cosines of a given vector and vice versa.

Exercise 10.2: This exercise has eleven questions that ask students to find the magnitude and direction of a given vector, as well as to perform vector operations such as addition, subtraction, and scalar multiplication.

Exercise 10.3: This exercise has seven questions that ask students to find the scalar and vector products of two given vectors.

Exercise 10.4: This exercise has six questions that ask students to find the scalar and vector triple products of given vectors.

Miscellaneous Exercise: This exercise has six questions that ask students to apply the concepts of vector algebra to solve problems in geometry and physics.


Overall, this chapter is essential for students who want to pursue higher education in Mathematics and Physics. It provides a solid foundation for students to understand the concepts of vectors and their applications in real-world problems.


Access NCERT Solutions for Class 12 Maths Chapter 10 – Vector Algebra

Exercise 10.1

1. Represent graphically a displacement of $40\;{\text{km}},{30^\circ }$ east of north.

Ans:


Displacement of $40\;{\text{km}},{30^\circ }$ east of nort


Here, the vector OP is representing the displacement of $40\;{\text{km}},{30^\circ }$ in East of North direction.


2. Classify the following measures as scalars and vectors.

i. $10\;{\text{kg}}$

Ans: $10\;{\text{kg}}$is a scalar quantity because it has only magnitude not direction.

ii. 2 meters north-west

Ans: 2 meters north-west is a vector quantity because it has both magnitude as well as direction.

iii. ${40^\circ }$

Ans: ${40^\circ }$ is a scalar quantity because it has only magnitude, not direction.

iv. $40$ watt

Ans: 40 watts is a scalar quantity because it has only magnitude, not direction.

v. ${10^{ - 19}}$coulomb

Ans: ${10^{ - 19}}$Coulomb is a scalar quantity because it has only magnitude not direction.

vi. $20\;{\text{m/}}{{\text{s}}^2}$

Ans: $20\;{\text{m/}}{{\text{s}}^2}$ is a vector quantity because it has both magnitude as well as direction.


3.  Classify the following as scalar and vector quantities:-

i. Time period

Ans: Time period is a scalar quantity because it has only magnitude.

ii. distance

Ans: Distance is a scalar quantity because it has only magnitude.

iii. force   

Ans: Force is a vector quantity because it has both magnitude as well as direction.

iv. velocity

Ans: Velocity is a vector quantity because it has both magnitude as well as direction.

v. work done

Ans: Work done is a scalar quantity because it has only magnitude.


4.  In Figure, identify the following vectors.


Vectors


i. Co-initial

Ans: Vectors $\vec a$ and $\vec d$ are co-initial.

ii. Equal   

Ans: Vectors $\vec b$ and $\vec d$ are equal.

iii. Collinear but not equal 

Ans: Vectors $\vec a$ and $\vec c$ are collinear but not equal.


5. Answer the following as true or false.

i.$\vec a$ and $ - \vec a$ and are collinear.

Ans: True

ii. Two collinear vectors are always equal in magnitude.

Ans: False

iii. Two vectors having same magnitude are collinear.

Ans: False

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

Ans: False


Exercise 10.2

1.  Compute the magnitude of the following vectors:-

$\vec a = \hat i + \hat j + \hat k\;;\;\vec b = 2\hat i - 7\hat j - 3\hat k\;;\;\vec c = \dfrac{1}{{\sqrt 3 }}\hat i + \dfrac{1}{{\sqrt 3 }}\hat j - \dfrac{1}{{\sqrt 3 }}\hat k$

Ans:  $|\vec a| = \sqrt {{{(1)}^2} + {{(1)}^2} + {{(1)}^2}}  = \sqrt 3 $

$|\vec b| = \sqrt {{{(2)}^2} + {{( - 7)}^2} + {{( - 3)}^2}} $

$ = \sqrt {4 + 49 + 9} $

$ = \sqrt {62} $

$|\vec c| = \sqrt {{{\left( {\dfrac{1}{{\sqrt 3 }}} \right)}^2} + {{\left( {\dfrac{1}{{\sqrt 3 }}} \right)}^2} - {{\left( {\dfrac{1}{{\sqrt 3 }}} \right)}^2}} $

$ = \sqrt {\dfrac{1}{3} + \dfrac{1}{3} + \dfrac{1}{3}}  = 1$

2. Write two different vectors having the same magnitude. 

Ans: $\vec a = (\hat i - 2\hat j + 3\hat k)$  and  $\vec b = (2\hat i + \hat j - 3\hat k)$.

$|\vec a| = \sqrt {{1^2} + {{( - 2)}^2} + {3^2}}  = \sqrt {1 + 4 + 9}  = \sqrt {14} $

$|\vec b| = \sqrt {{2^2} + {1^2} + {{( - 3)}^2}}  = \sqrt {4 + 1 + 9}  = \sqrt {14} $

But $\vec a \ne \vec b$

3. Write two different vectors having same direction. 

Ans:  $\vec p = (\hat i + \hat j + \hat k)$  and  $\vec q = (2\hat i + 2\hat j + 2\hat k)$.

The Direction Cosines of $\vec p$ are $a = \dfrac{1}{{\sqrt {{1^2} + {1^2} + {1^2}} }} = \dfrac{1}{{\sqrt 3 }},b = \dfrac{1}{{\sqrt {{1^2} + {1^2} + {1^2}} }} = \dfrac{1}{{\sqrt 3 }},c = \dfrac{1}{{\sqrt {{1^2} + {1^2} + {1^2}} }} = \dfrac{1}{{\sqrt 3 }}$

The Direction Cosines of $\vec q$ are$a = \dfrac{2}{{\sqrt {{2^2} + {2^2} + {2^2}} }} = \dfrac{2}{{2\sqrt 3 }} = \dfrac{1}{{\sqrt 3 }},b = \dfrac{2}{{\sqrt {{2^2} + {2^2} + {2^2}} }} = \dfrac{2}{{2\sqrt 3 }} = \dfrac{1}{{\sqrt 3 }},c = \dfrac{2}{{\sqrt {{2^2} + {2^2} + {2^2}} }} = \dfrac{2}{{2\sqrt 3 }} = \dfrac{1}{{\sqrt 3 }}$

The Direction Cosines of $\vec p$ and $\vec q$ are equal but $\vec p \ne \vec q$.


4. 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. 

Ans:  $2\hat i + 3\hat j = x\hat i + y\hat j \Rightarrow x = 2,y = 3$


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

Ans:  Let the points be ${\text{P}}(2,1)$ and ${\text{Q}}( - 5,7)$ 

$\overrightarrow {PQ}  = ( - 5 - 2)\hat i + (7 - 1)\hat j$

$ \Rightarrow \overrightarrow {PQ}  =  - 7\hat i + 6\hat j$

So, scalar components of required vector  are $ - 7$ and 6 and the vector components are $ - 7\hat i$ and $6\hat j$.


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

Ans:  $\vec a + \vec b + \vec c = (1 - 2 + 1)\hat i + ( - 2 + 4 - 6)\hat j + (1 + 5 - 7)\hat k{\text{ }} = 0\hat i - 4\hat j - 1\hat k{\text{ }} =  - \;4\hat j - \hat k{\text{ }}$


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

Ans: $|\vec a| = \sqrt {{1^2} + {1^2} + {2^2}}  = \sqrt {1 + 1 + 4}  = \sqrt 6 $

$\therefore \hat a = \dfrac{{\vec a}}{{|\vec a|}} = \dfrac{{\hat i + \hat j + 2\hat k}}{{\sqrt 6 }} = \dfrac{1}{{\sqrt 6 }}\hat i + \dfrac{1}{{\sqrt 6 }}\hat j + \dfrac{2}{{\sqrt 6 }}\hat k$


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

Ans:  $\overrightarrow {PQ}  = (4 - 1)\hat i + (5 - 2)\hat j + (6 - 3)k = 3\hat i + 3\hat j + 3k$

$|\overrightarrow {PQ} | = \sqrt {{3^2} + {3^2} + {3^2}}  = \sqrt {9 + 9 + 9}  = \sqrt {27}  = 3\sqrt 3 $

So, unit vector $ = \dfrac{{\overrightarrow {PQ} }}{{|\overrightarrow {PQ} |}} = \dfrac{{3\hat i + 3\hat j + 3\hat k}}{{3\sqrt 3 }} = \dfrac{1}{{\sqrt 3 }}\hat i + \dfrac{1}{{\sqrt 3 }}\hat j + \dfrac{1}{{\sqrt 3 }}\hat k$


9. 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 $\vec a + \vec b$. 

Ans: $\vec a + \vec b = (2 - 1)\hat i + ( - 1 + 1)\hat j + (2 - 1)\hat k = 1\hat i + 0\hat j + 1\hat k = \hat i + \hat k$

$|\vec a + \vec b| = \sqrt {{1^2} + {1^2}}  = \sqrt 2 $

So, unit vector $ = \dfrac{{(\vec a + \vec b)}}{{|\vec a + \vec b|}} = \dfrac{{\hat i + \hat k}}{{\sqrt 2 }} = \dfrac{1}{{\sqrt 2 }}\hat i + \dfrac{1}{{\sqrt 2 }}\hat k$


10. Find a vector in the direction of vector $5\hat i - \hat j + 2\hat k$ which has magnitude 8 units.

Ans: $\vec a = 5\hat i - \hat j + 2\hat k$

$|\vec a| = \sqrt {{5^2} + {{( - 1)}^2} + {2^2}}  = \sqrt {25 + 1 + 4}  = \sqrt {30} $

$\therefore \hat a = \dfrac{{\vec a}}{{|\vec a|}} = \dfrac{{5\hat i - \hat j + 2\hat k}}{{\sqrt {30} }}$

So, a vector in direction of  $5\hat i - \hat j + 2\hat k$ with magnitude 8 units is:$8\hat a = 8\left( {\dfrac{{5\hat i - \hat j + 2\hat k}}{{\sqrt {30} }}} \right) = \dfrac{{40}}{{\sqrt {30} }}\hat i - \dfrac{8}{{\sqrt {30} }}\hat j + \dfrac{{16}}{{\sqrt {30} }}\hat k$


11. 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.

Ans:  $\vec a = 2\hat i - 3\hat j + 4\hat k$ and $\vec b =  - 4\hat i + 6\hat j - 8\hat k$

$\vec b =  - 4\hat i + 6\hat j - 8\hat k =  - 2(2\hat i - 3\hat j + 4\hat k) =  - 2\vec a$

$\therefore \vec b = \lambda \vec a,\;\;\lambda  =  - 2$

So, the given vectors are collinear.


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

Ans: $\vec a = \hat i + 2\hat j + 3\hat k$.

$|\vec a| = \sqrt {{1^2} + {2^2} + {3^2}}  = \sqrt {1 + 4 + 9}  = \sqrt {14} $

So, the Direction cosines of $\vec a$ are $\left( {\dfrac{1}{{\sqrt {14} }},\dfrac{2}{{\sqrt {14} }},\dfrac{3}{{\sqrt {14} }}} \right)$


13.  Find the direction cosines of the vector joining the points ${\text{A}}(1,2, - 3)$ and ${\text{B}}( - 1, - 2,1)$ directed from A to b.

Ans: $\overrightarrow {{\text{AB}}}  = ( - 1 - 1)\hat i + ( - 2 - 2)\hat j + \{ 1 - ( - 3)\} \hat k$

$ \Rightarrow \overrightarrow {{\text{AB}}}  =  - 2\hat i - 4\hat j + 4\hat k$

$|\overrightarrow {{\text{AB}}} | = \sqrt {{{( - 2)}^2} + {{( - 4)}^2} + {4^2}}  = \sqrt {4 + 16 + 16}  = \sqrt {36}  = 6$

So, the Direction cosines of $\overrightarrow {{\text{AB}}} $ are $\left( { - \dfrac{2}{6}, - \dfrac{4}{6},\dfrac{4}{6}} \right) = \left( { - \dfrac{1}{3}, - \dfrac{2}{3},\dfrac{2}{3}} \right)$


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

Ans: $\vec a = \hat i + \hat j + \hat k$

$|\vec a| = \sqrt {{1^2} + {1^2} + {1^2}}  = \sqrt 3 $

So, the Direction Cosines of $\vec a$ are $\left( {\dfrac{1}{{\sqrt 3 }},\dfrac{1}{{\sqrt 3 }},\dfrac{1}{{\sqrt 3 }}} \right)$

$\cos \alpha  = \dfrac{1}{{\sqrt 3 }},\cos \beta  = \dfrac{1}{{\sqrt 3 }},\cos \gamma  = \dfrac{1}{{\sqrt 3 }}$

So, the vector is equally inclined to OX, OY, and ${\text{OZ}}$.


15. Find the position vector of a point ${\text{R}}$ 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 

Ans: $\overrightarrow {{\text{OP}}}  = \hat i + 2\hat j - \hat k$ and $\overrightarrow {{\text{OQ}}}  =  - \hat i + \hat j + \hat k$

 The position vector of ${\text{R}}$ is $\overrightarrow {{\text{OR}}}  = \dfrac{{2( - \hat i + \hat j + \hat k) + 1(\hat i + 2\hat j - \hat k)}}{{2 + 1}} = \dfrac{{( - 2\hat i + 2\hat j + 2\hat k) + (\hat i + 2\hat j - \hat k)}}{3}$

$ = \dfrac{{ - \;\hat i + 4\hat j + \hat k}}{3} =  - \dfrac{1}{3}\hat i + \dfrac{4}{3}\hat j + \dfrac{1}{3}\hat k$

ii. Externally.

Ans: The position vector of $R$ is $\overrightarrow {{\text{OR}}}  = \dfrac{{2( - \hat i + \hat j + \hat k) - 1(\hat i + 2\hat j - \hat k)}}{{2 - 1}} = ( - 2\hat i + 2\hat j + 2\hat k) - (\hat i + 2\hat j - \hat k)$

$ =  - 3\hat i + 3\hat k$


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

Ans: The position vector of $R$ is

$\overrightarrow {{\text{OR}}}  = \dfrac{{(2\hat i + 3\hat j + 4\hat k) + (4\hat i + \hat j - 2\hat k)}}{2} = \dfrac{{(2 + 4)\hat i + (3 + 1)\hat j + (4 - 2)\hat k}}{2}$

$ = \dfrac{{6\hat i + 4\hat j + 2\hat k}}{2} = 3\hat i + 2\hat j + \hat k$


17. Show that the points ${\text{A}},{\text{B}}$ and ${\text{C}}$ 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. 

Ans: 

$\overrightarrow {{\text{AB}}}  = \vec b - \vec a = (2 - 3)\hat i + ( - 1 + 4)\hat j + (1 + 4)\hat k =  - \hat i + 3\hat j + 5\hat k$

$\overrightarrow {{\text{BC}}}  = \vec c - \vec b = (1 - 2)\hat i + ( - 3 + 1)\hat j + ( - 5 - 1)\hat k =  - \hat i + 2\hat j - 6\hat k$

$\overrightarrow {{\text{CA}}}  = \vec a - \vec c = (3 - 1)\hat i + ( - 4 + 3)\hat j + ( - 4 + 5)\hat k = 2\hat i - \hat j + \hat k$

$|\overrightarrow {{\text{AB}}} {|^2} = {( - 1)^2} + {3^2} + {5^2} = 1 + 9 + 25 = 35$

$|\overrightarrow {{\text{BC}}} {|^2} = {( - 1)^2} + {( - 2)^2} + {( - 6)^2} = 1 + 4 + 36 = 41$

$|\overrightarrow {CA} {|^2} = {2^2} + {( - 1)^2} + {1^2} = 4 + 1 + 1 = 6$

$|\overrightarrow {{\text{AB}}} {|^2} + |\overrightarrow {{\text{CA}}} {|^2} = 35 + 6 = 41 = |\overrightarrow {{\text{BC}}} {|^2}$

So, ABC is a right angled triangle.


18. In triangle ABC, which of the following is not true?


Triangle ABC


a. $\overrightarrow {{\text{AB}}}  + \overrightarrow {{\text{BC}}}  + \overrightarrow {{\text{CA}}}  = \vec 0$

b. $\overrightarrow {{\text{AB}}}  + \overrightarrow {{\text{BC}}}  - \overrightarrow {{\text{AC}}}  = \vec 0$

c. $\overrightarrow {{\text{AB}}}  + \overrightarrow {{\text{BC}}}  - \overrightarrow {{\text{CA}}}  = \vec 0$

d. $\overrightarrow {{\text{AB}}}  - \overrightarrow {{\text{CB}}}  + \overrightarrow {{\text{CA}}}  = \vec 0$

Ans:


Triangle ABC


$\overrightarrow {{\text{AB}}}  + \overrightarrow {{\text{BC}}}  = \overrightarrow {{\text{AC}}} $

$ \Rightarrow \overrightarrow {{\text{AB}}}  + \overrightarrow {{\text{BC}}}  =  - \overrightarrow {{\text{CA}}} $

$ \Rightarrow \overrightarrow {{\text{AB}}}  + \overrightarrow {{\text{BC}}}  + \overrightarrow {{\text{CA}}}  = \vec 0$

$\overrightarrow {{\text{AB}}}  + \overrightarrow {{\text{BC}}}  = \overrightarrow {{\text{AC}}} $

$ \Rightarrow \overrightarrow {{\text{AB}}}  + \overrightarrow {{\text{BC}}}  - \overrightarrow {{\text{AC}}}  = \vec 0$

$\overrightarrow {{\text{AB}}}  + \overrightarrow {{\text{BC}}}  + \overrightarrow {{\text{CA}}}  = \vec 0$

$\overrightarrow {{\text{AB}}}  - \overrightarrow {{\text{CB}}}  + \overrightarrow {{\text{CA}}}  = \vec 0$

If $\overrightarrow {{\text{AB}}}  + \overrightarrow {{\text{BC}}}  - \overrightarrow {{\text{CA}}}  = \vec 0$

$\overrightarrow {{\text{AC}}}  = \overrightarrow {{\text{CA}}} $

$ \Rightarrow \overrightarrow {{\text{AC}}}  =  - \overrightarrow {{\text{AC}}} $

$ \Rightarrow \overrightarrow {{\text{AC}}}  + \overrightarrow {{\text{AC}}}  = \vec 0$

$ \Rightarrow 2\overrightarrow {{\text{AC}}}  = \vec 0$

$ \Rightarrow \overrightarrow {{\text{AC}}}  = \vec 0$, which is not true.

So, the equation given in option ${\text{C}}$ is False.

Hence, the correct answer is ${\text{C}}$.


19. If $\vec a$ and $\vec b$ are two collinear vectors, then which of the following are incorrect?

a. $\vec b = \lambda \vec a$, for some scalar $\lambda $

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 

Ans: If $\vec a$ and $\vec b$ are collinear vectors, they are parallel. $\vec b = \lambda \vec a($ scalar $\lambda )$

If $\lambda  =  \pm 1$, then $\vec a =  \pm \vec b$

If $\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k$ and $\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k$,

$\vec b = \lambda \vec a \Rightarrow {b_1}\hat i + {b_2}\hat j + {b_3}\hat k = \lambda \left( {a\hat i + {a_2}\hat j + {a_3}\hat k} \right)$

$ \Rightarrow {b_1}\hat i + {b_2}\hat j + {b_3}\hat k = \left( {\lambda {a_1}} \right)\hat i + \left( {\lambda {a_2}} \right)\hat j + \left( {\lambda {a_3}} \right)\hat k$

$ \Rightarrow {b_1} = \lambda {a_1},{b_2} = \lambda {a_2},{b_3} = \lambda {a_3}$

$ \Rightarrow \dfrac{{{b_1}}}{{{a_1}}} = \dfrac{{{b_2}}}{{{a_2}}} = \dfrac{{{b_3}}}{{{a_3}}} = \lambda $

Thus, respective components of $\vec a$ and $\vec b$ are proportional. 

But, $\vec a$ and $\vec b$ may have different directions. So, option D is incorrect. The correct answer is ${\text{D}}$.


Exercise 10.3

1. Find the angle between two vectors $\vec a$ and $\vec b$ with magnitudes $\sqrt 3 $ and 2, respectively having $\vec a \cdot \vec b = \sqrt 6 $

Ans: $|\vec a| = \sqrt 3 \;,\;|\vec b| = 2\;,\;\vec a\;.\;\vec b = \sqrt 6 $

$\therefore \sqrt 6  = \sqrt 3  \times 2 \times \cos \theta $

$ \Rightarrow \cos \theta  = \dfrac{{\sqrt 6 }}{{\sqrt 3  \times 2}}$

$ \Rightarrow \cos \theta  = \dfrac{1}{{\sqrt 2 }}$

$ \Rightarrow \theta  = \dfrac{\pi }{4}$


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

Ans: $|\vec a| = \sqrt {{1^2} + {{( - 2)}^2} + {3^2}}  = \sqrt {1 + 4 + 9}  = \sqrt {14} $

$|\vec b| = \sqrt {{3^2} + {{( - 2)}^2} + {1^2}}  = \sqrt {9 + 4 + 1}  = \sqrt {14} $

$\vec a\;.\;\vec b = (i - 2\hat j + 3\hat k)(3\hat i - 2\hat j + \hat k)$

$ = 1.3 + ( - 2)( - 2) + 3.1$

$ = 3 + 4 + 3$

$ = 10$

$\therefore 10 = \sqrt {14} \sqrt {14} \cos \theta $

$ \Rightarrow \cos \theta  = \dfrac{{10}}{{14}}$

$ \Rightarrow \theta  = {\cos ^{ - 1}}\left( {\dfrac{5}{7}} \right)$


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

Ans: $\vec a = \hat i - \hat j$ and $\vec b = \hat i + \hat j$

Projection of $\vec a$ on $\vec b$ is $\dfrac{1}{{\mid \vec b|}}(\vec a \cdot \vec b) = \dfrac{1}{{\sqrt {1 + 1} }}\{ 1.1 + ( - 1)(1)\}  = \dfrac{1}{{\sqrt 2 }}(1 - 1) = 0$


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

Ans: $\vec a = \hat i + 3\hat j + 7\hat k$ and $\vec b = 7\hat i - \hat j + 8\hat k$

Projection of $\vec a$ on $\vec b$ is $\dfrac{1}{{|\vec b|}}(\vec a \cdot \vec b) = \dfrac{1}{{\sqrt {{7^2} + {{( - 1)}^2} + {8^2}} }}\{ 1(7) + 3( - 1) + 7(8)\}  = \dfrac{{7 - 3 + 56}}{{\sqrt {49 + 1 + 64} }} = \dfrac{{60}}{{\sqrt {114} }}$


5. Show that each of the given three vectors is a unit vector, which are mutually perpendicular to each other. $\dfrac{1}{7}(2\hat i + 3\hat j + 6\hat k)\;\;,\;\;\dfrac{1}{7}(3\hat i - 6\hat j + 2\hat k)\;\;,\;\;\dfrac{1}{7}(6\hat i + 2\hat j - 3\hat k)$

Ans: $\vec a = \dfrac{1}{7}(2\hat i + 3\hat j + 6\hat k) = \dfrac{2}{7}\hat i + \dfrac{3}{7}\hat j + \dfrac{6}{7}\hat k$,

$\vec b = \dfrac{1}{7}(3\bar i - 6\hat j + 2\hat k) = \dfrac{3}{7}\hat i - \dfrac{6}{7}\bar j + \dfrac{2}{7}\hat k$

$\vec c = \dfrac{1}{7}(6\hat i + 2\hat j - 3\hat k) = \dfrac{6}{7}\hat i + \dfrac{2}{7}\hat j - \dfrac{3}{7}\hat k$

$|\vec a| = \sqrt {{{\left( {\dfrac{2}{7}} \right)}^2} + {{\left( {\dfrac{3}{7}} \right)}^2} + {{\left( {\dfrac{6}{7}} \right)}^2}}  = \sqrt {\dfrac{4}{{49}} + \dfrac{9}{{49}} + \dfrac{{36}}{{49}}}  = 1$

$|\vec b| = \sqrt {{{\left( {\dfrac{3}{7}} \right)}^2} + {{\left( { - \dfrac{6}{7}} \right)}^2} + {{\left( {\dfrac{2}{7}} \right)}^2}}  = \sqrt {\dfrac{9}{{49}} + \dfrac{{36}}{{49}} + \dfrac{4}{{49}}}  = 1$

$|\vec c| = \sqrt {{{\left( {\dfrac{6}{7}} \right)}^2} + {{\left( {\dfrac{2}{7}} \right)}^2} + {{\left( { - \dfrac{3}{7}} \right)}^2}}  = \sqrt {\dfrac{{36}}{{49}} + \dfrac{4}{{49}} + \dfrac{9}{{49}}}  = 1$

So, each of the vectors is a unit vector.

$\vec a\;.\;\vec b = \dfrac{2}{7} \times \dfrac{3}{7} + \dfrac{3}{7} \times \left( {\dfrac{{ - 6}}{7}} \right) + \dfrac{6}{7} \times \dfrac{2}{7} = \dfrac{6}{{49}} - \dfrac{{18}}{{49}} + \dfrac{{12}}{{49}} = 0$

$\vec b \cdot \vec c = \dfrac{3}{7} \times \dfrac{6}{7} + \left( {\dfrac{{ - 6}}{7}} \right) \times \dfrac{2}{7} + \dfrac{2}{7} \times \left( {\dfrac{{ - 3}}{7}} \right) = \dfrac{{18}}{{49}} - \dfrac{{12}}{{49}} - \dfrac{6}{{49}} = 0$

$\vec c\;.\;\vec a = \dfrac{6}{7} \times \dfrac{2}{7} + \dfrac{2}{7} \times \dfrac{3}{7} + \left( {\dfrac{{ - 3}}{7}} \right) \times \dfrac{6}{7} = \dfrac{{12}}{{49}} + \dfrac{6}{{49}} - \dfrac{{18}}{{49}} = 0$

So, given vectors are mutually perpendicular to each other.


6. Find $|\vec a|$ and $|\vec b|$, if $(\vec a + \vec b) \cdot (\vec a - \vec b) = 8$ and $|\overrightarrow a |\; = \;8|\vec b|$.

Ans: $(\vec a + \vec b) \cdot (\vec a - \vec b) = 8$

$ \Rightarrow \vec a\vec a - \vec a\;.\;\vec b + \vec b \cdot \vec a - \vec b\vec b = 8$

$ \Rightarrow |\vec a{|^2} - |\vec b{|^2} = 8$

$ \Rightarrow {(8|\vec b|)^2} - |\vec b{|^2} = 8$

$ \Rightarrow 64|\vec b{|^2} - |\vec b{|^2} = 8$

$ \Rightarrow 63|\vec b{|^2} = 8$

$ \Rightarrow |\vec b{|^2} = \dfrac{8}{{63}}$

$ \Rightarrow |\vec b| = \sqrt {\dfrac{8}{{63}}} $

$ \Rightarrow |\vec b| = \dfrac{{2\sqrt 2 }}{{3\sqrt 7 }}$

$|\vec a| = 8|\vec b| = \dfrac{{8 \times 2\sqrt 2 }}{{3\sqrt 7 }} = \dfrac{{16\sqrt 2 }}{{3\sqrt 7 }}$


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

Ans: $(3\vec a - 5\vec b) \cdot (2\vec a + 7\vec b)$

$ = 3\vec a \cdot 2\vec a + 3\vec a.7\vec b - 5\vec b \cdot 2\vec a - 5\vec b \cdot 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} + 11\vec a\vec b - 35|\vec b{|^2}$


8. Find the magnitude of two vectors $\vec a$ and $\vec b$, having the same magnitude and such that the angle between them is ${60^\circ }$ and their scalar product is $\dfrac{1}{2}$ 

Ans: Let $\theta $ be angle between $\vec a$ and $\vec b$.

$|\vec a| = |\vec b|,\vec a\;.\;\vec b = \dfrac{1}{2}$, and $\theta  = {60^\circ }$

$\therefore \dfrac{1}{2} = |\vec a|\vec b\mid \cos {60^\circ }$

$ \Rightarrow \dfrac{1}{2} = |\vec a{|^2} \times \dfrac{1}{2}$

$ \Rightarrow |\vec a{|^2} = 1$

$ \Rightarrow |\vec a| = |\vec b| = 1$


9. Find $|\vec x|$, if for a unit vector $\vec a\;,\;(\vec x - \vec a) \cdot (\vec x + \vec a) = 12$

Ans: $(\vec x - \vec a) \cdot (\vec x + \vec a) = 12$

$ \Rightarrow \vec x \cdot \vec x + \vec x\vec a - \vec a\vec x - \vec a \cdot \vec a = 12$

$ \Rightarrow |\vec x{|^2} - |\vec a{|^2} = 12$

$ \Rightarrow |\vec x{|^2} - 1 = 12$

$ \Rightarrow |\vec x{|^2} = 13$

$\therefore |\vec x| = \sqrt {13} $


10. 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 $\vec a + \lambda \vec b$ is perpendicular to $\vec c$, then find the value of $\lambda $.

Ans: $\vec a + \lambda \vec b = (2\hat i + 2\hat j + 3\hat k) + \lambda ( - \hat i + 2\hat j + \dot k) = (2 - \lambda )\hat i + (2 + 2\lambda )\hat j + (3 + \lambda )\hat k$

$(\vec a + \lambda \vec b) \cdot \vec c = 0$

$ \Rightarrow [(2 +  - )\hat i + (2 + 2\lambda )\hat j + (3 + \lambda )\hat k] \cdot (3\hat i + j) = 0$

$ \Rightarrow 3(2 - \lambda ) + (2 + 2\lambda ) + 0(3 + \lambda ) = 0$

$ \Rightarrow 6 - 3\lambda  + 2 + 2\lambda  = 0$

$ \Rightarrow  - \lambda  + 8 = 0$

$ \Rightarrow \lambda  = 8$


11. 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 $\vec b$.

Ans: $(|\vec a|\vec b + |\vec b|\vec a) \cdot (|\vec a|\vec b - |\vec b|\vec a)$

$ = |\vec a{|^2}\vec b\vec b - |\vec a||\vec b|\vec b\vec a + |\overrightarrow b ||a|\vec a\vec b - |\vec b{|^2}\vec a \cdot \vec a$

$ = |\vec a{|^2}|\overrightarrow b {|^2} - |\overrightarrow b {|^2}|\overrightarrow a {|^2}\;\; = \;\;0$


12. If $\vec a \cdot \vec a = 0$ and $\vec a\vec b = 0$, then what can be concluded above the vector $\vec b$ ? 

Ans: $\vec a \cdot \vec a = 0 \Rightarrow |\vec a{|^2} = 0 \Rightarrow |\vec a| = 0$

$\therefore \vec a$ is the zero vector. Thus, any vector $\vec b$ can satisfy $\vec a\;.\;\vec b = 0$.


13. If $\vec a,\vec b,\vec c$ are unit vectors such that $\vec a + \vec b + \vec c = 0$, find the value of $\vec a \cdot \vec b + \vec b \cdot \vec c + \vec c\;.\;\vec a$.

Ans: $|\vec a + \vec b + \vec c{|^2} = (\vec a + \vec b + \vec c) \cdot (\vec a + \vec b + \vec c) = |\vec a{|^2} + |\vec b{|^2} + |\vec c{|^2} + 2(\vec a\vec b + \vec b\vec c + \vec c \cdot \vec a)$

$ \Rightarrow 0 = 1 + 1 + 1 + 2(\vec a\vec b + \vec b \cdot \vec c + \vec c \cdot \vec a)$

$ \Rightarrow (\vec a\;.\;\vec b + \vec b \cdot \vec c + \vec c\;.\;\vec a) = \dfrac{{ - 3}}{2}$


14. If either vector $\vec a = \vec 0$ or $\vec b = \vec 0$, then $\vec a \cdot \vec b = 0.$ But the converse need not be true. Justify your answer with an example.

Ans: $\overrightarrow {\text{a}}  = 2\widehat {\text{i}} + 4\widehat {\text{j}} + 3\widehat {\text{k}}$ and $\overrightarrow b  = 3\widehat {\text{i}} + 3\widehat {\text{j}} - 6\widehat {\text{k}}$

$\vec a \cdot \vec b = 2.3 + 4.3 + 3( - 6) = 6 + 12 - 18 = 0$

$|{\text{a}}| = \sqrt {{2^2} + {4^2} + {3^2}}  = \sqrt {29} $

$\therefore \vec a \ne \vec 0$

$|\vec b| = \sqrt {{3^2} + {3^2} + {{( - 6)}^2}}  = 54$

$\therefore \vec b \ne \bar 0$

So, it is clear from the above example that the converse of the given statement need not be true.


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

Ans: $\overrightarrow {{\text{BA}}}  = \{ 1 - ( - 1)\hat i + (2 - 0)\hat j + (3 - 0)\hat k = 2\hat i + 2\hat j + 3\hat k$

$\overrightarrow {{\text{BC}}}  = \{ 0 - ( - 1)\} \hat i + (1 - 0)\hat j + (2 - 0)\hat k = \hat i + \hat j + 2\hat k$

$\overrightarrow {{\text{BA}}}  \cdot \overrightarrow {{\text{BC}}}  = (2\hat i + 2\hat \jmath  + 3\hat k) \cdot (\hat i + \hat \jmath  + 2\hat k) = 2 \times 1 + 2 \times 1 + 3 \times 2 = 2 + 2 + 6 = 10$

$|\overrightarrow {{\text{BA}}} | = \sqrt {{2^2} + {2^2} + {3^2}}  = \sqrt {4 + 4 + 9}  = \sqrt {17} $

$|\overrightarrow {BC} | = \sqrt {1 + 1 + {2^2}}  = \sqrt 6 $

$\therefore 10 = \sqrt {17}  \times \sqrt 6 \cos (\angle {\text{ABC}})$

$ \Rightarrow \cos (\angle {\text{ABC}}) = \dfrac{{10}}{{\sqrt {17}  \times \sqrt 6 }}$

$ \Rightarrow (\angle {\text{ABC}}) = {\cos ^{ - 1}}\left( {\dfrac{{10}}{{\sqrt {102} }}} \right)$


16. Show that the points ${\text{A}}(1,2,7),\;{\text{B}}(2,6,3)$ and ${\text{C}}(3,10, - 1)$ are collinear. 

Ans: $\overrightarrow {{\text{AB}}}  = (2 - 1)\hat i + (6 - 2)\hat j + (3 - 7)\hat k = \hat i + 4\hat j - 4\hat k$

$\overrightarrow {{\text{BC}}}  = (3 - 2)\hat i + (10 - 6)\hat \jmath  + ( - 1 - 3)\hat k = \hat i + 4\hat j - 4\hat k$

$\overrightarrow {{\text{AC}}}  = (3 - 1)\hat i + (10 - 2)\hat j + ( - 1 - 7)\hat k = 2\hat i + 8\hat j - 8\hat k$

$|\overrightarrow {AB} | = \sqrt {{1^2} + {4^2} + {{( - 4)}^2}}  = \sqrt {1 + 16 + 16}  = \sqrt {33} $

$|\overrightarrow {BC} | = \sqrt {{1^2} + {4^2} + {{( - 4)}^2}}  = \sqrt {1 + 16 + 16}  = \sqrt {33} $

$|\overrightarrow {AC} | = \sqrt {{2^2} + {8^2} + {8^2}}  = \sqrt {4 + 64 + 64}  = 2\sqrt {33} $

$|\overrightarrow {{\text{AC}}} | = |\overrightarrow {{\text{AB}}} | + |\overrightarrow {{\text{BC}}} |$

Hence, the given points are collinear.


17. 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.

Ans: $\overrightarrow {{\text{OA}}}  = 2\hat i - \hat j + \hat k\;,\;\overrightarrow {{\text{OB}}}  = \hat i - 3\hat j - 5\hat k\;,\;\overrightarrow {{\text{OC}}}  = 3\hat i - 4\hat j - 4\hat k$

$\overrightarrow {{\text{AB}}}  = (1 - 2)\hat i + ( - 3 + 1)\hat j + ( - 5 - 1)\hat k =  - \hat i - 2\hat j  - 6 \hat k$

$\overrightarrow {{\text{BC}}}  = (3 - 1)\hat i + ( - 4 + 3)\hat j + ( - 4 + 5)\hat k = 2\hat i - \hat j + \hat k$

$\overrightarrow {{\text{CA}}}  = (2 - 3)\hat i + ( - 1 + 4)\hat j + (1 + 4)\hat k =  - \hat i + 3\hat j + 5\hat k$

$|\overrightarrow {{\text{AB}}} | = \sqrt {{{( - 1)}^2} + {{( - 2)}^2} + {{( - 6)}^2}}  = \sqrt {1 + 4 + 36}  = \sqrt {41} $

$|\overrightarrow {BC} | = \sqrt {{2^2} + {{( - 1)}^2} + {1^2}}  = \sqrt {4 + 1 + 1}  = \sqrt 6 $

$|\overrightarrow {{\text{AC}}} | = \sqrt {{{( - 1)}^2} + {3^2} + {5^2}}  = \sqrt {1 + 9 + 25}  = \sqrt {35} $

$\therefore |\overrightarrow {BC} {|^2} + |\overrightarrow {AC} {|^2} = 6 + 35 = 41 = |\overrightarrow {AB} {|^2}$

Hence, $\Delta {\text{ABC}}$ is a right triangle.


18. If $\vec a$ is a nonzero vector of magnitude 'a' and $\lambda $ a nonzero scalar. then $\lambda \vec a$ is unit vector if

a. $\lambda  = 1$

b. $\lambda  =  - 1$

c. $a = |\lambda |$

d. $a = \dfrac{1}{{|\lambda |}}$

Ans: $|\lambda \vec a| = 1$

$ \Rightarrow |\vec a| = \dfrac{1}{{|\lambda |}}$

$ \Rightarrow a = \dfrac{1}{{|\lambda |}}$


Exercise 10.4

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:  We have, $\vec a = \hat i - 7\hat j + 7\hat k$  and  $\vec b = 3\hat i - 2\hat j + 2\hat k$

$\vec a \times \vec b = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ 1&{ - 7}&7 \\ 3&{ - 2}&2 \end{array}} \right|$

$ = \hat i( - 14 + 14) - \hat j(2 - 21) + \hat k( - 2 + 21) = 19\hat j + 19\hat k$

$\therefore |\vec a \times \vec b| = \sqrt {{{(19)}^2} + {{(19)}^2}}  = \sqrt {2 \times {{(19)}^2}}  = 19\sqrt 2 $


2.  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 $\vec b = \hat i + 2\hat j - 2\hat k$.

Ans:  $\vec a = 3\hat i + 2\hat j + 2\hat k$ and $\vec b = \hat i + 2\hat j - 2\hat k$

$\vec a + \vec b = 4\hat i + 4\hat j\;,\;\vec a - \vec b = 2\hat i + 4\hat k$

$(\vec a + \vec b) \times (\vec a - \vec b) = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ 4&4&0 \\ 2&0&4 \end{array}} \right| = \hat i(16) - \hat j(16) + \hat k( - 8) = 16\hat i - 16\hat j - 8\hat k$

$|(\vec a + \vec b) \times (\vec a - \vec b)| = \sqrt {{{16}^2} + {{( - 16)}^2} + {{( - 8)}^2}} $

$ = \sqrt {{2^2} \times {8^2} + {2^2} \times {8^2} + {8^2}} $

$ = 8\sqrt {{2^2} + {2^2} + 1}  = 8\sqrt 9  = 8 \times 3 = 24$

So, the unit vector is $ =  \pm \dfrac{{(\vec a + \vec b) \times (\vec a - \vec b)}}{{|(\vec a + \vec b) \times (\vec a - \vec b)\mid }} =  \pm \dfrac{{16\hat i - 16\hat j - 8\hat k}}{{24}}$

$ =  \pm \dfrac{{2\hat i - 2\hat j - \hat k}}{3} =  \pm \dfrac{2}{3}\hat i \mp \dfrac{2}{3}\hat j \mp \dfrac{1}{3}\hat k$


3.  If a unit vector $\vec a$ makes an angle $\dfrac{\pi }{3}$ with $\hat i,\dfrac{\pi }{4}$ with $\hat j$ and an acute angle $\theta $ with $\hat k$, then find $\theta $ and hence, the components of $\vec a$ 

Ans: $\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k$

$|\vec a|\; = 1$. $\cos \dfrac{\pi }{3} = \dfrac{{{a_1}}}{{|\vec a|}}$

$ \Rightarrow \dfrac{1}{2} = {a_1}$

$\cos \dfrac{\pi }{4} = \dfrac{{{a_2}}}{{|\vec a|}}$

$ \Rightarrow \dfrac{1}{{\sqrt 2 }} = {a_2}$

$\cos \theta  = \dfrac{{{a_3}}}{{|\vec a|}}$

$ \Rightarrow {a_3} = \cos \theta $

$ \Rightarrow \sqrt {a_1^2 + a_2^2 + a_3^2}  = 1$

$ \Rightarrow {\left( {\dfrac{1}{2}} \right)^2} + {\left( {\dfrac{1}{{\sqrt 2 }}} \right)^2} + {\cos ^2}\theta  = 1$

$ \Rightarrow \dfrac{1}{4} + \dfrac{1}{2} + {\cos ^2}\theta  = 1$

$ \Rightarrow \dfrac{3}{4} + {\cos ^2}\theta  = 1$

$ \Rightarrow {\cos ^2}\theta  = 1 - \dfrac{3}{4} = \dfrac{1}{4}$

$ \Rightarrow \cos \theta  = \dfrac{1}{2} \Rightarrow \theta  = \dfrac{\pi }{3}$

$\therefore {a_3} = \cos \dfrac{\pi }{3} = \dfrac{1}{2}$

So, $\theta  = \dfrac{\pi }{3}$ and components of $\vec a$ are $\left( {\dfrac{1}{2},\dfrac{1}{{\sqrt 2 }},\dfrac{1}{2}} \right)$


4.  Show that $(\vec a - \vec b) \times (\vec a + \vec b) = 2(\vec a \times \vec b)$

Ans: $(\vec a - \vec b) \times (\vec a + \vec b)$

$ = (a - b) \times \vec a + (a - b) \times b$

$ = a \times \vec a - \vec b \times \vec d + a \times \vec b - \vec b \times \vec b$

$ = 0 + \vec a \times \vec b + \vec a \times \vec b - 0$

$ = 2\;(\vec a \times \vec b)$


5. Find $\lambda $ and $\mu $ if $(2\hat i + 6\hat j + 27\hat k) \times (\hat i + \lambda \hat j + \mu \hat k) = \vec 0$

Ans: $(2\hat i + 6\hat j + 27\hat k) \times (\hat i + \lambda \hat j + \mu \hat k) = \vec 0$

$ \Rightarrow \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ 2&6&{27} \\ 1&\lambda &\mu \end{array}} \right| = 0\hat i + 0\hat j + 0\hat k$

$ \Rightarrow \hat i(6\mu  - 27\lambda ) - \hat j(2\mu  - 27) + \hat k(2\lambda  - 6) = 0\hat i + 0\hat j + 0\hat k$

$6\mu  - 27\lambda  = 0$

$2\mu  - 27 = 0$

$2\lambda  - 6 = 0$

$\lambda  = 3$

$\mu  = \dfrac{{27}}{2}$


6.  Given that $\vec a\;.\;\vec b = 0$ and $\vec a \times \vec b = 0$. What can you conclude about $\vec a$ and $\vec b$ ? 

Ans: $\vec a\;.\;\vec b = 0$

i. $|\vec a| = 0$ or $|\vec b| = 0$ or $\vec a \bot \vec b\quad ($ if $|\vec a| \ne 0$ and $|\vec b| \ne 0)$

$\vec a \times \vec b = 0$

ii. $|\overrightarrow a |\; = \;0\;{\text{or}}\;|\vec b|\; = \;0$ or  (if $|\vec a| \ne 0$ and $|\vec b| \ne 0)$

But $\overrightarrow a $ and $\overrightarrow b $ cannot be parallel and perpendicular at same time.

So, $|\vec a| = 0$ or $|\vec b| = 0$


7. Let the vectors $\vec a,\vec b,\vec c$ 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$

Ans: $(\vec b + \vec c) = \left( {{b_1} + {c_1}} \right)\hat i + \left( {{b_2} + {c_2}} \right)\hat j + \left( {{b_3} + {c_3}} \right)\hat k$

Now, $\vec a \times (\vec b + \vec c)\left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ {{a_1}}&{{a_2}}&{{a_3}} \\ {{b_1} + {c_1}}&{{b_2} + {c_2}}&{{b_3} + {c_3}} \end{array}} \right|$

$ = i\left[ {{a_2}\left( {{b_3} + {c_3}} \right) - {a_3}\left( {{b_2} + {c_2}} \right)} \right] - \hat j\left[ {{a_1}\left( {{b_3} + {c_3}} \right) - {a_3}\left( {{b_1} + {c_i}} \right)} \right] + \hat k\left[ {{a_1}\left( {{b_2} + {c_2}} \right) - {a_2}\left( {{b_1} + {c_1}} \right)} \right]$

$ = i\left[ {{a_2}{b_3} + {a_2}{c_3} - {a_3}{b_2} - {a_3}{c_2}} \right] + \hat j\left[ { - {a_1}{b_3} - {a_1}{c_3} + {a_3}{b_1} + {a_3}{c_1}} \right] + \hat k\left[ {{a_1}{b_2} + {a_1}{c_2} - {a_2}{b_1} - {a_2}{c_i}} \right]$

$\vec a \times \vec b = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ {{a_1}}&{{a_2}}&{{a_3}} \\ {{b_1}}&{{b_2}}&{{b_3}} \end{array}} \right|$

$ = i\left[ {{a_2}{b_3} - {a_3}{b_2}} \right] + \hat j\left[ {{b_1}{a_3} - a{b_3}} \right] + \hat k\left[ {{a_1}{b_2} - {a_2}h} \right]$

$\vec a \times \vec c = \left| {\begin{array}{*{20}{l}} {\hat i}&{\hat j}&{\hat k} \\ {{a_1}}&{{a_2}}&{{a_3}} \\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right|$

$ = \hat i\left[ {{a_2}{c_3} - {a_3}{c_2}} \right] + \hat j\left[ {{a_3}{c_1} - {a_1}{c_3}} \right] + \hat k\left[ {a{b_2} - {a_2}b} \right]$

$\left. {(\vec a \times \vec b) + (\vec a \times \vec c) = \hat i\left[ {{a_2}{b_3} + {a_2}{c_3} - {a_3}{b_2} - {a_3}{c_2}} \right] + \hat j{{[{b_1}{a_3} + {a_3}{c_1} - {a_1}{b_3} - {a_i}\} }_3}} \right]$

$ + \hat k\left[ {a,b, + {a_1}c, - a,{b_1} - a,{c_1}} \right]$

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

Hence, proved.


8. If either $\vec a = 0$ or $b = 0$, then $\vec a \times b = 0$.Is the converse true? Justify your answer with an example.

Ans: Let $\vec a = 2\hat i + 3\hat j + 4\hat k\;,\;\vec b = 4\hat i + 6\hat j + 8\hat k\;,\;\vec a \times \vec b = \vec 0$

$\vec a \times \vec b = \left| {\begin{array}{*{20}{l}} {\hat i}&{\hat j}&{\hat k} \\ 2&3&4 \\ 4&6&8 \end{array}} \right| = \hat i(24 - 24) - \hat j(16 - 16) + \hat k(12 - 12) = 0\hat i + 0\hat j + 0\hat k$

$|\vec a| = \sqrt {{2^2} + {3^2} + {4^2}}  = \sqrt {29} $

$\therefore \vec a \ne \vec 0$

$|\vec b| = \sqrt {{4^2} + {6^2} + {8^2}}  = \sqrt {116} $

$\therefore \vec b \ne \vec 0$

Hence, the converse of the statement need not be true.


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

 Ans: $\overrightarrow {{\text{AB}}}  = (2 - 1)\hat i + (3 - 1)\hat j + (5 - 2)\hat k = \hat i + 2\hat j + 3\hat k$

$\overrightarrow {{\text{BC}}}  = (1 - 2)\hat i + (5 - 3)\hat j + (5 - 5)\hat k =  - \hat i + 2\hat j$

Area $ = \dfrac{1}{2}|\overrightarrow {{\text{AB}}}  \times \overrightarrow {{\text{BC}}} |$

$\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{BC}}} = \left| {\begin{array}{*{20}{l}} {\hat i}&{\hat j}&{\hat k} \\ 1&2&3 \\ { - 1}&2&0 \end{array}} \right| = \hat i( - 6) - \hat j(3) + \hat k(2 + 2) = - 6\hat i - 3\hat j + 4\hat k$

$|\overrightarrow {{\text{AB}}}  \times \overrightarrow {{\text{BC}}} | = \sqrt {{{( - 6)}^2} + {{( - 3)}^2} + {4^2}}  = \sqrt {36 + 9 + 16}  = \sqrt {61} $

So, the area of $\Delta {\text{ABC}}$ is $\dfrac{{\sqrt {61} }}{2}$ sq units.


10.  Find the area of the parallelogram whose adjacent sides are determined by the vector $\vec a = \hat i - \hat j + 3\hat k$ and $\overrightarrow {\text{b}}  = 2\hat i - 7\hat j + \hat k$

Ans: $\vec a \times \overrightarrow {\text{b}} = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ 1&{ - 1}&3 \\ 2&{ - 7}&1 \end{array}} \right| = \hat i( - 1 + 21) - \hat j(1 - 6) + \hat k( - 7 + 2) = 20\hat i + 5\hat j - 5\hat k$

$\therefore |\vec a \times \vec b| = \sqrt {{{20}^2} + {5^2} + {5^2}}  = \sqrt {400 + 25 + 25}  = 15\sqrt 2 $

So, area of parallelogram is $15\sqrt 2 $ sq units


11. Let the vectors $\vec a$ and $\overrightarrow {\text{b}} $ be such that $|\vec a| = 3$ and $|\vec b| = \dfrac{{\sqrt 2 }}{3}$, then $\vec a \times \overrightarrow {\text{b}} $ is a unit vector, if the angle between $\vec a$ and $\vec b$is:

a. $\dfrac{\pi }{6}$

b. $\dfrac{\pi }{4}$

c. $\dfrac{\pi }{3}$

d. $\dfrac{\pi }{2}$

Ans:$|\vec a \times \vec b| = 1$

$ \Rightarrow ||a||\vec b|\sin \theta | = 1$

$ \Rightarrow |\vec a||\overrightarrow {\text{b}} ||\sin \theta | = 1$

$ \Rightarrow 3 \times \dfrac{{\sqrt 2 }}{3} \times \sin \theta  = 1$

$ \Rightarrow \sin \theta  = \dfrac{1}{{\sqrt 2 }}$

$ \Rightarrow \theta  = \dfrac{\pi }{4}$


12. Area of a rectangle having vertices ${\text{A}},{\text{B}},{\text{C}}$, and ${\text{D}}$ with position vectors $ - \hat i + \dfrac{1}{2}\hat j + 4\hat k,\hat i + \dfrac{1}{2}\hat j + 4\hat k,\hat i - \dfrac{1}{2}\hat j + 4\hat k$ and $ - \hat i - \dfrac{1}{2}\hat j + 4\hat k$ respectively is

a. $\dfrac{1}{2}$

b. 1

c. 2

d. 4

Ans: $\overrightarrow {{\text{AB}}}  = (1 + 1)\hat i + \left( {\dfrac{1}{2} - \dfrac{1}{2}} \right)\hat j + (4 - 4)\hat k = 2\hat i$

$\overrightarrow {{\text{BC}}}  = (1 - 1)\hat i + \left( { - \dfrac{1}{2} - \dfrac{1}{2}} \right)\hat j + (4 - 4)\hat k =  - \hat j$

$\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{BC}}} = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ 2&0&0 \\ 0&{ - 1}&0 \end{array}} \right| = \hat k( - 2) = - 2\hat k$

$|\overrightarrow {{\text{AB}}}  \times \overrightarrow {{\text{BC}}} | = \sqrt {{{( - 2)}^2}}  = 2$

So, the area of the required rectangle is 2 square units.


Miscellaneous Exercise

1. Write down a unit vector in XY-plane, making an angle of ${30^\circ }$ with the positive direction of x axis. 

Ans: Unit vector is $\vec r = \cos \theta \hat i + \sin \theta \hat j$, where $\theta $ is angle with positive ${\text{X}}$ axis. $\vec r = \cos {30^\circ }\hat i + \sin {30^\circ }\hat j = \dfrac{{\sqrt 3 }}{2}\hat i + \dfrac{1}{2}\hat j$


2. Find the scalar components and magnitude of the vector joining the points ${\text{P}}\left( {{x_1},{y_1},{z_1}} \right)$ and ${\text{Q}}\left( {{x_2},{y_2},{z_2}} \right)$

Ans: $\overrightarrow {{\text{PQ}}}  = \left( {{x_2} - {x_1}} \right)\hat i + \left( {{y_2} - {y_1}} \right)\hat j + \left( {{z_2} - {z_1}} \right)\hat k$

$|\overrightarrow {{\text{PQ}}} |\; = \;\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} $


3. A girl walks 4 km towards west, then she walks 3 km in a direction ${30^\circ }$ east of north and stops. Determine the girl's displacement from her initial point of departure. 

Ans:  


Girl's displacement from her initial point of departure


$\overrightarrow {{\text{OA}}}  =  - 4\hat i$

$\overrightarrow {{\text{AB}}}  = \hat i|\overrightarrow {{\text{AB}}} |\cos {60^\circ } + \hat j|\overrightarrow {{\text{AB}}} |\sin {60^\circ }$

$ = \hat i3 \times \dfrac{1}{2} + \hat j3 \times \dfrac{{\sqrt 3 }}{2}$

$ = \dfrac{3}{2}\hat i + \dfrac{{3\sqrt 3 }}{2}\hat j$

$\overrightarrow {{\text{OB}}}  = \overrightarrow {{\text{OA}}}  + \overrightarrow {{\text{AB}}} $

$ = ( - 4\hat i) + \left( {\dfrac{3}{2}\hat i + \dfrac{{3\sqrt 3 }}{2}j} \right)$

$ = \left( { - 4 + \dfrac{3}{2}} \right)\hat i + \dfrac{{3\sqrt 3 }}{2}\hat j$

$ = \left( {\dfrac{{ - 8 + 3}}{9}} \right)\hat i + \dfrac{{3\sqrt 3 }}{9}\hat j$

$ = \dfrac{{ - 5}}{2}\vec i + \dfrac{{3\sqrt 3 }}{2}\hat j$


4. If $\overrightarrow a  = \vec b + \vec c$, then is it true that $|\overrightarrow a | = |\vec b| + |\vec c|$ ? Justify your answer. 

Ans:  $\operatorname{In} \Delta {\text{ABC}},\overrightarrow {\;CB}  = \vec a,\;\overrightarrow {CA}  = b,\;\overrightarrow {AB}  = \vec c$

$\vec a = \vec b + \vec c$,  by triangle law of addition for vectors.

$|\vec a|\; < \;|\vec b| + |\vec c|$,  by triangle inequality law of lengths.

 Hence, it's not true that $|\vec a| = |\vec b| + |\vec c|$


5. Find the value of $x$ for which $x(\hat i + \hat j + \hat k)$ unit vector.

Ans: $|x(\hat i + \hat j + \hat k)| = 1$

$ \Rightarrow \sqrt {{x^2} + {x^2} + {x^2}}  = 1$

$ \Rightarrow \sqrt {3{x^2}}  = 1$

$ \Rightarrow \sqrt 3 x = 1$

$ \Rightarrow x =  \pm \dfrac{1}{{\sqrt 3 }}$


6.  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 $\vec b = \hat i - 2\hat j + \hat k$

Ans: $\vec c = \vec a + \vec b = (2 + 1)\hat i + (3 - 2)\hat j + ( - 1 + 1)\hat k = 3\hat i + \hat j$

$|\vec c| = \sqrt {{3^2} + {1^2}}  = \sqrt {9 + 1}  = \sqrt {10} $

$\therefore \hat c = \dfrac{{\vec c}}{{|\vec c|}} = \dfrac{{(3\hat i + \hat j)}}{{\sqrt {10} }}$

So, a vector of magnitude 5 and parallel to the resultant of $\vec a$ and $\vec b$ is $ \pm 5(\hat c) =  \pm 5\left( {\dfrac{1}{{\sqrt {10} }}(3\hat i + \hat j)} \right) =  \pm \dfrac{{3\sqrt {10} }}{2}\hat i \pm \dfrac{{\sqrt {10} }}{2}\hat j$


7.  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 $2\vec a - \vec b + 3\vec c$.

Ans: $2\vec a - \vec b + 3\vec c = 2(\hat i + \hat j + \hat k) - (2\hat i - \hat j + 3\hat k) + 3(\hat i - 2\hat j + \hat k)$

$ = 2\hat i + 2\hat j + 2\hat k - 2\hat i + j - 3\vec k + 3\hat i - 6\hat j + 3\hat k$

$ = 3\hat i - 3\hat j + 2\hat k$

$|2\vec a - \vec b + 3c| = \sqrt {{3^2} + {{( - 3)}^2} + {2^2}}  = \sqrt {9 + 9 + 4}  = \sqrt {22} $

Thus ,required unit vector is $\dfrac{{2\vec a - \vec b + 3\vec c}}{{|2\vec a - \vec b + 3\vec c|}} = \dfrac{{3\hat i - 3\hat j + 2\hat k}}{{\sqrt {22} }} = \dfrac{3}{{\sqrt {22} }}\hat i - \dfrac{3}{{\sqrt {22} }}\hat j + \dfrac{2}{{\sqrt {22} }}\hat k$


8. Show that the points ${\text{A}}(1, - 2, - 8),{\text{B}}(5,0, - 2)$ and ${\text{C}}(11,3,7)$ are collinear, and find the ratio in which B divides Ac. 

Ans: $\overrightarrow {{\text{AB}}}  = (5 - 1)\hat i + (0 + 2)\hat j + ( - 2 + 8)\hat k = 4\hat i + 2\hat j + 6\hat k$

$\overrightarrow {{\text{BC}}}  = (11 - 5)\hat i + (3 - 0)\hat j + (7 + 2)\hat k = 6\hat i + 3\hat j + 9\hat k$

$\overrightarrow {AC}  = (11 - 1)\hat i + (3 + 2)\hat j + (7 + 8)\hat k = 10\hat i + 5\hat j + 15\hat k$

$|\overrightarrow {{\text{AB}}} | = \sqrt {{4^2} + {2^2} + {6^2}}  = \sqrt {16 + 4 + 36}  = \sqrt {56}  = 2\sqrt {14} $

$|\overrightarrow {{\text{BC}}} | = \sqrt {{6^2} + {3^2} + {9^2}}  = \sqrt {36 + 9 + 81}  = \sqrt {126}  = 3\sqrt {14} $

$|\overrightarrow {AC} | = \sqrt {{{10}^2} + {5^2} + {{15}^2}}  = \sqrt {100 + 25 + 225}  = \sqrt {350}  = 5\sqrt {14} $

$\therefore \;\;|\overrightarrow {{\text{AC}}} | = |\overrightarrow {{\text{AB}}} | + |\overrightarrow {{\text{BC}}} |$

So, the points are collinear.

Let B divide AC in ratio $\lambda :1$. $\overrightarrow {{\text{OB}}}  = \dfrac{{\lambda \overrightarrow {{\text{OC}}}  + \overrightarrow {{\text{OA}}} }}{{(\lambda  + 1)}}$

$ \Rightarrow 5\hat i - 2\hat k = \dfrac{{\lambda (11\hat i + 3\hat j + 7\hat k) + (\hat i - 2\hat j - 8\hat k)}}{{\lambda  + 1}}$

$ \Rightarrow (\lambda  + 1)(5\hat i - 2\hat k) = 11\lambda \hat i + 3\lambda \hat j + 7\lambda \hat k + \hat i - 2\hat j - 8\hat k$

$ \Rightarrow 5(\lambda  + 1)\hat i - 2(\lambda  + 1)\hat k = (11\lambda  + 1)\hat i + (3\lambda  - 2)\hat j + (7\lambda  - 8)\hat k$

$ \Rightarrow \lambda  = \dfrac{2}{3}$

So, the required ratio is 2:3


9. 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. 

Ans: $\overrightarrow {{\text{OP}}}  = 2\vec a + \vec b,\overrightarrow {{\text{OQ}}}  = \overrightarrow {\text{a}}  - 3\overrightarrow {\text{b}} $

$\overrightarrow {OR}  = \dfrac{{2(2\vec a + \vec b) - (\vec a - 3\vec b)}}{{2 - 1}} = \dfrac{{4\vec a + 2\vec b - \vec a - 3\vec b}}{1} = 3\vec a + 5\vec b$

So, the position vector of ${\text{R}}$ is $3\overrightarrow {\text{a}}  + 5\overrightarrow {\text{b}} $

Position vector of midpoint of ${\text{RQ}} = \dfrac{{\overrightarrow {{\text{OQ}}}  + \overrightarrow {{\text{OR}}} }}{2}$

$ = \dfrac{{(a\sqrt 6 ) + (3\vec a + 5\bar b)}}{2}$

$ = 2\vec a + \vec b$

$ = \overrightarrow {OP} $

Thus, ${\text{P}}$ is midpoint of line segment ${\text{RQ}}$


10. 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. 

Ans:  Diagonal of a parallelogram is $\vec a + \vec b$ 

$\vec a + \vec b = (2 + 1)\hat i + ( - 4 - 2)\hat j + (5 - 3)\hat k = 3\hat i - 6\hat j + 2\hat k$

So, the unit vector parallel to diagonal is $\dfrac{{\vec a + \vec b}}{{|\vec a + \vec b|}} = \dfrac{{3\hat i - 6\hat j + 2\hat k}}{{\sqrt {{3^2} + {{( - 6)}^2} + {2^2}} }} = \dfrac{{3\hat i - 6\hat j + 2\hat k}}{{\sqrt {9 + 36 + 4} }} = \dfrac{{3\hat i - 6\hat j + 2\hat k}}{7} = \dfrac{3}{7}\hat i - \dfrac{6}{7}\hat j + \dfrac{2}{7}\hat k$

$\vec a \times \vec b = \left| {\begin{array}{*{20}{r}} {\hat i}&{\hat j}&k \\ 2&{ - 4}&3 \\ 1&{ - 2}&{ - 3} \end{array}} \right|$

$ = \hat i(12 + 10) - \hat j( - 6 - 5) + \bar k( - 4 + 4)$

$ = 22\hat i + 11\hat j$

$ = 11(2\hat i + \hat j)$

$\therefore |\vec a \times \vec b| = 11\sqrt {{2^2} + {1^2}}  = 11\sqrt 5 $

So, area of parallelogram is $11\sqrt 5 $ sq units


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

Ans: Let a vector be equally inclined to ${\text{OX}},{\text{OY}}$, and ${\text{OZ}}$ at an angle $\alpha $.

So, the Direction Cosines of the vector are $\cos \alpha ,\cos \alpha $and $\cos \alpha $.

${\cos ^2}\alpha  + {\cos ^2}\alpha  + {\cos ^2}\alpha  = 1$

$ \Rightarrow 3{\cos ^2}\alpha  = 1$

$ \Rightarrow \cos \alpha  = \dfrac{1}{{\sqrt 3 }}$

So, the DCs of the vector are $\dfrac{1}{{\sqrt 3 }},\dfrac{1}{{\sqrt 3 }},\dfrac{1}{{\sqrt 3 }}$.


12. Let $\vec a = \hat i + 4\hat j + 2\hat k$ and $\vec b = 3\hat i - 2\hat j + 7\hat k$ and $\vec c = 2\hat i - \hat j + 4\hat k$. Find a vector $\vec d$ which is

perpendicular to both $\vec a$ and $\vec b$ and $\vec c \cdot \vec d = 15$ 

Ans: $\vec d = {d_1}\hat i + {d_2}\hat \jmath  + {d_3}\hat k$

$\vec d\;.\;\vec a = 0 \Rightarrow {d_1} + 4{d_2} + 2{d_3} = 0$

$\vec d\;.\;\vec b = 0 \Rightarrow 3{d_1} - 2{d_2} + 7{d_3} = 0$

$\vec c \cdot \vec d = 15 \Rightarrow 2{d_1} - {d_2} + 4{d_3} = 15$

Solving these equations, we get ${d_1} = \dfrac{{160}}{3},{d_2} =  - \dfrac{5}{3},{d_3} =  - \dfrac{{70}}{3}$

$\therefore \vec d = \dfrac{{160}}{3}\hat i + \dfrac{5}{3}\widehat j + \dfrac{{70}}{3}\hat k = \dfrac{1}{3}(160\hat i + 5\hat j + 70\hat k)$


13. The scalar product of the vector $\hat i + \hat j + \hat k$ with a unit vector along the sum of vectors $2\hat i + 4\hat j - 5\hat k$ änd $\lambda \hat i + 2\hat j + 3\hat k$ is equal to one. Find the value of $\lambda $. 

Ans: $(2\hat i + 4\hat j - 5\hat k) + (\lambda \hat i + 2\hat j + 3\hat k) = (2 + \lambda )\hat i + 6\hat j - 2\hat k$

So, unit vector along $(2\hat i + 4\hat j - 5\hat k) + (\lambda \hat i + 2\hat j + 3\hat k)$ is $\left( {\dfrac{{(2 + \lambda )\hat i + 6\hat j - 2\hat k}}{{\sqrt {{\lambda ^2} + 4\lambda  + 44} }}} \right)$

$(\hat i + \hat j + \hat k) \cdot \left( {\dfrac{{(2 + \lambda )\hat i + 6\hat j - 2\hat k}}{{\sqrt {{\lambda ^2} + 4\lambda  + 44} }}} \right) = 1$

$ \Rightarrow \dfrac{{(2 + \lambda ) + 6 - 2}}{{\sqrt {{\lambda ^2} + 4\lambda  + 44} }} = 1$

$ \Rightarrow \sqrt {{\lambda ^2} + 4\lambda  + 44}  = \lambda  + 6$

$ \Rightarrow {\lambda ^2} + 4\lambda  + 44 = {(\lambda  + 6)^2}$

$ \Rightarrow {\lambda ^2} + 4\lambda  + 44 = {\lambda ^2} + 12\lambda  + 36$

$ \Rightarrow 8\lambda  = 8$

$ \Rightarrow \lambda  = 1$


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

Ans: $\vec a\;.\;\vec b = \vec b\;.\;\vec c = \vec c\;.\;\vec a = 0$

$|\vec a| = |\vec b| = |\vec c|$

Let $\vec a + \vec b + \vec c$ be inclined to $\vec a,\vec b,\vec c$ at angles ${\theta _1},{\theta _2},{\theta _3}$ respectively. $\cos {\theta _1} = \dfrac{{(\vec a + \vec b + \vec c) \cdot \vec a}}{{|\vec a + \vec b + \vec c||\vec a\mid }} = \dfrac{{\vec a \cdot \vec a + \vec b \cdot \vec a + \vec c \cdot \vec a}}{{|\vec a + \vec b + \vec c||\vec a\mid }} = \dfrac{{|\vec a{|^2}}}{{|\vec a + \vec b + \vec c||\vec a|}} = \dfrac{{|\vec a|}}{{|\vec a + \vec b + \vec c|}}$

$\cos {\theta _2} = \dfrac{{(\vec a + \vec b + \vec c)\vec b}}{{|\vec a + \vec b + \vec c||\vec b\mid }} = \dfrac{{\vec a\vec b + \vec b\vec b + \vec c\vec b}}{{|\vec a + \vec b + \vec c||\vec b\mid }} = \dfrac{{|\vec b{|^2}}}{{|\vec a + \vec b + \vec c||\vec b|}} = \dfrac{{|\vec b|}}{{|\vec a + \vec b + \vec c|}}$

$\operatorname{os} {\theta _3} = \dfrac{{(\vec a + \vec b + \vec c) \cdot \vec c}}{{|\vec a + \vec b + \vec c||\bar c|}} = \dfrac{{\vec a\vec c + \vec b\vec c + \vec c\vec c}}{{|\vec a + \vec b + \vec c||\vec c\mid }} = \dfrac{{|\vec c{|^2}}}{{|\vec a + \vec b + \vec c||\vec c|}} = \dfrac{{|\vec c|}}{{|\vec a + \vec b + \vec c|}}$

Since, $|\vec a| = |\vec b| = |\vec c|\; \Rightarrow \;\cos {\theta _1} = \cos {\theta _2} = \cos {\theta _3}$, So, ${\theta _1} = {\theta _2} = {\theta _3}$


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

Ans: $(\vec a + \vec b) \cdot (\vec a + \vec b) = |\vec a{|^2} + |\vec b{|^2}$

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

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

$ \Rightarrow 2\vec a\;.\;\vec b = 0$

$ \Rightarrow \vec a\;.\;\vec b = 0$

So $\vec a$ and $\vec b$ are perpendicular.


16. If $\theta $ is the angle between two vectors $\vec a$ and $\vec b$, then $\vec a\vec b \geqslant 0$ only when

a. $0 < \theta  < \dfrac{\pi }{2}$

b. $0 \leqslant \theta  \leqslant \dfrac{\pi }{2}$

c. $0 < \theta  < \pi $

d. $0 \leqslant \theta  \leqslant \pi $

Ans: $\therefore \vec a\vec b \geqslant 0$

$ \Rightarrow |\vec a||\vec b|\cos \theta  \geqslant 0$

$ \Rightarrow \cos \theta  \geqslant 0\quad \because [|\vec a| \geqslant 0$ and $|\vec b| \geqslant 0]$

$ \Rightarrow 0 \leqslant \theta  \leqslant \dfrac{\pi }{2}$

$\vec a \cdot \vec b \geqslant 0$ if $0 \leqslant \theta  \leqslant \dfrac{\pi }{2}$

So the right answer is B


17. Let $\vec a$ and $\vec b$ be two unit vectors and $\theta $ is the angle between them. Then $\vec a + \vec b$ is a unit vector if

a. $\theta  = \dfrac{\pi }{4}$

b. $\theta  = \dfrac{\pi }{3}$

c. $\theta  = \dfrac{\pi }{2}$

d. $\theta  = \dfrac{{2\pi }}{3}$

Ans: $|\vec a| = |\vec b| = 1$

$|\vec a + \vec b| = 1$

$ \Rightarrow (\vec a + \overrightarrow b )(\vec a + \overrightarrow b ) = 1$

$ \Rightarrow \vec a \cdot \vec a + \vec a\;.\;\vec b + \vec b\;.\;\vec a + \vec b\;.\;\vec b = 1$

$ \Rightarrow |\vec a{|^2} + 2\vec a\;.\;\vec b + |\overrightarrow b {|^2} = 1$

$ \Rightarrow {1^2} + 2|\vec a||\vec b|\cos \theta  + {1^2} = 1$

$ \Rightarrow 1 + 2.1.1\cos \theta  + 1 = 1$

$ \Rightarrow \cos \theta  =  - \dfrac{1}{2}$

$ \Rightarrow \theta  = \dfrac{{2\pi }}{3}$

So, $\vec a + \vec b$ is unit vector if $\theta  = \dfrac{{2\pi }}{3}$ 

The correct answer is D


18. The value of $\hat i.(\hat j \times \hat k) + \hat j \cdot (\hat i \times \hat k) + \hat k \cdot (\hat i \times \hat j)$ is

$\begin{array}{*{20}{l}} {{\text{ a. }}0}&{{\text{ b. }} - 1}&{{\text{ c. }}1}&{{\text{ d. }}3} \end{array}$

Ans:

$\hat i \cdot (\hat j \times \hat k) + \hat j \cdot (\hat i \times \hat k) + \hat k \cdot (\hat i \times \hat j)$

$ = \hat i\;.\;\hat i + \hat j \cdot ( - \hat j) + \hat k\;.\;\hat k$

$ = 1 - 1 + 1$

$ = 1$

The correct answer is C.


19. If $\theta $ is the angle between any two vectors $\vec a$ and $\vec b$, then $|\vec a\vec b| = |\vec a \times \vec b|$ when $\theta $ is equal to

a. 0 

b. $\dfrac{\pi }{4}$

c. $\dfrac{\pi }{2}$

d. n

Ans: $|\vec a\vec b| = |\vec a \times \vec b|$

$ \Rightarrow |\vec a|\vec b|\cos \theta  = |\vec a||\vec b\mid \sin \theta $

$ \Rightarrow \cos \theta  = \sin \theta $

$ \Rightarrow \tan \theta  = 1$

$ \Rightarrow \theta  = \dfrac{\pi }{4}$

The correct answer is B


NCERT Solutions for Class 12 Maths PDF Download

 

NCERT Solution Class 12 Maths of Chapter 10 All Exercises

Chapter 10 - Vector Algebra Exercises in PDF Format

Exercise 10.1

5 Questions & Solutions (5 Short Answers)

Exercise 10.2

19 Questions & Solutions (5 Short Answers, 14 Long Answers)

Exercise 10.3

18 Questions & Solutions (5 Short Answers, 13 Long Answers)

Exercise 10.4

12 Questions & Solutions (4 Short Answers, 8 Long Answers)


NCERT Solutions for Class 12 Maths Chapter 10 - Vector Algebra

NCERT Solutions for Class 12 Maths – Free PDF Download

Students who are in search of Vector Algebra class 12 CBSE NCERT Solutions need to open the link available at Vedantu website. Complete data and explanation of the topic will be available. This has been done for better understanding and clarity of the topic for students. Students will be able to understand the topic in a much easier and quicker way.


Important Points

  • A vector is defined as a quantity that has a magnitude as well as direction.

  • According to the triangle law of vector addition, “If two vectors are represented by two sides of a triangle taken in order, then their sum or resultant is given by the third side taken in opposite order”.

  • The dot product or the scalar product of any two vectors x and y having an angle θ between them is defined as x.y = |x||y|cos θ.

  • The vector or cross product of two given vectors p and q having an angle θ between them is defined as p x q = |p||q|sin θ.


Chapter 10 – Vector Algebra

NCERT Solutions for Class 12 Maths Chapter 10

10.1 Introduction

In Vector Algebra Class 12, you will be able to learn about the concept, operations and algebraic and geometric properties of vectors. There are various quantities connected with vectors like displacement, velocity acceleration, force, weight, momentum, electric field intensity, etc. You have the real-life application of the vector. For example, it is used to store positions, directions and velocities in games. By combining algebraic and geometric properties, you will be able to learn vital applicability of vectors in different areas.

 

10.2 Some Basic Concepts

Any quantity or object that has both magnitude and direction is known as a vector. You can take an example of a directed line segment AB, where the initial point is A from where the vector starts and the terminating point is B where it ends. The distance between the initial point and the terminating point is known as the magnitude of a vector. The direction of the arrow is represented by the arrow.

  • Positional Vector: In a set of three-dimensional coordinates, the distance between two points is called a position vector.

  • Direction Cosines: The cosine values of the angles made by the vector in a three-dimensional plane is called direction cosines.

 

10.3 Types of Vectors

Zero Vector: A vector with magnitude 0 is called a zero vector. Here the initial and terminating point of the vector co-occurs together and thus gives no direction.


Unit Vector: A vector whose magnitude is equivalent to its unit length i.e. 1 is known as a unit vector.


Coinitial Vectors: Two or more vectors initiating from the same point are known as coinitial vectors.


Collinear Vectors: Two or more vectors are referred to as collinear if they are parallel to the same line, irrespective of their magnitudes and directions.


Equal Vectors: Irrespective of the initial points, if two or more vectors have equal magnitude and directions, they are known as equal vectors.


Negative of a Vector: A vector with the same magnitude but a different direction that of a given vector is known as a negative vector of the given vector.


10.4 Addition of Vectors

For the addition of vectors, you have two laws:

1. Triangle Law of Vector Addition: According to this law, the initial point of one vector should coincide with the terminating point of the other vector.

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So it can be concluded that: \[\overrightarrow{AC}\]= \[\overrightarrow{AB}\]+ \[\overrightarrow{BC}\]

2. Parallelogram Law of Vector Addition: According to this law, two vectors \[\overrightarrow{a}\] and \[\overrightarrow{b}\] representing adjacent sides of a parallelogram, when added, results into the diagonal of the parallelogram.

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\[\overrightarrow{OA}\] + \[\overrightarrow{OB}\]= \[\overrightarrow{OC}\]

 

Properties of Vector Addition

1. Commutative Property: For any two vectors  \[\overrightarrow{a}\] and \[\overrightarrow{b}\],

\[\overrightarrow{a}\] + \[\overrightarrow{b}\] = \[\overrightarrow{b}\] + \[\overrightarrow{a}\]

2. Associative Property: For any three vector \[\overrightarrow{a}\],  \[\overrightarrow{b}\] and  \[\overrightarrow{c}\],

(\[\overrightarrow{a}\]  +  \[\overrightarrow{b}\]) +  \[\overrightarrow{c}\]  = \[\overrightarrow{a}\]  +  (\[\overrightarrow{b}\]  +  \[\overrightarrow{c}\] ) 

3. Additive Identity: For any vector a,

\[\overrightarrow{a}\]  +  \[\overrightarrow{0}\]  =  \[\overrightarrow{0}\]  + \[\overrightarrow{a}\]  =  \[\overrightarrow{a}\]

 

10.5 Multiplication of a Vector by a Scalar

Incase of multiplication of vector \[\overrightarrow{(a)}\] by a scalar say (\[\gamma\] ):

│\[\gamma\] │\[\overrightarrow{(a)}\]││ = | \[\overrightarrow{\gamma \;a}\] |

The direction of the vector \[\overrightarrow{\gamma \;a}\]  depends upon the value of “ \[\gamma\] ” .If “ \[\gamma\] ” is positive, the direction will be the same but if “ \[\gamma\] ” is negative, the direction will be opposite.

This can be related to the additive inverse where,

\[\overrightarrow{a}\] + \[\overrightarrow{(-a)}\] = \[\overrightarrow{(-a)}\] + \[\overrightarrow{a}\] = \[\overrightarrow{0}\]

 

10.5.1 Components of a Vector

r = x î + yĵ + z k̂

The above form of any vector is known as the component form of a vector in a 3D plane.

Where, 

OA  =  x î

OB  = yĵ

OC  = z k̂

Magnitude of such a vector can be calculated by using pythagoras theorem:

r =  x î + yĵ + z k̂

⇒|r| = √( x2 + y2  + z2)

 

10.5.2 Vector Joining Two Points

A vector obtained by joining two points P1 and P2 in a plane will be P1P2.

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In order to obtain magnitude (P1P2) you can apply the triangle law:

\[\overrightarrow{OP_{1}}\] + \[\overrightarrow{P_{1} \; P_{2}}\] =  \[\overrightarrow{OP_{2}}\]

(or)

\[\overrightarrow{P_{1} \; P_{2}}\] =  \[\overrightarrow{OP_{2}}\] - \[\overrightarrow{OP_{1}}\]

That 

P1 P2  = (x2 î + y2 ĵ + z2 k̂)  - (x1 î + y1 ĵ + z1 k̂)

= ( x2 – x1) î + ( y2 – y1 ) ĵ + ( z2 – z1 ) k̂

Thus, the above equation represents the vector P1P2. Its magnitude can be given by:

P1 P2  = √ [( x2−x1 )2 + ( y2−y1 )2 + ( z2−z1 )2]

 

10.5.3 Section Formula

Section formula is a method to find the position of a line segment that divides a vector externally or internally in a given ratio.

CASE 1: When PQ is divided internally by R

You can assume that PQ is divided by R in the ratio of m:n.

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\[\overrightarrow{OR}\] =  \[\frac{mb+na}{mn}\]

CASE 2: When PQ is divided externally by R

You can assume that PQ is divided by R in the ratio of m:n.

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\[\overrightarrow{OR}\] =  \[\frac{mb-na}{mn}\]

 

10.6 Product of Two Vectors

In this topic, you will understand the multiplication( product) of two vectors. Two vectors can be multiplied by two methods. First one is a Scalar (or dot) product where the result is a scalar, whereas in the second one is a Vector (or cross) product of two vectors where the result is a vector.

 

10.6.1 Scalar (or dot) Product of Two Vectors

In these two non zero vectors, a and b are multiplied with respect to cos which denotes the angle between both the vectors.

The scalar product can be written as:

a.b=|a||b| cosθ

Property 1: Distributivity of scalar product over addition.

Let  \[\overrightarrow{a}\], \[\overrightarrow{b}\] and \[\overrightarrow{c}\] be  any three vectors, then:

\[\overrightarrow{a}\] . ( \[\overrightarrow{b}\] + \[\overrightarrow{c}\] ) = \[\overrightarrow{a}\] . \[\overrightarrow{b}\] + \[\overrightarrow{a}\] . \[\overrightarrow{c}\]

Property 2:  Let a and b be any two vectors, and l be any scalar. Then:

\[\overrightarrow{(\gamma a)}\] . \[\overrightarrow{b}\] = \[\overrightarrow{(\gamma a)}\] . \[\overrightarrow{b}\] = \[\overrightarrow{\gamma}\] \[\overrightarrow{(a \; . \; b)}\] = \[\overrightarrow{a}\] . \[\overrightarrow{(\gamma b)}\]

 

10.6.2 Projection of a Vector on a Line

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When a vector \[\overrightarrow{AB}\] with a given directed line “l” makes an angle in an anticlockwise direction, the vector \[\overrightarrow{p}\] formed will be termed as the projection vector and its magnitude will be termed as the projection of the vector \[\overrightarrow{AB}\] on the directed line “l”.

 

10.6.3 Vector (or Cross) Product of Two Vectors

In this two non zero vectors \[\overrightarrow{a}\] and \[\overrightarrow{b}\] are multiplied with respect to sin which denotes the angle between both the vectors and ň is a unit vector perpendicular to both \[\overrightarrow{a}\] and \[\overrightarrow{b}\], in order to form a right-handed system wherein the right-handed system rotated from \[\overrightarrow{a}\] to \[\overrightarrow{b}\] moves in the direction of ň.

The vector product can be written as:

\[\overrightarrow{a \times b}\] = |\[\overrightarrow{a}\] | |\[\overrightarrow{b}\] | \[\overrightarrow{sin \theta}\]ň

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Key Features of NCERT Solutions for Class 12 Maths Chapter 10

NCERT Solutions for Class 10 Maths offered by Vedantu are the one which can guide you the best as they provide complete and comprehensive details of the topic. Not only for the understanding part but it also provides important practice questions made on CBSE pattern. You can understand the NCERT concept enriched with expert’s knowledge who writes these solutions. The key features to consider are:

  • You get a better mode to clarify your doubts as it offers in-depth knowledge in easy language.

  • It makes the concept easier by simplifying and classifying each and every concept of the topic.

  • Since it is made on the CBSE pattern, it gives you the idea of how to attempt the question paper.

  • After a  lot of research work, these solutions are made to build foundations for your higher studies.

FAQs on NCERT Solutions for Class 12 Maths Chapter 10 - Vector Algebra

1. What are the major topics of Class 12 Maths Chapter 10 as per the latest NCERT syllabus?

As per the latest syllabus of CBSE Class 12 Mathematics, following are some of the major topics covered in Chapter 10 Vector Algebra:

1. Introduction
2. Basic Concepts
  • Position Vector
  • Direction Cosines

3. Types of Vectors
  • Zero Vector

  • Unit Vector

  • Coinitial Vectors

  • Collinear Vectors

  • Equal Vectors

  • Negative of a Vector

4. Addition of Vectors
  • Properties of vector addition

5. Multiplication of a Vector by a Scalar
  • Components of a vector

  • Vector joining two points

  • Section formula

6. Product of Two Vectors

  • Scalar or dot product of two vectors

  • Projection of a vector on a line

  • Vector or cross product of two vectors

2. What is the difference between scalar and vector?

The main difference between a vector and a scalar quantity is that a vector quantity has a direction and a magnitude while a scalar has only magnitude. If a quantity has a direction associated with it, it is a vector quantity. However, the scalar quantity is not associated with the direction. Speed is a scalar quantity but velocity is a vector quantity as it specifies both a direction as well as a magnitude. In mathematical terms, a vector is an element of a vector space. Mathematically, we can demonstrate a vector as a directed line segment.

3. What are the important takeaways for a student from the 10th chapter of NCERT Class 12 Mathematics?

Students will understand how to differentiate between a scalar and vector quantity with the help of Class 12 Maths Chapter 10 Vector Algebra. The important things that a student will learn from this chapter are the magnitude and direction of a vector, direction cosines and direction ratios of a vector, types of vectors (equal, unit, zero, parallel and collinear vectors), position vector of a point, negative of a vector, components of a vector, etc. Students will also be able to perform addition of vectors, multiplication of a vector by a scalar and position vector of a point dividing a line segment in a given ratio. Students will get familiarized with the definition of Vectors, geometrical interpretation of vectors, properties and application of scalar (dot) product of vectors, vector (cross) product of vectors, the scalar triple product of vectors, etc.

4. Why should I refer to NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra?

Vector Algebra is an important branch of mathematics that deals with quantities having directions. The concepts of vector algebra are also useful in various topics of physics. Hence, students must pay attention to the chapter and solve all the problems given in the exercise. Students facing any doubt in the chapter can refer to NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra by Vedantu. These solutions include step-by-step answers to textbook questions. The expert in-house teachers at Vedantu provide the solutions. Hence, students can rely on them for exam preparation and revision. Students can perform better in exams with the help of NCERT Solutions as it will provide a better understanding of the chapter which is crucial for conceptual chapters like Vector Algebra.

5. What is vector algebra? 

Vector algebra is one of the important topics of algebra. It deals with the algebraic expressions of vector quantities. Vector algebra is an algebra where important symbols, letters, numbers and variables denote vector algebraic expressions. In Mathematics, algebraic operations are performed on vector spaces and vectors. This branch of mathematics has hypotheses and rules based on the behaviour and properties of vectors. On Vedantu, you can understand the various concepts of the chapter with some solved examples. 

6. What are the different types of vectors?

There are six types of vectors. These are- 

  • Zero vector- This type of vector always has a zero magnitude. 

  • Unit vector- A vector whose magnitude is always unity or one. It is denoted by a vector that is equal to the direction of the AB vector. 

  • Coinitial vector- The vectors that have the same initial points. 

  • Parallel or collinear vector- Two or more vectors are said to be collinear if they are parallel to the same line regardless of their directions and magnitudes. 

  • Equal vector- The vectors that have the same direction and magnitude irrespective of the position of their initial points. 

  • Negative of a vector- Vectors that have similar magnitude but have opposite directions. 

7. What are the observations of the scalar product of two vectors? 

The observations of the scalar product of two vectors are- 

  • The dot product of two vectors is a real number. 

  • If a and b vectors are non-zero vectors then, its dot product is zero only if a and b vectors are perpendicular to each other. 

  • If the angle between a and b vectors is zero then, its dot product is equal to the magnitude of the individual product of the vectors. 

  • If the angle between a and b vectors is 180 degrees then, its dot product is equal to the magnitude of the individual product of the vectors but having an opposite direction. 

8. What are the observations of the cross product of two vectors? 

The observations of the cross product of two vectors are as follows- 

  • The cross product of two vectors is always a vector quantity. 

  • If a and b vectors are not equal to zero vectors then, their cross product is equal to the product of the magnitude of individual vectors, sin angle between the two vectors and the normal vector. 

  • If the angle between two vectors, a and b is 90 degrees then sin 90 is equal to 1. Therefore, in this case the vector product is equal to the product of the magnitude of individual vectors and the normal vector. 

9. How can students receive decent marks in Chapter 10 of Class 12 Maths? 

Chapter 10 Vector Algebra is one of the important chapters of Class 12 Maths. Students must learn this chapter very well as it is crucial from the board exam point of view. So, to get good marks in this chapter students have to follow the given tips. 

  • Class 12 students have to take their classes seriously. 

  • They should make notes for this chapter for doing a proper revision. 

  • Students must practise the NCERT Solutions of Chapter 10 Vector Algebra to know the type of questions asked from this chapter in the exams.