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

Class 12 Math Chapter 10 Vector Algebra carries around 6–8 marks in the CBSE board exams and almost the same in the state board exams. As per notification, Vector Algebra is combined with 3D Geometry in the JEE Main syllabus, these two chapters carry a weightage of approximately 10–12% of the mathematics section. Vector Algebra and Three-Dimensional Theory include very important topics vector addition, vector multiplication; cross, and dot product, and many more. Students can check the key topics and important formulae below;

Class 12 Vector Algebra Key Topics

• Basics of Vectors: Representation of vectors in 2D, 3D.
• Types of Vectors: Unit, Position, Co-initial, collinear, and coplanar vectors
• Operations:  Addition, subtraction, and multiplication by a scalar
• Properties of vector addition and scalar multiplication and Scalar Product of Vectors
• Applications: Angle between two vectors, projection of one vector onto another
• Properties of cross-product: Geometric interpretation, Area of a parallelogram, Triangle, 
• Applications: Conditions for coplanarity of three vectors, Vector Equations of lines, and planes

Check out for Class 12 Vector Algebra Dropped Topics – 10.7 Scalar Triple Product, 10.7.1 Coplanarity of Three Vectors.

Class 12 Math Chapter 10 Vector Algebra Important Formulae for CBSE and Competitive Exams

Students can check the important topics below;

Basic Vector Operations

• Position Vector: A point $P(x, y, z)$in 3D space has a position vector:

$$\vec{A}=x\hat{i}+y\hat{j}+z\hat{k}$$

• Addition of Two Vectors: If $\vec{A} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$and $\vec{B} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$then:

$$\vec{A}+\vec{B}=(a_1+b_1)\hat{i}+(a_2+b_2)\hat{j}+(a_3+b_3)\hat{k}$$

• Magnitude of a Vector: For a vector $\vec{A} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$​ $$|\vec{A}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$$
  
•  Dot Product (Scalar Product): If  $\vec{A} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$and $\vec{B} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$, then;

$$\vec{A} \cdot \vec{B} = a_1b_1 + a_2b_2 + a_3b_3$$

• Angle between Two Vectors:

$$\vec{A}\cdot \vec{B}=\mathrm{\mid }\vec{A}\mathrm{\mid }\mathrm{\mid }\vec{B}\mathrm{\mid }\cos \theta$$

 

• Cross Product (Vector Product): If $\vec{A} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$and $\vec{B} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$, then;

$$\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$$

• Magnitude of Cross Product:

$$|\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \sin\theta$$

• Projection of a Vector: 

1. Projection of $\vec{A}$on $\vec{B}$:

   $$\text{Projection}=\frac{\vec{A}\cdot \vec{B}}{\mathrm{\mid }\vec{B}\mathrm{\mid }}$$

2. Vector Component of $\vec{A}$along $\vec{B}$:

   $$\text{Component}=\left(\frac{\vec{A}\cdot \vec{B}}{\mathrm{\mid }\vec{B}{\mathrm{\mid }}^2}\right)\vec{B}$$

• Collinearity of Two Vectors: Two vectors $\vec{A}$ and $\vec{B}$ are collinear if;

$$\vec{A} \times \vec{B} = \vec{0}$$

• Vector Equation of a Line

1. Vector form: A line passing through $\vec{A}$and parallel to $\vec{B}$:

   $$\vec{r}=\vec{A}+t\vec{B},t\in \mathbb{R}$$

2. Cartesian form: A line passing through $(x_1, y_1, z_1)$and parallel to $\vec{B} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$:

   $$\frac{x-x_1}{b_1}=\frac{y-y_1}{b_2}=\frac{z-z_1}{b_3}$$

Shiksha has compiled the complete exercise-wise solution of the Class 12 Vector Algebra NCERT PDF in one place prepared by our experts. Students can use Class 12 Vector Algebra Solutions to build a strong conceptual understanding of Vector and 3D geometry. The Vector Algebra Class 12 NCERT Solution PDF will be useful for both CBSE boards and competitive exams aspirants; JEE Main, CUSAT CAT, BITSAT, etc... Students can access the Vector Algebra Class 12 Solutions pdf download for free below; 

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

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.  $40km,30^0$ 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  $\stackrel{\to }{a}$ and  $\stackrel{\to }{d}$ are co initial same initial point.

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

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

 

Q5. Answer the following as true or false:

(i)   $\stackrel{\to }{a}$  and -  $\stackrel{\to }{a}$  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.

A.5. (i) True, as vector quantity  $\stackrel{\to }{a}$ and -  $\stackrel{\to }{a}$ 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.

Class 12 Chapter 10 Vector Algebra Exercise 10.2 focuses on advanced concepts such as the dot product (scalar product) and its applications in finding angles between vectors, the projection of one vector on another, and more. These concepts are not only crucial for board exams but also form the foundation for higher studies in physics, engineering, and applied mathematics. Vector Algebra Exercise 10.2 Solutions consist of 19 Questions. Students can check the complete Vector Algebra Exercise 10.2 solutions below

Class 12 Vector Algebra Exercise 10.2 Solutions

Q1 Compute the magnitude of the following vectors: $\stackrel{}{}$ $\mathrm{}$

 

Q2. Write two different vectors having same magnitude.

A.2. Two different vectors having same magnitude: -

(i)  $2\hat{i}+\hat{j}+3\hat{k}$

(ii)  $\hat{i}+3\hat{j}+2\hat{k}$

 

Q3. Write two different vectors having same direction.

A.3. Two different vectors having same directions: -

(i)  $\hat{i}+\hat{j}+\hat{k}$

(ii)  $2\hat{i}+2\hat{j}+2\hat{k}$

 

Q4. Find the values of x and y so that the vectors   $2\hat{i}+3\hat{j}andx\hat{i}+y\hat{j}$ are equal.

A.4. Note that two vector are equal only if their corresponding components are equal.

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

 

Q5.  Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (–5, 7).

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

shown as,

$\begin{array}{l}\stackrel{\to }{PQ}=(-5,-2)\hat{i}+(7,-1)\hat{j}\\ \stackrel{\to }{PQ}=-7\hat{i}+6\hat{j}\end{array}$

The scalar components are -7 and 6.

The vector components are -7i and 6j.

 

Q6. Find the sum of the vectors:   $\stackrel{\to }{a}=\hat{i}-2\hat{j}+\hat{k},\stackrel{\to }{b}=-2\hat{i}-4\hat{j}+5\hat{k}and\stackrel{\to }{c}=\hat{i}-6\hat{j}-7\hat{k}$  

A.6. The given vectors are

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

$\stackrel{\to }{b}=\widehat{-2i}+4\hat{j}+5\hat{k}$

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

The sum of the vector is

$\stackrel{\to }{a}+\stackrel{\to }{b}+\stackrel{\to }{c}=\left(a_1+a_2+a_3\right)\hat{i}+\left(b_1+b_2+b_3\right)\widehat{\stackrel{^}{j}+\left(c_1+c_2+c_3\right)\stackrel{^}{k}}$

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

 

Q7. Find the unit vector in the direction of the vector   $\stackrel{\to }{a}=\hat{i}+\hat{j}+2\hat{k}$

 

Q8.  Find the unit vector in the direction of the vector   $\stackrel{\to }{PQ}$  where P and Q. are the points (1, 2, 3) and (4, 5, 6) respectively.

A.8. Given,  $P\left(1,2,3\right)\&Q\left(4,5,6\right)$

So,

$\begin{array}{l}\stackrel{\to }{PQ}=(4-1)\hat{i}+(5-2)\hat{j}+(6-3)\hat{k}\\ =3\hat{i}+3\hat{j}+3\hat{k}\end{array}$

 

Q9. For given vectors   $a=2\hat{i}-\hat{j}+2\hat{k}andb=-\hat{i}+\hat{j}-\hat{k}$ , find the unit vector in the direction of    $\stackrel{\to }{a}+\stackrel{\to }{b}$

A9. Given,  $\stackrel{\to }{a}=2\hat{i}-\hat{j}+2\hat{k}\&\stackrel{\to }{b}=-\hat{i}+\hat{j}-\hat{k}$

The sum of given vectors is given by

 

Q11. Show that the vectors    $2\hat{i}-3\hat{j}+4\hat{k}and\stackrel{\to }{b}=-4\hat{i}+\widehat{6j}-8\hat{k}$  are collinear.

A.11. Let,  $\stackrel{\to }{a}=2\hat{i}-3\hat{j}+4\hat{k}\&\stackrel{\to }{b}=-4\hat{i}+\widehat{6j}-8\hat{k}$

It is seen that

$\begin{array}{l}\stackrel{\to }{b}=-4\hat{i}+\widehat{6j}-8\hat{k}\\ =-2\left(2\hat{i}-3\hat{j}+4\hat{k}\right)\\ =-2\stackrel{\to }{a}\\ \therefore \stackrel{\to }{b}=\lambda \stackrel{\to }{a}\end{array}$

Where,  $\lambda =-2$

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

 

Q12. Find the direction cosines of the vector  $\hat{i}+2\hat{j}+3\hat{k}.$

A.12. Let  $\stackrel{\to }{a}=\hat{i}+2\hat{j}+3\hat{k}$

 

Q13. Find the direction cosines of the vector joining the points A (1, 2, –3) and B (–1, –2, 1) directed from A to B.

A.13. Given, A(1,2,-3) and (-1,-2,1)

Now,

$\begin{array}{l}\left|\stackrel{\to }{AB}\right|=(-1-1)\hat{i}+(-2-2)\hat{j}+\left(1-(-3)\right)\hat{k}\\ =-2\hat{i}-4\hat{j}+4\hat{k}\end{array}$

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

Let, l, m, n be direction cosine,

$l=\frac{x}{\left|\stackrel{\to }{AB}\right|}=\frac{-2}{6}=\frac{-1}{3};m=\frac{y}{\left|\stackrel{\to }{AB}\right|}=\frac{-4}{6}=\frac{-2}{3};n=\frac{z}{\left|\stackrel{\to }{AB}\right|}=\frac{4}{6}=\frac{2}{3}$

Therefore, direction cosine of  $\stackrel{\to }{AB}$ are  $\left(\frac{-1}{3},\frac{-2}{3},\frac{2}{3}\right)$

 

Q14. Show that the vector   $\hat{i}+\hat{j}+\hat{k}$  is equally inclined to the axes OX, OY and OZ.

A.14. Here,

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

Then,

 

Q15. 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.

A.15. (i) The position vector of point R dividing the join of P and Q. internally in

the ratio 2:1 is,

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

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

A15. (ii)

$\begin{array}{l}\stackrel{\to }{OR}=\frac{2\left(-\hat{i}+\hat{j}+\hat{k}\right)-1\left(\hat{i}+\widehat{2j}-\hat{k}\right)}{2-1}\\ =-2\hat{i}+\widehat{2j}+2\hat{k}-\hat{i}-\widehat{2j}+\hat{k}\\ =-3\hat{i}+\hat{k}\end{array}$

 

Q16. Find the position vector of the mid-point of the vector joining the points P (2, 3, 4) and Q. (4, 1, – 2).

A.16. The Position vector of mid-point R of the vector joining point P (2,3,4) and Q (4,1,-2) is given by;

$\begin{array}{l}\stackrel{\to }{OR}=\frac{\left(2\hat{i}+3\hat{j}+4\hat{k}\right)+\left(4\hat{i}+\hat{j}-\widehat{2k}\right)}{2}\\ =\frac{(2+4)\hat{i}+(3+1)\hat{j}+(4-2)\hat{k}}{2}\\ =\frac{6\hat{i}+4\hat{j}+2\hat{k}}{2}=\frac{6\hat{i}}{2}+\frac{4\hat{j}}{2}+\frac{2\hat{k}}{2}\\ =3\hat{i}+2\hat{j}+\hat{k}\end{array}$

 

Q17.  Show that the points A, B and C with position vectors   $\stackrel{\to }{a}=3\hat{i}-4\hat{j}-4\hat{k},\stackrel{\to }{b}=2\hat{i}-\hat{j}+\hat{k}and\stackrel{\to }{c}=\hat{i}-3\hat{j}-5\hat{k}$   respectively form the vertices of a right angled triangle.

A.17. We have,

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

$\begin{array}{l}\stackrel{\to }{AB}=(2-3)\hat{i}+(-1-(-4))\hat{j}+(1-(-4))\hat{k}\\ =-\hat{i}+3\hat{j}+5\hat{k}\\ \stackrel{\to }{BC}=(1-2)\hat{i}+(-3-(-1))\hat{j}+(-5-1)\hat{k}\\ =-\hat{i}-2\hat{j}-6\hat{k}\\ \stackrel{\to }{CA}=(1-3)\hat{i}+(-3-(-4))\hat{j}+(-5-(-4))\hat{k}\\ =-2\hat{i}+\hat{j}-\hat{k}\end{array}$

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

Hence,

${\left|\stackrel{\to }{AB}\right|}^2+{\left|\stackrel{\to }{CA}\right|}^2=35+6=41={\left|\stackrel{\to }{BC}\right|}^2$

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

 

Q18. In triangle ABC (Fig. below), which of the following is not true:

A18. (A)  $\stackrel{\to }{AB}+\stackrel{\to }{BC}+\stackrel{\to }{CA}=\stackrel{\to }{0}$

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

$\begin{array}{l}\stackrel{\to }{AB}+\stackrel{\to }{BC}=\stackrel{\to }{AC}-----(1)\\ \stackrel{\to }{AB}+\stackrel{\to }{BC}=-\stackrel{\to }{CA}\end{array}$

$\stackrel{\to }{AB}+\stackrel{\to }{BC}+\stackrel{\to }{CA}=\stackrel{\to }{0}-----(2)$

So, (A) is true.

(B)  $\stackrel{\to }{AB}+\stackrel{\to }{BC}-\stackrel{\to }{AC}=0$

$\begin{array}{l}\stackrel{\to }{AB}+\stackrel{\to }{BC}=\stackrel{\to }{AC}\\ \stackrel{\to }{AB}+\stackrel{\to }{BC}-\stackrel{\to }{AC}=0\end{array}$

So, (B) is true.

(C)  $\stackrel{\to }{AB}+\stackrel{\to }{BC}-\stackrel{\to }{CA}=0$

$\begin{array}{l}\stackrel{\to }{AB}+\stackrel{\to }{BC}=\stackrel{\to }{CA}-----(3)\\ From,(1)\&(3),\\ \stackrel{\to }{AC}=\stackrel{\to }{CA}\\ \stackrel{\to }{AC}=-\stackrel{\to }{AC}\\ \stackrel{\to }{AC}+\stackrel{\to }{AC}=0\\ 2\stackrel{\to }{AC}=0\end{array}$

$\therefore$ The eQ.uation in alternative C  $\stackrel{\to }{AC}=\stackrel{\to }{0}$ , which is not true, is incorrect.

(D)  $\stackrel{\to }{AB}-\stackrel{\to }{CB}+\stackrel{\to }{CA}=\stackrel{\to }{0}$

$\begin{array}{l}From,eqn(2)wehave\\ \stackrel{\to }{AB}-\stackrel{\to }{CB}+\stackrel{\to }{CA}=0\end{array}$

The, equation given is alternative is D is true.

$\therefore$ The correct answer is C.

 

Q.19. If   $\stackrel{\to }{a}$  and   $\stackrel{\to }{b}$  are two collinear vectors, then which of the following are incorrect:

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

(B)   $\stackrel{\to }{a}$  = ±  $\stackrel{\to }{b}$

(C) The respective components of   $\stackrel{\to }{a}$  and   $\stackrel{\to }{b}$  are proportional.

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

A.19. We know,

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

So,

$\begin{array}{l}\stackrel{\to }{b}=\lambda \stackrel{\to }{a}\\ If,\lambda =\pm 1,then,\stackrel{\to }{a}=\pm \stackrel{\to }{b}\\ If,\stackrel{\to }{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}\\ \stackrel{\to }{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k},then\\ \stackrel{\to }{b}=\lambda \stackrel{\to }{a}\\ b_1\hat{i}+b_2\hat{j}+b_3\hat{k}=\lambda \left(a_1\hat{i}+a_2\hat{j}+a_3\hat{k}\right)\\ =\left(\lambda a_1\right)\hat{i}+\left(\lambda a_2\right)\hat{j}+\left(\lambda a_3\right)\hat{k}\\ b_1=\lambda a_1,b_2=\lambda a_2,b_3=\lambda a_3\\ \frac{b_1}{a_1}=\frac{b_2}{a_2}=\frac{b_3}{a_3}=\lambda \end{array}$

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

Thus, the statement given in D is incorrect.

The correct answer is D.

Class 12 Chapter 10 Vector Algebra Exercise 10.3 focuses on the geometric interpretation of the cross product of vectors, its role in determining the area of parallelograms and triangles, and its significance in three-dimensional space. These concepts help, students in understanding the true nature of vectors and 3D planes. Vector Algebra Exercise 10.3 Solutions involves the solution of 18 Questions. Students can check the complete exercise 10.3 solutions below;

Class 12 Vector Algebra Exercise 10.3 NCERT Solutions

Q1.  Find the angle between two vectors   $\stackrel{\to }{a}$  and   $\stackrel{\to }{b}$  with magnitude √3 and 2 respectively having   $\stackrel{\to }{a}$  .  $\stackrel{\to }{b}$  = √6

A.1. Given,

$\left|\stackrel{\to }{a}\right|=\surd 3$

 

, | b →
= 2 a n d a → . b → =√6

We have,

Now,

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

Also, we know,

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

A.3. Let,

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

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

$\therefore$ The projection of vector  $\stackrel{\to }{a}$ on  $\stackrel{\to }{b}$ is 0.

 

Q4. Find the projection of the vector   $\hat{i}+3\hat{j}+7\hat{k}$  on the vector   $7\hat{i}-\hat{j}+8\hat{k}$

A.4. Let,

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

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

 

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

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

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

 

Q6. Find   $\left|\stackrel{\to }{a}\right|$ and  $\left|\stackrel{\to }{b}\right|$ ,if  $\left(\stackrel{\to }{a}+\stackrel{\to }{b}\right).\left(\stackrel{\to }{a}-\stackrel{\to }{b}\right)=8$ and  $\left|\stackrel{\to }{a}\right|=8\left|\stackrel{\to }{b}\right|$

A.6.  $\left|\stackrel{\to }{a}\right|$ and  $\left|\stackrel{\to }{b}\right|$ ,if  $\left(\stackrel{\to }{a}+\stackrel{\to }{b}\right).\left(\stackrel{\to }{a}-\stackrel{\to }{b}\right)=8$ and  $\left|\stackrel{\to }{a}\right|=8\left|\stackrel{\to }{b}\right|$

$\begin{array}{l}\left(\stackrel{\to }{a}+\stackrel{\to }{b}\right).\left(\stackrel{\to }{a}-\stackrel{\to }{b}\right)=8\\ and\left|\stackrel{\to }{a}\right|=8\left|\stackrel{\to }{b}\right|\\ \left(\stackrel{\to }{a}+\stackrel{\to }{b}\right).\left(\stackrel{\to }{a}-\stackrel{\to }{b}\right)=8\\ \stackrel{\to }{a}.\stackrel{\to }{a}-\stackrel{\to }{a}.\stackrel{\to }{b}+\stackrel{\to }{b}.\stackrel{\to }{a}-\stackrel{\to }{b}.\stackrel{\to }{b}=8\\ {\left|\stackrel{\to }{a}\right|}^2-{\left|\stackrel{\to }{b}\right|}^2=8\\ {\left(8\left|\stackrel{\to }{b}\right|\right)}^2-{\left|\stackrel{\to }{b}\right|}^2=8\\ 64{\left|\stackrel{\to }{b}\right|}^2-{\left|\stackrel{\to }{b}\right|}^2=8\\ 63{\left|\stackrel{\to }{b}\right|}^2=8\\ \left|\stackrel{\to }{b}\right|=\surd 8/\surd 63\left(\therefore magnitudeofavectorisnon-negative\right)\\ \left|\stackrel{\to }{b}\right|=\frac{\mathrm{2\surd 2}}{\mathrm{3\surd 7}}\\ And\\ \left|\stackrel{\to }{a}\right|=8\left|\stackrel{\to }{b}\right|=\frac{2\times \mathrm{2\surd 2}}{\mathrm{3\surd 7}}=\frac{1\mathrm{6\surd 2}}{\mathrm{3\surd 7}}\end{array}$

 

Q7. Evaluate the product   $\left(3\stackrel{\to }{a}-5\stackrel{\to }{b}\right).\left(2\stackrel{\to }{a}+7\stackrel{\to }{b}\right).$

A.7.  $\left(3\stackrel{\to }{a}-5\stackrel{\to }{b}\right).\left(2\stackrel{\to }{a}+7\stackrel{\to }{b}\right).$

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

 

Q8. Find the magnitude of two vectors   $\stackrel{\to }{a}$  and  $\stackrel{\to }{b}$  having the same magnitude such that the angle between them is 60° and their scalar product is 1/2.

A.8. Let  $\theta$ be the angle between the vectors  $\left|\stackrel{\to }{a}\right|$ and  $\left|\stackrel{\to }{b}\right|$ .

It is given that  $\left|\stackrel{\to }{a}\right|=\left|\stackrel{\to }{b}\right|,\stackrel{\to }{a}.\stackrel{\to }{b}=\frac{1}{2}and\theta =60^{\circ }----(1)$

We know,  $\stackrel{\to }{a}.\stackrel{\to }{b}=\left|\stackrel{\to }{a}\right|\left|\stackrel{\to }{b}\right|cos\theta$

$\begin{array}{l}\therefore \frac{1}{2}=\left|\stackrel{\to }{a}\right|\left|\stackrel{\to }{a}\right|cos60^{\circ }(using(1))\\ \frac{1}{2}={\left|\stackrel{\to }{a}\right|}^2\times \frac{1}{2}\\ {\left|\stackrel{\to }{a}\right|}^2=1\\ \left|\stackrel{\to }{a}\right|=\left|\stackrel{\to }{b}\right|=1\end{array}$

$\therefore$ Magnitude of two vector=1

 

Q9. Find   $\left|\stackrel{\to }{x}\right|$  if for a unit vector   $\left(\stackrel{\to }{x}-\stackrel{\to }{a}\right).\left(\stackrel{\to }{x}-\stackrel{\to }{a}\right)=12$ .

A.9. $\begin{array}{l}\left(\stackrel{\to }{x}-\stackrel{\to }{a}\right).\left(\stackrel{\to }{x}-\stackrel{\to }{a}\right)=12\\ \stackrel{\to }{x}.\stackrel{\to }{x}+\stackrel{\to }{x}.\stackrel{\to }{a}-\stackrel{\to }{a}.\stackrel{\to }{x}-\stackrel{\to }{a}.\stackrel{\to }{a}=12\\ {\left|\stackrel{\to }{x}\right|}^2-{\left|\stackrel{\to }{a}\right|}^2=12\\ {\left|\stackrel{\to }{x}\right|}^2-1=12\left[\left|\stackrel{\to }{a}\right|=1asa\stackrel{\to }{a}isunitvector\right]\\ {\left|\stackrel{\to }{x}\right|}^2=12+1=13\\ \therefore \left|\stackrel{\to }{x}\right|=\surd 13\end{array}$

 

Q10. If   $\stackrel{\to }{a}=2\hat{i}+2\hat{j}+3\hat{k},\stackrel{\to }{b}=-\hat{i}+1\stackrel{\to }{j}+\stackrel{\to }{k}andc=3\hat{i}+\hat{j}$  are such that   $\stackrel{\to }{a}+\lambda \stackrel{\to }{b}$ is perpendicular to   $\stackrel{\to }{c}$  then find the value of λ

A.10. Given,

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

Now,

$\begin{array}{l}\stackrel{\to }{a}+\lambda \stackrel{\to }{b}=\left(2\hat{i}+2\hat{j}+3\hat{k}\right)+\lambda \left(-\hat{i}+2\hat{j}+\hat{k}\right)\\ =\left(2\hat{i}+2\hat{j}+3\hat{k}\right)+\left(-\lambda \hat{i}+2\lambda \hat{j}+\lambda \hat{k}\right)\\ =\left(2-\lambda \right)\hat{i}+\left(2+2\lambda \right)\hat{j}+\left(3+\lambda \right)\hat{k}\end{array}$

If  $\left(\stackrel{\to }{a}+\lambda \stackrel{\to }{b}\right)$ is perpendicular to  $\stackrel{\to }{c}$ , then  $\left(\stackrel{\to }{a}+\lambda \stackrel{\to }{b}\right).\stackrel{\to }{c}=0$

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

Therefore, the required value of  $\lambda$ is 8.

 

Q11. Show that   $\left|\stackrel{\to }{a}\right|\stackrel{\to }{b}+\left|\stackrel{\to }{b}\right|$  is perpendicular to   $\left|\stackrel{\to }{a}\right|\stackrel{\to }{b}-\left|\stackrel{\to }{b}\right|\stackrel{\to }{a}$  for any two non-zero vectors   $\stackrel{\to }{a}$  and   $\stackrel{\to }{b}$

A.11.  $\left(\left|\stackrel{\to }{a}\right|\stackrel{\to }{b}+\left|\stackrel{\to }{b}\right|\stackrel{\to }{a}\right).\left(\left|\stackrel{\to }{a}\right|\stackrel{\to }{b}-\left|\stackrel{\to }{b}\right|\stackrel{\to }{a}\right)$

$\begin{array}{l}=\left|\stackrel{\to }{a}\right|\stackrel{\to }{b}.\left|\stackrel{\to }{a}\right|\stackrel{\to }{b}-\left|\stackrel{\to }{a}\right|\stackrel{\to }{b}.\left|\stackrel{\to }{b}\right|\stackrel{\to }{a}+\left|\stackrel{\to }{b}\right|\stackrel{\to }{a}.\left|\stackrel{\to }{a}\right|\stackrel{\to }{b}-\left|\stackrel{\to }{b}\right|\stackrel{\to }{a}.\left|\stackrel{\to }{b}\right|\stackrel{\to }{a}\\ ={\left|\stackrel{\to }{a}\right|}^2\stackrel{\to }{b}.\stackrel{\to }{b}-{\left|\stackrel{\to }{b}\right|}^2\stackrel{\to }{a}.\stackrel{\to }{a}\\ ={\left|\stackrel{\to }{a}\right|}^2{\left|\stackrel{\to }{b}\right|}^2-{\left|\stackrel{\to }{b}\right|}^2{\left|\stackrel{\to }{a}\right|}^2\\ =0\end{array}$

$\therefore$ Therefore,  $\left|\stackrel{\to }{a}\right|\stackrel{\to }{b}+\left|\stackrel{\to }{b}\right|\stackrel{\to }{a}$ and  $\left|\stackrel{\to }{a}\right|\stackrel{\to }{b}-\left|\stackrel{\to }{b}\right|\stackrel{\to }{a}$ are perpendicular.

 

Q12. If and   $\stackrel{\to }{a}$ .  $\stackrel{\to }{a}$  = 0 and   $\stackrel{\to }{a}$ .  $\stackrel{\to }{b}$  = 0 , then what can be concluded about the vector   $\stackrel{\to }{b}$ ?

A.12. We know,

$\stackrel{\to }{a}.\stackrel{\to }{a}=0$ and  $\stackrel{\to }{a}.\stackrel{\to }{b}=0$

Now,

$\stackrel{\to }{a}.\stackrel{\to }{a}=0\Rightarrow {\left|\stackrel{\to }{a}\right|}^2\Rightarrow \left|\stackrel{\to }{a}\right|=0$

$\therefore$  $\stackrel{\to }{a}$ is a zero vector.

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

 

Q13. If   $\stackrel{\to }{a},\stackrel{\to }{b}$   and   $\stackrel{\to }{c}$ are unit vectors such that   $\stackrel{\to }{a}+\stackrel{\to }{b}+\stackrel{\to }{c}$ = 0 find the value of  $\stackrel{\to }{a}.\stackrel{\to }{b}+\stackrel{\to }{b}.\stackrel{\to }{c}+\stackrel{\to }{c}.\stackrel{\to }{a}$

A.13. $\left|\stackrel{\to }{a}+\stackrel{\to }{b}+\stackrel{\to }{c}\right|=\left(\stackrel{\to }{a}+\stackrel{\to }{b}+\stackrel{\to }{c}\right).\left(\stackrel{\to }{a}+\stackrel{\to }{b}+\stackrel{\to }{c}\right)$

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

 

Q14. If either vector   $\stackrel{\to }{a}=0or\stackrel{\to }{b}=0,then\stackrel{\to }{a}.\stackrel{\to }{b}=0$ . But the converse need not be true. Justify your answer with an example.

A14. Consider

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

Then,

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

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

 

Q15. 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   $\overline{BA}$ and  $\overline{BC}$ ]

A.15. Vertices of  $△ABC$ are given as

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

$\therefore △ABC$ is the angle between the vectors  $B\stackrel{\to }{A}$ and  $B\stackrel{\to }{C}$

 

Q16. Show that the points A (1, 2, 7), B (2, 6, 3) and C (3, 10, –1) are collinear.

A.16. Given, point are

 

Q17. Show that the vectors   $2\hat{i}-\hat{j}+\hat{k}and3\hat{i}-4\hat{j}-4\hat{k}$  form the vertices of a right angled triangle.

A.17. 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,

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

Now, vectors  $A\stackrel{\to }{B},B\stackrel{\to }{C}$ and  $A\stackrel{\to }{C}$ represents the sides of  $△ABC$ .

Hence,

 

Q18. If  $\stackrel{\to }{a}$ is a nonzero vector of magnitude ‘a’ and λ a nonzero scalar, then λ   $\stackrel{\to }{a}$ is unit vector if

(A) λ = 1

(B) λ = –1

(C) a = | λ 

|(D) a = 1/|λ|

A.18. Vector  $\lambda \stackrel{\to }{a}$ is a unit vector if  $\left|\lambda \stackrel{\to }{a}\right|=1$

Now,

$\begin{array}{l}\left|\lambda \stackrel{\to }{a}\right|=1\\ \left|\lambda \right|\left|\stackrel{\to }{a}\right|=1\\ \left|\stackrel{\to }{a}\right|=\frac{1}{\left|\lambda \right|}\left[\lambda \ne 0\right]\\ a=\frac{1}{\left|\lambda \right|}\left[\left|\stackrel{\to }{a}\right|=a\right]\end{array}$

$\therefore$ Therefore, vectar  $\lambda \stackrel{\to }{a}$ is a unit vector if a=  $\frac{1}{\left|\lambda \right|}$ .

Option (D)is correct.