# NCERT Solutions for Exercise 10.2 Class 12 Maths Chapter 10 - Vector Algebra

NCERT solutions for exercise 10.2 Class 12 Maths chapter 10 introduces a few more concepts of vectors. The questions of exercise 10.2 Class 12 Maths deals with the concepts like addition of vectors, magnitude of vectors, unit vectors etc. Also in NCERT solutions for Class 12 Maths chapter 10 exercise 10.2 questions related to collinear vectors and components of vectors. As the Class 12 Maths syllabus is concerned, it is important to practice Class 12 Maths chapter 10 exercise 10.2. Questions related to the Class 12th Maths chapter 10 exercise 10.2 can be expected for CBSE Class 12 Board Exams as well as competitive exams like JEE Mains. The chapter Vector Algebra exercises will be helpful for both Mathematics and Physics. To solve more problems of NCERT Class 12 Maths, students can also make use of NCERT Exemplar Solutions for Class 12 Maths.

Also practice-

• Vector Algebra Exercise 10.1

• Vector Algebra Exercise 10.3

• Vector Algebra Exercise 10.4

• Vector Algebra Miscellaneous Exercise

## Vector Algebra-Exercise: 10.2

Question:1 Compute the magnitude of the following vectors:

(1) $\vec a = \hat i + \hat j + \hat k$

Here

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

Magnitude of $\vec a$

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

Question:1 Compute the magnitude of the following vectors:

(2) $\vec b = 2 \hat i - 7 \hat j - 3 \hat k$

Here,

$\vec b = 2 \hat i - 7 \hat j - 3 \hat k$

Magnitude of $\vec b$

$\left | \vec b \right |=\sqrt{2^2+(-7)^2+(-3)^2}=\sqrt{62}$

Question:1 Compute the magnitude of the following vectors:

(3) $\vec c = \frac{1}{\sqrt 3 }\hat i + \frac{1}{\sqrt 3 }\hat j -\frac{1}{\sqrt 3 }\hat k$

Here,

$\vec c = \frac{1}{\sqrt 3 }\hat i + \frac{1}{\sqrt 3 }\hat j -\frac{1}{\sqrt 3 }\hat k$

Magnitude of $\vec c$

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

Question:2 Write two different vectors having same magnitude

Two different Vectors having the same magnitude are

$\vec a= 3\hat i+6\hat j+9\hat k$

$\vec b= 9\hat i+6\hat j+3\hat k$

The magnitude of both vector

$\left | \vec a \right |=\left | \vec b \right | = \sqrt{9^2+6^2+3^2}=\sqrt{126}$

Question:3 Write two different vectors having same direction.

Two different vectors having the same direction are:

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

$\vec b=2\hat i+4\hat j+6\hat k$

Question: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.

$2 \hat i + 3 \hat j$ will be equal to $x \hat i + y \hat j$ when their corresponding components are equal.

Hence when,

$x=2$ and

$y=3$

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

Let point P = (2, 1) and Q = (– 5, 7).

Now,

$\vec {PQ}=(-5-2)\hat i+(7-1)\hat j=-7\hat i +6\hat j$

Hence scalar components are (-7,6) and the vector is $-7\hat i +6\hat j$

Question:6 Find the sum of the vectors $\vec a = \hat i - 2 \hat j + \hat k , \vec b = -2 \hat i + 4 \hat j + 5 \hat k \: \: and\: \: \: \vec c = \hat i - 6 \hat j - 7 \hat k$

Given,

$\\ \vec a = \hat i - 2 \hat j + \hat k ,\\ \vec b = -2 \hat i + 4 \hat j + 5 \hat k \: \: and\: \: \: \\\vec c = \hat i - 6 \hat j - 7 \hat k$

Now, The sum of the vectors:

$\vec a +\vec b+\vec c = \hat i - 2 \hat j + \hat k + -2 \hat i + 4 \hat j + 5 \hat k + \hat i - 6 \hat j - 7 \hat k$

$\vec a +\vec b+\vec c = (1-2+1)\hat i +(-2+4-6) \hat j + (1+5-7)\hat k$

$\vec a +\vec b+\vec c =-4\hat j-\hat k$

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

Given

$\vec a = \hat i + \hat j + 2 \hat k$

Magnitude of $\vec a$

$\left |\vec a \right |=\sqrt{1^2+1^2+2^2}=\sqrt{6}$

A unit vector in the direction of $\vec a$

$\vec u = \frac{\hat i}{\left | a \right |} + \frac{\hat j}{\left | a \right |} +\frac{2\hat k}{\left | a \right |} =\frac{\hat i}{\sqrt{6}}+\frac{\hat j}{\sqrt{6}}+\frac{2\hat k}{\sqrt{6}}$

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

Given P = (1, 2, 3) and Q = (4, 5, 6)

A vector in direction of PQ

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

$\vec {PQ}=3\hat i+3\hat j +3\hat k$

Magnitude of PQ

$\left | \vec {PQ} \right |=\sqrt{3^2+3^2+3^2}=3\sqrt{3}$

Now, unit vector in direction of PQ

$\hat u=\frac{\vec {PQ}}{\left | \vec {PQ} \right |}=\frac{3\hat i+3\hat j+3\hat k}{3\sqrt{3}}$

$\hat u=\frac{\hat i}{\sqrt{3}}+\frac{\hat j}{\sqrt{3}}+\frac{\hat k}{\sqrt{3}}$

Question: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 $\vec a + \vec b$ .

Given

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

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

Now,

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

$\vec a + \vec b=\hat i+\hat k$

Now a unit vector in the direction of $\vec a + \vec b$

$\vec u= \frac{\vec a + \vec b}{\left |\vec a + \vec b \right |}=\frac{\hat i+\hat j}{\sqrt{1^2+1^2}}$

$\vec u= \frac{\hat i}{\sqrt{2}}+\frac{\hat j}{\sqrt{2}}$

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

Given a vector

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

the unit vector in the direction of $5 \hat i - \hat j + 2 \hat k$

$\vec u=\frac{5\hat i - \hat j + 2 \hat k}{\sqrt{5^2+(-1)^2+2^2}}=\frac{5\hat i}{\sqrt{30}}-\frac{\hat j}{\sqrt{30}}+\frac{2\hat k}{\sqrt{30}}$

A vector in direction of $5 \hat i - \hat j + 2 \hat k$ and whose magnitude is 8 =

$8\vec u=\frac{40\hat i}{\sqrt{30}}-\frac{8\hat j}{\sqrt{30}}+\frac{16\hat k}{\sqrt{30}}$

Question: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.

Let

$\vec a =2 \hat i -3 \hat j + 4 \hat k$

$\vec b=- 4 \hat i + 6 \hat j - 8 \hat k$

It can be seen that

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

Hence here $\vec b=-2\vec a$

As we know

Whenever we have $\vec b=\lambda \vec a$ , the vector $\vec a$ and $\vec b$ will be colinear.

Here $\lambda =-2$

Hence vectors $2 \hat i -3 \hat j + 4 \hat k$ and $- 4 \hat i + 6 \hat j - 8 \hat k$ are collinear.

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

Let

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

$\left |\vec a \right |=\sqrt{1^2+2^2+3^2}=\sqrt{14}$

Hence direction cosine of $\vec a$ are

$\left ( \frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}} ,\frac{3}{\sqrt{14}}\right )$

Question:13 Find the direction cosines of the vector joining the points A(1, 2, –3) and B(–1, –2, 1), directed from A to B.

Given

point A=(1, 2, –3)

point B=(–1, –2, 1)

Vector joining A and B Directed from A to B

$\vec {AB}=(-1-1)\hat i +(-2-2)\hat j+(1-(-3))\hat k$

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

$\left | \vec {AB} \right |=\sqrt{(-2)^2+(-4)^2+4^2}=\sqrt{36}=6$

Hence Direction cosines of vector AB are

$\left ( \frac{-2}{6},\frac{-4}{6},\frac{4}{6} \right )=\left ( \frac{-1}{3},\frac{-2}{3},\frac{2}{3} \right )$

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

Let

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

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

Hence direction cosines of this vectors is

$\left ( \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} \right )$

Let $\alpha$ , $\beta$ and $\gamma$ be the angle made by x-axis, y-axis and z- axis respectively

Now as we know,

$cos\alpha=\frac{1}{\sqrt{3}}$ , $cos\beta=\frac{1}{\sqrt{3}}$ $and\:cos\gamma=\frac{1}{\sqrt{3}}$

Hence Given vector is equally inclined to axis OX,OY and OZ.

Question:15 (1) Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are $i + 2 j - k$ and $- i + j + k$ respectively, in the ratio 2 : 1 internally

As we know

The position vector of the point R which divides the line segment PQ in ratio m:n internally:

$\vec r=\frac{m\vec b+n\vec a}{m+n}$

Here

position vector os P = $\vec a$ = $i + 2 j - k$

the position vector of Q = $\vec b=- i + j + k$

m:n = 2:1

And Hence

$\vec r = \frac{2(-\hat i+\hat j +\hat k)+1(\hat i+2\hat j-\hat k)}{2+1}=\frac{-2\hat i+2\hat j +2\hat k+\hat i+2\hat j-\hat k}{3}$

$\vec r = \frac{-2\hat i+2\hat j +2\hat k+\hat i+2\hat j-\hat k}{3}=\frac{-\hat i+4\hat j+\hat k}{3}$

$\vec r = \frac{-\hat i}{3}+\frac{4\hat j}{3}+\frac{\hat k}{3}$

Question:15 (2) 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 externally

As we know

The position vector of the point R which divides the line segment PQ in ratio m:n externally:

$\vec r=\frac{m\vec b-n\vec a}{m-n}$

Here

position vector os P = $\vec a$ = $i + 2 j - k$

the position vector of Q = $\vec b=- i + j + k$

m:n = 2:1

And Hence

$\vec r = \frac{2(-\hat i+\hat j +\hat k)-1(\hat i+2\hat j-\hat k)}{2-1}=\frac{-2\hat i+2\hat j +2\hat k-\hat i-2\hat j+\hat k}{1}$

$\vec r = -3\hat i +3\hat k$

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

Given

The position vector of point P = $2\hat i+3\hat j +4\hat k$

Position Vector of point Q = $4\hat i+\hat j -2\hat k$

The position vector of R which divides PQ in half is given by:

$\vec r =\frac{2\hat i+3\hat j +4\hat k+4\hat i+\hat j -2\hat k}{2}$

$\vec r =\frac{6\hat i+4\hat j +2\hat k}{2}=3\hat i+2\hat j +\hat k$

Question:17 Show that the points A, B and C with position vectors, $\vec a = 3 \hat i - 4 \hat j - 4 \hat k , \vec b = 2 \hat i - \hat j + \hat k \: \: and \: \: \: \vec c = \hat i - 3 \hat j - 5 \hat k$ , respectively form the vertices of a right angled triangle.

Given

the position vector of A, B, and C are

$\\\vec a = 3 \hat i - 4 \hat j - 4 \hat k ,\\\vec b = 2 \hat i - \hat j + \hat k \: \: and \\\vec c = \hat i - 3 \hat j - 5 \hat k$

Now,

$\vec {AB}=\vec b-\vec a=-\hat i+3\hat j+5\hat k$

$\vec {BC}=\vec c-\vec b=-\hat i-2\hat j-6\hat k$

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

$\left | \vec {AB} \right |=\sqrt{(-1)^2+3^2+5^2}=\sqrt{35}$

$\left | \vec {BC} \right |=\sqrt{(-1)^2+(-2)^2+(-6)^2}=\sqrt{41}$

$\left | \vec {CA} \right |=\sqrt{(2)^2+(-1)^2+(1)^2}=\sqrt{6}$

AS we can see

$\left | \vec {BC} \right |^2=\left | \vec {CA} \right |^2+\left | \vec {AB} \right |^2$

Hence ABC is a right angle triangle.

Question:18 In triangle ABC (Fig 10.18), which of the following is not true:

$A ) \overline{AB}+ \overline{BC}+ \overline{CA} = \vec 0 \\\\ B ) \overline{AB}+ \overline{BC}- \overline{AC} = \vec 0 \\\\ C ) \overline{AB}+ \overline{BC}- \overline{CA} = \vec 0 \\\\ D ) \overline{AB}- \overline{CB}+ \overline{CA} = \vec 0$

From triangles law of addition we have,

$\vec {AB}+\vec {BC}=\vec {AC}$

From here

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

also

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

Also

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

Hence options A,B and D are true SO,

Option C is False.

Question:19 If are two collinear vectors, then which of the following are incorrect:
(A) $\vec b = \lambda \vec a$ for some saclar $\lambda$
(B) $\vec a = \pm \vec b$
(C) the respective components of $\vec a \: \:and \: \: \vec b$ are not proportional
(D) both the vectors $\vec a \: \:and \: \: \vec b$ have same direction, but different magnitudes.

If two vectors are collinear then, they have same direction or are parallel or anti-parallel.
Therefore,
They can be expressed in the form $\vec{b}= \lambda \vec{a}$ where a and b are vectors and $\lambda$ is some scalar quantity.

Therefore, (a) is true.
Now,
(b) $\lambda$ is a scalar quantity so its value may be equal to $\pm 1$

Therefore,
(b) is also true.

C) The vectors and are proportional,
Therefore, (c) is not true.

D) The vectors and can have different magnitude as well as different directions.

Therefore, (d) is not true.

Therefore, the correct options are (C) and (D).

## More About NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.2

The NCERT Class 12 Math chapter 10 exercise 10.2 starts with the questions to find the magnitude of vectors. The 5th question of exercise 10.2 Class 12 Maths is related to the concept of position vector. The concepts of unit vectors are explained in questions 7 to 10. NCERT solutions for Class 12 Maths chapter 10 exercise 10.2 also goes through the topics like direction cosines, and section formula.

## Read Also| Vector Algebra Class 12 Chapter 10 Notes

Importance of NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.2

• The Class 12 Maths NCERT exercises are solved by math experts and can be used for preparations of various examinations.
• By practising the NCERT Solutions for Class 12 Maths chapter 10 exercise 10.2 students will be able to revise the concepts studied before the exercise and also will be able to clarify doubts.

## Also see-

• NCERT Exemplar Solutions Class 12 Maths Chapter 10

• NCERT Solutions for Class 12 Maths Chapter 10

## NCERT Solutions Subject Wise

• NCERT Solutions Class 12 Chemistry

• NCERT Solutions for Class 12 Physics

• NCERT Solutions for Class 12 Biology

• NCERT Solutions for Class 12 Mathematics

## Subject Wise NCERT Exemplar Solutions

• NCERT Exemplar Class 12 Maths

• NCERT Exemplar Class 12 Physics

• NCERT Exemplar Class 12 Chemistry

• NCERT Exemplar Class 12 Biology