# NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra

Two chapters vector algebra & three-dimensional geometry has 17% weightage in 12th board maths final examination. The CBSE NCERT solutions for class 12 maths chapter 10 vector algebra will help you to score good marks in the exam. The concepts of vector algebra are very useful in both maths and physics. In chapter 10 vector algebra total of 54 questions in 4 exercises and 30 solved examples are given in the NCERT textbook. These NCERT questions are prepared and explained in a detailed manner to help students in their board exam as well as in competitive exams. Before discussing this chapter in detail, you need to understand what is a vector quantity and what is a scalar quantity.

Vector Quantity- Quantity which involves both the value magnitude and direction. Vector quantities like weight, velocity, acceleration, displacement, force, momentum, etc.

Scalar Quantity- Quantity which involves only one value (magnitude) which is a real number. Scalar quantities like distance,  length, time, mass, speed, area, temperature, work, money, volume, voltage, density, resistance, etc.

The focus of class 12 maths chapter 10 vector algebra is on vector quantities. To develop a grip on the topic, students must try to solve the miscellaneous exercise also. In the NCERT solutions for class 12 maths chapter 10 vector algebra, you will find solutions to miscellaneous exercises too.

## Topics of NCERT class 12 maths chapter 10 Vector Algebra

10.1 Introduction

10.2 Some Basic Concepts

10.3 Types of Vectors

10.5 Multiplication of a Vector by a Scalar

10.5.1 Components of a vector

10.5.2 Vector joining two points

10.5.3 Section formula

10.6 Product of Two Vectors

10.6.1 Scalar (or dot) product of two vectors

10.6.2 Projection of a vector on a line

10.6.3 Vector (or cross) product of two vectors

## Solutions of NCERT for class 12 maths chapter 10 vector algebra-Exercise: 10.1

Represent graphically a displacement of 40 km,  east of north.

N,S,E,W are all 4 direction north,south,east,west respectively.

is displacement vector which

= 40 km.

makes an angle of 30 degree east of north  as shown in figure. Question:2 (1) Classify the following measures as scalars and vectors.

10Kg

10kg is a scalar quantity as it has only magnitude.

This is a vector quantity as it has both magnitude and direction.

Question:2 (3) Classify the following measures as scalars and vectors.

This is a scalar quantity as it has only magnitude.

This is a scalar quantity as it has only magnitude.

Question:2 (5) Classify the following measures as scalars and vectors.

This is a scalar quantity as it has only magnitude.

Question:2 (6) Classify the following measures as scalars and vectors.

This is a Vector quantity as it has magnitude as well as direction.by looking at the unit, we conclude that measure is acceleration which is a vector.

This is a scalar quantity as it has only magnitude.

(2)   distance

Distance is a scalar quantity as it has only magnitude.

(3)   force

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

Velocity is a vector quantity as it has both magnitude and direction.

(5)     work done

work done is a scalar quantity, as it is the product of two vectors. Since vector and vector  are starting from the same point, they are coinitial. Since Vector and Vector both have the same magnitude and same direction, they are equal.

(3) Collinear but not equal Since vector and vector  have the same magnitude but different direction, they are colinear and not equal.

True,    and   are collinear. they both are parallel to one line hence they are colinear.

False, because colinear means they are parallel to the same line but their magnitude can be anything and hence this is a false statement.

Question:5 Answer the following as true or false.

(3) Two vectors having same magnitude are collinear.

False, because any two non-colinear vectors can have the same magnitude.

Question:5 Answer the following as true or false.

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

False, because two colinear vectors with the same magnitude can have opposite  direction

## CBSE NCERT solutions for class 12 maths chapter 10 vector algebra-Exercise: 10.2

(1)

Here

Magnitude of

(2)

Here,

Magnitude of

(3)

Here,

Magnitude of

Two different Vectors having the same magnitude are

The magnitude of both vector

Two different vectors having the same direction are:

will be equal to  when their corresponding components are equal.

Hence when,

and

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

Now,

Hence scalar components are (-7,6) and the vector is

Question:6 Find the sum of the vectors

Given,

Now, The sum of the vectors:

Given

Magnitude of

A unit vector in the direction of

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

A vector in direction of PQ

Magnitude of PQ

Now, unit vector in direction of PQ

Given

Now,

Now a unit vector in the direction of

Given a vector

the unit vector in the direction of

A vector in direction of  and whose magnitude is 8 =

Question:11 Show that the vectors and   are collinear.

Let

It can be seen that

Hence here

As we know

Whenever we have , the vector  and  will be colinear.

Here

Hence  vectors and   are collinear.

Question:12 Find the direction cosines of the vector

Let

Hence direction cosine of  are

Given

point A=(1, 2, –3)

point B=(–1, –2, 1)

Vector joining A and B Directed from A to B

Hence Direction cosines of vector AB are

Let

Hence direction cosines of this vectors is

Let  and  be the angle made by x-axis, y-axis and z- axis respectively

Now as we know,

,

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

As we know

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

Here

position vector os P =  =

the position vector of Q  =

m:n = 2:1

And Hence

As we know

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

Here

position vector os P =  =

the position vector of Q  =

m:n = 2:1

And Hence

Given

The position vector of point P =

Position Vector of point Q =

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

Given

the position vector of A, B, and C are

Now,

AS we can see

Hence ABC is a right angle triangle.

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

Therefore, (a) is true.
Now,
(b)   is a scalar quantity so its value may be equal to

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).

## Question:1 Find the angle between two vectors  with magnitudes , respectively having .

Given

As we know

where  is the angle between two vectors

So,

Hence the angle between the vectors is .

Question:2 Find the angle between the vectors

Given two vectors

Now As we know,

The angle between two vectors  and  is given by

Hence the angle between

Let

Projection of vector  on

Hence, Projection of vector  on  is 0.

Let

The projection of  on   is

Hence, projection of vector  on  is

Given

Now magnitude of

Hence, they all are unit vectors.

Now,

Hence all three are mutually perpendicular to each other.

Question:6 Find  , if .

Given in the question

Since

So, answer of the question is

Question:7 Evaluate the product .

To evaluate the product

Given two vectors

Now Angle between

Now As we know that

Hence, the magnitude of two vectors

Question:9 Find   , if for a unit vector

Given in the question that

And we need to find

So the value of is

Given in the question is

and  is perpendicular to

and we need to find the value of ,

so the value of -

As   is perpendicular to

the value of ,

Given in the question that -

are two non-zero vectors

According to the question

Hence   is perpendicular to   .

Given in the question

Therefore  is a zero vector. Hence any vector   will satisfy

Given in the question

are unit vectors

and

and we need to find the value of

Let

we see that

we now observe that

Hence here converse of the given statement is not true.

Given points,

A=(1, 2, 3),

B=(–1, 0, 0),

C=(0, 1, 2),

As need to find Angle between

Hence angle between them ;

Answer- Angle between the vectors is

Given in the question

A=(1, 2, 7), B=(2, 6, 3) and C(3, 10, –1)

To show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear

As we see that

Hence point A, B , and C are colinear.

Given the position vector of A, B , and C are

To show that the vectors   form the vertices of a right angled triangle

Here we see that

Hence A,B, and C are the vertices of a right angle triangle.

Given   is a nonzero vector of magnitude ‘a’ and    a nonzero scalar

is a unit vector when

Hence the correct option is D.

## Solutions of NCERT for class 12 maths chapter 10 vector algebra-Exercise: 10.4

Question:1 Find

Given in the question,

and we need to find

Now,

So the value of is

Given in the question

Now , A vector which perpendicular to both  is

And a unit vector in this direction :

Hence Unit vector perpendicular to each of the vector  is .

Given in the question,

angle between  and  :

angle between  and

angle with  and :

Now, As we know,

Now components of  are:

Question:4 Show that

To show that

LHS=

As product of a vector with itself is always Zero,

As cross product of a and b is equal to negative of cross product of b and a.

= RHS

LHS is equal to RHS, Hence Proved.

Question:5 Find  and   if

Given in the question

and we need to find values of  and

From Here we get,

From here, the value of   and is

Given in the question

and

When , either  are perpendicular to each other

When  either are parallel to each other

Since two vectors can never be both parallel and perpendicular at same time,we conclude that

Given in the question

We need to show that

Now,

Now

Hence they are equal.

No, the converse of the statement is not true, as there can be two non zero vectors, the cross product of whose are zero. they are colinear vectors.

Consider an example

Here

Hence converse of the given statement is not true.

Given in the question

vertices A=(1, 1, 2), B=(2, 3, 5) and C=(1, 5, 5). and we need to find the area of the triangle

Now as we know

Area of triangle

The area of the triangle is square units

Given in the question

Area of parallelogram with adjescent side  and ,

The area of the parallelogram whose adjacent sides are determined by the vectors    and   is

Given in the question,

As given   is a unit vector, which means,

Hence the angle between two vectors is . Correct option is B.

(A)1/2

(B) 1

(C) 2

(D) 4

Given 4 vertices of rectangle are

Now,

Area of the Rectangle

Hence option C is correct.

## CBSE NCERT solutions for class 12 maths chapter 10 vector algebra-Miscellaneous Exercise

As we know

a unit vector in XY-Plane making an angle  with x-axis :

Hence for

Answer- the unit vector in XY-plane, making an angle of with the positive direction of x-axis is

Given in the question

And we need to finrd the scalar components and magnitude of the vector joining the points P and Q

Magnitiude of vector PQ

Scalar components are

As the girl walks 4km towards west

Position vector =

Now as she moves 3km in direction 30 degree east of north.

hence final position vector is;

No, if then we can not conclude that   .

the condition     satisfies in the triangle.

also, in a triangle,

Since, the condition  is contradicting with the triangle inequality, if  then we can not conclude that

Given in the question,

a unit vector,

We need to find the value of x

The value of x   is

Given two vectors

Resultant of  and :

Now, a unit vector in the direction of

Now, a unit vector of magnitude in direction of

Hence the required vector is

Given in the question,

Now,

let vector

Now, a unit vector in direction of

Now,

A unit vector parallel to

OR

Given in the question,

points A(1, – 2, – 8), B(5, 0, –2) and C(11, 3, 7)

now let's calculate the magnitude of the vectors

As we see that AB = BC + AC, we conclude that three points are colinear.

we can also see from here,

Point B divides AC in the ratio  2 : 3.

Given, two vectors

the point  R which divides line segment PQ in ratio 1:2 is given by

Hence position vector of R is .

Now, Position vector of the midpoint of RQ

which is the position vector of Point P . Hence, P is the mid-point of RQ

Given, two adjacent sides of the parallelogram

The diagonal will be the resultant of these two vectors. so

resultant R:

Now unit vector in direction of R

Hence unit vector along the diagonal of the parallelogram

Now,

Area of parallelogram

Hence the area of the parallelogram is .

Let a vector  is equally inclined to axis OX, OY  and OZ.

let direction cosines of this vector be

Now

Hence direction cosines are:

Given,

Let

now, since it is given that d is perpendicular to  and , we got the condition,

and

And

And

here we got 2 equation and 3 variable. one more equation will come from the condition:

so now we have three equation and three variable,

On solving this three equation we get,

,

Hence Required vector :

Let, the sum of vectors  and  be

unit vector along

Now, the scalar product of this with

squaring both the side,

Given

and

Now, let vector is inclined to  at  respectively.

Now,

Now, Since,

Hence vector   is equally inclined to  .

Given in the question,

are perpendicular and we need to prove that

LHS=

if   are perpendicular,

= RHS

LHS ie equal to RHS

Hence proved.

Given in the question

is the angle between two vectors

this will satisfy when

Hence option B is the correct answer.

Gicen in the question

be two unit vectors and   is the angle between them

also

Then is a unit vector if

Hence option D is correct.

Question:18 The value of   is

(A) 0

(B) –1

(C) 1

(D) 3

To find the value of

Hence option C is correct.

Given in the question

is the angle between any two vectors  and

To find the value of

Hence option D is correct.

## NCERT solutions for class 12 maths chapter wise

 chapter 1 Solutions of NCERT for class 12 maths chapter 1 Relations and Functions chapter 2 CBSE NCERT solutions for class 12 maths chapter 2 Inverse Trigonometric Functions chapter 3 NCERT solutions for class 12 maths chapter 3 Matrices chapter 4 Solutions of NCERT for class 12 maths chapter 4 Determinants chapter 5 CBSE NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability chapter 6 NCERT solutions for class 12 maths chapter 6 Application of Derivatives chapter 7 Solutions of NCERT for class 12 maths chapter 7 Integrals chapter 8 CBSE NCERT solutions for class 12 maths chapter 8 Application of Integrals chapter 9 NCERT solutions for class 12 maths chapter 9 Differential Equations chapter 10 Solutions of NCERT for class 12 maths chapter 10 Vector Algebra chapter 11 CBSE NCERT solutions for class 12 maths chapter 11 Three Dimensional Geometry chapter 12 NCERT solutions for class 12 maths chapter 12 Linear Programming chapter 13 Solutions of NCERT for class 12 maths chapter 13 Probability

## Benefits of NCERT solutions

• NCERT solutions are explained in a step-by-step manner, so it will be very easy for you to understand the concepts.

• NCERT Solutions for class 12 maths chapter 10 vector algebra will give you some new way to solve the problem.

• Performance in the 12th board exam plays a very important role in deciding the future, so you can get admission to a good college. Scoring good marks in the exam is now a reality with the help of these solutions of NCERT for class 12 maths chapter 10 vector algebra.

• To develop a grip on the concept, you should solve the miscellaneous exercise also. In CBSE NCERT solutions for class 12 maths chapter 10 vector algebra article, you will get a solution of miscellaneous exercise also.

Happy learning !!!