Class 12 NCERT Solutions - Mathematics Part ii – Chapter 10 – Vector Algebra Exercise 10.2

In the article, we will solve Exercise 10.2 from Chapter 10, “Vector Algebra” in the NCERT. Exercise 10.2 covers the basics of vectors like scalar and vector components of vectors, section formulas, Multiplication of a Vector by a scalar, etc.

Vector Algebra Formula to Solve Exercise 10.2

The basic vector algebra formulas to solve this exercise are mentioned below:

1. Magnitude of vector

let a vector ,

if

then

2.Section formula

(a) internally: let the position vector of a point R which divides the line joining P and Q in the ratio m:n internally.

where and

(b) Externally: let the position vector of a point R which divides the line joining P and Q in the ratio m:n externally.

3.Direction Cosine: let a vector

then Direction Cosines will be:

; ;

4. Unit Vector

let a vector

Unit vector

Exercise 10.2 Solutions

Q1. Compute the magnitude of the following vectors:

Solution:

using formula of magnitude of vector

Q2. Write two different vectors having same magnitude.

Solution:

An infinite number of possible answers. An example is given below:

let vectors as and

their magnitudes

we can see that but

Q3. Write two different vectors having same direction.

Solution:

An infinite number of possible answers. An example is given below:

let and

we are using Direction Cosines (D.C) to find direction of a vector

Direction Cosine of

and Direction Cosine of

This proved that direction of and is same whereas

Q4. Find the values of x and y so that the vectors and are equal

Solution:

As it is given that and are equal

i.e.

The corresponding components will also be equal. By Comparing corresponding components

we will get x = 2 and y = 3

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

Solution:

Let p(2 , 1) and q(-5 , 7) and position vector and is and respectively.

Therefore,

i.e.

So, the scalar components are -7 and 6 , and the vector components are and

Q6. Find the sum of the vectors and

Solution:

To find sum of vector we will add corresponding and components

so,

Q7. Find the unit vector in the direction of the vector

Solution:

Given that and We know that,

So,

Q8. Find the unit vector in the direction of vector , where P and Q are the points (1 , 2 , 3) and (4 , 5 , 6) respectively.

Solution:

Given P and Q are the points (1 , 2 , 3) and (4 , 5 , 6).

Therefore,

We know that

Therefore,

Q9. For given vectors , and , find the unit vector in the direction of the vector .

Solution:

Given,

and

Therefore,

and

We know that,

Unit vector =

Q10. Find a vector in the direction of vector which has magnitude 8 units.

Solution:

Given,

and

Therefore,

Thus, a vector parallel to with magnitude 8 units is

So,

Q11. Show that the vectors and are collinear.

Solution:

Given that,

and we know that,

When for two Collinear.

Where

So, the vectors and are Collinear.

Q12. Find the direction cosines of the vector .

Solution:

Let

and using magnitude concept of vector

So, Direction Cosine (D.C) of vector a will be

The D.C s of vector a are

Q13. Find the direction of the cosines of the vectors joining the points A(1 , 2 ,-3)and B(-1 , -2 , 1) directions from A to B.

Solution:

Let position vectors of point A and B are and respectively.

So,

i.e.

and

Therefore, The Direction Cosine of vector AB

Q14. Show that the vector is equally inclined to the axis OX, OY and OZ.

Solution:

Let

So,

Therefore Direction Cosine will be

Let , and be the angles formed by with the positive directions of x , y and z axes respectively.

So, , ,

All angles and are equal.

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

Q15. Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are and respectively, in the ratio 2:1

(i) Internally

(ii) Externally

Solution:

Let the position vectors of P and Q are and

(i) The position vector of R which divides the line joining two points P and Q internally in the ratio 2:1 is

using Section Formula(internally)

(ii) The position vector of R which divides the line joining two points P and Q externally in the ratio 2:1 is

using Section Formula(externally)

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

Solution:

Mid point divide the vector joining the points (2 , 3 , 4) and (4 , 1 , -2) in the ratio 1:1 internally.

The position vector of the mid-point R will be

Q.17: Show that the points A,B and C with position vectors, , and , respectively form the vertices of a right angled triangle.

Solution:

Position vectors of points A, B, and C are respectively given as:

, and

So,

Using magnitude of a vector formula,

As

Therefore, ABC is a right-angled triangle.

Q18. In triangle ABC (Fig 10.18), which of the following is not true:

(A)

(B)

(C)

(D)

Solution: (C)

Using Triangle law of addition,

So, the equation given in option A is true.

So, the equation given in option B is true.

and

Hence, the equation given in option D is true.

Thus, the correct option will be (C).

Q19. If a and b are two collinear vectors, then which of the following are incorrect?

(A) , for some scalar

(B)

(C) the respective components of a and b are proportional.

(D) both the vectors a and b have same direction, but different magnitudes.

Solution: (D)

If a and b are collinear vectors, they are parallel.

Therefore, for some scalar

i.e. option A is true.

If , then

option B is true.

and

Then,

Comparing the components of both the sides

we have

So, the respective components of vector a and b are proportional.

Thus, the correct option is D

Also, Check

Class 12 NCERT Solutions- Mathematics Part ii – Chapter 10 – Vector Algebra Exercise 10.1