NCERT solution class 12 chapter 4 Differential Equations exercise 4.2 mathematics part 2

EXERCISE 4.2

Question 1:

Compute the magnitude of the following vectors:

The given vectors are:

Question 2:

Write two different vectors having same magnitude.

Hence, are two different vectors having the same magnitude. The vectors are different because they have different directions.

Question 3:

Write two different vectors having same direction.

The direction cosines of are the same. Hence, the two vectors have the same direction.

Question 4:

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

The two vectors will be equal if their corresponding components are equal.

Hence, the required values of x and y are 2 and 3 respectively.

Question 5:

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

The vector with the initial point P (2, 1) and terminal point Q (–5, 7) can be given by,

Hence, the required scalar components are –7 and 6 while the vector components are

Question 6:

Find the sum of the vectors.

The given vectors are.

Question 7:

Find the unit vector in the direction of the vector.

The unit vector in the direction of vector is given by.

Question 8:

Find the unit vector in the direction of vector, where P and Q are the points

(1, 2, 3) and (4, 5, 6), respectively.

The given points are P (1, 2, 3) and Q (4, 5, 6).

Hence, the unit vector in the direction of  is

.

Question 9:

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

The given vectors are and.

Hence, the unit vector in the direction of is

a→+b→a→+b→=i^+k^2=12i⏜+12k⏜.

Question 10:

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

Hence, the vector in the direction of vector which has magnitude 8 units is given by,

Question 11:

Show that the vectorsare collinear.

.

Hence, the given vectors are collinear.

Question 12:

Find the direction cosines of the vector

Hence, the direction cosines of

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.

The given points are A (1, 2, –3) and B (–1, –2, 1).

Hence, the direction cosines of are

Question 14:

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

Therefore, the direction cosines of

Now, let αβ, and γbe the angles formed by with the positive directions of xy, and z axes.

Then, we have

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

Question 15:

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

(i) internally

(ii) externally

The position vector of point R dividing the line segment joining two points

P and Q in the ratio m: is given by:

1. Internally:

1. Externally:

Position vectors of P and Q are given as:

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

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

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).

The position vector of mid-point R of the vector joining points P (2, 3, 4) and Q (4, 1, – 2) is given by,

Question 17:

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

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

AB→2+CA→2=35+6=41=BC→2Hence, ABC is a right-angled triangle.

Question 18:

In triangle ABC which of the following is not true:

A.

B.

C.

D.

On applying the triangle law of addition in the given triangle, we have:

From equations (1) and (3), we have:

Hence, the equation given in alternative C is incorrect.

Question 19:

If are two collinear vectors, then which of the following are incorrect:

A. , for some scalar λ

B.

C. the respective components of are proportional

D. both the vectors have same direction, but different magnitudes

If are two collinear vectors, then they are parallel.

Therefore, we have:

(For some scalar λ)

If λ = ±1, then .

Thus, the respective components of are proportional.

However, vectors can have different directions.

Hence, the statement given in D is incorrect.