# Class 12 RD Sharma Solutions – Chapter 23 Algebra of Vectors – Exercise 23.5

Last Updated : 18 Mar, 2021

Solution:

We have,

### Question 2. If the position vector  of a point (12,n) is such that  , find the value(s).

Solution:

We have,

On squaring both sides,

### Question 3. Find a vector of magnitude 4 units which is parallel to the vector.

Solution:

Given,

Let is a vector parallel to Therefore, for any scalar

### Question 4. Express in terms of unit vectors (i)A = (4,-1),B = (1,3) (ii)A = (-6,3) , B = (-2,-5)

Solution:

(i) We have, A = (4,-1) B = (1,3) Position Vector of A = Position Vector of B = Now, Therefore, (ii) We have, A = (-6,3) B = (-2,-5) Position Vector of A = Position Vector of B = Now, Therefore,

### Question 5. Find the coordinates of the tip of the position vector which is equivalent to , where the coordinates of A and B are (-1,3) and (-2,1)

Solution:

We have,

A = (-1,3)

B = (-2,1)

Now,

Position Vector of

Position Vector of

Therefore,

Coordinate of the position vector

### Question 6. ABCD is a parallelogram. If the coordinates of A,B,C are (-2,-1), (3,0),(1,-2) respectively, find the coordinates of D.

Solution:

Here, A = (-2,-1)

B = (3,0)

C = (1,-2)

Let us assume D be (x , y).

Computing Position Vector of AB, we have,

= Position Vector of B – Position Vector of A

Comparing LHS and RHS of both,

5 = 1-x

x = -4

And,

1 = -2-y

y = -3

So, coordinates of D = (-4,-3).

### Question 7. If the position vectors of the points A(3,4), B(5,-6) and C(4,-1) are respectively, compute the value of .

Solution:

Computing the position vectors of all the points we have,

Now,

Computing the final value after substituting the values,

### Question 8. If be the position vector whose tip is (-5,3), find the coordinates of a point B such that , the coordinates of A being (-4,1).

Solution:

Given, Coordinate of A = (4,-1) Position vector of A = Position vector of Let coordinate of point B = (x, y) Position vector of B = Given that, Position vector of B – Position vector of A = \vec{a} Comparing the coefficients of LHS and RHS x – y = 5 x = 9 Also, y + 1 = 3 y = -1 So, coordinate of B = (9,-4)

### Question 9. Show that the points  form an isosceles triangle.

Solution:

So, the two sides AB and AC of the triangle ABC are equal.

Therefore, ABC is an isosceles triangle.

### Question 10. Find a unit vector parallel to the vector  .

Solution:

We have,

Let

Suppose  is any vector parallel to

, where Î» is any scalar.

Unit vector of

Therefore,

### Question 11. Find the components along the coordinate axes of the position vector of each of the following points :

(i) P(3,2)

(ii) Q(-5,1)

(iii) R(-11,-9)

(iv) S(4,-3)

Solution:

(i) Given, P = (3,2)

Position vector of P =

Component of P along x-axis =

Component of P along y-axis =

(ii) Given, Q = (-5,1)

Position vector of Q =

Component of Q along x-axis =

Component of Q along y-axis =

(iii) Given, R = (-11,-9)

Position vector of R =

Component of R along x-axis =

Component of R along y-axis =

(iv) Given, S = (4,-3)

Position vector of S =

Component of S along x-axis =

Component of S along y-axis =

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