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

• Last Updated : 18 Mar, 2021

Question 1. If the position vector of a point (-4,-3) be , find .

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 λ 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 = My Personal Notes arrow_drop_up