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Class 12 RD Sharma Solutions – Chapter 23 Algebra of Vectors – Exercise 23.3

Question 1.  Find the position vector of a point R which divides the line joining the two points P and Q with position vectors and   respectively in the ratio 1:2 internally and externally.

Solution:

The point R divides the line joining points P and Q in the ratio 1:2 internally.



The position vector of R =  

Point R divides the line joining P and Q in the ratio 1:2 externally.



The position vector of R = 

Question 2. Let   and  be the position vectors of the four distinct points A, B, C, D. If   then show that ABCD is a parallelogram.                                                                                                         

Solution: 

Given that  are the position vectors of the four distinct points A, B, C, D 

such that 

Given that, 

So, AB is parallel and equal to DC 

Hence, ABCD is a parallelogram.

Question 3. If  are the position vectors of A, B respectively, find the position vector of a point C in AB produced such that AC = 3AB and that a point D in BA produced such that  BD = 2BA.

Solution: 

Given that are the position vector of A and B

Let C be a point in AB produced such that AC = 3AB.

From the given data we can say that point C divides the line AB in

Ratio 3:2 externally. So, the position vector of point C can be written as

=  

D be a point in BA produced such that BD = 2BA

It is clear that point D divides the line in 1:2 externally. 

Then the position vector  can be written as

=  

Hence  and  

Question 4. Show that the four points A, B, C, D with position vectors   and  respectively such that  are coplanar. Also, find the position vector of the point of intersection of the lines AC and BD.

Solution:

Given that 

Sum of the coefficients on both sides of the given equation is 8

so, divide the equation by 8 on both the sides

It is clear that the position vector of a point P dividing Ac in the 

Ratio 3:5 is same as that of point P diving BD in the ratio 2:6.

Point P is common to AC and BD. Hence, P is the point of intersection of AC and BD.

Therefore, A, B, C and D are coplanar.

The position vector of point P can be written as 

 or 

Question 5:  Show that the four points P, Q, R, S with position vectors  and respectively such that are coplanar. Also, find the position vector of the point of intersection of the lines PR and QS.

Solution:

Given that 

Here  and 

are the position vectors of point P, Q, R, S

          -(1)

Sum of the coefficients on both the sides of the equation (1) is 11. 

So divide the equation (1) by 11 on both sides.

It shows that position vector of a point A dividing PR in the ratio of 6:5 and

 QS in the ratio 9:2. So A is the common point to PR and QS.

Therefore, P, Q, R and S are coplanar.

The position vector of point A is given by 

or 

Question 6: The vertices A, B, C of triangle ABC have respectively position vectors  with respect to a given origin O. Show that the point D where the bisector of   meets BC has position vector where . Hence deduce that the incentre I has position vector  where 

Solution: 

Let ABC be a triangle and the position vectors of A, B, C with respect to some origin say O be 

Let D be the point on BC where the bisector of  meets.

 be the position vector of D which divides BC internally in the ratio  

and where  

Thus, 

Therefore, by section formula, the position vector of D is given by

 

Let 

Incentre is the concurrent point of angle bisectors.

Thus, Incentre divides the line AD in the ratio  and 

the position vector of incentre is equal to

 


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