Question 3 & 4, Exercise 3.2

Solutions of Question 3 & 4 of Exercise 3.2 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

If r=ˆi9ˆj, a=ˆi+2ˆj and b=5ˆiˆj, determine the real number p and q such that r=pa+qb.

Given r=pa+qb. We put the values of r,a and b in the given equation. We get ˆi9ˆj=p(ˆi+2ˆj)+q(5ˆiˆj) ˆi9ˆj=(p+5q)ˆi+(2pq)ˆj. By comparing the coeffients of ˆi and ˆj, we have, p+5q=1(i) 2pq=9(ii) Multiply 2 by (i) and subtract (ii) from (i). We have 2p+10q=2+2p+q=+911q=11 q=1
Put the value of q in (i). We have, p+5(1)=1p=4 Hence we have p=4 and q=1.

If p=2ˆiˆj and q=xˆi+3ˆj, then find the value of x such that |p+q|=5.

We calculate p+q=2ˆiˆj+xˆi+3ˆj=(2+x)ˆi+2ˆj Thus |p+q|=(2+x)2+22=x2+4x+4+4=x2+4x+8 But we are given that |p+q|=5x2+4x+8=5x2+4x+8=25x2+4x17=0 This is quadratic equation with a=1, b=4 and c=17, so x=b±b24ac2a=4±424(1)(17)2(1)=4±842=4±2212=2±21 Thus x=2±21.