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.
Question 3
If →r=ˆi−9ˆj, →a=ˆi+2ˆj and →b=5ˆi−ˆj, determine the real number p and q such that →r=p→a+q→b.
Solution
Given
→r=p→a+q→b.
We put the values of →r,→a and →b in the given equation. We get
ˆi−9ˆj=p(ˆi+2ˆj)+q(5ˆi−ˆj)
⟹ˆi−9ˆj=(p+5q)ˆi+(2p−q)ˆj.
By comparing the coeffients of ˆi and ˆj, we have,
p+5q=1…(i)
2p−q=−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)=1⟹p=−4
Hence we have p=−4 and q=1.
Question 4
If →p=2ˆi−ˆj and →q=xˆi+3ˆj, then find the value of x such that |→p+→q|=5.
Solution
We calculate →p+→q=2ˆi−ˆj+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|=5⟹√x2+4x+8=5⟹x2+4x+8=25⟹x2+4x−17=0 This is quadratic equation with a=1, b=4 and c=−17, so x=−b±√b2−4ac2a=−4±√42−4(1)(−17)2(1)=−4±√842=−4±2√212=−2±√21 Thus x=−2±√21.
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