Question

Three non-zero non-collinear vectors $\hat{a},\hat{b}$ and $\hat{c}$ are such that $\hat{a}+3\hat{b}$ is collinear with $\hat{c}$, while $\hat{c}$ is $3\hat{b}+2\hat{c}$ collinear with â. Then $\hat{a}+3\hat{b}+2\hat{c}$ equals to

• A. 0
• B. $2\hat{a}$
• C. $3\hat{b}$
• D. $4\hat{c}$

Answer 1

Answer: (A)

Given, $a+3b$ is collinear with $c$.
$\therefore a+3b=\lambda c$
Or $a+3b-\lambda c=0...(i)$
And $3b+2c$ is collinear with $a$.
$\therefore 3b+2c=\mu a$
$3b+2c-\mu a=0...(ii)$
From Eqs. (i) and (ii), we get
$a+3b-\lambda c=3b+2c-\mu a$
On equating $c$, we get
$\lambda =-2$
On putting $\lambda =-2$ in Eq. (i), we get
$a+3b+2c=0$