Question

Let $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ be three non-zero vectors such that no two of these are collinear. If the vector $\overrightarrow{a}+2\overrightarrow{b}$ is collinear with $\overrightarrow{c}$ and $\overrightarrow{b}+3\overrightarrow{c}$ is collinear with $\overrightarrow{a}$ ($\lambda$ being some non-zero scalar), then $\overrightarrow{a}+2\overrightarrow{b}+6\overrightarrow{c}$ equals

• A. $\lambda \overrightarrow{c}$
• B. $\lambda \overrightarrow{b}$
• C. $\lambda \overrightarrow{a}$
• D. $\overrightarrow{0}$

Answer 1

Answer: (D)

$3\vec{a}+7\vec{b}=p\vec{c}$
$3\vec{b}+2\vec{c}=q\vec{a}$
$3\vec{b}+2\vec{c}=q\left(\frac{p\vec{c}-7\vec{b}}{3}\right)$
$9\vec{b}+6\vec{c}=qp\vec{c}-7q\vec{b}$
$\left(9+7q\right)\vec{b}\left(qp-6\right)$ is b and c is not collinear so
$q-9=79$
$q=-\frac{9}{7}$
$3\vec{b}+2\vec{c}=q\vec{a}$
$3\vec{b}+2\vec{c}=\frac{-9}{7}\vec{a}$
$21\vec{b}+14\vec{c}=-9\vec{a}$
$9\vec{a}+21\vec{b}+14\vec{c}=\vec{0}$