Question

Let $\overline{a},\overline{b},\overline{c}$ be three non-zero vectors whcih are pairwise non-collinear. If $\overline{a}+3\overline{b}$ is collinear with $\overline{c}$ and $\overline{b}+2\overline{c}$ is collinear with $\overline{a}$, then $\overline{a}+3\overline{b}+6\overline{c}$ is ______

Answer 1

Given,

$(\overrightarrow{a}+3\overrightarrow{b})$ is collinear with $\overrightarrow{c}$

$\therefore \overrightarrow{a}+3\overrightarrow{b}=\lambda \overrightarrow{c}$

$\Rightarrow \overrightarrow{c}=\frac{1}{\lambda }\overrightarrow{a}+\frac{3}{\lambda }\overrightarrow{b}$

$\overrightarrow{b}+2\overrightarrow{c}=t\overrightarrow{a}$

$\Rightarrow \overrightarrow{b}+2(\frac{\overrightarrow{a}}{\lambda }+\frac{3}{\lambda }\overrightarrow{b})=t\overrightarrow{a}$

$\Rightarrow \overrightarrow{b}+\frac{6\overrightarrow{b}}{\lambda }+\frac{2\overrightarrow{a}}{\lambda }-t\overrightarrow{a}=0$

$\Rightarrow \overrightarrow{b}(1+\frac{6}{\lambda })+\overrightarrow{a}(\frac{2}{\lambda }-t)=0$

$\therefore 1+\frac{6}{\lambda }=0$ $\frac{2}{\lambda }=t$

$\therefore \lambda =-6$ $\frac{2}{-6}=t$

$\therefore t=\frac{-1}{3}$

$\therefore \overrightarrow{a}+3\overrightarrow{b}=-6\overrightarrow{c}$

$\Rightarrow \overrightarrow{a}+3\overrightarrow{b}+6\overrightarrow{c}=\overrightarrow{0}$