Question

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

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

Answer 1

The correct option is D $0$

As $\overline{a}+2\overline{b}$ is collinear with $\overline{c}$
$\therefore \overline{a}+2\overline{b}=P\overline{c}....\left(i\right)$ and
$\overline{b}+3\overline{c}$ is collinear with $\overline{a}$
$\therefore \overline{b}+3\overline{c}=Q\overline{a}....\left(ii\right)$

Now by $\left(i\right)$ and $\left(ii\right),$ we have

$,\overline{a}-6\overline{c}=P\overline{c}-2Q\overline{a}$
$\Rightarrow \overline{a}(1+2Q)+\overline{c}(-6-P)=0$
$\Rightarrow 1+2Q=0$ and $-P-6=0$
$Q=-\frac{1}{2},P=-6$
Putting these value either in $\left(i\right)$ or in $\left(ii\right),$ we get,
$\overline{a}+2\overline{b}+6\overline{c}=0$