Question

Let $\vec{a}$, $\vec{b}$ and $\vec{c}$ be three non-zero vectors such that no two of these are collinear. If the vectors $\vec{a}+2\vec{b}$ is collinear with $\vec{c}$ and $\vec{b}+3\vec{c}$ is collinear with $\vec{a}$, then $\vec{a}+2\vec{b}+6\vec{c}$ equals to

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

Answer 1

Answer: (D)

If $\vec{a}+2\vec{b}$ is collinear with $\vec{c}$, then
$\vec{a}+2\vec{b}=t\vec{c}\dots \left(i\right)$
Also, if $\vec{b}+3\vec{c}$ is collinear with $\vec{a}$, then
$\vec{b}+3\vec{c}=\lambda \vec{a}\dots \left(ii\right)$
$\Rightarrow \vec{b}=\lambda \vec{a}-3\vec{c}$
On putting, this value in eq. $\left(i\right)$, we get
$\vec{a}+2\left(\lambda \vec{a}-3\vec{c}\right)=t\vec{c}$
$\vec{a}+2\lambda \vec{a}-6\vec{c}=t\vec{c}$
$\Rightarrow \left(\vec{a}-6\vec{c}\right)=t\vec{c}-2\lambda \vec{a}$
On comparing, we get
$1=-2\lambda$
$\lambda =-1/2$ and $-6=t$
$\Rightarrow t=-6$
From eq. $\left(i\right)$
$\vec{a}+2\vec{b}=-6\vec{c}$
$\Rightarrow \vec{a}+2\vec{b}+6\vec{c}=\vec{0}$