Question

$\vec{A}$ , $\vec{B}$ , $\vec{C}$ are three non-zero vectors; no two of them are parallel. If $\vec{A}+\vec{B}$ is collinear to $\vec{C}$ and $\vec{B}+\vec{C}$ is collinear to $\vec{A}$ , then $\vec{A}+\vec{B}+\vec{C}$ is equal to

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

Answer 1

Answer: (D)

Since, $\vec{A}+\vec{B}$ is collinear to $\vec{C}$ and $\vec{B}+\vec{C}$ is
collinear to $\vec{A}$.
$\therefore \vec{A}+\vec{B}=\lambda \vec{C}\text{ and }\vec{B}+\vec{C}=\mu \vec{A}$
where $\lambda$ and $\mu$ are scalars.
$\Rightarrow \vec{A}+\vec{B}+\vec{C}=(\lambda +1)\vec{C}$
and $\vec{A}+\vec{B}+\vec{C}=(\mu +1)\vec{A}$
$\Rightarrow (\lambda +1)\vec{C}=(\mu +1)\vec{A}$
If $\lambda \ne -1$, then
$(\lambda +1)\vec{C}=(\mu +1)\vec{A}$
$\vec{C}=\frac{\mu +1}{\lambda +1}\vec{A}$
$\Rightarrow \vec{C}$ and $\vec{A}$ are collinear.
This is a contradiction to the given condition
$\therefore \lambda =-1$
$\therefore \vec{A}+\vec{B}+\vec{C}=\vec{0}$