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 $\lambda \neq-1$, 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
$\therefore \lambda=-1$
$\therefore \vec{ A }+\vec{ B }+\vec{ C }=\vec{0}$