Question

The position vectors of the points $P$ and $Q$ with respect to the origin $O$ are $\overrightarrow{a}=\hat{i}+3\hat{j}-2\hat{k}$ and $\overrightarrow{b}=3\hat{i}-\hat{j}-2\hat{k}$, respectively. If $M$ is a point on $PQ$, such that $OM$ is the bisector of $POQ$, then $\overrightarrow{OM}$ is

• A. $2(\hat{i}-\hat{j}+\hat{k})$
• B. $2\hat{i}+\hat{j}-2\hat{k}$
• C. $2(-\hat{i}+\hat{j}-\hat{k})$
• D. $2(\hat{i}+\hat{j}+\hat{k})$

Answer 1

Answer: (B)

$2\hat{i}+\hat{j}-2\hat{k}$

Since $|\overrightarrow{OP}|=|\overrightarrow{OQ}|=\sqrt{14},\Delta$OPQ is isosceles.

Hence, the internal bisector $OM$ is perpendicular to $PQ$ and $M$ is the midpoint of $P$ and $Q$.

$\therefore \overrightarrow{OM}=\frac{1}{2}(\overrightarrow{OP}+\overrightarrow{OQ})$

$=2\hat{i}+\hat{j}-2\hat{k}$