Question

Let $\overrightarrow{AB}=3i+j-k$ and $\overrightarrow{AC}=i-j+3k$. If the point $P$ on the line segment $BC$ is equidistant from $AB$ and $AC$, then $\overrightarrow{p}$ is

• A. $2i-k$
• B. $i-k$
• C. $2i+k$
• D. None of these

Answer 1

The correct option is C $2i+k$

$\overrightarrow{AB}=3i+j-k;\overrightarrow{AC}=i-j+3k$

Observe that $|AB|=|AC|=\sqrt{11}$ units

$\therefore ABC$ is isosceles $P$ is the mid point of $BC$ (treating $A$ as origin)

$\overrightarrow{p}=\frac{1}{2}(4i+2k)=2i+k$