Question

$AB=3\hat{i}+\hat{j}-\hat{k}$ and $AC=\hat{i}-\hat{j}+3\hat{k}.$ If the point $P$ on the line segment $BC$ is equdistant from $AB$ and $AC,$ then $AP$ is

• A. $2\hat{i}-\hat{k}$
• B. $\hat{i}-2\hat{k}$
• C. $2\hat{i}+\hat{k}$
• D. None of these

Answer 1

Answer: (C)

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

A point equidistant from $AB$ and $AC$ is on the bisector of the$\angle BAC.$

A vector along the internal bisector of the $\angle BAC$

$=\frac{AB}{\left|AB\right|}+\frac{AC}{\left|AC\right|}$

$=\frac{3\hat{i}+\hat{j}+\hat{k}}{\sqrt{9+1+1}}+\frac{\hat{i}-\hat{j}+3\hat{k}}{\sqrt{1+1+9}}=\frac{1}{\sqrt{11}}(4\hat{i}+2\hat{k})$

$\therefore AP=t(2\hat{i}+\hat{k})$

$\therefore BP=AP-AB=t(2\hat{i}+\hat{k})-(3\hat{i}+\hat{j}-\hat{k})$

$=(2t-3)\hat{i}-\hat{j}+(t+1)k$

Also $BC=AC-AB=(\hat{i}-\hat{j}+3\hat{k})-(3\hat{i}+\hat{j}-\hat{k})$

$=-2\hat{i}-2\hat{j}+4\hat{k}.$

But $BP=sBC.$

$\therefore (2t-3)\hat{i}-\hat{j}+(t+1)k=s(-2\hat{i}-2\hat{j}+4\hat{k})$

$\therefore 2t-3=-2s,-1=-2s,t+1=4s$

$\therefore s=\frac{1}{2}$ and $t=1$ $\therefore AP=2\hat{i}+\hat{k}.$