The vertices of triangle $\mathrm{ABC}$ are $\mathrm{A} \equiv(3,0,0) ; \mathrm{B} \equiv(0,0,4)$; $\mathrm{C} \equiv(0,5,4)$. Find the position vector of the point in which the bisector of angle A meets $\mathrm{BC}$ is

• A

  $5 \hat{\mathrm{i}}+12 \hat{\mathrm{j}}$
• Correct answer
  B

  $\frac{5 \hat{\mathrm{i}}+12 \hat{\mathrm{k}}}{3}$
• C

  $\frac{5 \hat{\mathrm{i}}+12 \hat{\mathrm{j}}}{13}$
• D

  $\frac{5 \hat{\mathrm{i}}-12 \hat{\mathrm{j}}}{3}$

Solution

Let $\mathrm{AD}$ be the angle bisector of angle $\mathrm{A}$ which divides $\mathrm{BC}$ in the ratio $\mathrm{AB}: \mathrm{AC}$ Here $\mathrm{AB}=\sqrt{9+16}=\sqrt{25}$ and $$ \begin{aligned} & \mathrm{AC}=\sqrt{9+25+16} \\ & =\sqrt{50} \end{aligned} $$ $\therefore \mathrm{D}$ divides $\mathrm{BC}$ in the ratio $\sqrt{25}: \sqrt{50}$ i.e., $1: 2$ $\therefore$ Position vector of $\mathrm{D}=\frac{(4)(2) \hat{\mathrm{k}}+5 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}}{1+2}=\frac{5 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}}{3}$

Hence, the correct answer is option: 'B'.