Question

The position vector of $A$ and $B$ are $2\hat{i}+2\hat{j}+\hat{k}$ and $2\hat{i}+4\hat{j}+4\hat{k}$
The length of the internal bisector of $\angle BOA$ triangle $AOB$ is

• A. $\sqrt{\frac{136}{9}}$
• B. $\frac{\sqrt{136}}{9}$
• C. $\sqrt{\frac{217}{9}}$
• D. $\frac{20}{3}$

Answer 1

Answer: (A)

We have,
$OA=\left|2\hat{i}+2\hat{j}+\hat{k}\right|=3;OB=\left|2\hat{i}+4\hat{j}+4\hat{k}\right|=6$
Since the internal bisector divides opposite side in the
ratio of adjacent sides
$\therefore \frac{AC}{BC}=\frac{3}{6}=\frac{1}{2}$
where $OC$ is the bisector of $\angle BOA.$
$\therefore$ Position vector of $C$ is
$\frac{2\left(2\hat{i}+2\hat{j}+\hat{k}\right)+\left(2\hat{i}+4\hat{j}+4\hat{k}\right)}{2+1}=2\hat{i}+\frac{8}{3}\hat{j}+2\hat{k}$
$\therefore OC=\left|2\hat{i}+\frac{8}{3}\hat{j}+2\hat{k}\right|=\sqrt{\frac{136}{9}}$