Question

The position vectors 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 $\angle BOA$ of the triangle $AOB$ is

• A. $\sqrt{\frac{136}{9}}$
• B. $\sqrt{\frac{139}{9}}$
• C. $\frac{20}{3}$
• D. $\sqrt{\frac{217}{9}}$

Answer 1

Answer: (A)

$\sqrt{\frac{136}{9}}$

Here $OA=3,OB=6$

$\therefore$ Internal bisector $OD$ divides $AB$ in the ratio $1:2$

$\Rightarrow$ Position vector of $D$ is $(2,\frac{8}{3},2)$

$\therefore \left|OD\right|=\sqrt{4+\frac{64}{9}+4}=\sqrt{\frac{136}{9}}$