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 $?BOA$ 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)

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

Answer 2

The correct answer is A:$\sqrt{\frac{136}{9}}$
Given that;
Position vector of A & B are;
$2\hat i +2\hat j+\hat k$ &
$2\hat i+4\hat j+4\hat k$ respectively
$\therefore|OA|=\sqrt{2^2+2^2+1^2}=3units$
$|OB|=\sqrt{2^2+4^2+4^2}=6units$
Now let internal bisector OP divides AB with ratio $1\ratio 2$
$\therefore$ Position vector of D is $(2,\frac{8}{3},2)$
$\therefore |OD|=\sqrt{2^2+(\frac{8}{3})^2+2^2}$
$=\sqrt{\frac{136}{9}}$