Question

The position vectors of the $A$ and $B$ to $O$ are $2i+2j+k$ and $2i+4j+4k$.

Length of internal bisector of $\angle BOA$ of $△AOB$ is

• A. $\sqrt{\frac{136}{9}}$
• B. $\sqrt{\frac{136}{3}}$
• C. $\frac{20}{3}$
• D. $\frac{25}{3}$

Answer 1

The correct option is A

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

Explanation for the correct option:

Step-1: Solve for the required ratio of internal bisection

Given that the position vectors of the $A$ and $B$ to $O$ are $2i+2j+k$ and $2i+4j+4k$

$\Rightarrow OA=$$2i+2j+k$ and $OB=$$2i+4j+4k$

If $P=xi+yj+zk$ then $\left|P\right|=\sqrt{x^2+y^2+z^2}$

$\therefore$ $\left|OA\right|=\sqrt{2^2+2^2+1^2}=\sqrt{9}=3$

and $\left|OB\right|=\sqrt{2^2+4^2+4^2}=\sqrt{36}=6$

Let $OD$ be the angle bisector of $\angle BOA$

According to angle bisector theorem, the angle bisector of a triangle divides the opposite side into two parts that are proportional to the other two sides of the triangle.

$\Rightarrow \frac{AD}{DB}=\frac{\left|OA\right|}{\left|OB\right|}=\frac{3}{6}=\frac{1}{2}$

Thus point $D$ divides $AB$ in the ratio $1:2$ internally

Step-2: Solve for the length of angle bisector

We know that the position vector of $R$ which divides $\overrightarrow{PQ}$ in the ratio $a:b$ internally is given by $\frac{a·\overrightarrow{OQ}+b·\overrightarrow{OP}}{a+b}$

$\therefore$ The position vector of $D$$=$$\frac{1·\overrightarrow{OB}+2·\overrightarrow{OA}}{1+2}$

$\begin{array}{ccl}& =& \frac{1\left(2i+4j+4k\right)+2\left(2i+2j+k\right)}{3}\\ & =& \frac{6i+8j+6k}{3}\\ \Rightarrow \overrightarrow{OD}& =& 2i+\frac{8}{3}j+2k\end{array}$

$\therefore$Length of the angle bisector is $\left|OD\right|=\sqrt{2^2+{\left(\frac{8}{3}\right)}^2+2^2}$

$\begin{array}{ccc}& =& \sqrt{4+\frac{64}{9}+4}\\ & =& \sqrt{\frac{36+64+36}{9}}\\ \Rightarrow \left|OD\right|& =& \sqrt{\frac{136}{9}}\end{array}$

Hence, the correct option is option(A) i.e. $\sqrt{\frac{136}{9}}$