Question

If $OA=2\hat{i}+2\hat{j}+\hat{k},OB=2\hat{i}+4\hat{j}+4\hat{k}$ and the length of the internal bisector of $\angle BOA$ of triangle $AOB$ is $k$, then $9k^2=$

• A. $\sqrt{225}$
• B. 136
• C. 712
• D. 20

Answer 1

Answer: (B)

Given vectors
$OA=2i+2j+k$
and $OB=2\hat{i}+4\hat{j}+4\hat{k}$
$|OA|=\sqrt{4+4+1}=3=m(\text{ let })$ and
$|OB|=\sqrt{4+16+16}=6=n$ (let)
$\because$ The angle bisector of $\angle BOA$ intersect the side $AB$ at point $P$ in the ratio $3:6=1:2$ so
$OP=\frac{2(OA)+1(OB)}{3}$
$=\frac{6\hat{i}+8\hat{j}+6\hat{k}}{3}=2\hat{i}+\frac{8}{3}\hat{j}+2\hat{k}$
$\therefore |OP|=\sqrt{z^2+{\left(\frac{8}{3}\right)}^2+2^2}$
$=\sqrt{4+\frac{64}{9}+4}$
$=\sqrt{\frac{72+64}{9}}=\sqrt{\frac{136}{9}}=k$
$\therefore 9k^2=9\left(\frac{136}{9}\right)=136$