Question

If $O A =2 \hat{ i }+2 \hat{ j }+\hat{ k }, O B =2 \hat{ i }+4 \hat{ j }+4 \hat{ k }$ and the length of the internal bisector of $\angle B O A$ of triangle $A O B$ is $k$, then $9 k^{2}=$

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

Answer 1

Answer: (B)

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