Question

If $a$ and $b$ are respectively the internal and external bisectors of the angles between the vectors $-\hat{ i }+2 \hat{ j }-2 \hat{ k }$ and $3 \hat{ i }+4 \hat{ j }$ and $| a |=\frac{2}{3} \sqrt{6},| b |=\frac{2}{3} \sqrt{3}$, then one of the values of
$a - b$ is

• A. $\frac{1}{10}(-8 \hat{ i }+11 \hat{ j }-2 \hat{ k })$
• B. $\frac{2}{3}(-\hat{ i }+2 \hat{ j }-2 \hat{ k })$
• C. $\frac{1}{15}(9 \hat{ i }-11 \hat{ j }+3 \hat{ k })$
• D. $\frac{1}{12}(2 \hat{ i }-3 \hat{ j }-\hat{ k })$

Answer 1

Answer: (B)

We have,
$-\hat{ i }+2 \hat{ j }-2 \hat{ k }$ and $\hat{ i }+4 \hat{ j }$
$a$ and $b$ are the angle bisector of vectors
$ \therefore | a |=\lambda\left|\frac{3 \hat{ i }+4 \hat{ j }}{5}+\frac{-\hat{ i }+2 \hat{ j }-2 \hat{ k }}{3}\right| $
$=\lambda\left|\frac{4 \hat{ i }+22 \hat{ j }-10 \hat{ k } \mid}{15}\right| $
$| a |=\frac{2}{3} \sqrt{6}=\frac{\lambda}{15} \sqrt{16+484+100}$
$ \Rightarrow \lambda=1 $
Similarly $| b |=\mu\left[\frac{3 \hat{ i }+4 \hat{ j }}{5}-\frac{-\hat{ i }+2 \hat{ j }-2 \hat{ k }}{3}\right] $
$=\mu\left(\frac{14 \hat{ i }+2 \hat{ j }+10 \hat{ k }}{15}\right)$
$ | b |=\frac{2}{3} \sqrt{3}=\frac{\mu}{15} \sqrt{196+4+100} $
$ \Rightarrow \mu =1 $
$ \therefore a - b =\frac{4 \hat{ i }+2 \hat{ j }-10 \hat{ k }}{15}-\frac{14 \hat{ i }+2 \hat{ j }+10 \hat{ k }}{15} $
$=\frac{2}{3}(-\hat{ i }+2 \hat{ j }-2 \hat{ k }) $