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 $|b|=\mu \left[\frac{3\hat{i}+4\hat{j}}{5}-\frac{-\hat{i}+2\hat{j}-2\hat{k}}{3}\right]$
$=\mu \left(\frac{14\hat{i}+2\hat{j}+10\hat{k}}{15}\right)$
$|b|=\frac{2}{3}\sqrt{3}=\frac{\mu }{15}\sqrt{196+4+100}$
$\Rightarrow \mu =1$
$\therefore a-b=\frac{4\hat{i}+2\hat{j}-10\hat{k}}{15}-\frac{14\hat{i}+2\hat{j}+10\hat{k}}{15}$
$=\frac{2}{3}(-\hat{i}+2\hat{j}-2\hat{k})$