Question

The vector $c$, directed along the internal bisector of the angle between the vectors $a=7\hat{i}+4\hat{j}-4\hat{k}$ and $b=-2\hat{i}-\hat{j}+2\hat{k}$ with $\left|c\right|=5\sqrt{6}$, is

• A. $\frac{5}{3}(\hat{i}-7\hat{j}+2\hat{k})$
• B. $\frac{5}{3}(5\hat{i}+5\hat{j}+2\hat{k})$
• C. $\frac{5}{3}(\hat{i}+7\hat{j}+2\hat{k})$
• D. $\frac{5}{3}(-5\hat{i}+5\hat{j}+2\hat{k})$

Answer 1

The correct option is A $\frac{5}{3}(\hat{i}-7\hat{j}+2\hat{k})$

The required vector $c$ is given by,
$c=\lambda (a+b)$
$\Rightarrow c=\lambda (\frac{a}{\left|a\right|}+\frac{b}{\left|b\right|})$
$\Rightarrow c=\lambda (\frac{1}{9}(7\hat{i}-4\hat{j}-4\hat{k})+\frac{1}{3}(-2\hat{i}-\hat{j}+2\hat{k}))$
$\Rightarrow c=\frac{\lambda }{9}(\hat{i}-7\hat{j}+2\hat{k})$
$\Rightarrow \left|c\right|=\pm \frac{\lambda }{9}\sqrt{1+49+4}=\pm \frac{\lambda }{9}\sqrt{54}$
But $\left|c\right|=5\sqrt{6}$
$\Rightarrow \pm \frac{\lambda }{9}\sqrt{54}=5\sqrt{6}$
$\Rightarrow \lambda =\pm 15$
Hence, $c=\pm \frac{\lambda }{9}(\hat{i}-7\hat{j}+2\hat{k})=\pm \frac{5}{3}(\hat{i}-7\hat{j}+2\hat{k})$