Question

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

Answer: (C)

$\frac{5}{3}(\hat{i}+7\hat{j}+2\hat{k})$

The required vector $\overrightarrow{C}$ is given by

$\overrightarrow{C}=\lambda \left(\frac{\overrightarrow{a}}{\left|\overrightarrow{a}\right|}+\frac{\overrightarrow{b}}{\left|\overrightarrow{b}\right|}\right)$

$\Rightarrow \overrightarrow{C}=\lambda \left(\frac{1}{9}(7\hat{i}-4\hat{j}+4\hat{k})+\frac{1}{3}(-2\hat{i}-\hat{j}+2\hat{k})\right)$

$\Rightarrow \overrightarrow{C}=\frac{\lambda }{9}(\hat{i}-7\hat{j}+2\hat{k})\Rightarrow \left|\overrightarrow{C}\right|=\pm \frac{\lambda }{9}\sqrt{1+49+4}=\pm \frac{\lambda }{9}\sqrt{54}$

But $\left|\overrightarrow{C}\right|=5\sqrt{6}$ $\Rightarrow 5\sqrt{6}=\frac{\lambda }{9}\sqrt{54}\Rightarrow \lambda =\pm 15$

Hence, $\overrightarrow{C}=\pm \frac{15}{9}(\hat{i}-7\hat{j}+2\hat{k})$ $\Rightarrow \overrightarrow{C}=\pm \frac{5}{3}(\hat{i}-7\hat{j}+2\hat{k})$