Question

The vector $\overrightarrow{c}$ directed along the internal bisector of the angle between the vectors $\overrightarrow{a}=2\hat{i}-\hat{j}-2\hat{k}$ and $\overrightarrow{b}=2\hat{i}+2\hat{j}+\hat{k}$ with $\left|\overrightarrow{c}\right|=3\sqrt{2}$ is

• A. $4\hat{i}-\hat{j}-\hat{k}$
• B. $4\hat{i}-\hat{j}+\hat{k}$
• C. $4\hat{i}+\hat{j}-\hat{k}$
• D. $-4\hat{i}+\hat{j}+\hat{k}$

Answer 1

Answer: (C)

$4\hat{i}+\hat{j}-\hat{k}$

The vector along internal bisectors of angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is along $\frac{\overrightarrow{a}}{\left|\overrightarrow{a}\right|}+\frac{\overrightarrow{b}}{\left|\overrightarrow{b}\right|}$
$\therefore \frac{2\hat{i}-\hat{j}-2\hat{k}}{|2\hat{i}-\hat{j}-2\hat{k}|}+\frac{2\hat{i}+2\hat{j}+\hat{k}}{|2\hat{i}+2\hat{j}+\hat{k}|}$
$=\frac{4\hat{i}+\hat{j}-\hat{k}}{3}$ on simplification
$\therefore |2\hat{i}-\hat{j}-2\hat{k}|=\sqrt{4+1+4}=3$
$\Rightarrow \left|\overrightarrow{c}\right|=3\sqrt{2}$(given)
$\Rightarrow \overrightarrow{c}=4\hat{i}+\hat{j}-\hat{k}$