Question

If a vector $\overrightarrow{r}$ of magnitude $3\sqrt{6}$ is directed along the bisector of the angle between the vector $\overrightarrow{a}=7\hat{i}-4\hat{j}-4\hat{k}$ and $\overrightarrow{b}=-2\hat{i}-\hat{j}+2\hat{k}$ , then $\overrightarrow{r}$ is equal to

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

Answer 1

Answer: (A)

The reqd. vector $\vec{r}=t\left(\vec{a}+\vec{b}\right),t$ is a scalar.
or $\vec{r}=t\left[\frac{1}{9}\left(7\hat{i}-4\hat{j}-4\hat{k}\right)+\frac{1}{3}\left(-2\hat{i}-\hat{j}-2\hat{k}\right)\right]$
$=\frac{1}{9}\left(\hat{i}-7\hat{j}+2\hat{k}\right)$
Since $\left|\vec{r}\right|=3\sqrt{6}$
$\therefore {\left|\vec{r}\right|}^2=54$
$\Rightarrow \frac{t^2}{81}\left(1+49+4\right)=54$
$\Rightarrow t^2=81$
$\Rightarrow t=\pm 9$
Hence $\vec{r}=\pm \left(\hat{i}-7\hat{j}+2\hat{k}\right)$