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 $\vec{r}=\pm\left(\hat{i}-7\,\hat{j}+2\,\hat{k}\right)$