Question

If a vector $\overrightarrow{r}$ of magnitude $3\sqrt{6}$ is collinear with the bisector of the angle between the vectors $\overrightarrow{a}=7\hat{i}-4\hat{j}-4\hat{k}$ & $\overrightarrow{b}=-2\hat{i}-\hat{j}+2\hat{k}$ then $\overrightarrow{r}=$

• A. $\hat{i}-7\hat{j}+2\hat{k}$
• B. $\hat{i}+7\hat{j}-2\hat{k}$
• C. $\frac{13\hat{i}-\hat{j}-10\hat{k}}{\sqrt{5}}$
• D. $\hat{i}-7\hat{j}-2\hat{k}$

Answer 1

Answer: (A), (C)

$\hat{i}-7\hat{j}+2\hat{k}$
$\frac{13\hat{i}-\hat{j}-10\hat{k}}{\sqrt{5}}$

A vector $\overrightarrow{r}$ of magnitude $3\sqrt{6}$ is collinear with the bisector of the angle between the vectors $\overrightarrow{a}=7\hat{i}-4\hat{j}-4\hat{k}$ & $\overrightarrow{b}=-2\hat{i}-\hat{j}+2\hat{k}$
1.$\overrightarrow{r}=t(\hat{a}+\hat{b})$
$\Rightarrow \overrightarrow{r}=t(\frac{7\hat{i}-4\hat{j}-4\hat{k}}{9}+\frac{-2\hat{i}-\hat{j}+2\hat{k}}{3})$
$\Rightarrow \overrightarrow{r}=t\left(\frac{\hat{i}-7\hat{j}+2\hat{k}}{9}\right)$
but $|\overrightarrow{r}|=3\sqrt{6}$, simplifying for $t$ gives $t=9$
$\therefore \overrightarrow{r}=\hat{i}-7\hat{j}+2\hat{k}$
2.$\overrightarrow{r}=s(\hat{b}-\hat{a})$
$\Rightarrow \overrightarrow{r}=s(\frac{-2\hat{i}-\hat{j}+2\hat{k}}{3}-\frac{7\hat{i}-4\hat{j}-4\hat{k}}{9})$
$\Rightarrow \overrightarrow{r}=s\left(\frac{-13\hat{i}+\hat{j}-10\hat{k}}{9}\right)$
but $|\overrightarrow{r}|=3\sqrt{6}$, simplifying for $s$ gives $s=\frac{9}{\sqrt{5}}$
$\therefore \overrightarrow{r}=\frac{13\hat{i}-\hat{j}-10\hat{k}}{\sqrt{5}}$
Hence, options A and C.