Question

The distance between the parallel lines $\overrightarrow{r}=2\hat{i}+3\hat{j}-\hat{k}+\lambda (\hat{i}-\hat{j}+2\hat{k})$ and $\overrightarrow{r}=-3\hat{i}+4\hat{j}+\hat{k}+\mu (\hat{i}-\hat{j}+2\hat{k})$ is

• A. $2\sqrt{\frac{22}{3}}$
• B. $\sqrt{\frac{80}{3}}$
• C. $\frac{14}{\sqrt{6}}$
• D. $7$

Answer 1

The correct option is A $2\sqrt{\frac{22}{3}}$

$\overrightarrow{a_1}=2\hat{i}+3\hat{j}-\hat{k},\overrightarrow{a_2}=-3\hat{i}+4\hat{j}+\hat{k},$
$\overrightarrow{b}=\hat{i}-\hat{j}+2\hat{k},\overrightarrow{a_1}-\overrightarrow{a_2}=5\hat{i}-\hat{j}-2\hat{k}$
$\overrightarrow{b}\times (\overrightarrow{a_1}-\overrightarrow{a_2})=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 2\\ 5& -1& -2\end{array}\right|=4\hat{i}+12\hat{j}+4\hat{k}$
$\Rightarrow |\overrightarrow{b}\times (\overrightarrow{a_1}-\overrightarrow{a_2}\left)\right|=4\sqrt{11},\left|\overrightarrow{b}\right|=\sqrt{6}$
Distance between parallel lines= $\frac{\Rightarrow |\overrightarrow{b}\times (\overrightarrow{a_1}-\overrightarrow{a_2}\left)\right|}{\left|\overrightarrow{b}\right|}=\frac{4\sqrt{11}}{\sqrt{6}}=2\sqrt{\frac{22}{3}}$