Question

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

• A. $\sqrt{15}$
• B. $\frac{\sqrt{58}}{3}$
• C. $3\sqrt{2}$
• D. $\sqrt{26}$

Answer 1

The correct option is D $\sqrt{26}$

$\overrightarrow{a_1}=\hat{i}+2\hat{j}+3\hat{k},\overrightarrow{a_2}=2\hat{i}-\hat{j}-\hat{k},$
$\overrightarrow{b}=\hat{i}-\hat{j}+\hat{k},\overrightarrow{a_1}-\overrightarrow{a_2}=-\hat{i}+3\hat{j}+4\hat{k}$
$\overrightarrow{b}\times (\overrightarrow{a_1}-\overrightarrow{a_2})=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 1\\ -1& 3& 4\end{array}\right|=-7\hat{i}-5\hat{j}+2\hat{k}$
$\Rightarrow |\overrightarrow{b}\times (\overrightarrow{a_1}-\overrightarrow{a_2}\left)\right|=\sqrt{49+25+4}=\sqrt{78},\left|\overrightarrow{b}\right|=\sqrt{1+1+1}$
Distance between parallel lines= $\frac{\Rightarrow |\overrightarrow{b}\times (\overrightarrow{a_1}-\overrightarrow{a_2}\left)\right|}{\left|\overrightarrow{b}\right|}=\frac{\sqrt{78}}{\sqrt{3}}=\sqrt{26}$
Hence, distance between two parallel lines is $\sqrt{26}$