Question

Find the shortest distance between the following pairs of parallel lines.
$\overrightarrow{r}=(\hat{i}+\hat{j})+\lambda (2\hat{i}+\hat{j}-\hat{k})$ and $\overrightarrow{r}=(\hat{i}-\hat{j}+\hat{k})+\mu (-\hat{i}+\hat{j}-\hat{k})$

Answer 1

The equations of the given lines are

$\overrightarrow{r}=(\hat{i}+\hat{j})+\lambda (2\hat{i}+\hat{j}-\hat{k})$ and $\overrightarrow{r}=(\hat{i}-\hat{j}+\hat{k})+\mu (-\hat{i}+\hat{j}-\hat{k})$

It is known that the shortest distance between the lines $\overrightarrow{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$ and $\overrightarrow{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}$ is given by

$d=\left|\frac{(\overrightarrow{b_1}\times \overrightarrow{b_2}).(\overrightarrow{a_2}-\overrightarrow{a_1})}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}\right|$ ......$\left(1\right)$

Comparing the equations,

$\overrightarrow{a_1}=\hat{i}+\hat{j}$

$\overrightarrow{a_2}=\hat{i}-\hat{j}+\hat{k}$

$\overrightarrow{b_1}=2\hat{i}+\hat{j}-\hat{k}$

$\overrightarrow{b_2}=-\hat{i}+\hat{j}-\hat{k}$

$\overrightarrow{a_2}-\overrightarrow{a_1}=\hat{i}-\hat{j}+\hat{k}-\hat{i}-\hat{j}=-2\hat{j}+\hat{k}$

$\overrightarrow{b_1}\times \overrightarrow{b_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -1\\ -1& 1& -1\end{array}\right|$

$=\hat{i}(-1+1)-\hat{j}(-2-1)+\hat{k}(2+1)$

$=3\hat{j}+3\hat{k}$

$|\overrightarrow{b_1}\times \overrightarrow{b_2}|=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt{2}$

$\therefore d=\left|\frac{(\overrightarrow{b_1}\times \overrightarrow{b_2}).(\overrightarrow{a_2}-\overrightarrow{a_1})}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}\right|$

$=\left|\frac{(3\hat{j}+3\hat{k}).(-2\hat{j}+\hat{k})}{3\sqrt{2}}\right|$

$=\left|\frac{-6+3}{3\sqrt{2}}\right|$

$=\frac{1}{\sqrt{2}}$

$\therefore$ the shortest distance between the two lines is $\frac{1}{\sqrt{2}}$