Question

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

Answer 1

The equations of the given lines are
$\overrightarrow{r}=(\hat{i}+2\hat{j}+\hat{k})+\lambda (\hat{i}-\hat{j}+\hat{k})$ and
$\overrightarrow{r}=\widehat{2i}-\hat{j}-\hat{k}+\mu (\widehat{2i}+\hat{j}+2\hat{k})$
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,
$d=\left|\frac{(\overrightarrow{b_1}\times \overrightarrow{b_2})(\overrightarrow{a_1}-\overrightarrow{a_2})}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}\right|.....\left(1\right)$
Comparing the given equations, we obtain
$\overrightarrow{a_1}=\hat{i}+2\hat{j}+\hat{k}$
$\overrightarrow{b_1}=\hat{i}-\hat{j}+\hat{k}$
$\overrightarrow{a_2}=2\hat{i}-\hat{j}-\hat{k}$
$\overrightarrow{b_2}=2\hat{i}+\hat{j}+2\hat{k}$
$\overrightarrow{a_2}-\overrightarrow{a_1}=(2\hat{i}-\hat{j}-\hat{k})-(\hat{i}+2\hat{j}+\hat{k})=(\hat{i}-3\hat{j}-2\hat{k})$
$\overrightarrow{b_1}\times \overrightarrow{b_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{j}\\ 1& -1& 1\\ 2& 1& 2\end{array}\right|$
$\overrightarrow{b_1}\times \overrightarrow{b_2}=(-2-1)\hat{i}-(2-2)\hat{j}+(1+2)\hat{k}=-3\hat{i}+3\hat{k}$
$\Rightarrow$ $|\overrightarrow{b_1}\times \overrightarrow{b_2}|=\sqrt{{(-3)}^2+{\left(3\right)}^2}=\sqrt{18}=3\sqrt{2}$
Substituting all the values in equation (1), we obtain
$d=\left|\frac{(\widehat{3\hat{i}}+3\hat{k}).(\hat{i}-3\hat{j}-2\hat{k})}{3\sqrt{2}}\right|$
$\Rightarrow$ $d=\left|\frac{-3.1+3(-2)}{3\sqrt{2}}\right|$
$\Rightarrow$ $d=\left|\frac{-9}{3\sqrt{2}}\right|$
$\Rightarrow$ $d=\frac{3}{\sqrt{2}}=\frac{3\times \sqrt{2}}{\sqrt{2}\times \sqrt{2}}=\frac{3\sqrt{2}}{2}$
Therefore, the shortest distance between the two lines is $\frac{3\sqrt{2}}{2}$ units.