Question

Find the shortest distance between the lines $r=(2i-j)+\lambda (2i+j-3k)$ and $r=(i-j+2k)+\mu (2i+j-5k)$.

Answer 1

The equations of the givn lines are

$\overrightarrow{r_1}=(2\hat{i}-\hat{j})+\lambda (2\hat{i}+\hat{j}-3\hat{k})$

$\overrightarrow{r_2}=(\hat{i}-\hat{j}+2\hat{k})+\mu (2\hat{i}+\hat{j}-5\hat{k})$

Shortest between the lines $\overrightarrow{r_1}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$ and $\overrightarrow{r_2}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}$ is

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

where $\overrightarrow{a_1}=2\hat{i}-\hat{j}$, $\overrightarrow{b_1}=2\hat{i}+\hat{j}-3\hat{k}$,$\overrightarrow{a_2}=\hat{i}-\hat{j}+2\hat{k}$,$\overrightarrow{b_2}=2\hat{i}+\hat{j}-5\hat{k}$

$(\overrightarrow{b_1}\times \overrightarrow{b_2})=\left|\begin{array}{ccc}i& j& k\\ 2& 1& -3\\ 2& 1& -5\end{array}\right|$

$=\hat{i}(-5+3)-\hat{j}(-10+6)+\hat{k}(-5+3)$

$=-2\hat{i}+4\hat{j}-2\hat{k}$

$|\overrightarrow{b_1}\times \overrightarrow{b_2}|=\sqrt{4+16+4}=\sqrt{24}=2\sqrt{6}$

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

$d=\left|\frac{(-2\hat{i}+4\hat{j}-2\hat{k}).(-\hat{i}+2\hat{k})}{2\sqrt{6}}\right|$

$d=\left|\frac{2-4}{2\sqrt{6}}\right|=\frac{1}{\sqrt{6}}$

$\therefore$ the shortest distance between the two lines is $\frac{1}{\sqrt{6}}$units.