Question

Find the shortest distance between the lines $\overline{r}=(4\hat{i}-\hat{j})+\lambda (\hat{i}+2\hat{j}-3\hat{k})$ and $\overline{r}=(\hat{i}-\hat{n}j+2\hat{k})+\mu (\hat{i}+4\hat{j}-5\hat{k})$, where $\lambda$ and $\mu$ are parameters.

Answer 1

We know that the shortest distance between the line, $\overrightarrow{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$ and $\overrightarrow{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}$ is given as,
$d=\left|\frac{(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot (\overrightarrow{b_1}\times \overrightarrow{b_2})}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}\right|$
Here, $\overrightarrow{a_1}=4\hat{i}-\hat{j},\overrightarrow{a_2}=\hat{i}-\hat{j}+2\hat{k},\overrightarrow{b_1}=\hat{i}+2\hat{j}-3\hat{k}$
and $\overrightarrow{b_2}=\hat{i}+4\hat{j}-5\hat{k}$
Now, $\overrightarrow{a_2}-\overrightarrow{b_1}=(\hat{i}-\hat{j}+2\overrightarrow{k})-(4\overrightarrow{i}-\overrightarrow{j})=-3\overrightarrow{i}+0\overrightarrow{j}+2\overrightarrow{k}$
$=-3\overrightarrow{i}+2\overrightarrow{k}$
and $\overrightarrow{b_1}\times \overrightarrow{b_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& -3\\ 1& 4& -5\end{array}\right|$
$=\hat{i}(-10+12)-\hat{j}(-5+3)+\hat{k}(4-2)$$=2\hat{i}+2\hat{j}+2\hat{k}$
$\therefore (\overrightarrow{a_2}-\overrightarrow{a_1})\cdot (\overrightarrow{b_1}\times \overrightarrow{b_2})=(-3\hat{i}+2\hat{k})\cdot (2\hat{i}+2\hat{j}+2\hat{k})=-6+4=-2$
and $|\overrightarrow{b_1}\times \overrightarrow{b_2}|=\sqrt{(2{)}^2+(2{)}^2+(2{)}^2}=\sqrt{12}$
$\therefore$ Shortest distance, $d=\left|\frac{-2}{\sqrt{12}}\right|$
$=\frac{2}{2\sqrt{3}}=\frac{1}{\sqrt{3}}$ units.