Question

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

Answer 1

From the given equations, lets assume

$\overrightarrow{a_1}=(4\hat{i}-\hat{j}),\overrightarrow{b_1}=(\hat{i}+2\hat{j}-3\hat{k}),\overrightarrow{a_2}=(\hat{i}-\hat{j}+2\hat{k})\text{and}\overrightarrow{b_2}=(2\hat{i}+4\hat{j}-5\hat{k})Now,\overrightarrow{a_2}-\overrightarrow{a_1}=-3\hat{i}+0\hat{j}+2\hat{k}\overrightarrow{b_1}\times \overrightarrow{b_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& -3\\ 2& 4& -5\end{array}\right|=2\hat{i}-\hat{j}+0\hat{k}|\overrightarrow{b_1}\times \overrightarrow{b_2}|=\sqrt{(2{)}^2+(-1{)}^2}=\sqrt{5}$

$d=\frac{|(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot (\overrightarrow{b_1}\times \overrightarrow{b_2})|}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}$

$=\frac{|(-3\hat{i}+0\hat{j}+2\hat{k}).(2\hat{i}-\hat{j}+0\hat{k})|}{\sqrt{5}}$

$=\frac{|-6+0+0|}{\sqrt{5}}=\frac{6}{\sqrt{5}}$

or $\frac{6\sqrt{5}}{5}$ units.