Question

Find the shortest distance between 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

Given lines are $\overrightarrow{r}=(4\hat{i}-\hat{j})+\lambda (\hat{i}+2\hat{j}-3\hat{k})$ and $\overrightarrow{r}=(\hat{i}-\hat{j}+2\hat{k})+\mu (2\hat{i}+4\hat{j}-5\hat{k})$

Here $\overrightarrow{a_1}=4\hat{i}-\hat{j},\overrightarrow{b_1}=\hat{i}-2\hat{j}-3\hat{k}\text{and}\overrightarrow{a_2}=\hat{i}-\hat{j}+2\hat{k},\overrightarrow{b_2}=2\hat{i}+4\hat{j}-5\hat{k}$

So, $\overrightarrow{a_2}-\overrightarrow{a_1}=-3\hat{i}+2\hat{k},\overrightarrow{b_1}\times \overrightarrow{b_2}=\left[\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& -3\\ 2& 4& -5\end{array}\right]=2\hat{i}-\hat{j}$

Therefore, S.D $=\frac{|(\overrightarrow{a_2}-\overrightarrow{a_1}).(\overrightarrow{b_1}\times \overrightarrow{b_2})|}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}=\frac{|(-3\hat{i}+2\hat{k}).(2\hat{i}-\hat{j})|}{|2\hat{i}-\hat{j}|}=\frac{|-6|}{\sqrt{4+1}}=\frac{6}{\sqrt{5}}$ units

Hence required shortest distance is $\frac{6\sqrt{5}}{5}$ units.