Question

Find the shortest distance between the following pair of skew lines :
$\frac{x-1}{2}=\frac{2-y}{3}=\frac{z+1}{4}$ and $\frac{x+2}{-1}=\frac{y-3}{2}=\frac{z}{3}$.

Answer 1

$\underset{r}{\to }=\hat{i}+2\hat{j}-\hat{k}+\lambda (2\hat{i}-3\hat{j}+4\hat{k}),\underset{r}{\to }=-2\hat{i}+3\hat{j}+\mu (-\hat{i}+2\hat{j}+3\hat{k})\therefore \underset{a_1}{\to }=\hat{i}+2\hat{j}-\hat{k},\underset{b_1}{\to }=2\hat{i}-3\hat{j}+4\hat{k},\underset{a_2}{\to }=-2\hat{i}+3\hat{j},\underset{b_2}{\to }=-\hat{i}+2\hat{j}+3\hat{k}\Rightarrow \underset{a_2}{\to }-\underset{a_1}{\to }=-3\hat{i}+\hat{j}+\hat{k},\underset{b_1}{\to }\times \underset{b_2}{\to }=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -3& 4\\ -1& 2& 3\end{array}\right|=-17\hat{i}-10\hat{j}+\hat{k}\therefore \text{Shortest distance between skew lines}\frac{|(\underset{a_2}{\to }-\underset{a_1}{\to }).\underset{b_1}{\to }\times \underset{b_2}{\to }|}{|\underset{b_1}{\to }\times \underset{b_2}{\to }|}\Rightarrow =\frac{|(-3\hat{i}+\hat{j}+\hat{k}).(-17\hat{i}-10\hat{j}+\hat{k})|}{\sqrt{289+100+1}}=\frac{42}{\sqrt{390}}\text{units}$