Question

Find the shortest distance between two lines whose vector equations are $\overrightarrow{r}=(\hat{i}+2\hat{j}+3\hat{k})+\beta (\hat{i}-3\hat{j}+2\hat{k})$ and $\overrightarrow{r}(4\hat{i}+5\hat{j}+6\hat{k})+\mu (2\hat{i}+3\hat{j}+\hat{k})$.

Answer 1

Line:$\overrightarrow{r}=(\hat{i}+2\hat{j}+3\hat{k})+\beta (\hat{i}-3\hat{j}+2\hat{k})-\left(i\right)\overrightarrow{r}=(4\hat{i}+5\hat{j}+6\hat{k})+\mu (2\hat{i}+3\hat{j}+\hat{k})-\left(ii\right)$

$\overrightarrow{a}=\hat{i}+2\hat{j}+3\hat{k},\overrightarrow{b}=4\hat{i}+5\hat{j}+6\hat{k}$ are two points on Line $\left(i\right)$ and $\left(ii\right)$.

$\therefore \overrightarrow{AB}=3\hat{i}+3\hat{j}+3\hat{k}$

Direction of shortest distance is perpendicular to both lines.

$\therefore \overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 1\\ 1& -3& 2\end{array}\right|=9\hat{i}-3\hat{j}+9\hat{k}$

$\therefore$ Perpendicular distance$=\overrightarrow{AB}.\frac{\overrightarrow{n}}{\left|\overrightarrow{n}\right|}=\frac{(3\hat{i}+3\hat{j}+3\hat{k}).(9\hat{i}-3\hat{j}+9\hat{k})}{\sqrt{9^2+9^2+1^2}}=\frac{3}{\sqrt{19}}units$