Question

Find the distance between the lines l₁ and l₂ given by
$\overrightarrow{r}=\hat{i}+2\hat{j}-4\hat{k}+\lambda \left(2\hat{i}+3\hat{j}+6\hat{k}\right)\mathrm{and},\overrightarrow{r}=3\hat{i}+3\hat{j}-5\hat{k}+\mu \left(2\hat{i}+3\hat{j}+6\hat{k}\right)$

Answer 1

Given:
$$\begin{gathered} \overrightarrow{r}=\hat{i}+2\hat{j}-4\hat{k}+\lambda \left(2\hat{i}+3\hat{j}+6\hat{k}\right) \\ \overrightarrow{r}=3\hat{i}+3\hat{j}-5\hat{k}+\mu \left(2\hat{i}+3\hat{j}+6\hat{k}\right) \end{gathered}$$

These two lines pass through the points having position vectors $\overrightarrow{a_1}=\hat{i}+2\hat{j}-4\hat{k}\mathrm{and}\overrightarrow{a_2}=3\hat{i}+3\hat{j}-5\hat{k}$ and are parallel to the vector $\overrightarrow{b}=2\hat{i}+3\hat{j}+6\hat{k}$.

Now,
$\overrightarrow{a_2}-\overrightarrow{a_1}=2\hat{i}+\hat{j}-\hat{k}$
and
$$\begin{gathered} \left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)\times \overrightarrow{b}=\left(2\hat{i}+\hat{j}-\hat{k}\right)\times \left(2\hat{i}+3\hat{j}+6\hat{k}\right) \\ =\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -1\\ 2& 3& 6\end{array}\right| \\ =9\hat{i}-14\hat{j}+4\hat{k} \\ \Rightarrow \left|\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)\times \overrightarrow{b}\right|=\sqrt{9^2+{\left(-14\right)}^2+4^2} \\ =\sqrt{81+196+16} \\ =\sqrt{293} \\ \mathrm{and}\left|\overrightarrow{b}\right|=\sqrt{2^2+3^2+6^2} \\ =\sqrt{4+9+36} \\ =7 \end{gathered}$$

The shortest distance between the two lines is given by

$\frac{\left|\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)\times \overrightarrow{b}\right|}{\left|\overrightarrow{b}\right|}=\frac{\sqrt{293}}{7}$