Question
Find the shortest distance between the lines $I_1$ and $l_2$ whose vector equation are
$\begin{array}{l}
\vec{r}=\hat{i}+\hat{j}+\lambda(2 \hat{i}-\hat{j}+\hat{k}) \text { and } \\
\vec{r}=2 \hat{i}+\hat{j}-\hat{k}+\mu(3 \hat{i}-5 \hat{j}+2 \hat{k})
\end{array}$

Answer

Comparing (1) and (2) with $\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$ and $\vec{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}$ respectively
We get,
$\overrightarrow{a_1}=\hat{i}+\hat{j}, \overrightarrow{b_1}=2 \hat{i}-\hat{j}+\hat{k}, \overrightarrow{a_2}=2 \hat{i}+\hat{j}-\hat{k}$
and $\overrightarrow{b_2}=3 \hat{i}-5 \hat{j}+2 \hat{k}$
Therefore, $\overrightarrow{a_2}-\overrightarrow{a_1}=\hat{i}-\hat{k}$
and $\overrightarrow{b_1} \times \overrightarrow{b_2}=(2 \hat{i}-\hat{j}+\hat{k}) \times(3 \hat{i}-5 \hat{j}+2 \hat{k})$
$\begin{array}{l}
=\left|\begin{array}{ccc}
\hat{i} & \widehat{j} & \widehat{k} \\
2 & -1 & 1 \\
3 & -5 & 2
\end{array}\right| \\
=3 \hat{i}-\hat{j}+7 \hat{k}
\end{array}$
So, $\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|=\sqrt{9+1+49}$
$=\sqrt{59}$
Hence, the shortest distance between the given lines is given by
$\begin{aligned}
d & =\left|\frac{\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right) \cdot\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)}{\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|}\right| \\
& =\left|\frac{3-0+7}{\sqrt{59}}\right| \\
& =\frac{10}{\sqrt{59}} \text { unit }
\end{aligned}$

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