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Question
Find the shortest distance between the lines \[\overrightarrow{r} = 6 \hat{i} + 2 \hat{j} + 2 \hat{k} + \lambda\left( \hat{i} - 2 \hat{j} + 2 \hat{k} \right) \text{ and } \overrightarrow{r} = - 4 \hat{i} - \hat{k} + \mu\left( 3 \hat{i} - 2 \hat{j} - 2 \hat{k} \right)\]
Solution
\[\overrightarrow{r} = 6 \hat{i} + 2 \hat{j} + 2 \hat{k} + \lambda\left( \hat{i} - 2 \hat{j} + 2 \hat{k} \right) \text{ and } \overrightarrow{r} = - 4 \hat{i} - \hat{k} + \mu\left( 3 \hat{i} - 2 \hat{j} - 2 \hat{k} \right)\]
Comparing the given equations with the equations
\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \text{ and } \overrightarrow{r} = \overrightarrow{a_2} + \mu \vec{b_2}\]
we get ,
\[\overrightarrow{a_1} = 6 \hat{i} + 2 \hat{j} + 2 \hat{k} \]
\[ \overrightarrow{a_2} = - 4 \hat{i} - \hat{k} \]
\[ \overrightarrow{b_1} = \hat{i} - 2 \hat{j} + 2 \hat{k} \]
\[ \overrightarrow{b_2} = 3 \hat{i} - 2 \hat{j} - 2 \hat{k} \]
\[ \therefore \overrightarrow{a_2} - \overrightarrow{a_1} = - 10 \hat{i} - 2 \hat{j} - 3 \hat{k} \]
\[\text { and } \overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & - 2 & 2 \\ 3 & - 2 & - 2\end{vmatrix}\]
\[ = 8 \hat{i} + 8 \hat{j} + 4 \hat{k} \]
\[ \Rightarrow \left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right| = \sqrt{8^2 + 8^2 + 4^2}\]
\[ = \sqrt{64 + 64 + 16}\]
\[ = \sqrt{144}\]
\[ = 12\]
\[\text { and } \left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) = \left( - 10 \hat{i} - 2 \hat{j} - 3 \hat{k} \right) . \left( 8 \hat{i} + 8 \hat{j} + 4 \hat{k} \right)\]
\[ = - 80 - 16 - 12\]
\[ = - 108\]
The shortest distance between the lines
\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \text{ and }\overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2}\] is given by
\[d = \left| \frac{\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right)}{\left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right|} \right|\]
\[ \Rightarrow d = \left| \frac{- 108}{12} \right|\]
\[ = 9\]
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