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Question
Find the shortest distance between the following pairs of parallel lines whose equations are: \[\overrightarrow{r} = \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) + \lambda\left( \hat{i} - \hat{j} + \hat{k} \right) \text{ and } \overrightarrow{r} = \left( 2 \hat{i} - \hat{j} - \hat{k} \right) + \mu\left( - \hat{i} + \hat{j} - \hat{k} \right)\]
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Solution
The vector equations of the given lines are
\[\overrightarrow{r} = \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) + \lambda\left( \hat{i} - \hat{j} + \hat{k} \right) . . . (1)\]
\[ \overrightarrow{r} = \left( 2 \hat{i} - \hat{j}- \hat{k} \right) + \mu\left( - \hat{i} + \hat{j} - \hat{k} \right)\]
\[ = \left( 2 \hat{i} - \hat{j} - \hat{k} \right) - \mu\left( \hat{i} - \hat{j} + \hat{k} \right) . . . (2)\]
These two lines pass through the points having position vectors \[\overrightarrow{a_1} = \hat{i} + 2 \hat{j} + 3 \hat{k} \text{ and } \overrightarrow{a_2} = 2 \hat{i} - \hat{j} - \hat{ k} \] and are parallel to the vector \[\overrightarrow{b} = \hat{ i} - \hat{j} + \hat{k} \]
Now,
\[\overrightarrow{a_2} - \overrightarrow{a_1} = \hat{i} - 3 \hat{j} - 4 \hat{k} \]
and
\[\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) \times \overrightarrow{b} = \left( \hat{i} - 3 \hat{j} - 4 \hat{k} \right) \times \left( \hat{i} - \hat{j} + \hat{k} \right)\]
\[ = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & - 3 & - 4 \\ 1 & - 1 & 1\end{vmatrix}\]
\[ = - 7 \hat{i} - 5 \hat{j} + 2 \hat{k} \]
\[ \Rightarrow \left| \left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) \times \overrightarrow{b} \right| = \sqrt{\left( - 7 \right)^2 + \left( - 5 \right)^2 + 2^2}\]
\[ = \sqrt{49 + 25 + 4}\]
\[ = \sqrt{78}\]
The shortest distance between the two lines is given by
\[\frac{\left| \left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) \times \overrightarrow{b} \right|}{\left| \overrightarrow{b} \right|} = \frac{\sqrt{78}}{\sqrt{3}} = \sqrt{26}\]
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