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Question
Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( 1 - t \right) \hat{i} + \left( t - 2 \right) \hat{j} + \left( 3 - t \right) \hat{k} \text{ and } \overrightarrow{r} = \left( s + 1 \right) \hat{i} + \left( 2s - 1 \right) \hat{j} - \left( 2s + 1 \right) \hat{k} \]
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Solution
\[\overrightarrow{r} = \left( 1 - t \right) \hat{i} + \left( t - 2 \right) \hat{j} + \left( 3 - t \right) \hat{k} \text{ and } \vec{r} = \left( s + 1 \right) \hat{i} + \left( 2s - 1 \right) \hat{j} - \left( 2s + 1 \right) \hat{k} \]
The vector equations of the given lines can be re-written as
\[\overrightarrow{r} = \hat{i} - 2 \hat{j} + 3 \hat{k} + t\left( - \hat{i} + \hat{j} - \hat{k}\right) \text{ and } \overrightarrow{r} = \hat{i} - \hat{j} - \hat{k} + s\left( \hat{i} + 2 \hat{j} - 2 \hat{k} \right)\]
Comparing the given equations with the equations \[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \text{ and } \overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2}\]
we get ,
\[\overrightarrow{a_1} = \hat{i} - 2 \hat{j} + 3 \hat{k} \]
\[ \overrightarrow{a_2} = \hat{i} - \hat{j} - \hat{k} \]
\[ \overrightarrow{b_1} = - \hat{i} + \hat{j} - \hat{k} \]
\[ \overrightarrow{b_2} = \hat{i} + 2 \hat{j} - 2 \hat{k} \]
\[ \therefore \overrightarrow{a_2} - \overrightarrow{a_1} = \hat{j} - 4 \hat{k} \]
\[\text{ and } \overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ - 1 & 1 & - 1 \\ 1 & 2 & - 2\end{vmatrix}\]
\[ = - 3 \hat{j} - 3 \hat{k} \]
\[ \Rightarrow \left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right| = \sqrt{\left( - 3 \right)^2 + \left( - 3 \right)^2}\]
\[ = \sqrt{9 + 9}\]
\[ = 3\sqrt{2}\]
\[\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) = \left( \hat{j} - 4 \hat{k} \right) . \left( - 3 \hat{j} - 3 \hat{k} \right)\]
\[ = - 3 + 12\]
\[ = 9\]
The shortest distance between the line
\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \text{ and } \overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2}\] is given by
\[d = \left| \frac{\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right)}{\left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right|} \right|\]
\[ = \left| \frac{9}{3\sqrt{2}} \right|\]
\[ = \frac{3}{\sqrt{2}}\]
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