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Question
Find the shortest distance between the following pairs of lines whose vector equations are: \[\vec{r} = 3 \hat{i} + 8 \hat{j} + 3 \hat{k} + \lambda\left( 3 \hat{i} - \hat{j} + \hat{k} \right) \text{ and } \vec{r} = - 3 \hat{i} - 7 \hat{j} + 6 \hat{k} + \mu\left( - 3 \hat{i} + 2 \hat{j} + 4 \hat{k} \right)\]
Solution
\[\vec{r} = 3 \hat{i} + 8 \hat{j} + 3 \hat{k} + \lambda\left( 3 \hat{i} - \hat{j} + \hat{k} \right) \text{ and } \vec{r} = - 3 \hat{i} - 7 \hat{j} + 6 \hat{k} + \mu\left( - 3 \hat{i} + 2 \hat{j} + 4 \hat{k} \right)\] Comparing the given equations with the equations
\[\vec{r} = \vec{a_1} + \lambda \vec{b_1} \text{ and } \vec{r} = \vec{a_2} + \mu \vec{b_2}\]
\[\vec{a_1} = 3 \hat{i} + 8 \hat{j} + 3 \hat{k} \]
\[ \vec{a_2} = - 3 \hat{i} - 7 \hat{j} + 6 \hat{k} \]
\[ \vec{b_1} = 3 \hat{i} - \hat{j} + \hat{k} \]
\[ \vec{b_2} = - 3 \hat{i} + 2 \hat{j} + 4 \hat{k} \]
\[ \therefore \vec{a_2} - \vec{a_1} = - 6 \hat{i} - 15 \hat{j} + 3 \hat{k} \]
\[and \vec{b_1} \times \vec{b_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 3 & - 1 & 1 \\ - 3 & 2 & 4\end{vmatrix}\]
\[ = - 6 \hat{i} - 15 \hat{j} + 3 \hat{k} \]
\[ \Rightarrow \left| \vec{b_1} \times \vec{b_2} \right| = \sqrt{\left( - 6 \right)^2 + \left( - 15 \right)^2 + 3^2}\]
\[ = \sqrt{36 + 225 + 9}\]
\[ = \sqrt{270}\]
\[\left( \vec{a_2} - \vec{a_1} \right) . \left( \vec{b_1} \times \vec{b_2} \right) = \left( - 6 \hat{i} - 15 \hat{j} + 3 \hat{k} \right) . \left( - 6 \hat{i} - 15 \hat{j} + 3 \hat{k} \right)\]
\[ = 36 + 225 + 9\]
\[ = 270\]
The shortest distance between the lines
\[\vec{r} = \vec{a_1} + \lambda \vec{b_1} \text{ and } \vec{r} = \vec{a_2} + \mu \vec{b_2}\] is given by
\[d = \left| \frac{\left( \vec{a_2} - \vec{a_1} \right) . \left( \vec{b_1} \times \vec{b_2} \right)}{\left| \vec{b_1} \times \vec{b_2} \right|} \right|\]
\[ = \frac{270}{\sqrt{270}}\]
\[ = \sqrt{270}\]
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