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Question
Find the shortest distance between the following pairs of lines whose cartesian equations are : \[\frac{x - 1}{2} = \frac{y + 1}{3} = z \text{ and } \frac{x + 1}{3} = \frac{y - 2}{1}; z = 2\]
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Solution
The equations of the given lines are
\[\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 0}{1} . . . (1) \]
\[\frac{x + 1}{3} = \frac{y - 2}{1} = \frac{z - 2}{0} . . . (2)\]
Since line (1) passes through the point (1,-1, 0) and has direction ratios proportional to 2, 3, 1, its vector equation is
\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \]
\[Here, \]
\[ \overrightarrow{a_1} = \hat{i} - \hat{j} + 0 \hat{k} \]
\[ \overrightarrow{b_1} = 2 \hat{i} + 3 \hat{j} + \hat{k} \]
Also, line (2) passes through the point ( -1, 2, 2) and has direction ratios proportional to 3, 1, 0. Its vector equation is
\[\overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2} \]
\[Here, \]
\[ \overrightarrow{a_2} = - \hat{i} + 2 \hat{j} + 2 \hat{k} \]
\[ \overrightarrow{b_2} = 3 \hat{i} + \hat{j} + 0 \hat{k} \]
Now,
\[\overrightarrow{a_2} - \overrightarrow{a_1} = - 2 \hat{i} + 3 \hat{j} + 2 \hat{k} \]
\[\text{ and } \overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 3 & 1 & 0\end{vmatrix}\]
\[ = - \hat{i} + 3 \hat{j} - 7 \hat{k} \]
\[ \Rightarrow \left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right| = \sqrt{\left( - 1 \right)^2 + 3^2 + \left( - 7 \right)^2}\]
\[ = \sqrt{1 + 9 + 49}\]
\[ = \sqrt{59}\]
\[\text{ and } \left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) = \left( - 2 \hat{i} + 3 \hat{j} + 2 \hat{k} \right) . \left( - \hat{i} + 3 \hat{j} - 7 \hat{k} \right)\]
\[ = 2 + 9 - 14\]
\[ = - 3\]
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
\[d = \left| \frac{\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right)}{\left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right|} \right|\]
\[ = \left| \frac{- 3}{\sqrt{59}} \right|\]
\[ = \frac{3}{\sqrt{59}}\]
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