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Question
Find the shortest distance between the following pairs of lines whose cartesian equations are : \[\frac{x - 1}{- 1} = \frac{y + 2}{1} = \frac{z - 3}{- 2} \text{ and } \frac{x - 1}{1} = \frac{y + 1}{2} = \frac{z + 1}{- 2}\]
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Solution
\[\frac{x - 1}{- 1} = \frac{y + 2}{1} = \frac{z - 3}{- 2} . . . (1) \]
\[\frac{x - 1}{1} = \frac{y + 1}{2} = \frac{z + 1}{- 2} . . . (2)\]
Since line (1) passes through the point (1, -2, 3) and has direction ratios proportional to -1,1-2,its vector equation is
\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \]
\[Here, \]
\[ \overrightarrow{a_1} = \hat{i} - 2 \hat{j}+ 3 \hat{k} \]
\[ \overrightarrow{b_1} = - \hat{i} + \hat{j} - 2 \hat{k} \]
Also, line (2) passes through the point (1, -1, -1 ) and has direction ratios proportional to 1, 2,-2.Its vector equation is
\[\overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2} \]
\[Here, \]
\[ \overrightarrow{a_2} = \hat{i} - \hat{j} - \hat{k} \]
\[ \overrightarrow{b_2} = \hat{i} + 2 \hat{j} - 2 \hat{k} \]
Now,
\[\overrightarrow{a_2} - \overrightarrow{a_1} = \hat{j} - 4 \hat{k} \]
\[\text{ and } \overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ - 1 & 1 & - 2 \\ 1 & 2 & - 2\end{vmatrix}\]
\[ = 2 \hat{i} - 4 \hat{j} - 3 \hat{k} \]
\[ \Rightarrow \left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right| = \sqrt{2^2 + \left( - 4 \right)^2 + \left( - 3 \right)^2}\]
\[ = \sqrt{4 + 16 + 9}\]
\[ = \sqrt{29}\]
\[\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \vec{b_1} \times \overrightarrow{b_2} \right) = \left( \hat{j} - 4 \hat{k} \right) . \left( 2 \hat{i} - 4 \hat{j} - 3 \hat{k} \right)\]
\[ = - 4 + 12\]
\[ = 8\]
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
\[d = \left| \frac{\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right)}{\left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right|} \right|\]
\[ = \left| \frac{8}{\sqrt{29}} \right|\]
\[ = \frac{8}{\sqrt{29}}\]
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