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Question
Find the shortest distance between the following pairs of lines whose cartesian equations are: \[\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} and \frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 5}{5}\]
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Solution
The equations of the given lines are
\[\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} \] ...........(1)
\[\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 5}{5}\] .............(2)
Since line (1) passes through the point (1, 2, 3) and has direction ratios proportional to 2, 3, 4, its vector equation is \[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \]
\[Here, \]
\[ \overrightarrow{a_1} = \hat{i} + 2 \hat{j} + 3 \hat{k} \]
\[ \overrightarrow{b_1} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \]
Also, line (2) passes through the point (2, 3, 5) and has direction ratios proportional to 3, 4, 5.
Its vector equation is \[\overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2} \]
\[\text{ Here } , \]
\[ \overrightarrow{a_2} = 2 \hat{i} + 3 \hat{j} + 5 \hat{k} \]
\[ \overrightarrow{b_2} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \]
Now,
\[\overrightarrow{a_2} - \overrightarrow{a_1} = \hat{i} + \hat{j} + 2 \hat{k} \]
\[\text{ and }\overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{vmatrix}\]
\[ = - \hat{i} + 2 \hat{j} - \hat{k} \]
\[ \Rightarrow \left| \vec{b_1} \times \vec{b_2} \right| = \sqrt{\left( - 1 \right)^2 + 2^2 + \left( - 1 \right)^2}\]
\[ = \sqrt{1 + 4 + 1}\]
\[ = \sqrt{6}\]
\[\text{ and } \left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) = \left( \hat{i} + \hat{j} + 2 \hat{k} \right) . \left( - \hat{i} + 2 \hat{j} - \hat{k} \right)\]
\[ = - 1 + 2 - 2\]
\[ = - 1\]
The shortest distance between the lines \[\vec{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \text{ and } \overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2}\] is given by
\[d = \left| \frac{\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right)}{\left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right|} \right|\]
\[ = \left| \frac{- 1}{\sqrt{6}} \right|\]
\[ = \frac{1}{\sqrt{6}}\]
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