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Question
By computing the shortest distance determine whether the following pairs of lines intersect or not: \[\frac{x - 1}{2} = \frac{y + 1}{3} = z \text{ and } \frac{x + 1}{5} = \frac{y - 2}{1}; z = 2\]
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Solution
\[\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 0}{1} and \frac{x + 1}{5} = \frac{y - 2}{1} = \frac{z - 2}{0}\]
Since the first line passes through the point (1, -1, 0) and has direction ratios proportional to 2, 3, 1, its vector equation is
\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} . . . (1) \]
\[\text{ Here } , \]
\[ \overrightarrow{a_1} = \hat{i} - \hat{j} + 0 \hat{k} \]
\[ \overrightarrow{b_1} = 2 \hat{i} + 3 \hat{j} + \hat{k} \]
Also, the second line passes through the point ( -1, 2, 2) and has direction ratios proportional to 5, 1, 0. Its vector equation is
\[\overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2} . . . (2) \]
\[\text{ Here }, \]
\[ \overrightarrow{a_2} = - \hat{i} + 2 \hat{j} + 2 \hat{k} \]
\[ \overrightarrow{b_2} = 5 \hat{i} + \hat{j} + 0 \hat{k} \]
Now,
\[\overrightarrow{a_2} - \overrightarrow{a_1} = - 2 \hat{i} + 3 \hat{j} + 2 \hat{k} \]
\[\text{ and } \overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 5 & 1 & 0\end{vmatrix}\]
\[ = - \hat{i} + 5 \hat{j} - 13 \hat{k} \]
\[\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) = \left( - 2 \hat{i} + 3 \hat{j} + 2 \hat{k} \right) . \left( - \hat{i} + 5 \hat{j} - 13 \hat{k} \right)\]
\[ = 2 + 15 - 26\]
\[ = - 9\]
\[\text{ We observe } \]
\[\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) \neq 0\]
\[\text{ Thus, the given lines do not intersect } .\]
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