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Question
By computing the shortest distance determine whether the following pairs of lines intersect or not: \[\frac{x - 5}{4} = \frac{y - 7}{- 5} = \frac{z + 3}{- 5} \text{ and } \frac{x - 8}{7} = \frac{y - 7}{1} = \frac{z - 5}{3}\]
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Solution
\[\frac{x - 5}{4} = \frac{y - 7}{- 5} = \frac{z + 3}{- 5} \text{ and } \frac{x - 8}{7} = \frac{y - 7}{1} = \frac{z - 5}{3}\] Since the first line passes through the point (5, 7,-3) and has direction ratios proportional to 4, -5,-5 its vector equation is
\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} . . . (1) \]
\[\text{ Here } , \]
\[ \overrightarrow{a_1} = 5 \hat{i} + 7 \hat{j} - 3 \hat{k} \]
\[ \overrightarrow{b_1} = 4 \hat{i} - 5 \hat{j} - 5 \hat{k}\]
Also, the second line passes through the point (8, 7, 5) and has direction ratios proportional to 7, 1, 3.
Its vector equation is
\[\overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2} . . . (2) \]
\[\text{ Here } , \]
\[ \overrightarrow{a_2} = 8 \hat{i} + 7 \hat{j} + 5 \hat{k} \]
\[ \overrightarrow{b_2} = 7 \hat{i} + \hat{j} + 3 \hat{k} \]
Now,
\[\overrightarrow{a_2} - \vec{a_1} = 3 \hat{i} + 8 \hat{k} \]
\[\text{ and } \overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 4 & - 5 & - 5 \\ 7 & 1 & 3\end{vmatrix}\]
\[ = - 10 \hat{i} - 47 \hat{j} + 39 \hat{k} \]
\[\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) = \left( 3 \hat{i} + 8 \hat{k} \right) . \left( - 10 \hat{i} - 47 \hat{j} + 39 \hat{k} \right)\]
\[ = - 30 + 312\]
\[ = 282\]
\[\text{ We observe }\]
\[\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) \neq 0\]
\[\text{ Thus, the given lines do not intersect } .\]
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