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Question
Show that the lines \[\frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z}{1} \text { and }\frac{x}{1} = \frac{y}{2} = \frac{z}{3}\] are perpendicular to each other.
Solution
We have
\[\frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z}{1} \text {and} \frac{x}{1} = \frac{y}{2} = \frac{z}{3}\]
These equations can be re-written as
\[\frac{x - 5}{7} = \frac{y - \left( - 2 \right)}{- 5} = \frac{z - 0}{1} . . . (1) \]
\[\frac{x - 0}{1} = \frac{y - 0}{2} = \frac{z - 0}{3} . . . (2)\]
\[\therefore \overrightarrow{m_1} = \text{ Vector parallel to line } (1) = 7 \hat{i} - 5 \hat{j} + \hat{k} \]
\[ \overrightarrow{m_2} = \text{ Vector parallel to line } (2) = \hat{i} + 2 \hat{j} + 3 \hat{k} \]
Now,
\[\overrightarrow{m_1} . \overrightarrow{m_2} = \left( 7 \hat{i} - 5 \hat{j} + \hat{k} \right) . \left( \hat{i}+ 2 \hat{j} + 3 \hat{k} \right)\]
\[ = 7 - 10 + 3\]
\[ = 0\]
Hence, the given two lines are perpendicular to each other.
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