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Question
Find the distance of the point (2, 4, −1) from the line \[\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{- 9}\]
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Solution
We know that the distance d from point P to the line l having equation \[\overrightarrow{r} = \overrightarrow{a} + \lambda \overrightarrow{b}\] is given by \[d = \frac{\left| \overrightarrow{b} \times \overrightarrow{PQ} \right|}{\left| \overrightarrow{b} \right|}\] where Q is any point on the line l . The equation of the given line is \[\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{- 9}\] Let P(2, 4, −1) be the given point. Now, in order to find the required distance we need to find a point Q on the given line. The given line passes through the point (−5, −3, 6). So, take this point as the required point Q(−5, −3, 6). Also, the line is parallel to the vector
\[\overrightarrow{b} = \hat{i} + 4 \hat{j} - 9 \hat{k} \]
Now ,
\[\overrightarrow{PQ} = \left( - 5 \hat{i} - 3 \hat{j} + 6 \hat{k} \right) - \left( 2 \hat{i} + 4 \hat{j} - \hat{k} \right) = - 7 \hat{i} - 7 \hat{j} + 7 \hat{k} \]
\[\therefore \overrightarrow{b} \times \overrightarrow{PQ} = \begin{vmatrix}\hat{ i} & \hat{j} & \hat{k} \\ 1 & 4 & - 9 \\ - 7 & - 7 & 7\end{vmatrix} = - 35 \hat{i} + 56 \hat{j} + 21 \hat{k} \]
\[ \Rightarrow \left| \overrightarrow{b} \times \overrightarrow{PQ} \right| = \sqrt{\left( - 35 \right)^2 + {56}^2 + {21}^2} = \sqrt{1225 + 3136 + 441} = \sqrt{4802} = 49\sqrt{2}\]
Let d be the required distance.
\[\therefore d = \frac{\left| \vec{b} \times \overrightarrow{PQ} \right|}{\left| \overrightarrow{b} \right|} = \frac{49\sqrt{2}}{\sqrt{1 + 16 + 81}} = \frac{49\sqrt{2}}{\sqrt{98}} = \frac{49\sqrt{2}}{7\sqrt{2}} = 7\]
Thus, the distance of the given point from the given line is 7 units.
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