Question

Find the distance of the point (2, 4, −1) from the line  \[\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{- 9}\] 

Answer 1

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.