Question

Find the distance of the point P(−1, −5, −10) from the point of intersection of the line joining the points A(2, −1, 2) and B(5, 3, 4) with the plane  \[x - y + z = 5\] . 

 

Answer 1

The equation of the line passing through the points A(2, −1, 2) and B(5, 3, 4) is given by

\[\frac{x - 2}{5 - 2} = \frac{y - \left( - 1 \right)}{3 - \left( - 1 \right)} = \frac{z - 2}{4 - 2}\]
\[\text{ Or }   \frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{2}\]

The coordinates of any point on the line 

\[\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{2} = \lambda\left( \text{say} \right)\] are 

\[\left( 3\lambda + 2, 4\lambda - 1, 2\lambda + 2 \right)\]

 .........(1)

If it lies on the plane

\[x - y + z = 5\] , then 

\[3\lambda + 2 - \left( 4\lambda - 1 \right) + 2\lambda + 2 = 5\]
\[ \Rightarrow \lambda + 5 = 5\]
\[ \Rightarrow \lambda = 0\]

Putting

\[\lambda = 0\] in (1), we get (2, −1, 2) as the coordinates of the point of intersection of the given line and plane.

∴ Required distance = Distance between points (−1, −5, −10) and (2, −1, 2)

\[= \sqrt{\left( 2 + 1 \right)^2 + \left( - 1 + 5 \right)^2 + \left( 2 + 10 \right)^2}\]
\[ = \sqrt{9 + 16 + 144}\]
\[ = \sqrt{169}\]
\[ = 13 \text{ units} \]