Question

Find the distance of the point P(3, 4, 4) from the point, where the line joining the points A(3, −4, −5) and B(2, −3, 1) intersects the plane 2x + y + z = 7.   

Answer 1

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

\[\frac{x - 3}{2 - 3} = \frac{y - \left( - 4 \right)}{- 3 - \left( - 4 \right)} = \frac{z - \left( - 5 \right)}{1 - \left( - 5 \right)}\]
\[\text{ Or }  \frac{x - 3}{- 1} = \frac{y + 4}{1} = \frac{z + 5}{6}\]

The coordinates of any point on the line 

\[\frac{x - 3}{- 1} = \frac{y + 4}{1} = \frac{z + 5}{6} = \lambda\left( \text{ say }    \right)\] are \[\left( - \lambda + 3, \lambda - 4, 6\lambda - 5 \right)\]..........................(1)

If it lies on the plane 2x + y + z = 7, then 

\[2\left( - \lambda + 3 \right) + \left( \lambda - 4 \right) + \left( 6\lambda - 5 \right) = 7\]
\[ \Rightarrow 5\lambda - 3 = 7\]
\[ \Rightarrow 5\lambda = 10\]
\[ \Rightarrow \lambda = 2\] 

Putting

\[\lambda = 2\]  in (1), we get (1, −2, 7) as the coordinates of the point of intersection of the given line and plane.

∴ Required distance = Distance between points (3, 4, 4) and (1, −2, 7)

\[= \sqrt{\left( 3 - 1 \right)^2 + \left( 4 + 2 \right)^2 + \left( 4 - 7 \right)^2}\]
\[ = \sqrt{4 + 36 + 9}\]
\[ = \sqrt{49}\]
\[ = 7 \text{ units} \]