Question

Find the distance of the point (1, -5, 9) from the plane

\[x - y + z =\] 5  measured along the line \[x = y = z\]  . 

 

Answer 1

The equation of line parallel to the line x = y = z and passing through the point (1, −5, 9) is

\[\frac{x - 1}{1} = \frac{y + 5}{1} = \frac{z - 9}{1}\]      .......(1)

Any point on this line is of the form (k + 1, k − 5, k + 9).

If (k + 1, k − 5, k + 9) be the point of intersection of line (1) and the given plane, then

(k + 1) − (k − 5) + (k + 9) = 5

⇒ k = −10

So, the point of intersection of line (1) and the given plane is (−10 + 1, −10 − 5, −10 + 9) i.e. (−9, −15, −1).

∴ Required distance = Distance between (1, −5, 9) and (−9, −15, −1) = 

\[\sqrt{\left( 1 + 9 \right)^2 + \left( - 5 + 15 \right)^2 + \left( 9 + 1 \right)^2} = \sqrt{3 \times {10}^2} = 10\sqrt{3}\]  units .