Question

Find the shortest distance of a point (2, 5, −3) from the plane − r. (6 − i − 3 − j + 2 − k) = 1?

Answer 1

Point  (2, 5, -3)
Plane     r . (6 i - 3 j + 2 k) = 1
             r = x i + y j + z k

 So The equation of the plane is :  6 x - 3 y + 2 z = 1

Shortest (perpendicular distance of P from the plane is:
    =  | 6 * 2 - 3 * 5 + 2 * (-3) | / √( 6²+ (-3)² + 2²)
    =  9 / √49