Calculate the disance between to parallel planes?

Question

If Pi1: 2x-6y-z+12=0 and Pi2: -6x+18y+3z+10=0

Calculate the distance between Pi and P2

Answer 1

One way is to choose a point in each plane, compute the vector from one of those points to the other, and find the magnitude of the component of this vector in the direction of a vector normal to either plane.

A point P in π1 is (0,2,0). A point Q in π2 is (0,0,-10/3). Then
vector PQ = 0i -2j -(10/3)k. We can read off from the equation for π1 that a vector N normal to π1 is given by

N = 2i - 6j - k

The component of PQ in the direction of N is given by

(PQ • N) / | N |

where | N | is the length, or magnitude, of N.

The length of N is √[2² + (-6)² + (-1)²] = √41. So the distance between π1 and π2 is

| PQ • N | / √41 = [(0)(2) + (-2)(-6) + (-10/3)(-1)]/ √41
= (46/3) / √41 = (46√41)/123