Consider two plans P1: ax + by + cz + e = 0 and P2: ax + by +cz + f = 0. Show that the distance between P1 and P2 is given by d = l e-f l / sqrt(a^2 + b^2 +c^2)
* any idea on what to do.
Look at reply #9 on this thread.
Because the planes are parallel, use any point on either plane find the distance to the other plane.
here is my solution:
l Ap + Bq + Cr + D l / sqrt(A^2 + B^2 + C^2)
point: (p, q, r)
n = P1: Ax + By + Cz + e = 0 and P2: Ax + By + Cz + f = 0
Isolate e and f and get:
e = - (Ax1 + By2 + Cz3)
- f = Ax1 + By2 + Cz3
Distaance between Plane 1 and 2 have the same exact distance as (p1, q1, r1) due to the fact its on the same plane.
l Ap + Bq + Cr + D l / sqrt(A^2 + B^2 + C^2)
D = l -f + e l / sqrt(A^2 + B^2 + C^2)
D = l e - f l / sqrt(A^2 + B^2 + C^2)