What is the distance between the parallel planes: ax+by+cz=d1 and ax+by+cz=d2?

Answer 1

Well, I did it but there is probably an easier way
A point in plane 1 is (0,0,d1/c)
A line through that point perpendicular to the plane is:
(x-0)/a = (y-0)/b = (z-d1/c)/c = t to make it parametric in t
then
x = at
y = bt
z = ct + d1/c
Where does that line hit the second plane?
a(at) + b(bt) +c(ct +d1/c) = d2
so
t = (d2-d1)/(a^2+b^2+c^2)
and at that point in that second plane
x2 = a t
y2 = b t
z2 = ct + d1/c
get (x2-0), (y2-0) , (z2-d1/c)
for distance formula
then
D = sqrt[(at)^2 + (bt)^2 + (ct)^2]
D = t sqrt (a^2+b^2+c^2)
D = [(d2-d1)/(a^2+b^2+c^2)]sqrt (a^2+b^2+c^2)
D = (d2-d1)/sqrt (a^2+b^2+c^2)