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

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    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)

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