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September 20, 2014

September 20, 2014

Posted by **sophie** on Sunday, May 22, 2011 at 6:11pm.

- math -
**Damon**, Sunday, May 22, 2011 at 7:03pmWell, 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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