There's one step in the derivation that I don't understand. I'll explain some of the derivation and then explain which step I don't understand.
Begin with scalar equation of plane:
A(x-x0) + B(y-yo) + C(z-zo) = 0
Divide this equation by C and
let a = -A/C and
let b = -B/C
It can now be written as:
z-zo = a(x-xo) + b(y-yo)
The intersection of this equation with the plane y = yo is:
z-zo = a(x-xo),
which is recognized as the equation of a line with slope a.
The rest of the derivation is irrelevant. I don't understand how, in step 2, letting a = -A/C makes the variable 'a' the slope of a line in the xz-plane. A and C are defined in the scalar equation of a plane, respectively, as the x and z component of a vector that's orthogonal to the plane that its equation defines.
It appears to say that dividing the x by the z component defines the slope of a line in the xz-plane, but shouldn't it be dividing the z by the x component for that to be true? Right now it seems equivalent as to defining the slope in the xy-plane as x/y, but we all know it is actually y/x. This is what I'm confused about.
Oh my! I think I might understand it now. since C/A is the slope of the line ORTHOGONAL to the line in the tangent plane in the xz-plane, then to find the slope of the line in the tangent plane, it takes the negative reciprocal of C/A to get -A/C!
If you understand this question, could you verify that this is the correct explanation for letting a = -A/C?