compute slope of a 3D plane

I have a set of (X,Y,Z) points representing different planar features. I need to calculate the slope of each plane using normal vectors. i think slope is given by the angle between normal vector (NV) of each plane and NV of imaginary horizontal plane. Assume, the plane equation that I use is; Ax+By+c=z. Then i guess the normal vector of my plane开发者_Go百科 is (a,b, -1). For my plane equation, what should be the equation of imaginary horizontal plane? I think equation of horizontal plane is z=c. Hence, the normal vector is (0,0,-1). Is this correct? Then the angle between my plane and the horizontal plane is; cos^(-1)⁡〖(a.0+b.0+(-1).1)/(√(〖a1〗^2+〖b1〗^2+〖c1〗^2 ).√(0^2+0^2+1^2 ))〗

Is that correct? please comment me and give me the correct equation.

Yes, that's mostly correct, but you've made some small mistakes substituting into the expression for the angle. The angle is cos^{-1} [(a * 0 + b * 0 + (-1) * (-1) / (√{a^2 + b^2 + (-1)^2} * √{0^2+0^2+(-1)^2}] = cos^{-1}(1/√{a^2 + b^2 + 1})