compute slope of a 3D plane

问题:

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 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.

回答1:

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})