How to efficiently rotate and translate a plane in 3D
I have a plane defined by a normal (n) and a distance (d) (from the origin). I would like to transform it into a new system. The long way is like this: 1) multiply distance (d) with normal (n) resulting in a a vector (p) 2) rotate (R) and translate (v) the vector (p) to get (p') 3) normalize (p') to get the normal 4) use another algorithm to find the smallest distance (d') between the new plane and the origin
I haven't tried this but I guess it should work. QUestion: Isn't there a faster way to get n' and d'? If the translation (v) is 0 than I can skip 4). But if it is not 0? Is there an easier way to get the new d'?
Answer
You need to be careful because normals don't necessarily transform like points do, and the distance is the perpendicular distance to the origin, so you have to compute d'= d + n.v. If all you're doing is translation and rotation, then you can rotate the normal and compute a new perpendicular distance. But, if you're scaling your axes differently, or doing a general projective transformation, then you need to treat things differently.
The way that works for everything is to use homogeneous coordinates, so all your transforms are 4x4 matrices, and both your points and your planes are 4-vectors:
point p=(x,y,z) -> homogeneous (x,y,z,1), equiv. to (x*W, y*W, z*W, W)
plane q=[n=(a,b,c), d] -> homogeneous [a,b,c,d], equiv. to [a*K, b*K, c*K, d*K)
-> point p is on plane q iff: p.q=0 (using homogeneous coords, as above)
Generally, you will multiply all your transformation matrices into one 4x4 matrix T, and use that matrix on every point, to determine its final transformed position. The trick is, you need to use the inverse transpose of T to transform your plane coordinates. From the following, you can see that this preserves incidence between points and planes:
point p' = T p
plane q' = (T^-1)^t q
-> point p' is on plane q' when: p'.q'=0
then, note: p'.q' = p^t T^t (T^-1)^t q = p^t q = p.q
so: p'.q'=0 whenever p.q=0