3d plane

Format:

1. Ax + By + Cz + D = 0;

2. dot(n, p) + d = 0,   d = - dot(n, p)          //p = {x, y, z}

3. dot(L, P) = 0            // L = [n, d],  //P = {x, y, z, 1}, the same as p in homogenous space

Proof:

3:  

dot(L, P) = dot(n,p) + d = 0   //according to 2

Relationship with arbitary point p:

if      dot(n, p) + d = 0 ;         p is on plane

if      dot(n, p) + d > 0 ;         p is on positive half plane (positive means the same direction as normal)

if      dot(n, p) + d < 0 ;         p is on negative half space

distance from p to plane:        abs(dot(n,p) + d) / length(n)

distance from origin:  abs(d) / length(n)

proof:

assume arbitary point p0 on plane

dot(n,p0p) = dot(n,p) - dot(n,p0)  = dot(n,p) + d,            // we can get proof for second and thirdconclusion by this

distance will be po0 project onto n, which will be:

      abs(dot(n, p0p) )/ length(n)

Transfrom

We can transform n and an arbitary point to get our new plane, have nothing to say about this way.

We can also transfrom L=[n,d] directory to get out new plane, the result is

L' = inverse(transpose(T))L

proof:

        dot(L, P) = 0 ===> transpose(L) P = 0 // matrix version of dot， here we use column vector

                               ===> transpose(L')P' = 0

                               ===> transpose(L')TP = 0

                               ===>transppose(L) = transpose(L')T

                                ===>L  = transpose(T)L'

                                ===>L' = inverse(transpose(T))

Or just think  L is 4d normal.