4x4 Matrix Transformation of the Implicit Form of the Plane

Kenneth E. Hoff III
7/13/98

Given a plane as a 4-vector [ A B C D ] whose components are the coefficients of the implicit form of the plane Ax+By+Cz+D where (A,B,C) is the normal to the plane (possibly not normalized) and D is the negative distance from the origin to the plane along the normal (in units of the normal's length if not normalized), and a 4x4 transformation matrix M, we wish to transform the plane.

The simplest method would be to convert the implicit form of the plane to the point-normal form. We already have the normal as (A,B,C) and we can obtain a point on the plane as follows:

PtOnPlane = -D*(A,B,C)

This is simple since -D is the negative distance from the origin to the plane along the normal (in units of the normal, if not normalized).

So now we have the point-normal form as 1 point on the plane and 1 vector that is the plane's normal. We can perform two transformations as follows:

   NewNormal N' = M * Normal
NewPtOnPlane P' = M * PtOnPlane

Through substitution and expansion we get:

[ N'x ] = M * [ A ]
[ N'y ]       [ B ]
[ N'z ]       [ C ]
[  0  ]       [ 0 ]

[ P'x ] = M * [ -AD ]
[ P'y ]       [ -BD ]
[ P'z ]       [ -CD ]
[  1  ]       [  1  ]

We can get the implicit form [ A' B' C' D' ] for the xformed plane as follows:

Normalize the normal N' (if desired)
[ A' B' C' ] = [ N'x N'y N'z ]
D' = -(P' dot N')

We should be able to remove the conversion step by expanding the terms:

[ A' ]   [ M00*A + M01*B + M02*C ]
[ B' ] = [ M10*A + M11*B + M12*C ]
[ C' ]   [ M20*A + M21*B + M22*C ]

[ D' ] = -( (M00*-AD + M01*-BD + M02*-CD + M03) * (M00*A + M01*B + M02*C) +
            (M10*-AD + M11*-BD + M12*-CD + M13) * (M10*A + M11*B + M12*C) +
            (M20*-AD + M21*-BD + M22*-CD + M23) * (M20*A + M21*B + M22*C) )
       = -( (-D * (M00*A + M01*B + M02*C) + M03) * A' +
            (-D * (M10*A + M11*B + M12*C) + M13) * B' +
            (-D * (M20*A + M21*B + M22*C) + M23) * C' )
       = -( (-DA' + M03) * A' + 
            (-DB' + M13) * B' + 
            (-DC' + M23) * C' )

This gives an efficient and compact form for the planar xform without renormalization of the normal. We could then normalize as follows:

L = sqrt(A'²+B'²+C'²)
[ A B C D ] = [ A/L B/L C/L D/L ]

We could incorporate the normalization into the calculation more efficiently by reordering the operations of D' to compute (A'²+B'²+C'²):

D' = -(-DA' + M03) * A' + (-DB' + M13) * B' + (-DC' + M23) * C'
   = -(-DA'² + A'M03 + -DB'² + B'M13 + -DC'² + C'M23)
   = -(-D*(A'² + B'² + C'²) + (A'M03 + B'M13 + C'M23))
   = D*(A'² + B'² + C'²) - (A'M03 + B'M13 + C'M23)

So now we can compute an intermediate value L2 (length squared) before computing D' and normalization:

L2 = A'² + B'² + C'²D' = D*L2 - (A'M03 + B'M13 + C'M23)
L = sqrt(L2)
[ A B C D ] = [ A'/L B'/L C'/L D'/L ]

<OR MORE EFFICIENTLY>

L2 = A'² + B'² + C'²L = sqrt(L2)
[ A B C ] = [ A'/L B'/L C'/L ]
D' = D*L2 - (A*M03 + B*M13 + C*M23)

Note that this assumes that the 4x4 matrix does not include a projection, a shear, or a nonuniform scaling. What does it mean to project an infinite plane?