Find plane normal and origin from homogeneous matrix

[mathieu]

Given a 3 x 4 matrix, how do I find the plane intersecting the three points expressed in homogeneous coordinate ?

Matrix is:

| a b c d |
| e f g h |
| i j k l |

Where X1=(a,b,c,d); X2=(e,f,g,h) and X3=(i,j,k,l) are 3 distinct points expressed in homogeneous coordinates. Those 3 points in space, define a plane such as Ax+By+Cz+D=0, (A,B,C,D constants not all zero). How do I find those A,B,C,D values ?

[Scott Hemphill]

Add a fourth row:

| a b c d |
| e f g h |
| i j k l |

| x y z w |

Where the {x,y,z,w} are symbolic, and compute the determinant of the
matrix. The coefficients of {x,y,z,w} are {A,B,C,D}.

Scott
--
Scott Hemphill hemp...@alumni.caltech.edu
"This isn't flying. This is falling, with style." -- Buzz Lightyear

[Nobody]

| b c d | | a c d | | a b d | | a b c |
A = | f g h | B = -| e g h | C = | e f h | D = -| e f g |
| j k l | | i k l | | i j l | | i j k |

Where |...| indicates the determinant.

This follows from Cramer's rule, and the fact that the plane equation is
just a 4D dot-product (Ax+By+Cz+Dw=0).

Also: provided that D is non-zero, you can set D=1 and A,B,C will be the
result of inverting the left-hand 3x3 submatrix and multiplying by
-(d,h,l)^T. If D=0, the 3x3 submatrix won't have an inverse (its
determinant is -D).