Let vectors \( \vec {AB} \) and \( \vec {AC} \) be defined by their endpoints as follows:
\( \vec {AB} \; = \; \lt B_x - A_x , B_y - A_y , B_z - A_z \gt \)
\( \vec {AC} \; = \; \lt C_x - A_x , C_y - A_y , C_z - A_z \gt \)
The cross product of vectors \( \vec {AB} \) and \( \vec {AC} \) is orthogonal to the plane defined by the three points \( A \), \( B \) and \( C \)
The cross product of vectors \( \vec {AB} \) and \( \vec {AC} \) is given by
\( \vec{AB} \times \vec{AC} \; = \; \lt AB_y \cdot AC_z - AB_z \cdot AC_y , AB_z \cdot AC_x - AB_x \cdot AC_z , AB_x \cdot AC_y-AB_y \cdot AC_x \gt \)
Let \( M \) be any point on the plane defined by its coordinates as follows
\( M \; = \; ( x , y , z ) \)
Let vector \( \vec {AM} \) be defined by
\( \vec {AM} \; = \; \lt x - A_x, y -A_y , z - A_z \gt \)
For a point \( M = (x,y,z) \) to be on the plane defined by the three points \( A \), \( B \) and \( C \), we need to have the dot product of the vectors \( \vec {AM} \) and \( \vec{AB} \times \vec{AC} \) equal to zero. Hence the equation of the plane:
\( \vec {AM} \cdot (\vec{AB} \times \vec{AC}) = 0 \)
Substitute the components and rewrite the above equation as
\( (x - A_x) \cdot (AB_y \cdot AC_z - AB_z \cdot AC_y) + (y -A_y) \cdot (AB_z \cdot AC_x - AB_x \cdot AC_z) + (z - A_z) \cdot (AB_x \cdot AC_y-AB_y \cdot AC_x) = 0 \)
Let
\( a \; = \; (AB_y \cdot AC_z - AB_z \cdot AC_y) \)
\( b \; = \; (AB_z \cdot AC_x - AB_x \cdot AC_z) \)
\( c \; = \; (AB_x \cdot AC_y-AB_y \cdot AC_x) \)
\( d \; = \; - A_x \cdot a - A_y \cdot b - A_z \cdot c \)
Rewrite the equation of the plane as:
\( a x + by + cz + d = 0 \)
