Let the plane specified by three points belonging to it K(xk;yk;zk), M(xm;ym;zm), P(xp;yp;zp). To find the normal vector to make the equation of this plane. Mark an arbitrary point lying on the plane, with the letter L, I want her to have coordinates (x;y;z). Now consider the three vectors PK, PM and PL, they lie on the same plane (coplanar), so their scalar triple product is zero.
Find the coordinates of vectors PK, PM, and PL:
PK = (xk-xp, yk-yp;zk-zp)
PM = (xm-xp, ym-yp;zm-zp)
PL = (x-xp, y-yp z-zp)
Mixed product of these vectors is equal to the determinant presented in the figure. This determinant should be calculated to find the equation for the plane. The calculation of the mixed product for a particular case, see example.
Let a plane defined by three points K(2;1;-2), M(0;0;-1) and P(1;8;1). You want to find a normal vector to the plane.
Take an arbitrary point L with coordinates (x;y;z). Compute the vectors PK, PM, and PL:
PK = (2-1;1-8;-2-1) = (1;-7;-3)
PM = (0-1;0-8;-1-1) = (-1;-8;-2)
PL = (x-1;y-8;z-1)
Make the determinant for the scalar triple product of vectors (it is on the picture).
Now decompose the determinant along the first row, and then count the value of the determinant of size 2 by 2.
Thus, the equation of plane is 10x + 5y - 15z - 15 = 0 or what is the same, -2x + y - 3z - 3 = 0. It is easy to determine the vector normal to plane: n = (-2;1;-3).