Normal vector of the plane (or a plane normal) call the vector perpendicular to this plane. One of ways to set the plane is the indication of coordinates of its normal and point lying on the plane. If the plane is set by Ax+By+Cz+D=0 equation, then to it the vector with coordinates is normal (A; B; C). In other cases for calculation of a normal vector it is necessary to work.

## Instruction

1. Let the plane be set by three K points belonging to it (xk; yk; zk), M (xm; ym; zm), P (xp; yp; zp). To find normal vector, we will work out the equation of this plane. Designate any point lying on the plane, letter L, let it will have coordinates (x; y; z). Now consider three vectors of PK, PM and PL, they lie on one plane (komplanarna) therefore their mixed work is equal to zero.

2. Find coordinates of vectors of 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) the Mixed work of these vectors will be equal to the determinant presented in the drawing. This determinant should be calculated to find the equation for the plane. You watch calculation of the mixed work for a specific case in an example.

3. Primerpust the plane is set by three points of K(2;1;-2), M (0;0;-1) and P(1;8;1). It is required to find a normal vector of the plane. Take any point of L with coordinates (x; y; z). Calculate vectors of 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 determinant for the mixed work of vectors (it in the drawing).

4. Now spread out determinant on the first line, and then count values of determinants of size 2 on 2. Thus the plane equation - 10x + 5y - 15z - 15 = 0 or that the same, - 2x + y - 3z - 3 = 0. From here it is easy to define a vector of a normal to the plane: n = (-2;1;-3).