The Cartesian equation of a plane P is ax + by + cz + d = 0, where a, b, c are the coordinates of the normal vector vec n = ( (a), (b), (c) )
Let A, B and C be three noncolinear points, A, B, C in P
Note that A, B and C define two vectors vec (AB) and vec (AC) contained in the plane P. We know that the cross product of two vectors contained in a plane defines the normal vector of the plane.
Now, we can use an example to illustrate the solution. Assume that the coordinates of the three points are the following:
A(1,2,3), B(-2,1,0) and C(0,3,2)
From the coordinates of the points A, B and C we can find the vectors vec (AB) and vec (AC):
vec (AB) = (x_b - x_a)*hat i + (y_b - y_a)*hat j + (z_b - z_a)*hat k
vec (AC) = (x_c - x_a)*hat i + (y_c - y_a)*hat j + (z_c - z_a)*hat k
where hat i, hat j and hat k are the unit vectors on the cartesian axes of coordinates Ox, Oy and Oz.
After plugging in the values of the coordinates, we have:
vec (AB) = (-2 - 1)*hat i + (1-2)*hat j + (0-3)*hat k
So, vec (AB) = -3 hat i - hat j -3 hat k
vec (AC) = (0-1)*hat i + (3-2)* hat j + (2-3)*hat k.
So, vec (AC) = -hat i + hat j - hat k
Next, we'll find the normal vector from the cross product of vec (AB) and vec (AC)
vec n = vec (AB) x vec (AC) = | (hat i,hat j,hat k),(-3,-1,-3),( -1,1,-1)| = 4 hat i - 4 hat k
Therefore, the equation of the plane is 4x-4z+d=0. In order to find d, we can plug in the coordinates of point A:
d= 4z - 4x => d = 4*3 - 4*1 = 8
So, the cartesian equation of the plane is 4x-4z+8=0.
