Recall that a plane is defined by three non-collinear points and the general form for a plane is, Ax + By + Cz + D = 0 where {A, B, C} is a set of direction numbers for a line normal or perpendicular to the plane.
Determine the set of direction numbers for the line normal to the plane. Write an equation in A, B, C and D for each of the three given points. For example if given points P1 (1,0,1 ), P2 (-1,-2, 1) and P3 (2,-2,2) write three equations:
P1 (1,0,1) yields A + C + D = 0
P2 (-1,-2,1) yields -A - 2B + C + D = 0
P3 (2,-2,2) yields 2A - 2B + 2C + D = 0.
Solve for A, B, C, and D using linear algebra. Let P1P2 and P1P3 represent vectors u and v respectively so:
u = (-1-1) i + (-2-0) j + (1-1) k = -2i - 2j
v = (2-1) i + (-2-0) j + (2-1) k = i -2j + k
Since u and v lie on the plane their cross product is perpendicular to the plane. Solving u x v yields,
u x v = [ -2i -2j ] x [ i -2j +k] = -2i + 0j +4k - 0i +2j + 2k = -2i + 2j +6k
The set of direction numbers perpendicular to the plane is therefore { -2, 2, 6 }.
Write the equation of the plane using the direction numbers and one of the three given points. For example with direction numbers {-2, 2, 6} and point (-1,-2,1) write, -2 ( x + 1) +2 (y +2) + 6( z - 1) = 0.
-2 ( x + 1) +2 (y +2) + 6 ( z - 1) = 0
-2x - 2 +2y + 4 + 6z - 6 = 0
-2x +2y + 6z - 4 = 0
-x + y + 3z - 2 = 0
-x + y + 3z = 2
Plug the values of point in question into the derived equation to determine if it is on the plane. If plugging in the point yields a true statement, the point is on the same plane. In the example the equation for the plane was -x + y +3z = 2. Considering a point with the coordinates, ( 1, 0, 1) , to determine if it is on the same plane plug into the derived equation:
-1 + 0 + 3 (1) = 2
-1 + 0 + 3 = 2
2 = 2
This is a true statement therefore this point is on the same plane.