Find an equation for the plane through the points (5,3,4),(−1,0,2),(0,2,−1).

Question

Find the equation.

The plane is: ______ ?

Answer 1

The standard form of the equation of a plane is:

            Ax + By + Cz + D = 0

             Where A, B, and C are from the vector Normal to the plane: N = [A, B, C]

With N and a point on the plane, (x0, y0, z0), the equation of the plane comes from simplifying:

             A(x – x0) + B (y – y0) + C( z – z0) = 0

Let’s start by Naming the Points:

             P (5, 3, 4)

             Q (-1, 0, 2)

             R (0, 2, -1)

Now, to find N, the vector normal to the plane, we must find a vector normal to any pair of vectors in the plane. To do this, we take the cross product of 2 vectors in the plane. Let’s use the vectors PQ and PR:

-         Vector PQ = [ Qx – Px , Qy – Py , Qz – Pz]

[ -1 -5, 0 – 3, 2 – 4]

[ -6, -3, -2 ]

-         Vector PR = [ Rx – Px , Ry – Py , Rz – Pz]

[ 0 - 5, 2 – 3, -1 – 4]

[ -5, -1. -5]

-         PQ x PR = ( (-3 x -5) – (-2 x -1) ) i – ( (-6 x -5) – (-2 x -5) ) j + ( ( -6 x -1) – ( -3 x -5)) k

(15 – 2) i – (30 – 10) j + (6 -15) k

13 i – 20 j – 9 k or [ 13, -20, -9] (A = 13, B = -20, C = 9)

Now we multiply the coefficients of this vector as shown in the opening paragraph:

A(x – x0) + B (y – y0) + C( z – z0) = 0. For (x0, y0, z0), I use point P (5, 3, 4):

-         Plane = 13 (x – 5) – 20 ( y – 3) – 9 (z – 4) = 0

13 x – 65 - 20 y + 60 – 9 z + 36 = 0

13 x – 20y – 9z – 65 + 60 + 36 = 0

13 x – 20y – 9z + 31 = 0