Find the Cartesian Equation of a Plane

The Cartesian equation of a plane in space is often very helpful when attempting to find its characteristics.

Steps

Find a normal vector. Sometimes this will be given, but often you will have to figure it out from other givens. For the purposes of this article, the normal vector is defined as n = < A, B, C> where n is a vector, and A, B, and C are scalars.
Find a point on the line. Usually this is a given, although occasionally you will have to come up with one based on the equation of a line. If you are given parametric equations, the easiest way to do this is to find the point where t = 0. For the purposes of this article, the point is defined as P(x₁, y₁, z₁)
Plug these values into the equation for a plane. A(x-x₁)+B(y-y₁)+C(z-z₁)=0
Simplify the equation. This will yield: Ax-Ax₁+By-By₁+Cz-Cz₁=0. Remembering that x₁, y₁, and z₁ are scalars, you can add them together and then add the opposite of that number to the other side of the equation. This value is known as D. The equation will now be in the standard Cartesian form of Ax+By+Cz=D

Because of the nature of the Cartesian Equation, a vector normal to the plane can be found directly from it. If the equation is in the standard form Ax+By+Cz=D, the normal vector is <a, b,="" c="">.
Why this works: Suppose the normal vector is AB, defined by <a, b,="" c="">. Also suppose the point A is coincident with the plane and that A(x, y, z), and again, the point is P(x₁, y₁, z₁). If you were to find the components of the vector PA, you would get (x-x₁)i + (y-y₁)j + (z-z₁)k. Since the dot product of two perpendicular vectors is equal to zero, and the vector PA is on the plane, PA must be normal to AB and taking the dot product of the two is equal to zero. By definition, therefore, A(x-₁)+B(y-y₁)+C(z-z₁)=0