Vector Equations of Planes

Representing planes with vector equations – extension of Vector Equations of Lines

Intuition

The general idea here is that a plane is described by two vectors:

• Vector 1: Formed by two points lying on the plane ($\vec{r}-\overrightarrow{r_0}$)
• Vector 2: Normal vector

Below are just different forms expressing these 2 vectors.

Suppose we have a normal vector that is orthogonal to the plane $\vec{n}=⟨a,b,c⟩$ For points $P_0=(x_0,y_0,z_0)$ and $P=(x,y,z)$ on the plane, let $\overrightarrow{r_0}$ and $\vec{r}$ be position vectors for $P_0$ and $P$. Then, vector $\vec{r}-\overrightarrow{r_0}$ lies on the plane.

Vector Equation of Plane

Since $\vec{n}$ is normal, by the properties of dot product we have:

$$\begin{array}{r}\vec{n}\cdot (\vec{r}-\overrightarrow{r_0})=0\\ A(x-x_0)+B(y-y_0)+C(z-z_0)=0\\ ⟨A,B,C⟩\cdot ⟨x-x_0,y-y_0,z-z_0⟩=0\end{array}$$

We can also use: $\vec{n}\cdot \vec{r}=\vec{n}\cdot \overrightarrow{r_0}$

Scalar Equation of Plane

A more useful form:

$$\begin{array}{rl}⟨a,b,c⟩\cdot (⟨x,y,z⟩-⟨x_0,y_0,z_0⟩)& =0\\ ⟨a,b,c⟩\cdot ⟨x-x_0,y-y_0,z-z_0⟩& =0\\ a(x-x_0)+b(y-y_0)+c(z-z_0)& =0\\ ax+by+cz& =d\end{array}$$

Here, $d=a{x}_0+b{y}_0+c{z}_0$. This equation gives us the normal vector easily because $\vec{n}=⟨a,b,c⟩$