Sometimes it is more appropriate to utilize what is known as the **vector form** of the equation of plane.

# Vector Form Equation of a Plane

Let $\vec{n} = (a, b, c)$ be a normal vector to our plane $\Pi$, that is $\Pi \perp \vec{n}$. Instead of using just a single point from the plane, we will instead take a vector that is parallel from the plane. Let $P(x, y, z)$ and $P_0(x_0, y_0, z_0)$ be two points on the plane. It follows that $\vec{P_0P} \perp \vec{n}$.

We will define the vector $\vec{r_0}$ to have its initial point at the origin and terminal point at $P_0$. We will also define $\vec{r}$ to have its initial point at the origin as well, but instead have its terminal point at $P$. It thus follows that $\vec{P_0P} = \vec{r} – \vec{r_0}$. Furthermore, we note that $\vec{r} – \vec{r_0}$ is parallel to the plane $\Pi$ and perpendicular to $\vec{n}$. From the dot product we get that:

(1)

Definition: Let $\Pi$ be a plane in $\mathbb{R}^3$. Then the Vector Form Equation of the plane $\Pi$ is $0 = \vec{n} \cdot (\vec{r} – \vec{r_0}) = 0$ where $\vec{n} = (a, b, c)$ is any normal vector to $\Pi$, $\vec{r} = \vec{OP}$ and $\vec{r_0} = \vec{OP_0}$. |

# Example 1

**Determine the vector form equation of a plane $\Pi$ given that the points $P(-5, 2, 8)$ and $Q(2, 3, 3)$ lie on the plane and the vector $\vec{n} = (4, 4, -1)$ is perpendicular to $\Pi$.**

To solve this, we must determine what our vectors $\vec{r}$ and $\vec{r_0}$ are. We note that if $O$ is the origin, then $\vec{OP} = \vec{r_0} = (-5, 2, 8)$ and $\vec{OQ} = \vec{r} = (2, 3, 3)$. Plugging these values and our norm $\vec{n}$ into the form we obtain:

(2)