Equation of Plane/Vector Form

Theorem

Let $P$ be a plane which passes through a point $C$ whose position vector relative to the origin $O$ is $\mathbf c$.

Let $\mathbf p$ be the vector perpendicular to $P$ from $O$.

Let $\mathbf r$ be the position vector of an arbitrary point on $P$.

Then $P$ can be represented by the equation:

$\mathbf p \cdot \paren {\mathbf r - \mathbf c} = 0$

where $\cdot$ denotes dot product.

Proof

It is seen that $\mathbf r - \mathbf c$ lies entirely within the plane $P$.

As $P$ is perpendicular to $\mathbf p$, it follows that $\mathbf r - \mathbf c$ is perpendicular to $\mathbf p$.

Hence by Dot Product of Perpendicular Vectors:

$\mathbf p \cdot \paren {\mathbf r - \mathbf c} = 0$

$\blacksquare$

Also presented as

The vector form of the equation of the plane can also be presented in the form:

$\mathbf r \cdot \mathbf n = p$

where:

$\mathbf r$ is the position vector of an arbitrary point on $P$

$\mathbf n$ is the unit normal vector to $P$

$p$ is the length of the normal vector to $P$ through the origin.

Sources

• 1951: B. Hague: An Introduction to Vector Analysis (5th ed.) ... (previous) ... (next): Chapter $\text {II}$: The Products of Vectors: $2$. The Scalar Product