Point-Normal Form of a Plane

In 2-space, a line can algebraically be expressed by simply knowing a point that the line goes through and its slope. This can be expressed in the form $y - y_0 = m(x - x_0)$. In 3-space, a plane can be represented differently. We will still need some point that lies on the plane in 3-space, however, we will now use a value called the normal that is analogous to that of the slope.

Point-Normal Form of a Plane

We must first define what a normal is before we look at the point-normal form of a plane:

Definition: A Normal Vector usually denoted $\vec{n} = (a, b, c)$, is a vector that is perpendicular to a plane $\Pi$. That is $\vec{n} \perp \Pi$.

Suppose that the points $P_0 = (x_0, y_0, z_0)$ and $P = (x, y, z)$ lay on the plane $\Pi$. The vector $\vec{P_0P}$ runs along this plane. Now let $\vec{n} = (a, b, c)$ be a normal of this plane. It thus follows that $\vec{P_0P} \perp \vec{n}$. From the dot product, we know that $\vec{P_0P} \cdot \vec{n} = 0$, which when expanded gives us the point-normal form of a plane:

\begin{align} 0 = \vec{P_0P} \cdot \vec{n}\\ 0 = (x - x_0, y - y_0, z - z_0) \cdot (a, b, c) \\ \: 0 = a(x - x_0) + b(y - y_0) + c(z - z_0) \end{align}
Definition: Let $\Pi$ be a plane in $\mathbb{R}^3$. Then the Point-Normal Form Equation of the plane $\Pi$ is $0 = \vec{n} \cdot \vec{P_0 P} = (a, b, c) \cdot (x - x_0, y - y_0, z - z_0)$ where $\vec{n}$ is any normal vector to $\Pi$ and $\vec{P_0}$ is any point on $\Pi$.

Often times we expand the last formula out to get $ax - ax_0 + by - by_0 + cz - cz_0 =$, and we let $d = -ax_0 - by_0 - cz_0$ to get $ax + by + cz + d = 0$.

Example 1

Find the equation of a plane that passes through the point $P(-2, 3, 4)$ and is perpendicular to the vector $\vec{n} = (1, 3, -7)$.

We can solve this question by inputting what we know into the formula, that is:

\begin{align} a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \\ (x + 2) +3(y - 3) -7(z - 4) = 0 \end{align}

We are done as we have an equation of a plane that satisfies our conditions. Alternatively we could make this equation neater and write it is $x + 3y - 7z + 21 = 0$.

Example 2

Determine the normal vector of the following equation of a plane $2x + 3y -6z + 3 = 0$.

The general equation of a plane is in the form $ax + by + cz + d = 0$. From this we can easily pull the normal to be $\vec{n} = (a, b, c) = (2, 3, -6)$.

Example 3

Determine the point-normal form of a plane that goes through the points $P(1, 4, 2)$, $Q(-10, 4, 3)$, and $R(2, 2, 4)$.

It is important to recognize that we will need both a single point and the normal vector to determine the point-normal form of this line. We already have a point given to us, in fact, we have three! We can use either as it does not matter as long as both lie on the plane (and both do according to the question). Hence, we will need to find a normal vector. We can do this with the use of the cross product, however, we will need to find two parallel vectors first. We can easily do this with utilization of the points given. Let's use the vector formed from the cross product of $\vec{PQ} = (-11, 0, 1)$ and $\vec{PR} = (1, -2, 2)$

\begin{align} \quad \vec{PQ} \times \vec{PR} = \vec{n} = (2, 23, 22) \end{align}

Thus we can now input our normal and one of the three points into our point-normal form a line to obtain $2(x - 1) + 23(x - 4) + 22(x - 2) = 0$.

Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License