Equations of Planes in Three Dimensional Space

# Equations of Planes in Three Dimensional Space

We will now look at equations of planes in $\mathbb{R}^3$. There are three forms of planes that we will look at.

 Definition: An equation in the form $Ax + By + Cz + D = 0$ represents the Standard Form Equation of a plane in $\mathbb{R}^3$.

For example, the equation $2x + 3y + z - 1 = 0$ represents a plane in $\mathbb{R}^3$. The point $(1, 1, -4)$ lies on this plane for example because $(1, 1, -4) \in S = \{ (x, y, z) \in \mathbb{R}^3 : 2x + 3y + z - 1 = 0 \}$ where $S$ is the set of points form $\mathbb{R}^3$ that are contained on this plane.

Now the standard form equation of the plane is the simplest form for planes. We will now look at two other forms a plane can be in, but before we do, we will need the following definition.

 Definition If $\Pi$ is a plane in $\mathbb{R}^3$, then if a vector $\vec{n} = (a, b, c)$ is such that $\vec{n} \perp \Pi$ then $\vec{n}$ is said to be a Normal Vector of The Plane $\Pi$.

We are now ready to look at the second different form a plane can be in.

 Definition: Let $\Pi$ be a plane and let $P_0(x_0, y_0, z_0)$ be any point on $\Pi$, and let $\vec{n} = (a, b, c)$ be a normal vector to $\Pi$. Then the Point-Normal Form Equation of $\Pi$ is $\vec{n} \cdot \vec{P_0P} = 0$.

We note that the point-normal form equation of the plane consists of all vectors $\vec{P_0P}$ that are perpendicular to the normal $\vec{n}$. Since the point $P_0(x_0, y_0, z_0)$ is on the plane to begin with, the vector $\vec{P_0P}$ is only the plane $\Pi$ only if $P(x, y, z)$ is also on the plane (because if not, then $\vec{n} \cdot \vec{P_0P} \neq 0$).

There is one more form of a plane that we will look at that is similar to the point-normal form.

 Definition: Let $\Pi$ be a plane and let $P_0(x_0, y_0, z_0)$ and $P(x, y, z)$ be points on the plane. Let $\vec{r_0}$ be the position vector with terminal point $P_0$ and let $\vec{r}$ be the position vector with terminal point $P$. Then the vector $r - r_0$ is parallel to $\Pi$. Let $\vec{n} = (a, b, c)$ be a normal vector to $\Pi$. The Vector Form Equation of $\Pi$ is $\vec{n} \cdot (\vec{r} - \vec{r_0}) = 0$.

We will now look at some examples regarding equations of planes in $\mathbb{R}^3$.

## Example 1

Determine the equation of the plane that passes through $(1, 1, 1)$ and has the normal vector $\vec{n} = (1, 2, 3)$.

The most convenient form to write this plane in is point-normal form as $(1, 2, 3) \cdot (x - 1, y - 1, z - 1) = 0$. We can expand this equation to get the general form of this plane as follows:

(1)
\begin{align} (1, 2, 3) \cdot (x-1, y-1, z-1) = 0 \\ 1(x-1) + 2(y-1) + 3(z-1) = 0 \\ x - 1 + 2y - 2 + 3z - 3 = 0 \\ x + 2y + 3z -6 = 0 \end{align}

## Example 2

Write the point-normal form equation of the plane $2x + 4y + 7z - 2 = 0$.

To write $2x + 4y + 7z - 2 = 0$ in point-normal form we must have a point on the plane and a normal vector to this plane. We can immediately pick up the normal vector for this plane to be $\vec{n} = (2, 4, 7)$. Now we just need a point on the plane. The point $(1, 0, 0)$ is on this plane, and so $(2, 4, 7) \cdot (x - 1, y, z) = 0$ represents the vector-form equation of $2x + 4y + 7z - 2 = 0$.

To verify this, all we need to do is compute the dot product $(2, 4, 7) \cdot (x - 1, y, z)$ as follows:

(2)
\begin{align} (2, 4, 7) \cdot (x - 1, y, z) = 0 \\ 2(x- 1) + 4(y) + 7(z) = 0 \\ 2x + 4y + 7z - 2 = 0 \end{align}

## Example 3

Determine an equation for the plane that passes through $P(6, 2, 3)$, $Q(4, 5, 6)$, and $R(1, 8, 9)$.

We need to find a point on this plane and a normal to this plane. We've been given three points, so all we actually need is a normal vector. To construct a normal vector, consider the vectors $\vec{PQ} = (-2, 3, 3)$ and $\vec{PR} = (-5, 6, 6)$. These vectors both lie on the plane because their initial and terminal points lie on the plane. If we take the cross product of these vectors, we will obtain a vector that is perpendicular to both $\vec{PQ}$ and $\vec{PR}$ and hence a vector that is perpendicular to the plane.

(3)
\begin{align} \quad \quad \vec{PQ} \times \vec{PQ} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -2 & 3 & 3\\ -5 & 6 & 6 \end{vmatrix} = \begin{vmatrix} 3 & 3\\ 6 & 6 \end{vmatrix} \vec{i} - \begin{vmatrix} -2 & 3\\ -5 & 6 \end{vmatrix} \vec{j} + \begin{vmatrix} -2 & 3\\ -5 & 6 \end{vmatrix} \vec{k} = 0\vec{i} -3 \vec{j} + 3 \vec{k} = (0, -3, 3) \end{align}

Therefore $\vec{n} = (0, -3, 3)$ will do. So $(0, -3, 3) \cdot (x - 6, y - 2, z - 3) = 0$ represents the plane that passes through $P$, $Q$, and $R$.