# Equations of planes

A plane is a flat, two-dimensional surface that extends to infinity. You can have a plane in $\displaystyle \mathbb{R}^{2}$, but we don’t talk about that often because
it’s not very interesting. There is only one plane in two-space, the *xy* plane. However, in three-space, there are $\displaystyle \infty$ planes.

Consider a point, defined by a tuple of coordinates. The point has zero dimensions because there is only *one* point. Now consider a line—a line is just a set of $\displaystyle \infty$ points that happen to lie in a straight line (yes, this is a circular definition). The line is one-dimensional and has one degree of freedom; it is controlled by the free parameter *t* in

$\displaystyle \vec{r} = \vec{r}_{0} + t \vec{m} , \qquad t \in \mathbb{R}$.

Now consider a plane. A plane another set of $\displaystyle \infty$ points, and these ones all lie on a flat surface. You could also think of it as $\displaystyle \infty$ parallel lines. The
plane is two-dimensional and has two degrees of freedom; it is controlled by the free parameters *s* and *t* in

$\displaystyle \vec{r} = \vec{r}_{0} + s \vec{u} + t \vec{v} , \qquad s , t \in \mathbb{R}$.

That’s the vector equation of a plane. As long as $\displaystyle \vec{u}$ and $\displaystyle \vec{v}$ are non-collinear, this allows $\displaystyle \vec{r}$ to move around anywhere on the plane by choosing different values for the free parameters.

We are trying to generalize ideas about sets of points in different dimensions. It is easier to see the natural progression of adding degrees of freedom with a drawing—but don’t forget that these are really two-dimensional representations:

The arrows on the line indicates that it extends to infinity in both directions. You may be tempted to think of the plane as a quadrilateral, but it is most definitely not. The plane doesn’t really have edges—it’s just hard to draw it without them. The same thing goes for the solid: it’s not a cube. It extends to infinity, so there is only one solid in three-space: the *xyz* solid.

As I mentioned before, there is a single plane in $\displaystyle \mathbb{R}^{2}$ but many planes in $\displaystyle \mathbb{R}^{3}$. We can generalize this idea to all the geometrical figures (sets of points) that we’ve seen so far:

Dimensions | Space | Many | One |
---|---|---|---|

0 | $\displaystyle \mathbb{R}^{0}$ | — | point |

1 | $\displaystyle \mathbb{R}^{1}$ | point | line |

2 | $\displaystyle \mathbb{R}^{2}$ | line | plane |

3 | $\displaystyle \mathbb{R}^{3}$ | plane | solid |

4 | $\displaystyle \mathbb{R}^{4}$ | solid | hypersolid |

Let’s look at the equation of the plane again:

$\displaystyle \vec{r} = \vec{r}_{0} + s \vec{u} + t \vec{v} , \qquad s , t \in \mathbb{R}$.

We can substitute components like we did for the equation of the line:

$\displaystyle \left [ x , y , z \right ] = \left [ x_{0} , y_{0} , z_{0} \right ] + s \left [ u_{1} , u_{2} , u_{3} \right ] + t \left [ v_{1} , v_{2} , v_{3} \right ]$.

From this, we can get the parametric equations of the plane:

$\displaystyle x = x_{0} + s u_{1} + t v_{1} , \quad y = y_{0} + s u_{2} + t v_{2} , \quad z = z_{0} + s u_{3} + t v_{3}$.

Another way to describe a plan is with a point and a *normal vector*. The normal vector is perpendicular to the plane, and we get it by crossing the plane’s two direction vectors:

$\displaystyle \vec{n} = \vec{u} \times \vec{v}$.

Given this normal vector and a position vector $\displaystyle \vec{r}_{0}$, every point $\displaystyle \vec{r}$ on the plane satisfies the equation

$\displaystyle \vec{n} \cdot \left ( \vec{r} - \vec{r}_{0} \right ) = 0$.

This works because the dot product of perpendicular vectors is always zero, and the vector $\displaystyle \vec{r} - \vec{r}_{0}$ is parallel to the plane. You can use points instead of position vectors if you wish: $\displaystyle \vec{n} \cdot \overrightharpoon{A P} = 0$, where A is the initial point and P represents all other points on the plane. Suppose we replace the vectors in that equation with components:

$\displaystyle \left [ a , b , c \right ] \cdot \left ( \left [ x , y , z \right ] - \left [ x_{0} , y_{0} , z_{0} \right ] \right ) = 0$.

We can simplify this to give us

$\displaystyle a \left ( x - x_{0} \right ) + b \left ( y - y_{0} \right ) + c \left ( z - z_{0} \right ) = 0$.

When we actually have numbers for $\displaystyle \vec{r}_{0}$, we usually write this as

$\displaystyle a x + b y + c z + d = 0$,

where
$\displaystyle d = - \left ( a x_{0} + b y_{0} + c z_{0} \right )$.
This is called the *standard form* of the equation of a plane. It is just like the standard form of a line in $\displaystyle \mathbb{R}^{2}$, but with an added term
for the third dimension. The standard form of a plane is also useful because the normal vector is in plain sight. Given a normal vector
$\displaystyle \vec{n} = \left [ a , b , c \right ]$
and a point P, we can jump straight to the standard form equation by substituting the components of
$\displaystyle \vec{n}$
into the equation above. We still don’t know *d*, so we need to substitute in the point and solve for it.

You’ll notice that the value of *d* and the normal vector are independent. Since *d* doesn’t affect the direction of the plane, it must affect something else. If we set it to zero, the plane passes through the origin. If we set it to one, the shortest distance between the origin and the plane is one. The value of *d* effectively moves the plane away from the origin in the direction of the normal vector.

The normal vector is useful for another reason: we can use it to find the angle between two planes. If we have planes with normal vectors $\displaystyle \vec{n}_{1}$ and $\displaystyle \vec{n}_{2}$, the acute angle separating them is given by

$\displaystyle \theta_{\text{acute}} = \cos^{- 1}{\left ( \frac{\left \lvert \vec{n}_{1} \cdot \vec{n}_{2} \right \rvert}{\left \lvert \vec{n}_{1} \right \rvert \left \lvert \vec{n}_{2} \right \rvert} \right )}$.