Any flat surface, such as a wall, table top or stiff piece of cardboard can be thought of as representing part of a plane. Consider a piece of cardboard with a point $P$ marked on it. One can take a nail and stick it into the cardboard at $P$ such that the nail is perpendicular to the cardboard; see Figure 10.6.1

This nail provides a “handle” for the cardboard. Moving the cardboard around moves $P$ to different locations in space. Tilting the nail (but keeping $P$ fixed) tilts the cardboard. Both moving and tilting the cardboard defines a different plane in space. In fact, we can define a plane by: 1) the location of $P$ in space, and 2) the direction of the nail.

The previous section showed that one can define a line given a point on the line and the direction of the line (usually given by a vector). One can make a similar statement about planes: we can define a plane in space given a point on the plane and the direction the plane “faces” (using the description above, the direction of the nail). Once again, the direction information will be supplied by a vector, called a normal vector, that is orthogonal to the plane.

What exactly does “orthogonal to the plane” mean? Choose any two points $P$ and $Q$ in the plane, and consider the vector $\vv{PQ}\text{.}$ We say a vector $\vec n$ is orthogonal to the plane if $\vec n$ is perpendicular to $\vv{PQ}$ for all choices of $P$ and $Q\text{;}$ that is, if $\vec n\cdot \vv{PQ}=0$ for all $P$ and $Q\text{.}$

This gives us way of writing an equation describing the plane. Let $P=(x_0,y_0,z_0)$ be a point in the plane and let $\vec n = \la a,b,c\ra$ be a normal vector to the plane. A point $Q = (x,y,z)$ lies in the plane defined by $P$ and $\vec n$ if, and only if, $\vv{PQ}$ is orthogonal to $\vec n\text{.}$ Knowing $\vv{PQ} = \la x-x_0,y-y_0,z-z_0\ra\text{,}$ consider: \begin{align*} \vv{PQ}\cdot\vec n \amp = 0\\ \la x-x_0,y-y_0,z-z_0\ra\cdot \la a,b,c\ra \amp =0\\ a(x-x_0)+b(y-y_0)+c(z-z_0) \amp =0\\ \end{align*} Equation (10.6.3) defines an implicit function describing the plane. More algebra produces: \begin{align*} ax+by+cz \amp = ax_0+by_0+cz_0.\\ \end{align*} The right hand side is just a number, so we replace it with $d\text{:}$ \begin{align*} ax+by+cz \amp = d .\\ \end{align*} As long as $c\neq 0\text{,}$ we can solve for $z\text{:}$ \begin{align*} z \amp = \frac1c(d-ax-by). \end{align*}

Equation (eq_planes3) is especially useful as many computer programs can graph functions in this form. Equations <<Unresolved xref, reference "eq_planes1"; check spelling or use "provisional" attribute>> and (eq_planes2) have specific names, given next.

##### Definition10.6.2Equations of a Plane in Standard and General Forms

The plane passing through the point $P=(x_0,y_0,z_0)$ with normal vector $\vec n=\la a,b,c\ra$ can be described by an equation with standard form \begin{equation*} a(x-x_0)+b(y-y_0)+c(z-z_0) =0; \end{equation*} the equation's general form is \begin{equation*} ax+by+cz = d. \end{equation*}

A key to remember throughout this section is this: to find the equation of a plane, we need a point and a normal vector. We will give several examples of finding the equation of a plane, and in each one different types of information are given. In each case, we need to use the given information to find a point on the plane and a normal vector.

##### Example10.6.3Finding the equation of a plane.

Write the equation of the plane that passes through the points $P=(1,1,0)\text{,}$ $Q = (1,2,-1)$ and $R = (0,1,2)$ in standard form.

Solution

We have just demonstrated the fact that any three non-collinear points define a plane. (This is why a three-legged stool does not “rock;” it's three feet always lie in a plane. A four-legged stool will rock unless all four feet lie in the same plane.)

##### Example10.6.5Finding the equation of a plane.

Verify that lines $\ell_1$ and $\ell_2\text{,}$ whose parametric equations are given below, intersect, then give the equation of the plane that contains these two lines in general form. \begin{equation*} \ell_1: \begin{array}{ccc} x\amp =\amp -5+2s \\ y\amp =\amp 1+s \\ z\amp =\amp -4+2s \end{array} \qquad\qquad \ell_2: \begin{array}{ccc} x \amp =\amp 2+3t\\ y\amp =\amp 1-2t \\ z\amp =\amp 1+t \end{array} \end{equation*}

Solution
##### Example10.6.7Finding the equation of a plane

Give the equation, in standard form, of the plane that passes through the point $P=(-1,0,1)$ and is orthogonal to the line with vector equation $\vec \ell(t) = \la -1,0,1\ra + t\la 1,2,2\ra\text{.}$

Solution
##### Example10.6.9Finding the intersection of two planes

Give the parametric equations of the line that is the intersection of the planes $p_1$ and $p_2\text{,}$ where: \begin{gather*} p_1: x-(y-2)+(z-1) =0\\ p_2: -2(x-2)+(y+1)+(z-3)=0 \end{gather*}

Solution
##### Example10.6.11Finding the intersection of a plane and a line

Find the point of intersection, if any, of the line $\ell(t) = \la 3,-3,-1\ra +t\la-1,2,1\ra$ and the plane with equation in general form $2x+y+z=4\text{.}$

Solution

# Subsection10.6.1Distances

Just as it was useful to find distances between points and lines in the previous section, it is also often necessary to find the distance from a point to a plane.

Consider Figure 10.6.13, where a plane with normal vector $\vec n$ is sketched containing a point $P$ and a point $Q\text{,}$ not on the plane, is given. We measure the distance from $Q$ to the plane by measuring the length of the projection of $\vv{PQ}$ onto $\vec n\text{.}$ That is, we want: $$\snorm{\text{ proj } _{\,\vec n}\,{\vv{PQ}}} = \snorm{\frac{\vec n\cdot \vv{PQ}}{\vnorm n^2}\vec n} = \frac{\abs{\vec n\cdot \vv{PQ}}}{\vnorm n} \label{eq_plane_dist}\tag{10.6.1}$$

Equation (10.6.1) is important as it does more than just give the distance between a point and a plane. We will see how it allows us to find several other distances as well: the distance between parallel planes and the distance from a line and a plane. Because Equation (10.6.1) is important, we restate it as a Key Idea.

##### Key Idea10.6.14Distance from a Point to a Plane

Let a plane with normal vector $\vec n$ be given, and let $Q$ be a point. The distance $h$ from $Q$ to the plane is \begin{equation*} h = \frac{\abs{\vec n\cdot \vv{PQ}}}{\vnorm n}, \end{equation*} where $P$ is any point in the plane.

##### Example10.6.15Distance between a point and a plane

Find the distance bewteen the point $Q = (2,1,4)$ and the plane with equation $2x-5y+6z=9\text{.}$

Solution

We can use Key Idea 10.6.14 to find other distances. Given two parallel planes, we can find the distance between these planes by letting $P$ be a point on one plane and $Q$ a point on the other. If $\ell$ is a line parallel to a plane, we can use the Key Idea to find the distance between them as well: again, let $P$ be a point in the plane and let $Q$ be any point on the line. (One can also use Key Idea 10.5.13.) The Exercise section contains problems of these types.

These past two sections have not explored lines and planes in space as an exercise of mathematical curiosity. However, there are many, many applications of these fundamental concepts. Complex shapes can be modeled (or, approximated) using planes. For instance, part of the exterior of an aircraft may have a complex, yet smooth, shape, and engineers will want to know how air flows across this piece as well as how heat might build up due to air friction. Many equations that help determine air flow and heat dissipation are difficult to apply to arbitrary surfaces, but simple to apply to planes. By approximating a surface with millions of small planes one can more readily model the needed behavior.

# Subsection10.6.2Exercises

Terms and Concepts

In the following exercises, give any two points in the given plane.

In the following exercises, give the equation of the described plane in standard and general forms.

In the following exercises, give the equation of the line that is the intersection of the given planes.

In the following exercises, find the point of intersection between the line and the plane.

In the following exercises, find the given distances.