V8 Equation of a plane

Open image

Learn how to find the equation of a plane (a 2-dimensional space):
a) through three points or
b) given a normal (line at right angles) and a point on the plane or
c) given a parallel plane and a point on the plane.

(See also Linear graphs)

A plane is a subset of three dimensional space. In non mathematical terms you can think of it as a flat surface that extends infinitely in two directions. It may be defined as

the surface that goes through three points or
the surface containing a point and having a fixed normal vector.

Cartesian Equation of a Plane

The above diagram represents a part of a plane. On the plane are two points, $P(x,y,z)$ and $P_{0}(x_{0},y_{0},z_{0})$ .

The vector $\overrightarrow{P_{0}P}=(x-x_{0})\hat{i}+(y-y_{0})\hat{j}+(z-z_{0})\hat{k}$.

Also illustrated is a normal vector to the plane (that is a vector at right angles to the plane) represented by $\overrightarrow{N}=a\hat{i}+b\hat{j}+c\hat{k}$, where $a,b$ and $c$ are constants.

Since $\overrightarrow{N}$ is perpendicular to $\overrightarrow{P_{0}P}$ their dot product must equal zero,11 The definition of the dot product of two vectors $\vec{a}$ and $\vec{b}$ is \[\begin{align*} \vec{a}\cdot\vec{b} & =\left|\vec{a}\right|\left|\vec{b}\right|\cos\theta \end{align*}\] where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$. If $\vec{a}$ and $\vec{b}$ are perpendicular, $\theta=90^{\circ}$. Hence $\cos\left(\theta\right)=\cos\left(90^{\circ}\right)=0$ and so \[\begin{align*} \vec{a}\cdot\vec{b} & =0. \end{align*}\] that is to say $\overrightarrow{N}.\overrightarrow{P_{0}P}=0$ therefore:

\[\begin{align*} \left(a\hat{i}+b\hat{j}+c\hat{k}\right).\left((x-x_{0})\hat{i}+(y-y_{0})\hat{j}+(z-z_{0})\hat{k}\right) & =0 \end{align*}\]

\[\begin{align*} a(x-x_{0})+b(y-y_{0})+c(z-z_{0}) & =0. \end{align*}\]

This equation defines the plane in Cartesian coordinates.

This equation may be rearranged: \[\begin{align*} a(x-x_{0})+b(y-y_{0})+c(z-z_{0}) & =0\\ ax+by+cz & =ax_{0}+by_{0}+cz_{0}\\ & =d \end{align*}\] where $d$ is a constant. In fact, the general equation of a plane with the normal vector $\overrightarrow{N}=a\hat{i}+b\hat{j}+c\hat{k}$ is \[\begin{align*} ax+by+cz & =d. \end{align*}\]

For example,
1. $\;2x-3y+z=6$ is a plane.
2.$\;x+y=4$ is a plane.
3.$\;z=x-2y+3$ is a plane. To see this rearrange it to get the Cartesian form of the plane, $x-2y-z=-3$.

Example 1

If a plane has the normal vector $\overrightarrow{N}=\hat{i}+2\hat{j}-5\hat{k}$ and contains the point $(3,4,1)$ then we can say that the equation of the plane is: \[\begin{align*} 1(x-3)+2(y-4)-5(z-1) & =0\\ x-3+2y-8-5z+5 & =0\\ x+2y-5z & =3+8-5\\ x+2y-5z & =6 \end{align*}\]

Compare this equation with the normal vector, $\overrightarrow{N}$. You will notice that the coefficients of $\hat{i}$, $\hat{j}$ and $\hat{k}$ for the normal vector are the same as the coefficients of $x,y$ and $z$ in the equation of the plane.

Example 2

Find the equation of the plane normal to the vector $6\hat{i}-5\hat{j}+\hat{k}$ that passes through the point $(1,2,3)$.

The equation of the plane will be 22 Obviously there are an infinite number of parallel planes that are perpendicular to the given vector. For instance $6x-5y+z=-1$ , $6x-5y+z=0$ , $6x-5y+z=6$ , $6x-5y+z=11$ etc. are all perpendicular to the vector $6\hat{i}-5\hat{j}+\hat{k}$ but only one of these, $6x-5y+z=-1$ , will pass through the given point. \[\begin{align*} 6(x-1)-5(y-2)+1(z-3) & =0\\ 6x-6-5y+10+z-3 & =0\\ 6x-5y+z & =6-10+3\\ 6x-5y+z & =-1 \end{align*}\]

Not only can we find the equation of a plane, given a point and a normal vector, but we can also find the equation of the normal vector given the equation of a plane.

For example the plane $3x+4y-z=9$ has a normal vector $\overrightarrow{N}=3\hat{i}+4\hat{j}-\hat{k}$.

Example 3

Find the equation of the plane that passes through the point $(2,1,-5)$ and is parallel to the plane $z=2x+3y-4$ .

Since the plane is parallel to $z=2x+3y-4$ , it will have the same normal vector.

The equation $z=2x+3y-4$ can be written as $2x+3y-z=4$ , therefore its normal vector will be $2\hat{i}+3\hat{j}-\hat{k}$ .

So the equation of the plane we wish to find will pass through the point $(2,1,-5)$ and have a normal vector $2\hat{i}+3\hat{j}-\hat{k}$ .

\[\begin{align*} 2(x-2)+3(y-1)-1(z+5) & =0\\ 2x-4+3y-3-z-5 & =0\\ 2x+3y-z & =4+3+5\\ 2x+3y-z & =12 \end{align*}\]

Example 4

Find the equation of the plane that contains the three points $A(1,1,1)$ , $B(2,4,3)$ and $C(3,2,1)$ .

First find the cross product of the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ as this will give a vector normal to the plane.

Then use this normal vector and any one of the three given points to find the equation of the plane.

$\overrightarrow{AB}\times\overrightarrow{AC}=($ $1\hat{i}+3\hat{j}+2\hat{k})\times(2\hat{i}+\hat{j})$

$=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k}\\ 1 & 3 & 2\\ 2 & 1 & 0 \end{array}\right|=-2\hat{i}+4\hat{j}-5\hat{k}$

So the plane has the normal vector $\overrightarrow{N}=-2\hat{i}+4\hat{j}-5\hat{k}$ and contains the point $(1,1,1)$33 Either of the other two points could also have been used

Therefore the equation of the plane is

\[\begin{align*} -2(x-1)+4(y-1)-5(z-1) & =0\\ -2x+2+4y-4-5z+5 & =0\\ -2x+4y-5z & =-2+4-5\\ -2x+4y-5z & =-3\\ \textrm{or $2x-4y+5z$ } & =3. \end{align*}\]

Exercises

What are the vectors normal to the following planes,

$2x+3y+7z=13$
$z=x+3y-9.$
Answer: a) $2\hat{i}+3\hat{j}+7\hat{k}\qquad$b) $\hat{i}+3\hat{j}-\hat{k}$.

Find the equation of the plane that contains the point $(2,3,1)$ and has the normal vector $4\hat{i}+3\hat{j}-2\hat{k}.$
Answer: $4x+3y-2z=15$.
Find the equation of the plane that contains the point $(5,-3,2)$ and has the normal vector $\hat{i}-9\hat{j}-4\hat{k}.$
Answer: $x-9y-4z=24$.
Find the equation of the plane that contains the point $(1,-1,0)$ and is parallel to the plane $x-3y+2z=0.$
Answer: $x-3y+2z=4$.
Find the equation of the plane that contains the points $(1,2,3)$ , $(2,1,1)$ and $(-3,0,4).$
Answer: $5x-7y+6z=9$ .