## V8 Equation of a plane

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.

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

1. the surface that goes through three points or

2. 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,1 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*}

So

\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 2 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 $$(1,1,1)$$3 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

1. What are the vectors normal to the following planes,
1. $$2x+3y+7z=13$$

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

1. 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: $$4x+3y-2z=15$$.

2. 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: $$x-9y-4z=24$$.

3. Find the equation of the plane that contains the point $$(1,-1,0)$$ and is parallel to the plane $$x-3y+2z=0.$$
Answer: $$x-3y+2z=4$$.

4. Find the equation of the plane that contains the points $$(1,2,3)$$ , $$(2,1,1)$$ and $$(-3,0,4).$$
Answer: $$5x-7y+6z=9$$ .

What's next... V9 Intersecting planes

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