3D Coordinate Geometry – Equation of a Plane

A plane is a flat, two-dimensional surface that extends infinitely far. A plane is the two-dimensional analog of a point (zero dimensions), a line (one dimension), and three-dimensional space. A plane in three-dimensional space has the equation

$ax + by + cz + d=0,$

where at least one of the numbers $$a, b,$$ and $$c$$ must be non-zero. A plane in 3D coordinate space is determined by a point and a vector that is perpendicular to the plane.

This wiki page is dedicated to finding the equation of a plane from different given perspectives.

## Introduction

A plane in 3D coordinate space is determined by a point and a vector that is perpendicular to the plane. Let $$P_{0}=(x_{0}, y_{0}, z_{0} )$$ be the point given, and $$\overrightarrow{n}$$ the orthogonal vector. Also, let $$P=(x,y,z)$$ be any point in the plane, and $$r$$ and $$r_{0}$$ the position vectors of points $$P$$ and $$P_{0},$$ respectively. Now, if we let $$\overrightarrow{n}=(a,b,c) ,$$ then since $$\overrightarrow{P_{0}P}$$ is perpendicular to $$\overrightarrow{n},$$ we have

\begin{align} \overrightarrow{P_{0}P} \cdot \overrightarrow{n} &= (\overrightarrow{r}-\overrightarrow{r_{0}}) \cdot \overrightarrow{n} \\ &= (x-x_{0}, y-y_{0}, z-z_{0}) \cdot (a, b, c) \\ &= a(x-x_{0}) + b(y-y_{0}) + c(z-z_{0} )\\ &=0. \end{align}

We can also write the above equation of the plane as

$ax+by+cz+d = 0,$

where $$d= -(ax_{0} + by_{0} + cz_{0}) .$$

This does not quite work if one of $$a, b, c$$ is zero. In that case the vector is parallel to one of the coordinate planes. Say $$c = 0$$ then the vector is parallel to the $$xy$$-plane and the equation of the required plane is $$a(x-x_{0}) + b(y-y_{0}) = 0$$ which is of course a straight line in the $$xy$$ plane and $$z$$ is unrestricted. Similar arguments apply if two of $$a, b, c$$ are zero.

Another way to think of the equation of the plane is as a flattened parallelepiped. A flattened parallelepiped, made of three vectors $$\vec{a} = \left \langle x_{1}, y_{1}, z_1 \right \rangle , \vec{b} = \left \langle x_2, y_2, z_2 \right \rangle, \vec{c} = \left \langle x_3, y_3, z_3 \right \rangle$$, has volume 0. We can use the scalar triple product to compute this volume:

$0 = \vec{a} \cdot \big(\vec{b} \times \vec{c}\big),$

where $$\big(\vec{b} \times \vec{c}\big)$$ gives the vector that is normal to the plane.

Let’s say that the endpoints of $$\big(\vec{b} \times \vec{c}\big)$$ are $$( x, y, z )$$ and $$(x_0, y_0, z_0 )$$ and the components of $$\vec{a}$$ are $$\left \langle a, b, c \right \rangle$$. Then by taking the dot product, we get the equation of a plane, which is

$0 = a(x-x_0) + b(y-y_0) + c(z-z_0).$

Here is a problem to try:

## Parallel to the Coordinate Planes

The equation of a plane which is parallel to each of the $$xy$$-, $$yz$$-, and $$zx$$-planes and going through a point $$A=(a,b,c)$$ is determined as follows:

1) The equation of the plane which is parallel to the $$xy$$-plane is $$z=c .$$
2) The equation of the plane which is parallel to the $$yz$$-plane is $$x=a .$$
3) The equation of the plane which is parallel to the $$zx$$-plane is $$y=b.$$

Here is an example based on the above:

What is the equation of the plane which passes through the point $$B=(4,1,0)$$ and is parallel to the $$yz$$-plane?

Since the $$x$$-coordinate of $$B$$ is 4, the equation of the plane passing through $$B$$ parallel to the $$yz$$-plane is

$x=4. \ _\square$

Try the following problem:

## Normal Vector and a Point

If we know the normal vector of a plane and a point passing through the plane, the equation of the plane is established.

Thus, the equation of a plane through a point $$A=(x_{1}, y_{1}, z_{1} )$$ whose normal vector is $$\overrightarrow{n} = (a,b,c)$$ is

$a(x-x_{1}) + b(y-y_{1}) + c(z-z_{1}) = 0 .$

Check out the following examples:

If a plane is passing through the point $$A=(1,3,2)$$ and has normal vector $$\overrightarrow{n} = (3,2,5),$$ then what is the equation of the plane?

The equation of the plane which passes through $$A=(1,3,2)$$ and has normal vector $$\overrightarrow{n} = (3,2,5)$$ is

\begin{align} 3(x-1) + 2(y-3) + 5(z-2) &= 0 \\ 3x – 3 + 2y – 6 + 5z – 10 &= 0 \\ 3x + 2y + 5z – 19 &=0. \ _\square \end{align}

If a plane is passing through the point $$A=(5,6,2)$$ and has normal vector $$\overrightarrow{n} = (-1,3,-7),$$ then what is the equation of the plane?

The equation of the plane which passes through the point $$A=(5,6,2)$$ and has normal vector $$\overrightarrow{n} = (-1,3,-7)$$ is

\begin{align} -1(x-5) + 3(y-6) -7(z-2) &= 0 \\ -x+5+3y-18-7z+14 &= 0 \\ -x+3y-7z+1 &=0. \ _\square \end{align}

Try the following problem:

## Passing through Three Points

When we know three points on a plane, we can find the equation of the plane by solving simultaneous equations.

Let $$ax+by+cz+d=0$$ be the equation of a plane on which there are the following three points: $$A=(1,0,2), B=(2,1,1),$$ and $$C=(-1,2,1).$$ Then the equation of the plane is established as follows:

We already have the equation of the plane with 4 unknown constants:

$ax + by + cz +d = 0. \qquad (1)$

We also get the following 3 equations by substituting the coordinates of $$A, B,$$ and $$C$$ into $$(1):$$

\begin{align} a \cdot 1 + b \cdot 0 + c \cdot 2 + d &= 0 \\ a \cdot 2 + b \cdot 1 + c \cdot 1 + d &= 0 \\ a \cdot (-1) + b \cdot 2 + c \cdot 1 +d &= 0, \end{align}

which gives $$b=3a, c=4a, d=-9a. \qquad (2)$$

Substituting $$(2)$$ into $$(1) ,$$ we have

\begin{align} ax + 3ay + 4az -9a &= 0 \\ x + 3y + 4z – 9 &=0. \end{align}

Hence, the equation of the plane passing through the three points $$A=(1,0,2), B=(2,1,1),$$ and $$C=(-1,2,1)$$ is

$x + 3y + 4z – 9 =0 .$

Using this method, we can find the equation of a plane if we know three points. Here are a couple of examples:

If a plane is passing through the three points $$A=(0,0,2), B=(1,0,1),$$ and $$C=(3,1,1) ,$$ then what is equation of the plane?

Let the equation of the plane be $$ax+by+cz+d=0. \qquad (1)$$

Then since this plane includes the three points $$A=(0,0,2), B=(1,0,1),$$ and $$C=(3,1,1) ,$$ we have

\begin{align} a \cdot 0 + b \cdot 0 + c \cdot 2 + d &= 0 \\ a \cdot 1 + b \cdot 0 + c \cdot 1 + d &= 0 \\ a \cdot 3 + b \cdot 1 + c \cdot 1 +d &= 0, \end{align}

which gives $$b=-2a, c=a, d=-2a. \qquad (2)$$

Substituting $$(2)$$ into $$(1) ,$$ we have

\begin{align} ax + -2ay + az -2a &= 0 \\ x -2y + z – 2 &=0. \end{align}

Hence, the equation of the plane passing through the three points $$A=(0,0,2), B=(1,0,1)$$ and $$C=(3,1,1)$$ is

$x -2y + z – 2 =0. \ _\square$

If a plane is passing through the three points $$A=(3,1,2), B=(6,1,2),$$ and $$C=(0,2,0) ,$$ then what is the equation of the plane?

Let the equation of the plane be $$ax+by+cz+d=0. \qquad (1)$$

Then since this plane includes the three points $$A=(0,0,2), B=(1,0,1),$$ and $$C=(3,1,1) ,$$ we have

\begin{align} a \cdot 3 + b \cdot 1 + c \cdot 2 + d &= 0 \\ a \cdot 6 + b \cdot 1 + c \cdot 2 + d &= 0 \\ a \cdot 0 + b \cdot 2 + c \cdot 0 +d &= 0, \end{align}

which gives $$a=0, c=\frac{1}{2}b, d=-2b . \qquad (2)$$

Substituting $$(2)$$ into $$(1) ,$$ we have

\begin{align} 0x + -by + \frac{1}{2}bz -2b &= 0 \\ x -y + \frac{1}{2}z – 2 &=0 \\ 2x – 2y +z-4 &=0. \end{align}

Hence, the equation of the plane passing through the three points $$A=(0,0,2), B=(1,0,1),$$ and $$C=(3,1,1)$$ is

$2x – 2y +z-4 =0. \ _\square$

Try the following problem:

## Problem Solving

This section is dedicated to improve your problem-solving skills through several problems to try.

You are watching: 3D Coordinate Geometry – Equation of a Plane. Info created by GBee English Center selection and synthesis along with other related topics.