# 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 analogue of a point(zero dimensions), a line(one dimension) and three-dimensional space. A plane in three-dimensional space has the equation

\(\qquad \qquad \qquad \qquad ax + by + cz + d=0,\)

where at least one of the numbers \(a, b\) and \( c\) must be non-zero. A plane in the 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.

#### Contents

## Introduction

A plane in the 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 3D space, and \( r\) and \(r_{0} \) the position vectors of the 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 \( d= -(ax_{0} + by_{0} + cz_{0}) .\)

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:

\[0 = \vec{a} \cdot (\vec{b} \times \vec{c}) \]

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

Let's say that the endpoints of \( \vec{b} \times \vec{c} \) 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, then 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.

## See Also

**Cite as:**3D Coordinate Geometry - Equation of a Plane.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/3d-coordinate-geometry-equation-of-a-plane/