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.

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## 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.