there exists a point $A$ with coordinates $(x_1,y_1,z_1)$ on the plane.
a vector $\boldsymbol{n}$ that is perpendicular to the plane. This vector is called the norm vector.

Visually, we have the following scenario:

Here, $R$ is another point on the plane with coordinates $(x,y,z)$. The difference between points $A$ and $R$ is that $A$ is assumed to be known, that is, $(x_1,y_1,z_1)$, are some fixed scalar values whereas $R$ can vary.

We know from this sectionlink that the vector pointing from $A$ to $R$, which we denote as $\boldsymbol{v}$, can be expressed as:

$$\boldsymbol{v}= \begin{pmatrix} x-x_1\\ y-y_1\\ z-z_1 \end{pmatrix} = \begin{pmatrix}x\\y\\z\end{pmatrix}- \begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix} = \boldsymbol{r}- \boldsymbol{a} $$

Where $\boldsymbol{r}$ and $\boldsymbol{a}$ are the position vectors pointing to $R$ and $A$ respectively.

From theoremlink, because $\boldsymbol{v}$ is perpendicular to the normal vector $\boldsymbol{n}$, their dot product must equal zero:

$$\begin{equation}\label{eq:NoUqoRBzM8vBpDGtLsK} \boldsymbol{n}\cdot\boldsymbol{v}=0 \end{equation}$$

Let's substitute the vectors into \eqref{eq:NoUqoRBzM8vBpDGtLsK} and simplify:

$$\begin{equation}\label{eq:Wy6nQjTBTpQ1iS7mdnO} \begin{aligned}[b] \begin{pmatrix} a\\b\\c \end{pmatrix}\cdot \begin{pmatrix} x-x_1\\ y-y_1\\ z-z_1 \end{pmatrix}&=0\\ a(x-x_1)+b(y-y_1)+c(z-z_1)&=0\\ ax+by+cz&=ax_1+by_1+cz_1 \end{aligned} \end{equation}$$

Here, the right-hand side is some constant because we assume that we are given:

the coordinates of $A$, that is, $(x_1,y_1,z_1)$.
the norm vector $\boldsymbol{n}$, that is, $a$, $b$ and $c$.

We now formally define two important concepts:

the vector equation of a plane.
the cartesian equation of a plane.

Definition.

Vector equation of a plane

Suppose we are given that a plane:

passes through the point $A$ represented by a position vector $\boldsymbol{a}$.
has a normal vector $\boldsymbol{n}$.

Let $R$ be some other point on the plane represented by a position vector $\boldsymbol{r}$. The so-called vector equation of a plane is given by:

$$\begin{equation}\label{eq:YeBAPpdg7dJHKpjIrsU} \boldsymbol{n}\cdot{\boldsymbol{v}}=0 \end{equation}$$

Where $\boldsymbol{v}$ is a vector pointing from $A$ to $R$ defined as:

$$\boldsymbol{v}= \begin{pmatrix} x-x_1\\ y-y_1\\ z-z_1 \end{pmatrix} = \boldsymbol{r}- \boldsymbol{a}$$

Using this, the vector equation of a plane \eqref{eq:YeBAPpdg7dJHKpjIrsU} is also sometimes expressed as:

$$\begin{equation}\label{eq:gczzhneSgMN2ws4fZwP} \boldsymbol{n}\cdot(\boldsymbol{r}-\boldsymbol{a})=0 \end{equation}$$

In some textbooks, the vector equation is equivalently defined as:

$$\boldsymbol{n}\cdot\boldsymbol{r}= \boldsymbol{n}\cdot\boldsymbol{a}$$

Where $\boldsymbol{n}\cdot{\boldsymbol{a}}$ can be thought of as some constant since the components in $\boldsymbol{n}$ and $\boldsymbol{a}$ are assumed to be known.

Example.

Finding the vector equation of a plane

Suppose a plane passes through the point $(2,4,3)$ and has the following normal vector:

$$\boldsymbol{n}= \begin{pmatrix} 5\\3\\1 \end{pmatrix}$$

Find the vector equation of the plane.

Solution. The plane is parallel to the following vector:

$$\boldsymbol{v}= \begin{pmatrix} x-2\\y-4\\z-3 \end{pmatrix}$$

The vector equation of the plane is:

$$\begin{pmatrix} 5\\3\\1 \end{pmatrix}\cdot \begin{pmatrix} x-2\\y-4\\z-3 \end{pmatrix}=0$$

Equivalently, the vector equation of the plane can be expressed as:

$$\begin{align*} \begin{pmatrix} 5\\3\\1 \end{pmatrix}\cdot \begin{pmatrix} x\\y\\z \end{pmatrix}&= \begin{pmatrix}5\\3\\1\end{pmatrix} \cdot \begin{pmatrix}2\\4\\3\end{pmatrix}\\ &=10+12+3\\ &=25 \end{align*}$$

If we let vector $\boldsymbol{r}$ be the position vector of any point $(x,y,z)$ on the plane, then the vector equation of the plane can also be compactly expressed as:

$$\begin{equation}\label{eq:n9Yy7FDRoaVTgoiGuk6} \boldsymbol{n}\cdot\boldsymbol{r}=25 \end{equation}$$

We can also compute the dot product to get:

$$5x+3y+z=25$$

This is called the Cartesian or the general equation of the plane.

Definition.

Cartesian (general) equation of the plane

Suppose we are given that a plane passes through the point $(x_1,y_1,z_1)$ and has the normal vector $\boldsymbol{n}$ below:

$$\boldsymbol{n}= \begin{pmatrix} a\\ b\\ c \end{pmatrix}$$

The Cartesian or the general equation of the plane is:

$$ax+by+cz \;{\color{blue}=}\; ax_1+by_1+cz_1 \;{\color{blue}=}\; d$$

Where the right-hand side is to be treated as some fixed constant, say $d$.

Example.

Finding the Cartesian equation of a plane given a point and a normal vector

Find the equation of the plane that passes through the point $(1,2,3)$ and has the normal vector below:

$$\boldsymbol{n}= \begin{pmatrix} 5\\3\\4 \end{pmatrix}$$

Solution. The cartesian equation of the plane is:

$$\begin{align*} 5x+3y+4z&=(5)(1)+(3)(2)+(4)(3)\\ &=23 \end{align*}$$

Example.

Finding the normal vector of a plane

Consider the following equation of a plane:

$$3x+2y+4z=20$$

Determine a vector that is perpendicular to this plane.

Solution. The normal vector is the coefficients of the cartesian equation of the plane:

$$\boldsymbol{n}= \begin{pmatrix} 3\\2\\4 \end{pmatrix}$$

Note that any scalar multiple of this normal vector will also be perpendicular to the plane.

Example.

Converting the cartesian form to the vector form

Consider the following general form of a plane:

$$2x+y+3z=5$$

Determine the vector equation of the plane.

Solution. The normal vector of the plane and the position vector of any point on the plane is:

$$\boldsymbol{n}= \begin{pmatrix} 2\\1\\3 \end{pmatrix},\;\;\;\;\; \boldsymbol{r}= \begin{pmatrix} x\\y\\z \end{pmatrix}$$

The vector equation of the plane is:

$$\boldsymbol{n} \cdot \boldsymbol{r}=5$$

Practice problems

Suppose a plane passes through the point $(4,2,5)$ and has the following normal vector:

$$\boldsymbol{n}= \begin{pmatrix} 1\\3\\2 \end{pmatrix}$$

Find the vector equation of the plane, that is, find $\boldsymbol{n}\cdot{\boldsymbol{r}}$.

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We apply the vector equation directly

$$\begin{align*} \boldsymbol{n}\cdot\boldsymbol{r}&= \begin{pmatrix}1\\3\\2\end{pmatrix} \cdot \begin{pmatrix}4\\2\\5\end{pmatrix}\\ &=4+6+10\\ &=20 \end{align*}$$

Find the Cartesian equation of the plane that contains the point $(2,1,3)$ and has the following normal vector:

$$\boldsymbol{n}= \begin{pmatrix} 2\\4\\3 \end{pmatrix}$$

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We can start by finding the vector equation of the plane:

$$\begin{align*} \boldsymbol{n}\cdot\boldsymbol{r}&= \begin{pmatrix}2\\4\\3\end{pmatrix} \cdot \begin{pmatrix}2\\1\\3\end{pmatrix}\\ &=4+4+9\\ &=17 \end{align*}$$

We then expand the dot product on the left-hand side to get the Cartesian equation of the plane:

$$2x+4y+3z=17$$

Published by Isshin Inada

Edited by 0 others

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