Math Planes Vector equation of a plane

# Vector equation of a plane

Alternatively, a plane can be described using a point and a vector perpendicular to the plane, called normal vector.

The vector equation of a plane has the following form:

$\text{E: } (\vec{x} - \vec{a}) \cdot \vec{n}=0$
• $\vec{a}$ is the position vector
• $\vec{n}$ is the normal vector

## Parametric equation → Vector equation

i

### Method

1. Adopt the position vector from the parametric equation
2. Calculate the normal vector
3. Insert position vector and normal vector
i

### Hint

The normal vector can be calculated with both the scalar product and the cross product. The calculation with the cross product is somewhat easier and faster, whereas the formula of the scalar product is much easier to remember.

### Example

$\text{E: } \vec{x} = \color{green}{\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}} + r \cdot \color{blue}{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}}$ $+ s \cdot \color{blue}{\begin{pmatrix} 1 \\ 5 \\ 2 \end{pmatrix}}$

1. #### Position vector

$\vec{a}=\color{green}{\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}}$
2. #### Normal vector

##### Option 1

Since both direction vectors are perpendicular to the normal vector $\vec{n}=\begin{pmatrix} x \\ y \\ z \end{pmatrix}$, the scalar product must result in zero.

1. $\begin{pmatrix} x \\ y \\ z \end{pmatrix}\cdot\color{blue}{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}} = 0$
2. $\begin{pmatrix} x \\ y \\ z \end{pmatrix}\cdot\color{blue}{\begin{pmatrix} 1 \\ 5 \\ 2 \end{pmatrix}} = 0$

The scalar product can now be calculated.

1. $1x+1y=0$
2. $1x+5y+2z= 0$

II.-I.

$4y+2z=0$

Choose $z$ freely, e.g. $z=4$

$4y+8=0\quad|-8$
$4y=-8\quad|:4$
$y=-2$

Calculate $x$ using I. (insert $y$)

$x+y=0$
$x-2=0\quad|+2$
$x=2$

$\vec{n}=\begin{pmatrix} 2 \\ -2 \\ 4 \end{pmatrix}$

##### Option 2

Now we only calculate the cross product of the vectors.

$\vec{n}$ $=\color{blue}{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}} \times \color{blue}{\begin{pmatrix} 1 \\ 5 \\ 2 \end{pmatrix}}$ $=\begin{pmatrix} 1\cdot2 - 0\cdot5 \\ 0\cdot1 - 1\cdot2 \\ 1\cdot5 - 1\cdot1 \end{pmatrix}$ $=\begin{pmatrix} 2 \\ -2 \\ 4 \end{pmatrix}$
3. #### Insert

$\text{E: } (\vec{x} - \vec{a}) \cdot \vec{n}=0$

$\text{E: } \left(\vec{x} - \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}\right) \cdot \begin{pmatrix} 2 \\ -2 \\ 4 \end{pmatrix}=0$

## Vector equation → parametric equation

i

### Method

1. Adopt the position vector of the vector equation
2. Use scalar product to determine the direction vector
3. Insert position vector and direction vector
i

### Hint

It can sometimes be easier to first transform it into the coordinate form and then into the parametric form.

### Example

$\text{E: } \left(\vec{x} - \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}\right) \cdot \begin{pmatrix} 2 \\ -2 \\ 4 \end{pmatrix}=0$

1. #### Position vector

$\vec{a}=\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}$
2. #### Direction vector

Using the normal vector we can determine both direction vectors needed for the parametric equation.

1. Direction vector

A vector must be found with which the scalar product results in zero.

$\begin{pmatrix} 2 \\ -2 \\ 4 \end{pmatrix}\cdot\color{blue}{\begin{pmatrix} \, \\ \, \\ \, \end{pmatrix}} = 0$

It is particularly easy to replace the first coordinate with 0 and then swap the other two coodinates and to change a sign.

$\begin{pmatrix} 2 \\ \color{red}{-2} \\ \color{red}{4} \end{pmatrix}\cdot\begin{pmatrix} 0 \\ \color{blue}{-4} \\ \color{blue}{-2} \end{pmatrix} = 0$

$\vec{u}=\begin{pmatrix} 0 \\ -4 \\ -2 \end{pmatrix}$

2 Direction vector

$\begin{pmatrix} 2 \\ -2 \\ 4 \end{pmatrix}\cdot\color{blue}{\begin{pmatrix} \, \\ \, \\ \, \end{pmatrix}} = 0$

Here the last coordinate needs to be replaced with 0 and the other two coordinates need to be switched and a sign is changed.

$\begin{pmatrix} \color{red}{2} \\ \color{red}{-2} \\ 4 \end{pmatrix}\cdot\begin{pmatrix} \color{blue}{-2} \\ \color{blue}{-2} \\ 0 \end{pmatrix} = 0$

$\vec{v}=\begin{pmatrix} -2 \\ -2 \\ 0 \end{pmatrix}$

3. #### Insert

$\text{E: } \vec{x} = \vec{a} + r \cdot \vec{u} + s \cdot \vec{v}$

$\text{E: } \vec{x} = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} + r \cdot \begin{pmatrix} 0 \\ -4 \\ -2 \end{pmatrix}$ $+ s \cdot \begin{pmatrix} -2 \\ -2 \\ 0 \end{pmatrix}$