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:

• $\vec{a}$ is the position vector
• $\vec{n}$ is the normal vector

Parametric equation → Vector equation

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

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

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

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