Math Planes Cartesian equation

# Cartesian equation of a plane

Planes also have a third form of representation, namely the cartesian equation.

$\text{E: } ax+by+cz=d$

$a, b, c, d \in \mathbb{R}$

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### Hint

The equations of the coordinate planes $E_{xy}: z=0$, $E_{xz}: y=0$, $E_{yz}: x=0$ are special cases of the cartesian equation.

## Vector equation → cartesian equation

The cartesian equation is obtained by multiplying the vector equation using the scalar product.

### Example

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

1. #### Substitute $\vec{x}$ and summarize

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

$\text{E: }$ $\begin{pmatrix} x-2 \\ y-1 \\ z-1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -2 \\ 4 \end{pmatrix}=0$

2. #### Scalar product

Use the scalar product of the two vectors on the left side of the equation:

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

$(x-2)\cdot2 + (y-1)\cdot(-2)$ $+ (z-1)\cdot4$ $=0$

$2x-4-2y+2+4z-4$ $=0$

$2x-2y+4z-6=0 \,\, |+6$

$2x-2y+4z=6$

3. #### Cartesian equation

$\text{E: } 2x-2y+4z=6$

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### Remember

With the cartesian form $\text{E: } ax+bx+cz=d$ a normal vector can always be read off directly:

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

## Cartesian equation → normal equation

Now we only need a random position vector because the normal vector can be read off.

$\text{E: } 2x-2y+4z=6$

### Example

1. #### Normal vector

The required normal vector can be read from the coefficients.

$\vec{n}=\begin{pmatrix} 2 \\ -2 \\ 4 \end{pmatrix}$
2. #### Position vector: find a point

It is particularly easy to choose an intercept point. To do this, all coordinates except one are set to 0.

You can immediately see that $A(3|0|0)$ is in the plane:

$2\cdot3-2\cdot0+4\cdot0=6$
$6=6$

$\vec{a}=\begin{pmatrix} 3 \\ 0 \\ 0 \end{pmatrix}$
3. #### Insert

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

$\text{E: } \left(\vec{x} - \begin{pmatrix} 3 \\ 0 \\ 0 \end{pmatrix}\right) \cdot \begin{pmatrix} 2 \\ -2 \\ 4 \end{pmatrix}$ $=0$

## Cartesian equation → parametric equation

First you search for any three points in the plane and then you set up the parametric equation.

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### Method

1. Find three points
2. Set up parametric equation

## Parametric equation → cartesian equation

Here you should take the detour via the normal equation:

parametric equation → vector equation → cartesian equation