Math ↠ Planes ↠ Cartesian equation

Cartesian equation of a plane

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

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

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

i

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

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

!

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.

i

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