Distance from point to plane

The Hesse normal form can be used to calculate the distance between a point $P$ and a plane $E$.

For this we insert the position vector $\vec{p}$ of the point into the Hesse normal form. The distance $d$ is then obtained.

$d = |(\vec{p} - \vec{a}) \cdot \vec{n_0}|$

Since you get negative values for points below the plane, but there is no negative distance, you need the amount.

If $d=0$ results, the point lies on the plane (point on the plane).

For the distance of a point to the plane, you first set up the Hesse normal form and then insert the point.

$A(1|2|1)$

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

Set up Hesse normal form

$|\vec{n}|=\sqrt{2^2+(-2)^2+4^2}$ $=\sqrt{24}$

$\vec{n_0}= \frac{\vec{n}}{|\vec{n}|}$ $=\begin{pmatrix} 2/\sqrt{24} \\ -2/\sqrt{24} \\ 4/\sqrt{24} \end{pmatrix}$

$\text{E: } \left(\vec{x} - \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}\right) \cdot \begin{pmatrix} 2/\sqrt{24} \\ -2/\sqrt{24} \\ 4/\sqrt{24} \end{pmatrix}=0$
Insert point

$\vec{p}=\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$

$d=$ $\left|\left(\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} - \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}\right) \cdot \begin{pmatrix} 2/\sqrt{24} \\ -2/\sqrt{24} \\ 4/\sqrt{24} \end{pmatrix} \right|$ $=\left|\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 2/\sqrt{24} \\ -2/\sqrt{24} \\ 4/\sqrt{24} \end{pmatrix} \right|$ $=|-\frac4{\sqrt{24}}|$ $\approx0.82$