Planes

Normal form

Let a plane be in its parameter form and rearrange it:
\begin{align} \vec{e}&=\vec{s}+k\vec{a}+t\vec{b}\ \ \ (k,t\in\mathbb{R})\\ \vec{e}-\vec{s}&=k\vec{a}+t\vec{b}\\ (\vec{e}-\vec{s})\cdot(\vec{a}\times\vec{b})&=(k\vec{a}+t\vec{b})\cdot(\vec{a}\times\vec{b})\\ [\vec{e}-\vec{s},\vec{a},\vec{b}]&=0 \end{align} Written es determinant: \begin{equation} \begin{vmatrix}e_1-s_1&e_2-s_2&e_3-s_3\\a_1&a_2&a_3\\b_a&b_2&b_3\end{vmatrix} = 0 \end{equation}

Hesse normal form

The normal vector $\vec{n}=\vec{a}{\small\times}\vec{b}$ is perpendicular to $\vec{a}$ and $\vec{b}$. The Hess normal form can now be derived: \begin{align} (\vec{e}-\vec{s})\vec{n}&=0\\ \vec{e}\vec{n}&=\vec{s}\vec{n}\\ \vec{e}\vec{n_0}&=d \end{align} $$\text{where}\ \ \ \vec{n_0}=\frac{\vec{n}}{|\vec{n}|}\ \ \ \text{and}\ \ \ \vec{d}=\vec{s}\vec{n_0}$$ Note: $|\vec{d}|$ is the distance between the plane and the origin.