Hyperplane

Definition

Vector Equation for a hyperplane

$f(\mathbf{x})={\beta }_0+{\mathbf{\beta }}^{⊺}\mathbf{x}=0$

Facts

Suppose the dimension of $\mathbf{x}$ is $p$

• $p=1$: point in $1$-dimensional space.
• $p=2$: line in $2$-dimensional space.
• $p=3$: plane in $3$-dimensional space.
• $p>3$: hyperplane in $p$-dimensional space.

$\forall {\mathbf{x}}_1,{\mathbf{x}}_2\in H,{\mathbf{\beta }}^{⊺}({\mathbf{x}}_1-{\mathbf{x}}_2)=0\Rightarrow \mathbf{\beta }\perp H$

$\mathbf{\beta }$ is perpendicular to the surface $H$

$\forall \mathbf{x}\in {\mathbb{R}}^{p+1},d_s(\mathbf{x},H)={\mathbf{\beta }}^{\ast ⊺}(\mathbf{x}-{\mathbf{x}}_0)=\frac{1}{||\mathbf{\beta }||}({\mathbf{\beta }}^{⊺}\mathbf{x}+{\beta }_0-\underset{=0}{\underbrace{({\beta }_0+{\mathbf{\beta }}^{⊺}{\mathbf{x}}_0)}})=\frac{1}{||\mathbf{\beta }||}({\mathbf{\beta }}^{⊺}\mathbf{x}+{\beta }_0)=\frac{f(\mathbf{x})}{||f^{\prime }(\mathbf{x})||}$ where $d_s(\cdot )$ is a signed distance, ${\mathbf{\beta }}^{\ast }:=\mathbf{\beta }/||\mathbf{\beta }||$, and ${\mathbf{x}}_0\in H$