An hyperplane in 3D

L. El Ghaoui

An hyperplane in 3D

Consider an affine set of dimension $2$ in $\mathbb{R}^3$ , which we describe as the set of points $x \in \mathbb{R}^3$ such that there exists two parameters $\lambda_1, \lambda_2$ such that

$x=\left(\begin{array}{c} 3 \lambda_1-4 \lambda_2+4 \\ \lambda_1 \\ \lambda_2 \end{array}\right)=\left(\begin{array}{l} 4 \\ 0 \\ 0 \end{array}\right)+\lambda_1\left(\begin{array}{l} 3 \\ 1 \\ 0 \end{array}\right)+\lambda_2\left(\begin{array}{c} -4 \\ 0 \\ 1 \end{array}\right) .$

The set ${\bf H}$ can be represented as a translation of a linear subspace: ${\bf H} = x_0 + {\bf L}$ , with

$x_0 : = \left(\begin{array}{l} 4 \\ 0 \\ 0 \end{array}\right),$

and ${\bf L}$ the span of the two independent vectors

$u: = \left(\begin{array}{l} 3 \\ 1 \\ 0 \end{array}\right), v: = \left(\begin{array}{c} -4 \\ 0 \\ 1 \end{array}\right).$

Thus, the set ${\bf H}$ is of dimension $2$ in $\mathbb{R}^3$ , hence it is an hyperplane. In $\mathbb{R}^3$ , hyperplanes are ordinary planes.

We can find a representation of the hyperplane in the standard form

${\bf H} = \{x: a^T(x-x_0)=0\}.$

We simply find $a$ that is orthogonal to both $u$ and $v$ . That is, we solve the equations

$0 = a^Tu = 3a_1+a_2, \hspace{0.05in} 0 = a^Tv = -4a_1+a_3$

The above leads to $a = (a_1, -3a_1, 4a_1)$ . Choosing for example $a_1 = 1$ leads to $a = (1, -3, 4)$ .

The hyperplane ${\bf H}$ can be expressed as $x_0 + {\bf span}(u,v)$ , where $x_0$ is a particular element, and $u,v$ are two independent vectors. The set ${\bf H}$ is represented in light blue; it is a translation of the corresponding span ${\bf L} = {\bf span}(u,v)$ . Any point $x \in {\bf H}$ is such that $x- x_0$ belongs to ${\bf L}$ . Thus we can represent the hyperplane as the set of points such that $x- x_0$ is orthogonal to $a$ , where $a$ is any vector orthogonal to both $u,v$ .

An hyperplane in 3D

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