# Unique set of vectors normal to a hyperplane

1. Oct 14, 2012

### Bipolarity

Let's say we have a point-normal representation of a space:
$$n \cdot P_{0}P = 0$$ where n is a vector $<a_{1},a_{2}...a_{n}>$ and $P_{0}$ is a point through which the space passes and P is the set of all points contained in the space.

In $ℝ^{2}$, the point-normal representation defines a line.
In $ℝ^{3}$, the point-normal representation defines a plane.
In $ℝ^{n}$, the point-normal representation defines a hyperplane, or (n-1) dimensional affine space.

It can be shown that this affine space can be represented in component form as:
$$a_{1}x_{1} + a_{2}x_{2} + ... + a_{n}x_{n} = b$$
where b is a constant.

My question is essentially asking about the converse of the representation done above: Can we go from the component-wise representation to the point-normal representation?
In other words, can we show that the only vectors normal to the space determined by $$a_{1}x_{1} + a_{2}x_{2} + ... + a_{n}x_{n} = b$$ are vectors of the form
$t<a_{1},a_{2}...a_{n}>$ where t is a scalar parameter?

BiP

2. Oct 15, 2012

### chiro

Hey Bipolarity.

You could use the identity that <tx,y> = t<x,y> for general inner products and if <x,y> = 0 then <tx,y> = t<x,y> = 0 for all x and y in the inner product space.