User:Dionyziz/Hyperplanes in N dimensions

This "proof" is not really a proof. It's just an idea I had, and I couldn't actually work out that well.

In some given N-dimensional space, with $$N \in \mathbb{N}, N >= 1$$, we traditionally conveniently define an affine (N - 1)-Hyperplane P using N distinct points A which belong to our target hyperplane.

i.e. Let $$N \in \mathbb{N}, N >= 1$$ and $$\vec{A_i} \in \mathbb{R}^{N} \forall i \in \delta = [1, N] \cap \mathbb{N}$$ and $$\vec{A_i} \not= \vec{A_j} \forall i, j \in \delta, i \not= j$$

Then the hyperplane can be easily defined. We do so by performing N - 1 exterior products on field $$\mathbb{R}^k \forall k \in \delta$$ (you can remove item one from delta for clarity and only apply exterior products on $$\delta - \{1\}$$, since a 1-dimensional exterior product doesn't affect the operands) between the first and all subsequent k-vectors. For example, for the second dimension, $$\vec{B_{i - 1}} = \vec{A_i} - \vec{A_1} \forall i \in \delta - { 1 }$$, and move on into reducing our result to only one N-vector, which now defines our hyperplane.

However, this method, even though it is simple and easy to perform, has the disadvantage that N points are used, while they are redundant. In fact, only one point that belongs to the given hyperplane can be used to define it. We can do so by picking that point carefully so that when it is projected to N-1 dimensions, its distance from the origin defines the N-1 angles needed to define our hyperplane (of course with the exception of hyperplanes that are parallel to our 0-frame, as, in that case, we wouldn't be able to define the distance as it would be static for all points). As hyperplanes are affine, the distance can range through R and it is up to our convenience to pick the desired point.

In an attempt to define a simple system conforming to this methodology, we could, for instance, take a distance equal to the angle, in radians, of the hyperplane in each dimension (therefore using distances only in $$[0, \pi]$$).