Is it really impossible to define a plane with less than two coordinates? So we all (should) know about the 3 point plane definition. Any 3 non collinear points define a plane.
Most of us also know the 2 point definition. 2 different points define a normal (the change between them) and have a position. Use the midpoint as the point for the plane to lie on, and use the normal as the direction that is perpendicular from the plane.
Well, I might be crazy, but I think it is possible to can define any plane that doesn't touch the origin with just one point.
Use the difference between the point and the origin (0,0,0) as the normal, and use the point itself as the position of the plane. Like this you can define any plane that doesn't touch the origin.
Is this a valid mathematical way to define a plane, and has it been done before? Are there any flaws with my logic?
 A: It's possible to define a plane using only one point. how? consider any desired bijection between $\mathbb R^2$ and $\mathbb R^6$, now take the planr that would be defined by the corresponding three points in $\mathbb R^6$. In fact you even have some points left over, since for example $(0,0,0,0,0,0)$ doesn't normally define a plane, however since a bijection is surjective there is in fact one point for each plane, so no worries.
On the other hand, this makes no sense, so perhaps we should use three points to define a plane since there is a geometric intuition behind how to do so. 
A: What you've noticed is that the parameter space of planes in $\mathbb{R}^3$ is three-dimensional.
See the wikipedia page discussing the  affine grassmannian, which notes that the parameter space of $k$-dimensional planes in $\mathbb{R}^n$, called the affine Grassmannian $\operatorname{Graff}_k(\mathbb{R}^n)$, is $(n-k)(k+1)$ dimensional. In this case, that is $(3-2)(2+1) = 3$ dimensional.
A: Once you have fixed the origin of the space in which you’re working, you can represent any plane using just three numbers.
You store a unit vector $\mathbf N$ and a number $d$. The plane these data represent is the one with normal $\mathbf N$ passing through the point $d\mathbf N$. If $\mathbf N = (a,b,c)$, then the plane has equation $ax+by+cz=d$.
A unit vector can be represented by two numbers, so the total is three numbers.
This is very similar to your idea, but it works for any plane, including ones that pass through the origin.
