Describing a $m$-dimensional plane of $\Bbb{R^n}$ with a $m$-dimensional subspace of $\Bbb{R^n}$

Question

I am given a definition of a plane as per the following : A $m$-dimensional plane in $\Bbb{R^n}$ is the set $$T=x_0+E:=\{x_0+y\in\Bbb{R^n}:y\in E\}$$

where $E$ is a $m$-dimensional subspace of $\Bbb{R^n}$, and $x_0$ is a point of the plane. I don't see how this definition of a plane fits in with the standard $ax+bx+cz = d$ equation of a plane. I'm trying to see how the plane $$x+y+z = 1$$

can be written in terms of the above definition. This equation describes a $2$-dimensional plane in $\Bbb{R^3}$, so $m=2$. Then the set $E$ must also be $2$-dimensional, hence it is also a plane. What exactly is this $E$, for the plane given by equation above?

Answer 1

$z=1-y-x$ so that $(x, y, z)=(x, y, 1-y-x) = (0, 0, 1)+x(1, 0, -1)+y(0, 1, -1)$. The vectors $(1, 0, -1)$ and $(0, 1, -1)$ span a 2 dimensional subspace E of $\Bbb{R^3}$. E defines a plane through the origin and your plane is a parallel one through $(0,0,1)$

Answer 2

Consider the case of $E$ hyperplane, $i.e.$ codimension 1. The hyperplane is definded by $$E=\{y\in\mathbb{R}^n,\,a_{1}y_{1}+\dots+a_{n}y_n=0\}=\{y\in\mathbb{R}^n,\,\langle y,a\rangle=0\}=a^{\perp}.$$

Let then $T=x_{0}+E$, hence for $x\in T$, $x=x_{0}+y$, we have $$\langle x,a\rangle=\langle x_0,a\rangle+\langle y,a\rangle=\langle x_0,a\rangle+0=\langle x_0,a\rangle,$$ which is the other definition if you define $d:=\langle x_0,a\rangle$.

For codimension greater than $k>1$, remark that you have $k$ equation, that is the intersection of $k$ hyperplanes.