Intersection of parallel affine subspaces Given two affin parallel subspaces, let's say $ A$ and $B$ , $dim(A) = dim(B) = m $, I'm trying to prove that both of them have the same set of points at infinity and that set is a non empty projective subspace of $ P_{\mathbb{K}}^n$ with dimension $ \phantom{1} m - 1 $
Intuition tells me that it is probably something simple to prove by induction so in my attempt I have decided to start with the simple case of two parallel lines. Since they are parallel, they have the same direction vector and therefore they intersect at the same point at infinity, which is a non-empty projective subspace of dimension $0$.
However, I am stuck at this point since two parallel planes do not necessarily have to be generated by two parallel vectors, perhaps I am giving the problem the wrong approach. Any hints or suggestions?
 A: One can be totally explicit:
Hint We can choose affine coordinates $(x^1, \ldots x^n)$ so that one of the planes has the form
$$x^{m + 1} = \cdots = x^{n - 1} = 0, \quad x^n = a$$ for some $a$. Since the second plane is parallel to the first, it must have the form
$$x^{m + 1} = \cdots = x^{n - 1} = 0, \quad x^n = b$$ for some $b$. If we projectivize these varieties with coordinates $[x^0 : \cdots : x^n]$, what are the defining (homogeneous) equations for the projective varieties? What is their intersection?
A: $\newcommand{\mysetminus}{\raise.3ex\smallsetminus}$
I want to make a few considerations that does not use coordinates.
Let ${\cal A}(E)$ be an affine space over the $\mathbb K$-vector space $E$. The injection
$$ J : {\cal A}(E)\; \longrightarrow\; {\cal P}(E\times\mathbb K)$$$$P\;\longmapsto\; \langle(P-O,1)\rangle, $$
where $O$ is any fixed point of ${\cal A}(E)$ and ${\cal P}(E\times \mathbb K)$ is the projective space over the vector space $E\times\mathbb K$, immerges the given affine space into its projective completion.
The points of
$$ {\mathcal A}(E)^\infty:={\mathcal P}(E\times\mathbb K)\mysetminus J({\mathcal A}(E)) $$
are the points at infinity of ${\mathcal A}(E).$ This construction depending on the choice of the point $O$, is not canonical, but the resulting set doesn't depends on it, as it is easy to prove that
$$ {\cal A}(E)^\infty  =  {\cal P}(E\times\{0\}). $$
That being said, let us consider an affine subspace $\;S=P+\vec F\;$ of $\;{\mathcal A}(E)$, where $\vec F$ is a vector subspace of $E$. The projective completion $\tilde S$ of $S\;$ in $\;{\cal P}(E\times \mathbb K)$ is by definition the projective subspace of this latter generated  by the set $J(S)$, and one can to prove that:
$$ \tilde S = (P+\vec F)^\tilde{}= J(P+\vec F)\cup {\mathcal P}(\vec F\times\{0\}), $$
so the set of the points at infinity of $S$ are given by
$$ (P+\vec F)^\infty = {\mathcal P}(\vec F\times\{0\}). $$
We may now conclude: $A=Q+\vec F_A$ and $B=R+\vec F_B$ being parallel and of the same dimension, we deduce that $\vec F_A=\vec F_B=:\vec F$. The preceeding considerations then imply
$$ A^\infty = B^\infty = {\mathcal P}(\vec F\times\{0\}) $$
and obviously $\dim({\mathcal P}(\vec F\times\{0\}))=\dim(\vec F)-1=m-1.$
