[proofplan]
We prove that the vectors parallel to $B$ are exactly the elements of $W$. First, using the difference characterization of the direction space, every difference of two points of $B=p+W$ lies in $W$, and every vector in $W$ occurs as such a difference. Then we use the equality $\operatorname{dir}(B)=W$ to rewrite $B$ with any point $r \in B$ as base point.
[/proofplan]
[step:Identify all differences of points of $B$]
For an affine subspace $C \subset A$, its direction space is
\begin{align*}
\operatorname{dir}(C):=\{a-a' : a,a' \in C\} \subset V.
\end{align*}
We show that this set is $W$ when $C=B$.
Let $u \in \operatorname{dir}(B)$. Then there exist points $b,b' \in B$ such that $u=b-b'$. Since $B=p+W$, there exist vectors $w,w' \in W$ such that
\begin{align*}
b=p+w
\end{align*}
and
\begin{align*}
b'=p+w'.
\end{align*}
By the affine-space action law,
\begin{align*}
u=(p+w)-(p+w')=w-w'.
\end{align*}
Since $W$ is a linear subspace of $V$, it is closed under subtraction, so $w-w' \in W$. Hence $\operatorname{dir}(B) \subset W$.
Conversely, let $w \in W$. Since $0 \in W$, both $p+w$ and $p=p+0$ belong to $B$. Therefore
\begin{align*}
(p+w)-p=w,
\end{align*}
so $w \in \operatorname{dir}(B)$. Hence $W \subset \operatorname{dir}(B)$.
Combining the two inclusions gives
\begin{align*}
\operatorname{dir}(B)=W.
\end{align*}
[guided]
The direction space records exactly which vectors can be obtained by subtracting one point of the affine subspace from another. For this affine subspace,
\begin{align*}
B=p+W=\{p+w:w\in W\},
\end{align*}
so every point of $B$ is obtained by starting at the base point $p$ and translating by a vector in $W$.
First take an arbitrary vector $u \in \operatorname{dir}(B)$. By the definition of direction space, there are points $b,b' \in B$ such that
\begin{align*}
u=b-b'.
\end{align*}
Because $b$ and $b'$ lie in $p+W$, there are vectors $w,w' \in W$ with
\begin{align*}
b=p+w
\end{align*}
and
\begin{align*}
b'=p+w'.
\end{align*}
The affine action is compatible with vector subtraction in the model [vector space](/page/Vector%20Space), so subtracting these two points cancels the common base point $p$ and leaves
\begin{align*}
b-b'=(p+w)-(p+w')=w-w'.
\end{align*}
Since $W$ is a linear subspace, it is closed under subtraction. Thus $w-w' \in W$, and therefore $u \in W$. This proves
\begin{align*}
\operatorname{dir}(B) \subset W.
\end{align*}
For the reverse inclusion, take an arbitrary vector $w \in W$. The point $p+w$ lies in $B$ by definition. Also $p$ lies in $B$ because $0 \in W$ and $p=p+0$. Their difference is
\begin{align*}
(p+w)-p=w.
\end{align*}
Thus $w$ occurs as a difference of two points of $B$, so $w \in \operatorname{dir}(B)$. This proves
\begin{align*}
W \subset \operatorname{dir}(B).
\end{align*}
The two inclusions give
\begin{align*}
\operatorname{dir}(B)=W.
\end{align*}
[/guided]
[/step]
[step:Rebase the affine subspace at an arbitrary point]
Since $W$ is a linear subspace and $\operatorname{dir}(B)=W$, the set $\operatorname{dir}(B)$ is a linear subspace of $V$.
Let $r \in B$. Since $B=p+W$, there exists $w_0 \in W$ such that
\begin{align*}
r=p+w_0.
\end{align*}
We prove that $B=r+W$. If $x \in r+W$, then there exists $w \in W$ such that
\begin{align*}
x=r+w=(p+w_0)+w=p+(w_0+w).
\end{align*}
Since $W$ is closed under addition, $w_0+w \in W$, so $x \in p+W=B$. Hence $r+W \subset B$.
Conversely, if $x \in B$, then there exists $w_1 \in W$ such that
\begin{align*}
x=p+w_1.
\end{align*}
Using $r=p+w_0$, we have
\begin{align*}
x=p+w_1=(p+w_0)+(w_1-w_0)=r+(w_1-w_0).
\end{align*}
Since $W$ is closed under subtraction, $w_1-w_0 \in W$, so $x \in r+W$. Hence $B \subset r+W$.
Therefore
\begin{align*}
B=r+W.
\end{align*}
Using $\operatorname{dir}(B)=W$, this becomes
\begin{align*}
B=r+\operatorname{dir}(B).
\end{align*}
This proves the stated consequence for every $r \in B$.
[/step]
