I'm a bit confused about when vectors are parallel. I was trying to prove the following theorem:
Let $f(\vec{x})=A\vec{x}+\vec{b}$ be an affine transformation. Then $f$ preserves the property of parallelism among lines
In the solution they said that
Suppose that $l\,:\, \vec{p}_{1}+t\vec{u}_{1}$ and $m\,:\,\vec{p}_{2}+t\vec{u}_{2}$ when $t\in\mathbb{R}$ are parallel lines. Then $\vec{u}_{2}=k\vec{u}_{1}$ for some $k\in\mathbb{R}$.
Then they showed that:
$$ T\left(\vec{p}_{1}+t\vec{u}_{1}\right)=A\left(\vec{p}_{1}+t\vec{u}_{1}\right)+\vec{b}=\left(A\vec{p}_{1}+\vec{b}\right)+t\left(A\vec{u}_{1}\right)\\ T\left(\vec{p}_{2}+t\vec{u}_{2}\right)=A\left(\vec{p}_{2}+t\vec{u}_{2}\right)+\vec{b}=\left(A\vec{p}_{2}+\vec{b}\right)+\left(tA\vec{u}_{2}\right)=\left(A\vec{p}_{2}+\vec{b}\right)+tk\left(A\vec{u}_{1}\right) $$ So if $\vec{q}_{1}\triangleq A\vec{p}_{1}+\vec{b}$, $\vec{v}\triangleq A\vec{u}_{1}$ and $\vec{q}_{2}\triangleq A\vec{p}_{2}+\vec{b}$, you get: $$ T\left(\vec{p}_{2}+t\vec{u}_{2}\right)=\vec{q}_{2}+t\left(k\vec{v}\right) $$ This means that $\vec{q}_{1}+t\vec{v}_{1}$ and $\vec{q}_{2}+kt\vec{v}$ are parallel.
My question is - why $l$ and $m$ are parallel? Why $\vec{q}_{1}+t\vec{v}_{1}$ and $\vec{q}_{2}+kt\vec{v}$ are parallel? What make those kind of vectors of form $\vec{a}+t\vec{b}$ parallel?
Edit: The solution (with changes to names of variables) is located here (page 260).