I am trying to understand, geometrically, a plot of the quotient of $\mathbb{R}^3$ quotiented by some line $L$, through the origin. The Wikipedia article states;
Similarly, the quotient space for $\mathbb{R}^3$ by a line through the origin can again be represented as the set of all co-parallel lines, or alternatively be represented as the vector space consisting of a plane which only intersects the line at the origin.
$L$ is a one-dimensional subspace, so this quotient space must surely be two-dimensional. However, a set of lines, as in the case of quotienting $\mathbb{R}^2$ by a line $L$, is one-dimensional. I'm having trouble both understanding this and visualizing the result.
As another matter, I don't want to just take for granted that the equivalence classes are sets of "co-parallel" lines, but my attempt to prove this by picking an arbitrary point $(x,y,z)$ in $\mathbb{R}^3$ and considering the set of $(a,b,c)$ such that $(x,y,z) - (a,b,c) \in L$ didn't produce a workable result (in fact, it produced a plane, rather than a line).
Consider a line $L$ passing throw origin, as a vector subspace of $\mathbb{R}^3$,
so the quotient space will be the space of equivalent relations that is geometrically a plan perpendicular to $L$, which is a $2$-dimensional vector subspace in $\mathbb{R}^3$.