with a single point (the 0-vector), we have no lines. thus a trivial vector space won't work.

in one dimension, we have no planes, so a vector space of dimension 1 won't work.

in two dimensions, we have just one plane (the entire space), so every line in the space is a subset of the (only) plane, and so intersects it at every point of the line.

in three dimensions, we now have many planes and lines to choose from. so here, the challenge is to show that if a line is not parallel to a plane, it must intersect the plane. this is a bit tricky, as you have to think, what does it MEAN for a line to be "parallel to a plane"? it might be useful to define a parallel line as one lying in a parallel plane to our given plane. can you think of a way to create a non-parallel plane out of your given line, created from lines parallel to the given line, such that EVERY ONE of the parallel lines intersects the given plane?

in four dimensions (say x,y,z,w) consider the xy-plane and the w-axis. these intersect at the origin (0,0,0,0). clearly the w-axis is perpendicular to the xy-plane, so it is NOT parallel. now shift the xy-plane 1 unit up along the z-axis. prove the plane:

P = {(x,y,1,0): x,y in R} and the line L = {(0,0,0,w): w in R} have no intersection

(if you like you can write P = s(1,0,0,0) + t(0,1,0,0) + (0,0,1,0), and L = u(0,0,0,1) + (0,0,0,0)).