What do these sets have in common? In $\mathbb{R}^2$ specify a set of vectors, that has many elements as possible, so that no vector is a constant multiple of another vector in the set.
What do such sets have in common?
In $\mathbb{R}^3$ do the same

I asked both of these earlier, but people are acting like it is a HW question (giving me "hints" etc.) but I'm not looking for HW help. I made these questions up.
The answers are the equivalence classes of fractions for $\mathbb{R}^2$ and for $\mathbb{R}^3$ the equivalence classes of 3-part ordered ratios.
But now that you have heard that answer are there any more interesting answers?
 A: An easy way to do this is to start with a circle resp. a sphere and delete one of each pair of antipodal vectors.  Any answer can be obtained from this answer by replacing any point with a scalar multiple of itself.  I don't know what else there is to say.
A: The intuitive way to answer this, for me, is the following.  Let's start with $\mathbb{R}^2$.  We're going to imagine the vectors as points in $\mathbb{R}^2-{(0,0)}$.  Once you choose one vector, you have basically blocked off the line through the origin that contains it.  So you can choose one vector for every line through the origin.  The same goes for $\mathbb{R}^3$.  There isn't much more to say about this, but if you require the set to be connected, so that each vector you choose is "close" to other vectors in your set, you've got an embedding of the "real projective plane", $\mathbb{RP}^{n-1}$, in $\mathbb{R}^n$.  The real projective plane is (equivalently), the space of lines through the origin, the sphere modulo the map sending $x$ to $-x$, or $\mathbb{R}^n$ modulo scalar multiplication.  The last definition is just what you said about equivalence classes of ordered pairs/triples.  Oh, and $\mathbb{RP}^2$ is what you get if you glue a circle to the single edge of a Moebius strip!
