Consider the vectors $r^1 = (0,2,-1)$, $r^2 = (-1,0,2)$, and $r^3 = (2,-1,0)$. Two properties of these vectors that interest us here are: 1) The $i$'th coordinate of $r^i$ is 0, and 2) The intersection of the convex hull of these three vectors and the nonnegative orthant in $R^3$ is the unit simplex.
Which are the triplets of vectors in $R^3$ that satisfy these two properties? Using a pencil and paper one can easily convince oneself that the answer is the following: either (in the description of the rays below, $x \geq 0$) a) $r^1$ lies on the ray $(0,-x,1+x)$, $r^2$ lies on the ray $(1+x,0,-x)$, and $r^3$ lies on the ray $(-x,1+x,0)$; or b) $r^1$ lies on the ray $(0,1+x,-x)$, $r^2$ lies on the ray $(-x,0,1+x)$, and $r^3$ lies on the ray $(1+x,-x,0)$.
Question: what happens in $R^4$, that is, what are the possible configurations for $r^1,\cdots,r^4$ so that the two properties above are satisfied? what happens in $R^n$?
