12:42 AM
"vectors don't always generate isometries" Agreed. "isometries aren't always generated by vector reflection" Disagree: a reflection is defined by orthogonal projection onto a hyperplane; for such a projection to exist, the orthogonal complement of the hyperplane must intersect the hyperplane trivially, and hence must be a line; any non-zero vector on this line represents the reflection in the Clifford algebra.
As for products of potentially null vectors: I do not immediately have an answer, and I want to think about this more, but here is an idea I am currently following. Let $v$ be null; then we can show that $X \mapsto vXv$ is (exterior algebra) grade-preserving. We can also show that $vwv = 2(w\cdot v)v$ when $w$ is a vector, and for any $X, Y$ that $v(X\wedge Y)v = (vXv)\wedge\hat Y + \hat X\wedge(vYv)$ where $X, Y$ are arbitrary multivectors and e.g. $\hat X$ is grade involution.
Now suppose that $V$ is a product of vectors and that $v'$ is another null vector such that $v\cdot v' = 1$. Then $\eta = v + v'$ is not null. Consider then that $vV\eta = vVv + vVv'$. If we had some condition under which either $vVv = 0$ or $vVv' = 0$, then we could get something like $vV\eta = vVv'$ so then $vV = vVv'\eta^{-1}$; this would be a method for reducing the RHS to a product of a smaller number of vectors.
If $X$ is a blade containing $v$, i.e. $v\wedge X = 0$, then $vXv = 0$. A blade is a product of orthogonal vectors, so this shows that any product of the form $vXv'(v+v')$ can be reduced to $vX = v\mathbin\rfloor X$.
