Number of diagonals perpendicular to given diagonal of the unit n-cube

Question

I think I know how to do this problem but just want to check that my reasoning is correct. "Let $U$ denote the unit cube in $\mathbb{R}^n$, and let $D$ be a given diagonal of $U$. How many other diagonals of $U$ are perpendicular to $D$?" I am unsure how to proceed with the 1 and 0 dimensional case, but in 2 dimensions that answer is 1. I conjecture that for an n-dimensional cube, if the diagonal traverses $m_o$, an odd number of dimensions, the number is zero, and if it traverses $m_e$, an even number of dimensions the number is the binomial coefficient $\binom{m_e}{m_e/2}$. We write the m dimensional diagonal as $(d_1,d_2,...d_m)$, with $d_i=\pm1$. By rotation, all diagonals are similar to $(1,1,..1)$ m times, so we need only find the solution for that particular diagonal. We look for another m dimensional vector such that its dot product with $(1,1,...1)$ is zero. For m odd, this is clearly impossible. For m even, half of the coordinates must be negative 1's and the other positive ones, so you have $\binom{m_e}{m_e/2}$ options. Does this reasoning make sense? Also, what should I do for the 0 and 1 dimensional cases? Sorry, wasn't sure what to tag this as either. Thanks.