# Number of diagonals in $n$-hypercube perpendicular to the diagonal.

If $v_0, v_1,\ldots, v_n$ are such vectors, that:

1. $(v_j-v_0)$ is perpendicular to $(v_i-v_0)$ for any different $i,j=1,\ldots,n$
2. $||v_i-v_0||=a,$ for any $i=1,\ldots,n$

Then $n$-hypercube in $\Bbb R^n$ is defined this way:
$K:= \{v \in \Bbb R^n \mid (v-v_0) = \alpha_1 (v_1-v_0) + \alpha_2 (v_2-v_0) + \ldots. +\alpha_n (v_n-v_0) ;\ \alpha_i \in [0,1] \}$
and the set of dots in $K$:
$W:= \{v \in \Bbb R^n \mid (v-v_0) = \alpha_1 (v_1-v_0) + \alpha_2 (v_2-v_0) + \ldots +\alpha_n (v_n-v_0) ;\ \alpha_i \in \{0,1\} \}$

How can I find the number of diagonals in $n$-hypercube, which are perpendicular to one diaginal?

Without loss of generality this fixed 'one diagonal' might as well be the vector joining $(0, 0, \cdots, 0)$ and $(1, 1, \cdots, 1)$. The direction of this vector is simply $(1, 1, \cdots, 1)$.
Any other diagonal will be a vector joining $(t_1, t_2, \cdots, t_n)$ and $(1-t_1, 1-t_2, \cdots, 1-t_n)$, where $t_1, t_2, \cdots, t_n \in \{0,1\}$. Why? Because to get from point point of the hypercube to the opposite point you need to invert all of the coordinates. The direction of this vector is $(1-2t_1, 1-2t_2, \cdots, 1-2t_n)$.
These two vectors are perpendicular if and only if their dot product is zero, i.e. if and only if $$(1-2t_1) + (1-2t_2) + \cdots + (1-2t_n) = n-2(t_1 + t_2 + \cdots + t_n) = 0$$
So you need to find the number of possible values in $0,1$ that $t_1, t_2, \cdots, t_n$ can take so that they satisfy the above equation. (And divide your answer by $2$, because this takes into account the direction of the diagonal and we don't care about direction.)
For instance when $n=2$ you just have a square; given one diagonal there is only one diagonal perpendicular to it. And sure enough there are two solutions to the equation $2-2(t_1+t_2)=0$, namely $(t_1,t_2)=(1,0)$ and $(0,1)$; and there are $\frac{2}{2}$ diagonals.