Orthogonality in $N$-dimensional spherical coordinates I have a question about orthogonality expressed in spherical coordinates in an $N$-dimensional space.
Given a unitary vector whose coordinates are : $$ \vec{V}  : (\phi_1, \phi_2, \cdots, \phi_{N-1})$$
What condition does a vector (let's say unitary to simplify) need to satisfy to be orthogonal to $\vec{V}$? In other words, what's the expression (in spherical coordinates) of the hyperplane orthogonal to $\vec{V}$ and containing the origin? Thanks in advance.
 A: Let's say the vector we are looking for is $x$. Suppose we write $x$ and your vector $v$ in cartesian coordinates, then for orthogonality  we have
$$
0 = \sum_{i=1}^N v_i x_i
$$
where $x_i$ and $v_i$ are the cartesian components.
The standard definition of $N$-dimensonal spherical coordinates can be found at https://en.wikipedia.org/wiki/N-sphere
Let's denote the angles of $v$ by $\phi$ (as you already did) and the angles of $x$ by $\alpha$. 
Use those definitions to replace the cartesian coordinates of $x$ and $v$ to get
$$
0 = \sum_{i=1}^N v_i x_i = \cos(\phi_1)\cos(\alpha_1) +  \sin(\phi_1)\cos(\phi_2)\sin(\alpha_1)\cos(\alpha_2) + \cdots + \prod_{i=1}^{N-1}\sin(\phi_i)\sin(\alpha_i)
$$  
This is not a very nice formula. So in general, what one would do is to first rotate the coordinate system in such a way that the vector $v$ is along one axis, i.e. let all $\phi_i = 0$. Then the formula simplifies to 
$
0 = \cos(\phi_1)\cos(\alpha_1) = \cos(\alpha_1)$
which just says $\alpha_1 = \pi/2$ and all other $\alpha_i$ can be freely chosen.
