The problem: Given unit vectors $\mathbf{u}, \mathbf{v} \in \mathbb{R}^{n}$ with $\mathbf{u} \cdot \mathbf{v} = 0$, consider $p_n = \|\mathbf{u} \odot \mathbf{v}\|_2$ or the norm of Hadamard product of the two. What is the maximum of $p_n$?
In $\mathbb{R}^2$, the problem reduces to $[\cos \theta, \sin\theta] \odot [\sin \theta, -\cos\theta] = \frac{1}{2} [\sin 2\theta, -\sin 2\theta]$. And the maximum is therefore $\frac{\sqrt{2}}{2}$. In $\mathbb{R}^3$ using spherical coordinates, it seems the maximum is again $\frac{\sqrt{2}}{2}$ (and since $\mathbb{R}^2$ can be embedded in $\mathbb{R}^3$, $p_3$ can't be less than $\frac{\sqrt{2}}{2}$).
But I was wondering if the conclusion is extendable to $\mathbb{R}^{n}$ and if there's a more intuitive way of seeing it rather than blindly following polar/spherical parametrization.
Working on a rotationally invariant unit sphere, we are free to choose a nice simple u-vector $$u = \frac{1}{{\sqrt n}}$$ and to construct the v-vector via Gram-Schmidt from an arbitrary $x\ne \lambda u$ $$\eqalign{ y &= x - u\,(u^Tx) \\ v &= \frac{y}{\|y\|} \\ }$$ With these choices the length function is radically simplified. $$\eqalign{ p_n^2 &= \|u\odot v\|^2 \\ &= \frac{1}{n}\|v\|^2 \\ &= \frac{1}{n} \\ }$$ and we conclude that $$p_n = \frac{\sqrt n}{n}$$