Vector placement for optimal projections

Question

Let $n_1,\cdots,n_m$ be unit vectors arbitrarily chosen in $\mathbb{R}^3$. Then, there exist some optimal unit vector $n$ such that $$|n^Tn_i|\geq \delta(m)>0 \quad \forall i=1,2,\cdots,m$$ I am looking for a lower bound on $\delta(m)$ for all possible configurations of $n_1,\cdots,n_m$.

For example if $m=2$, then it appears that $\delta(2)\geq \sin(\pi/4)$ (the worst case is when they are normal). Obviously, $\delta$ decreases as $m$ increases.

Edit: I am describing an attempt that appears promising. The set of all points that are normal to each vector $n_i$ define a great circle (GC) on the unit sphere. These are the points to be avoided by $n$. Thus there are $m$ such GCs. Assume now that we add these GCs one by one. Each one added can intersect every one of the existing ones in 2 points at most. Those intersections define new regions on the sphere surface. Thus the upper bound on the number of regions is $2+2+2\cdot 2+\cdots+2\cdot (m-1)=m^2-m+2$. Selecting the vector $n$ to lie on the center of the region that has the maximum distance from each edges is the optimal placement for $n$. A fact that can be exploited is that since the overall area is $4\pi$ there exists one region with area greater than or equal to $4\pi/(m^2-m+2)$.

For the case of $m=3$ vectors the situation is more simple. In this case the regions are 8 spherical triangles and we can prove that there exists one with lengths greater than or equal to $\pi/2$ on all edges.

Choosing the center of this triangle as the direction of $n$ yields the desired solution.

Answer 1

We can widen your great circles by $\phi$ on each side to make a band around the sphere. If we make $\phi$ small enough the total area of the bands will be less than $4 \pi$ and there must be at least one direction that is more than $\phi$ from being normal to every unit vector. That will give the lower bound you seek. The band comprises the whole sphere less two spherical caps of angle $\frac \pi 2 - \phi=\theta$ (to match the $\theta$ in the Wikipedia article) so its area is $4\pi \cos \theta$ We want $m 4\pi cos \theta \lt 4 \pi$ or $\cos \theta \lt \frac 1m$ We can therefore guarantee that $\delta(m) \gt \frac 1m$. You can improve this some by considering the overlap of the bands, which reduces the covered area and would allow increasing $\cos \theta$

Answer 2

This issue is a "maximin" (maximize a minimum value) issue.

Such an optimal vector always exists, due to the fact that

• the sphere $S$ is a compact subset of $\mathbb R^3$ and

• function $$f: \ (x,y,z) \ \to \ min_i(|a_ix+b_iy+c_iz|)$$

is continuous,

thus reaches its maximum for some $(x,y,z)=(x_0,y_0,z_0)$.

I fear that an exact result is out of reach because it is connected to the difficult issue of optimal placement of $m$ points on the (unit) sphere for a general $m$.