Sums of unit vectors contained in a half-space

Question

Consider $n$ unit vectors $\{v_1,...,v_n\}$ with $v_i\in \mathbb{R}^3$. Now define

$\text{H}(w):=\{w'\in\mathbb{R}^3 \ | \ (w',w)>0\}, \ w\in\mathbb{R}^3$

(where $(\cdot,\cdot)$ is the standard scalar product in $\mathbb{R}^3$), i.e. the strict halfspace identified by the vector $w$. Assume $v_i\in\text{H}(w) \ \forall i=1,...,n$. For $n=2$, this implies

$v_1+v_2\in \text{H}(v_1)\cap\text{H}(v_2)$.

Is it true that

$\sum_i^n v_i\in \bigcap_{i}^n\text{H}(v_i)$

for $n>2$? If the answer is yes, how can one prove it?

Answer 1

Let $w=(0,1,0)$,

Let $\theta \in (0, \frac{\pi}2)$ $$v_1= (\cos\theta, \sin \theta,0)$$

$$v_2=(-\cos\theta, \sin \theta,0)$$ $$v_3= (\cos\frac{\theta}2, \sin \frac{\theta}2,0)$$

We can check that $v_i \in H(w)$.

$$\sum_{i=1}^3 v_i = (\cos \frac{\theta}2,2\sin \theta + \sin \frac{\theta}2 ,0)$$

and \begin{align}\lim_{\theta\to 0^+}\langle \sum_{i=1}^3 v_i, v_2\rangle&=\lim_{\theta\to 0^+}\left(-\cos \theta\cos \frac{\theta}2+2\sin^2\theta+\sin \theta \sin \frac{ \theta}2\right) \\ &=-1\end{align}

In particular, if we let $\theta=0.01$,

$$v_1 = (\cos(0.01),\sin(0.01),0),$$$$ v_2 = (-\cos(0.01), \sin(0.01), 0), $$$$v_3=(\cos(0.005),\sin(0.005),0).$$

We can check that $v_1+v_2+v_3 \notin H(v_2).$

octave:3> w=[0,1,0]
w =

   0   1   0

octave:4> v1 = [cos(0.01), sin(0.01), 0]
v1 =

   0.99995   0.01000   0.00000

octave:5> v2 = [-cos(0.01), sin(0.01), 0]
v2 =

  -0.99995   0.01000   0.00000


octave:7> v3 = [cos(0.005), sin(0.005), 0]
v3 =

   0.99999   0.00500   0.00000

octave:8> (v1+v2+v3)*v2'
ans = -0.99969