3-dimensional vectors satisfying certain equalities Question: Are there 5 distinct vectors $u,v,w,x,y \in \mathbb{R}^3$, all on the unit sphere (i.e. $||u||=||v||=||w||=||x||=||y||=1$), such that:
$||u+v+x||=||u+v+y||=||u+w+x||=||u+w+y||=1$
?
Also, what is the answer if the vectors are taken from $\mathbb{R}^d$ for $d>3$? 
Let me explain my motivation for this question. My actual question is the following: What is the largest subset $A \subseteq \{1,...,m\}$ such that there is no non-trivial solution to the system:
$x_1 + x_5 = x_6 + x_8$,
$x_2 + x_5 = x_6 + x_9$,
$x_3 + x_5 = x_7 + x_8$,
$x_4 + x_5 = x_7 + x_9$,
with $x_1,...,x_9 \in A$. The reason I'm interested in such sets is that they may be helpful for a construction in graph theory. 
One way to tackle such questions is to think of vectors instead of integer numbers (integers may be mapped to vectors by considering their representation in some base. This is done in the famous Behrend construction of a dense set $A \subseteq \{1,...,m\}$ without a non-trivial solution for $x_1+x_2=2x_3$). Any direction for my question would be appreciated.   
 A: The answer depends on what is meant with "distinct".
If you allow the vectors $u,v,w,x,y$ to be colinear (for instance v = -u),
then there are many solutions. For example:
$u = (1,0,0)$, $v = (-1,0,0)$, $w=(0,1,0)$, $x=(0,-1,0)$ and $y = (-\frac{1}{2},-\frac{1}{2}, \frac{1}{\sqrt{2}})$. 
On the other hand, fix distinct $u,v,w\in S^2$ and let us assume that $u \neq - v$ and $u \neq - w$.
Claim: If $x$ and $y$ in $S^2$ are distinct and satisfy
 $$ (*) \quad \Vert u+v+x\Vert =  \Vert u+v+y\Vert =  \Vert u+w+x\Vert =  \Vert u+w+y\Vert = 1 $$ then either $x=-u$ or $y = -u$. 
One can see this by considering a reformulation: Write $v' = -v$ and $w' = -w$ and $x' = x+u$ and $y' = y+ u$.
If $x,y$ satisfy (*), then $x', y' \in \mathbb{R}^3$ are points of distance one from the three distinct points $u,v',w' \in S^2$.
Consider the sphere $S$ of radius one around $u$. The spheres of radius one around $v'$ and $w'$ intersect $S$ in two distinct (since $v \neq w$, $u \neq v'$ and $u \neq w'$!) circles. However, two distinct circles intersect in at most 2 points and so there is a unique choice (up to order) for $x'$ and $y'$. But since $u,v',w'$ lie on the unit circle, we know that 0 is of distance one from all three points. We conclude that either $x'=0$ or $y'=0$ and hence $x = -u$ or $y = -u$.
As the geometric argument above indicates, in dimension $d\geq 4$ there are again many solutions (if you restrict your question to 5 vectors). For example:
$u=(1,0,0,0)$, $v=(0,1,0,0)$, $w=(0,0,1,0)$, $x = \frac{1}{2}(-1,-1,-1,1)$ and $y = \frac{1}{2}(-1,-1,-1,-1)$.
