Can we always find such a vector? Suppose we have a finite $s$-dimensional grid $J\subset\mathbb{Z}^{s}$ containing $0_{s}$. 
Let $n_{i}\in\mathbb{Z}^{s}$, $i=1,\ldots,N$ be the vectors with ending points the points of the grid. 
Can we always find a vector $u\in\mathbb{R}^{s}$ such that all dot products $n_{i}\cdot u$, $i=1,\ldots, N$ are distinct?
My intuition leads me to believe the answer is yes, since I tried to find counterexamples in $1$ and $2$ dimensions but failed. I haven't come up with a solid proof though. 
A re-formulation of the problem would be proving that there exists a vector $u\in\mathbb{R}^{s}$ such that 
$$(n_{i}-n_{j})\cdot u\neq 0$$ for all $i\neq j$. Given that the grid contains $N$ vectors, the number of vectors $n_{i}-n_{j}$ for $i\neq j$ is $(N-1)!$. 
Any pointing to the right direction would be welcome.  
 A: Let a finite (or even countable) number of pairwise different vectors ${\bf n}_i$ be given. For $i\ne j$ one has ${\bf n}_i\cdot{\bf u}={\bf n}_j\cdot{\bf u}$ iff ${\bf u}$ is lying in the hyperplane $\>H_{ij}\!:\>({\bf n}_i-{\bf n}_j)\cdot{\bf u}=0$. There are at most countable many forbidden hyperplanes $H_{ij}$, and these hyperplanes do not fill all of space. It follows that there are plenty of vectors ${\bf u}$ satisfying your desires.
A: After rethinking the problem I'm returning with a new proposition.
One of the solutions could be if you select components of this vector from the set of values being square roots of primes $\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{7}, \dots$ etc $\dots$ let name this set $\mathbb{S}$. 
From this question and comments with links provided by Jyrki Lahtonen follows that there is no possibility that any vector with components from $\mathbb{S}$ and vector with components from $\mathbb{Z}$ can have their dot product equal $0$ unless components of integer vector are equal $0$. 
Note also that you can use values from the sets which look like, for example   $(\sqrt{3}- \sqrt{2}), (\sqrt{13}- \sqrt{11}),\dots,(\sqrt{73}- \sqrt{71}),\dots,(\sqrt{313}- \sqrt{311}) \dots $ and other appropriate  combinations with rational coefficients.
Selecting appropriately them you can obtain a vector which is very close (but not equal) to any predefined form, for example to vector $(1,1,1,..,1)^T$.
