fast way to calculate normal to set of vectors with $\pm$1 entries Say I have a set of $(n-1)$ linearly independent vectors $\mathbf{v}_i$ of dimension $n$ with entries $\pm1$. I am interested in finding the $n-$dimensional vector $\mathbf{u} $which is normal to the hyperplane spanned by the $\mathbf{v}_i$. In other words, $\mathbf{u}$ is orthogonal to each of the $\mathbf{v}_i$, where $\mathbf{u}$ is unique up to multiplicative constant. 
I have two questions:
1) Is it true that $\mathbf{u}$ is also a vector with entries $(0, \pm 1)$ with an arbitrary pre-factor (e.g. $\mathbf{u}=\alpha[-1,1,0]$)? Can this be proven?
2) I know that $\mathbf{u}$ can be calculated with standard linear algebraic techniques (e.g. Gaussian elimination), but if property 1) is true, is there a computationally faster trick to getting the answer?
Thank you for your help!
 A: There is a six-dimensional counterexample:


*

*$(+1,+1,+1,+1,+1,+1)$

*$(-1,-1,+1,+1,+1,+1)$

*$(+1,+1,-1,-1,+1,+1)$

*$(+1,+1,+1,-1,-1,+1)$

*$(+1,-1,+1,-1,+1,-1)$


The normal to the linear span of these five vectors is $(2,-2,-1,1,-1,1)$.
A: To recap: we are given $n-1$ linearly independent vectors in ${\bf R}^n$ with $\pm 1$ entries. The original Q1 asked, in effect, if there is always a vector with $\pm 1$ entries that is orthogonal to all of the given ones. A counter examples is provided for $n=3$ by taking
$$
v_1 = e_1+e_2+e_3\quad,\quad v_2=e_1+e_2-e_3
$$
since the normal vector to span($v_1$,$v_2$) must be a multiple of $e_1-e_2$. (Here I am using $e_1,e_2,\dots$ to denote the standard o.n. basis vectors of ${\bf R}^n$.)
A modified version of Q1 now relaxes the requirement on the normal vector, so that one merely requires it to have entries in $\{-1,0,1\}$. This also fails (Adam Goucher pointed our a counterexample while I was writing this answer) and in fact we can generate a family of examples for all $n\geq 4$ as follows. Let $u= e_1+\dots + e_n$ iinside ${\bf R}^n$ and consider the vectors $u-2e_1,\dots, u-2e_{n-1}$. A little bit of manipulation shows that a normal vector to this family must have the form $ae_1+\dots + ae_{n-1}+be_n$ where $(n-2)a+b=0$, and so it is not possible for such a vector to have entries in $\{-1,0,1\}$ when $n\geq 4$.
It's not clear whether the OP is interested in Q2 even when Q1 has a negative answer, and it's not something I have any immediate intuition for.
