Find a vector NOT perpendicular to a given set of vectors 
Let us suppose to have a finite set of vectors $S=\{v_1,\ldots,v_m\}$ in $\mathbb{R}^n$ (with $m \gg n$ in general). I need to find a vector $x \in \mathbb{R}^n$ that is NOT perpendicular to any vector in $S$. The existence of such a vector $x$ is guaranteed, moreover almost every vector in $\mathbb{R}^n$ satisfies this property. But I need to find an algorithm to determine a vector with these properties, I cannot close my eyes and choose.

Any ideas?
EDIT1: in my case, the vectors in $S$ have some "symmetries" in the sense that they are generated by permutation and change of signs of a few vectors in $S$
EDIT2: $v_1+\ldots+v_m=0 \in \mathbb{R}^n$
EDIT3: I simplified the solution proposed by Tom Collinge. We define a sequence $\{x_i\}_{i=1}^m$ such that $x_m=x$ is what we are looking for. First define $x_1=v_1$, so $x_1 \cdot v_1 \ne 0$ and they are not perpendicular (obviously). Then, recursively, define for $k \in \{2,\ldots,m\}$
$$
x_k=\begin{cases}
x_{k-1}+2v_k & \text{if }x_{k-1}=-v_k\\
x_{k-1}+v_k & \text{otherwise}
\end{cases}.
$$
By construction we have that $x_k \cdot v_i \ne 0$ for $i \le k$, then $x_m \cdot v_i \ne 0$ for all $v_i \in S$, so $x=x_m$ is not perpendicular to every element in $S$. Can it works?

EDIT4: As pointed out, the algorithm proposed in EDIT3 does not work


You probably want to have non orthogonal vectors to get out of numerical problems with dividing numbers by zero, right?
If that is the case, try to maximize that value instead!
So solve the problem
$$b_{opt}=\max_{b}\left(\min_i(v_i^Tb)\right)$$
This should have a nice solution that you can always work with.
