The 8 vectors to be made non-collinear Consider the set of $8$ vectors $V=\{ai+bj+ck:a,b,c \in \{-1,1\}\}$. How can I choose three non-collinear vectors from $V$?
My try:
Let there be three vectors
\begin{align*}
V_1&=a_1i+b_1j+c_1k,\\
V_2&=a_2i+b_2j+c_2k,\\
V_3&=a_3i+b_3j+c_3k.
\end{align*}
Now for $V_1,V_2,V_3$ to be collinear determinant of their $x,y,z$ components should be zero. But what next? What conditions should I apply?
 A: If you visualize the set of vectors, you will notice that they form the vertices of a cube. Since no three vertices of the cube are collinear, any choice of $V_1,V_2,V_3$ as three disctinct vectos of the set will do.
Let $V_1,V_2,V_3$ be any three distinct vectors of the set $V$.
Note that $V_2-V_1$ is a nonzero vector with components $\in\{-2,0,2\}$. Likewise, $V_3-V_1$ and $V_3-V_2$ are nonzero vectors with components $\in\{-2,0,2\}$. 
The only way for $V_3-V_1$ to be a multiple of $V_2-V_1$ which is equivalent to collinearity of $V_1,V_2,V_3$) is therefore that $V_3-V_1=\pm(V_2-V_1)$. But "$+$" leads to $V_3=V_2$, contrary to assumption. So $V_3-V_1=-(V_2-V_1)$ and hence $V_3-V_2=2\cdot (V_1-V_2)$ is a nonzero vector with components in $\{-4,0,4\}$. But we already know that its components are $\in\{-2,0,2\}$, so they must be zero, i.e. again $V_3=V_2$ contrary to assumption.
A: The cross product of any two of the three vectors you choose cannot be zero.  With the constraints you have, this in essence means that none of the vectors you choose can be the same as, or the negative of, any of the others.
So $(1,1,1), (1,1,-1), (1,-1,1)$ would work, for example.
