Q:

Finding a basis where the vectors are not orthogonal

Consider the vector space $\mathbb{R}^3$ and the subspace $S$ of dimension $1$ given by $$S = span[v_1], \ \ v_1 = \begin{pmatrix}
1\\
1\\
1\\
\end{pmatrix}$$
  Determine a basis $\{v_2,v_3\}$ of the subspace $S^\perp$ where the vectors $v_2$ and $v_3$ are not orthogonal.

Attmepted solution - We see that by definition $S^\perp$ has dimension $2$ comprising all vectors $w\in\mathbb{R}^3$ such that $w^T v_1 = 0$ We need to construct $v_2$ and $v_3$ such that $v_2^T v_3\neq 0$. 
I am sort of rusty with linear algebra. I was just wondering whether the elements of the vectors $v_2,v_3$ need to only have $-1$ or $1$ meaning that I could set $$v_2 = \begin{pmatrix}
1\\
-1\\
0\\
\end{pmatrix} \ \ \ \ v_3 = \begin{pmatrix}
1\\
0\\
-1\\
\end{pmatrix}$$
Could I have numbers that are not $1$ or $-1$ inside these vectors and still satisfy the conditions?

A:

There is nothing wrong with your vectors. But there are infinitely many other solutions. For instance, the two vectors
$$
v_2 = \begin{pmatrix}
1000\\
-500\\
-500\\
\end{pmatrix} \ \ \ \ v_3 = \begin{pmatrix}
1000\\
-700\\
-300\\
\end{pmatrix}
$$
work just as fine.

