The question came directly seeking for counterexample of the following Lemma :
$\textbf{Lemma} :$ $(V_{\mathbb{R}},\phi)$ is anisotropic $\iff \phi$ is definite.
If we restrict our view over rationals, i think although $\Leftarrow$ still holds, the other implication does not.
So the question, a little generalized would be
Exists a non-degenerate inner product of $\mathbb{Q}^2$ such that doesn't admit normalized orthogonal basis, anisotropic but not definite?
I think the answer is affirmative,but it was quite difficult to me. I tried a matrix of the following form : $$\begin{pmatrix}a & 1 \\ 1 & b\end{pmatrix}$$ With the condition that $\begin{cases} \phi(a,a) > 0 \\ \phi(b,b) < 0 \\ ab \ne 1 \end{cases}$
But i ended up with nothing, especially because of anisotropic condition :
$$\begin{pmatrix}x & y\end{pmatrix}\begin{pmatrix}a & 1 \\ 1 & b\end{pmatrix} \begin{pmatrix} x \\ y\end{pmatrix} = 0$$
Which gives the following conic $ax^{2}+2xy+by^{2} = 0$ over $\mathbb{Q}$.
Any solution, help or tip would be appreciated.