Dimension 3 quadratic forms are isotropic iff they are zero in the Witt ring?

Question

In the paper Quadratic Forms over $\mathbb C[t_1,t_2]$ by K. H. Kim and W. Roush (https://core.ac.uk/download/pdf/82242112.pdf), they mentioned on page 77 that over $\mathbb C(t_1,t_2)$, forms of dimension 3 are isotropic if and only if they are zero in the Witt ring. I haven't understood this point. Suppose that we have an isotropic form $q=<1>\perp<-1>\perp<a>$ where $a$ is a non-trivial element in $\mathbb C(t_1,t_2)^\times/\mathbb C(t_1,t_2)^{\times 2}$, then the class of $q$ in the Witt ring $W(\mathbb C(t_1,t_2))$ is determined by its anisotropic part and $q\sim <a>$ which shouldn't be zero in the Witt ring (if I'm not wrong)?