Existence of an isotropic vector orthogonal to a given one in a vector space over a finite field.

Question

Suppose $V$ is a vector space over a finite field $F$ with a quadratic form $N:V \rightarrow F$ and associated non-degenerate symmetric bilinear inner product $f : V\times V \rightarrow F$. Let $v \in V$ be an arbitrary vector. Is it always possible to find an isotropic vector $u\in V$ which is orthogonal to $v$? That is, $N(u) = 0$ and $f(v,u) = 0$. If $v$ is isotropic, then it is orthogonal to itself, however I do not see whether the statement is true for an arbitrary non-isotropic vector $v$.

Answer 1

No. For example take a two-dimensional vector space $V$. There are nondegenerate isotropic inner products on this, for example the "hyperbolic plane" coming from the quadratic form $xy$. Anyway in such a $V$, if $v$ has $N(v)\ne0$ then we cannot have a nonzero isotropic vector $u$ orthogonal to $v$. For $u$ and $v$ would be linearly independent, and so span the space $V$. Then $u$ would be orthogonal to the whole of $V$, contradicting its nondegeneracy.

None of this depends on the field $F$, save I suppose for it not being of characteristic two.