Question about polynomials over finite fields

The condition that $A$ and $B$ are nonempty disjoint can be replaced with the condition that they are distinct, as we may just replace $A$ and $B$ with $A \cap (A - B)$ and $B \cap (A- B)$ respectively.

Let $q = |\mathbb F|$.

Let $A$ and $B$ be two random linear subspaces of the space of polynomials of degree $\leq d$ each of codimension $k$. As long as $0< k < { d+2 \choose 2}$, with high probability $A$ and $B$ are distinct. As long as $A$ and $B$ intersect transversely the space of polynomials vanishing on $\ell$, the number of elements of $A$ and $B$ taking a given fixed value on $\ell$ is $ q^{ {d+2 \choose 2} - k - (d+1)}$.

Equivalently, this happens when the perpendicular spaces of $A$ and $B$ intersect only at $0$ the space of linear forms on degree $d$ polynomaials that factor through reduction to $\ell$. Each nonzero linear form has a probability of $$\frac{(q^k-1) }{ q^{ {d+2 \choose 2}} -1}$$ of being in the perpendicular subspace, and there are at most $(q^{d+1}-1)$ nontrivial linear forms that factor through restriction to each of $q (q+1)$ lines, so as long as

$$ q(q+1) (q^{d+1}-1) \frac{(q^k-1) }{ q^{ {d+2 \choose 2}} -1}< 1$$

with high probability $A$ and $B$ are transverse to all these linear subspaces. This happens when $k<{d+2 \choose 2} - d-3$. So it looks to me like even random $d+4$-dimensional subspaces will do the job.