Skew planes in $\mathbb{A}^4$

Question

Can there be two skew planes in $\mathbb{A}^4$?

By this I mean two disjoint planes $\pi_1,\pi_2\subset\mathbb{A}^4$ such that their underlying direction vector spaces only intersect at zero.

Answer 1

No, two such skew planes cannot exist:

Your planes are (like all affine planes) of the form $\pi_1=a+P_1, \pi_2=b+P_2$ for some $a,b\in \mathbb A^4$ and some complementary vectorial planes $P_1, P_2$.
Write $a=a_1+a_2\in P_1 \oplus P_2, b=b_1+b_2\in P_1 \oplus P_2$ and notice that $\pi_1=a+P_1=a_2+P_1,\pi_2=b+P_2=b_1+P_2$ (since $a_1+P_1=P_1$, etc.).
Then your affine planes $\pi_1, \pi_2$ are not disjoint since $$b_1+a_2\in \pi_1\cap \pi_2 $$