Suppose we have two arbitrary vectors $v_1$ and $v_2$ in $\mathbb F_p ^{2p-1}$ ($p$ is an odd prime number, and $\mathbb F_p$ is the finite field with $p$ elements):
Now is there any nonzero vector $w$ contains only $0$ and $1$ s.t the inner product of $v_1.w=v_2.w=0$?
For example if $v_1=\underbrace{(2,0,2,0\cdots,2)}_{2p-1}$ and $v_2=\underbrace{(0,1,0,1\cdots,0)}_{2p-1}$ then we can put $w=v_1=\underbrace{(1,0,1,0\cdots,1)}_{2p-1}$ and then $v_1.w=2p=0$ in $\mathbb F_p$ and $v_2.w=0$.
It's easy to see that when $v_1$ and $v_2$ consist of only $0$ and $1$, then always there exists such a $w$, but in the general case, I have been trying some tricks from linear algebra and combinatorics without any success!
And I should say that in $\mathbb F_p ^{2p-2}$ there is a counterexample:
$v_1=\underbrace{(1,0,1,0\cdots,0)}_{2p-2}$, $v_2=\underbrace{(0,1,0,1\cdots,1)}_{2p-2}$
Hint: It can be proved by use of Chevalley–Warning theorem and $p-1$th power in $2p-1$ variable, to get only $0$ and $1$.