Let's say I have two vectors, $$v_1 = (f(q),f(q),0,1)$$ $$v_2 = (f(q),f(q),1,0)$$ for some parameter $q$. For each $q$, these vectors span a plane. In particular there is a vector $$v_0 = (0,0,-1,1)$$ which always lies in that plane. I would like to find the vector perpendicular to $v_0$ which lies in the plane (for all $q$).
Note that adding $v_1$ and $v_2$ together produces the vector $$ w = \langle 2\,f(q), 2\,f(q), 1, 1\rangle $$ and this vector lives in the plane of interest.
Dotting $w$ with $v_0$ gives $$ w\cdot v_0 = \langle 2\,f(q), 2\,f(q), 1, 1\rangle\cdot \langle 0, 0, -1, 1\rangle = 0 $$ Hence $w$ lives in the plane and is orthogonal to $v_0$.
Side Note. In this problem we're interested in finding certain vectors orthogonal to $v_0=\langle 0, 0, -1, 1\rangle$. Every vector orthogonal to $v_0$ is of the form $w=\langle x_1, x_2, x_3, -x_3\rangle$. So, the first two coordinates of $w$ can be chosen "freely" and the last two coordinates need to be related by negation. Looking at the two vectors $v_1$ and $v_2$, we see that our $w$ can be created by adding.