# Find a vector in the plane spanned by two 4D vectors

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.
• Thanks! Any advice to spot something like that? I suppose that since I took the difference to get $v_0$ it would stand to reason trying the sum. But it seems these vectors are not orthogonal. I'm used to thinking in 3D, 4D vectors screw with me. – Kai Jul 30 at 2:33
• Did you mean $f(q)$ where you wrote $f(1)$? – J. W. Tanner Jul 30 at 2:37