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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 $q$).

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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.

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  • $\begingroup$ 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. $\endgroup$ – Kai Jul 30 at 2:33
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    $\begingroup$ @Kai I added a "side note" that hopefully addresses your question. $\endgroup$ – Brian Fitzpatrick Jul 30 at 2:36
  • $\begingroup$ Did you mean $f(q)$ where you wrote $f(1)$? $\endgroup$ – J. W. Tanner Jul 30 at 2:37
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    $\begingroup$ @J.W.Tanner yep. $\endgroup$ – Brian Fitzpatrick Jul 30 at 2:40

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