Say we have a vector v=$[3\ 0\ 4]$. Find a 3x3 orthogonal matrix Q such that only the second component of Qv is nonzero and such that this component is also positive. Is Q unique?
I tried applying a combination of sin and cos in the ith and jth positions that turn into zeros on Qv but it isn't quite working. Is this even the right way to do it?
Nah, just complete the basis. Consider an orthogonal basis $Q=[u,v,w]^T$ where v is just the normalized version of your v. You see I already constructed a matrix $Q$ that has $v$ at the second place, so $Qv$ already satisfies the requirement of $Qv=(0,a,0)$.
Now just find any pair of orthogonal vectors $u$ and $w$ that are also orthogonal to $v$. There is an infinite number of possibilities (you can rotate $u$ and $w$ around $v$ without any change of behaviour along $v$). You can find them in many ways, one of them being the Gram-Schmidt orthogonalization. Just start with two independent vectors, say (0,1,0) and (0,0,1). The first is already orthogonal to $v$, so you must just orthogonalize the last one and you are done.