Integer multiple of vector so angle is between $\pi/2$ and $\pi/3$ Given two vectors $u,v\in \mathbb{R}^2$ I wish to find rational $q$ so the angle $\theta
$ between $u$ and $v+qu$ satisfies
$$\frac{\pi}{3}\leq \theta \leq \frac{\pi}{2} $$
My first approach was inspired by the Gram–Schmidt process, letting $k=\left\lfloor\frac{\langle u,v\rangle}{\langle v,v\rangle}\right\rfloor v$, but nothing seems to come out of this.
fix: I changed $k$ from integer to rational, had miswritten originally
 A: Hint:
For the angle of two vectors $u, v$, we have
$$\cos \theta=\frac{\langle u,v\rangle}{\|u\|\,\|v\|}$$
and you want $\;0\le \cos\theta\le\frac12$.
A: First rotate the coordinate system such that $u=(a,0)$ with $a>0$. Then scale it such that $v=(b,\pm1)$. (If $u$ and $v$ are parallel, you're sunk in any case).
You now want $0 < b+ka < \tan\frac\pi6 = \frac{1}{\sqrt 3}$. There might be zero, one, or multiple integral $k$ that satisfy this.
A: Given three vectors $u$, $v$, and $w = v + ku \in \mathbb{R}^2$.
Based on the scalar product of two vector you can write that:
$u\cdot w = |u|\cdot|w|\cdot cos(\theta) = u_x\cdot w_x + u_y\cdot w_y$
So,
$$cos(\theta) = {u_x\cdot w_x + u_y\cdot w_y\over|u|\cdot|w|}$$
Now, replacing $w$ to $v+ku$:
$${u_x\cdot (v_x + ku_x) + u_y\cdot (v_y+ ku_y)\over|u|\cdot|v + ku|} $$
$${u_xv_x + ku_x^2 + u_yv_y+ ku_y^2\over|u|\cdot|v + ku|} $$
Then, rewrite the condition replacing $\theta$ by the above argument:
$$cos({\pi\over3}) \leq {u_xv_x + ku_x^2 + u_yv_y+ ku_y^2\over|u|\cdot|v + ku|} \leq cos({\pi\over2}) $$
Now you are able to solve it and obtain k.
