Integer multiple of vector so angle is between $\pi/2$ and $\pi/3$

Question

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

Answer 1

Hint:

For the angle of two vectors $u, v$, we have $$\cos \theta=\frac{\langle u,v\rangle}{\|u\|\,\|v\|}$$ and you want $\;0\le \cos\theta\le\frac12$.

Answer 2

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.

Answer 3

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.