Let $\vec{u}(\theta) = \left<\cos{\theta}, \sin{\theta} \right>$
Given the formula $t\times(\vec{u}(\theta) - \vec{K_1})=\vec{K_2}$ with $\vec{K_1}, \vec{K_2} \in \mathbb{R}^2$, $\theta \in [0, 2\pi)$ and $t \in \mathbb{R}^+$, how can I find a value of $\theta$ that minimizes $t$?
Note that $\vec{K_1}$ and $\vec{K_2}$ have values that guarantee infinitely many solutions to the equation above.
PS. I'm new to this topic and learn on my own, so I couldn't really come up with a good title. Feel free to suggest a better one, I'll gladly change it.
Take the derivative of the vector equation (using the chain rule), and find $\frac{{\rm d}t}{{\rm d}\theta}=0$.
$$ {\rm d} t \, ( \vec{u} - \vec{K}_1) + t \, \left( \frac{\partial \vec{u}}{\partial \theta} {\rm d}\theta \right) = 0 $$
$$ {\rm d}t = \frac{ t \, \frac{\partial \vec{u}}{\partial \theta}}{\vec{K}_1 - \vec{u} } {\rm d}\theta =0 $$
which has the solution:
$$ t\,\left< -\sin \theta, \cos \theta \right> = 0$$
or $$t=0$$