$\pi-2\pi$ angle between two 3D vectors

Question

I need to calculate the angle between two 3D vectors. There are plenty of examples available of how to do that but the result is always in the range $0-\pi$. I need a result in the range $\pi-2\pi$.

Let's say that $\vec x$ is a vector in the positive x-direction and $\vec y$ is a vector in the positive y-direction and $\vec z$ is a reference vector in the positive z-direction. $\vec z$ is perpendicular to both $\vec x$ and $\vec y$. Would it then be possible to calculate the angle between $\vec x$ and $\vec y$ and get a result in the range $\pi-2\pi$?

The angle value should be measured counter clockwise. I have not been able to figure out how to do that. I am no math guru but I have basic understanding of vectors at least. Thank you very much for the help!

Answer 1

You need to define a perpendicular axis first. Otherwise clockwise/counterclockwise don't make sense. For example one vector is along $x$, one along $y$. Is the angle $90^\circ$ or $270^\circ$? Depends if your axis is $z$ or $-z$.

Let's assume that you define the angle using the scalar and vector products, and none of the two vectors is null. Then $$\sin\theta=\frac{(\vec v_1\times\vec v_2)\cdot\hat d}{|\vec v_1||\vec v_2|}\\\cos\theta=\frac{\vec v_1\cdot\vec v_2}{|\vec v_1||\vec v_2|}$$ Here $\hat d$ is a unit vector in a direction perpendicular to the plane $(\vec v_1,\vec v_2)$. The choice of $\hat d$ will impact the value of the angle (the quadrant in which it lies). You can then use the $\mathrm{atan2}$ function to give an angle between $-\pi$ and $\pi$. If you really want, you can change that to go from $0$ to $2\pi$

Answer 2

The cross product $v_1 \times v_2$ will be either positively or negatively proportional to the unit vector along your specified z-axis. You can interpret the proportionality constant as $\sin \theta |v_1| |v_2|$. Together with the value of $\cos \theta$ from dot product this determines a unique $\theta \in [0, 2\pi)$.

This may be slightly unpleasant computationally. You can also just find the angle in $[0, \pi]$ and then compute the determinant of 3 by 3 matrix with columns $v_1, v_2, z$; if this determinant is negative then take $2\pi-\theta$, otherwise keep $\theta$.