So I want to calculate the relative uncertainty in an angle given two vectors with a given relative uncertainty. For the angle calculation I'm using the following formula. $$\cos{\alpha} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$$ where: $$ \vec{a} = \begin{bmatrix} x_a\\ y_a\\ z_a \end{bmatrix} \; \text{and, } \; \vec{b} = \begin{bmatrix} x_b\\ y_b\\ z_b \end{bmatrix}$$
When I represent the vectors with the relative uncertainty I get the following:
$$ \vec{a} = \begin{bmatrix} x_{a} \pm x_{a_{error}}\\ y_{a} \pm y_{a_{error}}\\ z_{a} \pm z_{a_{error}} \end{bmatrix} \; \vec{b} = \begin{bmatrix} x_{b} \pm x_{b_{error}}\\ y_{b} \pm y_{b_{error}}\\ z_{b} \pm z_{b_{error}} \end{bmatrix}$$
I tried to calculate the relative error in the enumerator and the denominator, but at the end I just got the same error for the enumerator and denominator and since $cos^{-1}(1) = 0$, I'm guessing I have done some sort of mistake. I guess I'm missing some knowledge on how I'm supposed to do this.
When I did the calculation I calculated the relative error for $\vec{a} \cdot \vec{b}$ and then for $|\vec{a}||\vec{b}|$. For both of them I got that the relative error is
$$x_{a_{error}} + x_{b_{error}} + y_{a_{error}} + y_{b_{error}} + z_{a_{error}} + z_{b_{error}}$$
I'm familiar with the basic rules when calculating with uncertainty, but it is still a new topic for me and I'm a but confused about how I should approach this problem. I can add the whole calculation as well, but I think I'm missing some basic knowledge. Since the result for me don't make any sence. I though that since I'm working with relative uncertainty I could use the basic rules when multiplying, deviding etc for the relative uncertainty.
Thank you for the help!