13. Vector $\vec{a}=(1, \sin \theta), \vec{b}=(\cos \theta, \sqrt{3}), \theta \in R$, then the range of $|\vec{a}-\vec{b}|$ is
$[1,3]$ $\qquad$
Answer:
$$
\begin{aligned}
|\vec{a}-\vec{b}| & =\sqrt{(1-\cos \theta)^{2}+(\sin \theta-\sqrt{3})^{2}}=\sqrt{5-2(\cos \theta-\sqrt{3} \sin \theta)} \\
& =\sqrt{5-4 \sin \left(\frac{\pi}{6}-\theta\right)} \text {, its maximum value is } 3 \text {, minimum value is } 1 \text {, the range of values is } [1,3]。
\end{aligned}
$$