Problem 51

Since the equation of the circle is $(x-\cos\beta)^{2} + (y+\sin\beta)^{2} = \frac{1}{2}$,

the center of the circle is at $(\cos\beta, -\sin\beta)$, and the radius is $\frac{\sqrt{2}}{2}$.

The distance $d$ from the center of the circle to the line $x\cos\alpha - y\sin\alpha + \frac{1}{2} = 0$ is $d = |\cos\alpha\cos\beta + \sin\alpha\sin\beta + \frac{1}{2}| = |\cos(\alpha-\beta) + \frac{1}{2}|$.

Since $\overrightarrow{m} = (2\cos\alpha, 2\sin\alpha)$ and $\overrightarrow{n} = (3\cos\beta, 3\sin\beta)$, and the angle between $\overrightarrow{m}$ and $\overrightarrow{n}$ is 60°,

then $2 \times 3 \times \cos60° = 6\cos\alpha\cos\beta + 6\sin\alpha\sin\beta$,

which implies $\cos\alpha\cos\beta + \sin\alpha\sin\beta = \frac{1}{2}$,

therefore, $d = |\frac{1}{2} + \frac{1}{2}| = 1 > \frac{\sqrt{2}}{2}$,

hence, the correct answer is $\boxed{\text{D}}$.

From the given equations of the line $x\cos\alpha - y\sin\alpha + \frac{1}{2} = 0$ and the circle $(x-\cos\beta)^{2} + (y+\sin\beta)^{2} = \frac{1}{2}$, we can easily derive the expression for the distance $d$ from the center of the circle to the line. Then, by using the vectors $\overrightarrow{m} = (2\cos\alpha, 2\sin\alpha)$ and $\overrightarrow{n} = (3\cos\beta, 3\sin\beta)$, and knowing the angle between $\overrightarrow{m}$ and $\overrightarrow{n}$ is 60°, we can calculate the value of $d$ and compare it with the radius of the circle to find the answer.

This problem is of medium difficulty. It examines the knowledge of scalar product operations of planar vectors and the positional relationship between a line and a circle. If the distance from the center of the circle to the line is $d$, and the radius of the circle is $r$, then: ① when $d r$, the circle and the line are separate.