Zero-based vector means origin-based vector

$\cos \theta = \dfrac{{\bf V_1} \cdot {\bf V_2}}{V_1 \cdot V_2}$

$\cos \theta = \dfrac{-\sqrt{3} \left( 2\sqrt{3} \right) + 1(2)}{\sqrt{\left( -\sqrt{3} \right)^2 + 1^2} \cdot \sqrt{\left( 2\sqrt{3} \right)^2 + 2^2}}$

$\cos \theta = -\frac{1}{2}$

$\theta = 120^\circ$ ← Answer: [ D ]