If \( \vec{a} \cdot \vec{b}=-\frac{1}{\sqrt{3}}|\vec{a} \times \vec{b}| \) then angle between vector \( \vec{a} \) and vector \( \vec{b} \) is : \( \begin{array}{llll}\text { (A) } \frac{\pi}{3} & \text { (B) } \frac{\pi}{6} & \text { (C) } \frac{\pi}{4} & \text { (D) } \frac{2 \pi}{3}\end{array} \)

To solve for the angle between the vectors \( \vec{a} \) and \( \vec{b} \), we start by recalling the formulas for the dot product and the magnitude of the cross product of two vectors. The dot product is given by: \[ \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos \theta \] where \( \theta \) is the angle between the vectors. The magnitude of the cross product is given by: \[ |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}| \sin \theta \] Given the equation \( \vec{a} \cdot \vec{b} = -\frac{1}{\sqrt{3}}|\vec{a} \times \vec{b}| \), we can substitute the formulas for dot product and cross product into this equation: \[ |\vec{a}||\vec{b}| \cos \theta = -\frac{1}{\sqrt{3}} |\vec{a}||\vec{b}| \sin \theta \] Assuming \( |\vec{a}| \) and \( |\vec{b}| \) are not zero, we can divide both sides by \( |\vec{a}||\vec{b}| \): \[ \cos \theta = -\frac{1}{\sqrt{3}} \sin \theta \] Rearranging this gives: \[ \frac{\cos \theta}{\sin \theta} = -\frac{1}{\sqrt{3}} \] This simplifies to: \[ \cot \theta = -\frac{1}{\sqrt{3}} \] Taking the inverse cotangent gives: \[ \theta = \tan^{-1}(-\sqrt{3}) \] The angle whose cotangent is \(-\frac{1}{\sqrt{3}}\) corresponds to: \[ \theta = \frac{2\pi}{3} \text{ (in the second quadrant, where tangent is negative)} \] Thus, the angle between vector \( \vec{a} \) and vector \( \vec{b} \) is: \[ \boxed{\frac{2\pi}{3}} \]