1. **State the problem:** We need to find the moment of the force \(\vec{F} = (10, \frac{\pi}{3})\) acting at point \(A(\sqrt{3}, 2)\) about the origin \(O\). 2. **Understand the force vector:** The force is given in polar form \((F, \theta)\) where \(F=10\) and \(\theta=\frac{\pi}{3}\). 3. **Convert force to Cartesian components:** \[ F_x = F \cos \theta = 10 \cos \frac{\pi}{3} = 10 \times \frac{1}{2} = 5 \] \[ F_y = F \sin \theta = 10 \sin \frac{\pi}{3} = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \] So, \(\vec{F} = (5, 5\sqrt{3})\). 4. **Position vector of point A:** \[ \vec{r} = (\sqrt{3}, 2) \] 5. **Moment (torque) about origin:** The moment \(\vec{M} = \vec{r} \times \vec{F}\) (cross product in 2D gives a vector along \(\hat{k}\)): \[ M_z = x F_y - y F_x \] Substitute values: \[ M_z = \sqrt{3} \times 5\sqrt{3} - 2 \times 5 = 5 \times 3 - 10 = 15 - 10 = 5 \] 6. **Direction:** The moment vector is along \(\hat{k}\), so \[ \vec{M} = 5 \hat{k} \] **Final answer:** \(5 \hat{k}\) which corresponds to option (b).