1. **Problem statement:** Given vectors $a$ and $b$ with magnitudes $\|a\| = \sqrt{3}$, $\|b\| = 2$, and the angle between them $\widehat{(a,b)} = 150^{0}$, find the area of the parallelogram formed by vectors $u = a + b$ and $v = 2a - b$. 2. **Recall the formula for the area of a parallelogram:** The area is given by the magnitude of the cross product of the two vectors: $$\text{Area} = \|u \times v\|$$ 3. **Express $u \times v$ using $a$ and $b$: ** $$u \times v = (a + b) \times (2a - b) = a \times 2a - a \times b + b \times 2a - b \times b$$ 4. **Simplify terms:** - $a \times a = 0$ because the cross product of any vector with itself is zero. - $b \times b = 0$ for the same reason. - $a \times b$ and $b \times a$ are related by $b \times a = - (a \times b)$. So, $$u \times v = 0 - a \times b + 2 b \times a - 0 = - a \times b + 2(- a \times b) = - a \times b - 2 a \times b = -3 a \times b$$ 5. **Magnitude of $u \times v$: ** $$\|u \times v\| = \|-3 a \times b\| = 3 \|a \times b\|$$ 6. **Calculate $\|a \times b\|$: ** Recall that $$\|a \times b\| = \|a\| \|b\| \sin \theta$$ where $\theta = 150^{0}$. Calculate $\sin 150^{0}$: $$\sin 150^{0} = \sin (180^{0} - 30^{0}) = \sin 30^{0} = \frac{1}{2}$$ Therefore, $$\|a \times b\| = \sqrt{3} \times 2 \times \frac{1}{2} = \sqrt{3}$$ 7. **Calculate the area:** $$\text{Area} = 3 \times \sqrt{3} = 3\sqrt{3}$$ **Final answer:** The area of the parallelogram is $3\sqrt{3}$, which corresponds to option (b).