1.Show that $\nabla \times (\nabla \times \vec{A}) = \nabla(\nabla \cdot \vec{A}) - \nabla^2 \vec{A}$.
Sol: Using the vector triple product identity $\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b}(\vec{a}\cdot\vec{c}) - (\vec{a}\cdot\vec{b})\vec{c}$ with operator $\nabla$ gives $\nabla \times (\nabla \times \vec{A}) = \nabla(\nabla \cdot \vec{A}) - (\nabla \cdot \nabla)\vec{A} = \nabla(\nabla \cdot \vec{A}) - \nabla^2 \vec{A}$.
2.Evaluate $\oint_C \vec{F} \cdot d\vec{r}$ where $\vec{F} = (x - y)\hat{i} + (x + y)\hat{j}$ around the unit circle $C$ using Green's Theorem.
Sol: $\oint_C (P dx + Q dy) = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx dy$. Here $\frac{\partial Q}{\partial x} = 1, \frac{\partial P}{\partial y} = -1 \implies 1 - (-1) = 2$. Integral = $2 \times \text{Area of unit circle} = 2\pi$.
3.Prove that if $\vec{F}$ is a conservative force field ($\nabla \times \vec{F} = 0$), then $\vec{F} = -\nabla U$ for some scalar potential function $U$.
Sol: Since curl of any gradient is identically zero ($\nabla \times (\nabla U) = 0$), irrotational field $\nabla \times \vec{F} = 0$ implies $\vec{F}$ can be represented as negative gradient of a scalar potential $U(\vec{r}) = -\int_{\vec{r}_0}^{\vec{r}} \vec{F} \cdot d\vec{r}$.
