1. **Problem statement:** Given the vector equation $\mathbf{v} = \mathbf{w} \times \mathbf{r}$, where $\mathbf{w}$ is a constant vector and $\mathbf{r}$ is the position vector, prove that $\mathbf{w} = \nabla \times \mathbf{u}$ for some vector field $\mathbf{u}$. 2. **Recall the curl definition:** The curl of a vector field $\mathbf{u}$ is defined as $$\nabla \times \mathbf{u} = \left( \frac{\partial u_z}{\partial y} - \frac{\partial u_y}{\partial z}, \frac{\partial u_x}{\partial z} - \frac{\partial u_z}{\partial x}, \frac{\partial u_y}{\partial x} - \frac{\partial u_x}{\partial y} \right).$$ 3. **Construct the vector field $\mathbf{u}$:** Let $$\mathbf{u} = \frac{1}{2} \mathbf{w} \times \mathbf{r}.$$ 4. **Calculate the curl of $\mathbf{u}$:** Using the vector calculus identity for curl of a cross product with a constant vector $\mathbf{w}$, $$\nabla \times (\mathbf{w} \times \mathbf{r}) = \mathbf{w} (\nabla \cdot \mathbf{r}) - (\mathbf{w} \cdot \nabla) \mathbf{r}.$$ Since $\mathbf{w}$ is constant, $\nabla \cdot \mathbf{r} = 3$ (divergence of position vector), and $(\mathbf{w} \cdot \nabla) \mathbf{r} = \mathbf{w}$ (directional derivative of $\mathbf{r}$ along $\mathbf{w}$), we get $$\nabla \times (\mathbf{w} \times \mathbf{r}) = 3 \mathbf{w} - \mathbf{w} = 2 \mathbf{w}.$$ 5. **Therefore,** $$\nabla \times \mathbf{u} = \nabla \times \left( \frac{1}{2} \mathbf{w} \times \mathbf{r} \right) = \frac{1}{2} \nabla \times (\mathbf{w} \times \mathbf{r}) = \frac{1}{2} \times 2 \mathbf{w} = \mathbf{w}.$$ 6. **Conclusion:** We have shown that for $\mathbf{u} = \frac{1}{2} \mathbf{w} \times \mathbf{r}$, $$\mathbf{w} = \nabla \times \mathbf{u},$$ which proves the statement.