Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 16 / a

The fixed vectors have constant components, so $\mathbf{a}\cdot \mathbf{b}$ is constant, $\nabla \cdot \mathbf{a}=0$ and $\nabla \cdot \mathbf{r}=3$. Hence the divergence is

The curl of the constant vector field $\mathbf{a}\times \mathbf{b}$ vanishes. The curl of the position field $\mathbf{r}=(x,y,z)$ also vanishes, since all its cross-coordinate partial derivatives are zero. Thus

The vector triple product rewrites the field as $\mathbf{b}(\mathbf{a}\cdot \mathbf{r})-\mathbf{r}(\mathbf{a}\cdot \mathbf{b})$. Its first term has divergence $\mathbf{b}\cdot \nabla (\mathbf{a}\cdot \mathbf{r})=\mathbf{a}\cdot \mathbf{b}$, while the second contributes $3\mathbf{a}\cdot \mathbf{b}$. Therefore

Again write the field as $\mathbf{b}(\mathbf{a}\cdot \mathbf{r})-\mathbf{r}(\mathbf{a}\cdot \mathbf{b})$. The second term has zero curl. Applying the product rule to the first gives $\nabla (\mathbf{a}\cdot \mathbf{r})\times \mathbf{b}=\mathbf{a}\times \mathbf{b}$. Thus