Find the curl of $\mathrm a \times ( \mathrm b \times \mathrm r)$

Question

I'm trying to find the curl of $a \times (b \times r)$ where $a$, $b$ are constant vectors and $r = (x,y,z)$.

I've worked through the problem to achieve the following answer:

$$(\vec a \cdot \vec r) (\nabla \times \vec b) + \nabla (\vec a \cdot \vec r) \times \vec b - (\vec a \cdot \vec b) (\nabla \times \vec r) - \nabla (\vec a \cdot \vec b) \times \vec r$$

I know that the final answer is, $a \times b$, however I'm unsure of the steps in between to get there. Help would be greatly appreciated.

Answer 1

\begin{align*} \mathbf{a} \times (\mathbf{b} \times \mathbf{r}) &= (\mathbf{a \cdot r}) \mathbf{b}-(\mathbf{a \cdot b}) \mathbf{r} \\[5pt] \nabla \times [\mathbf{a} \times (\mathbf{b} \times \mathbf{r})] &= \nabla \times [(\mathbf{a \cdot r}) \mathbf{b}]- (\mathbf{a \cdot b}) \nabla \times \mathbf{r} \\[5pt] &= \begin{vmatrix} \mathbf{e}_{x} & \mathbf{e}_{y} & \mathbf{e}_{z} \\ \partial_{x} & \partial_{y} & \partial_{z} \\ (\mathbf{a \cdot r}) b_x & (\mathbf{a \cdot r}) b_y & (\mathbf{a \cdot r}) b_z \\ \end{vmatrix} - \mathbf{0} \\[5pt] &= \sum_{xyz} \mathbf{e}_{x} \left \{ \frac{\partial}{\partial y} [(\mathbf{a\cdot r}) b_{z}]- \frac{\partial}{\partial z} [(\mathbf{a\cdot r}) b_{y}] \right \} \\[5pt] &= \sum_{xyz} \, \mathbf{e}_{x} (a_{y} b_{z}-a_{z} b_{y}) \\[5pt] &= \mathbf{a\times b} \end{align*}

Answer 2

To solve this kind of problems usually I use levi-civita symbol.

$(a\times b)_{k}=\varepsilon_{ijk}a_{i}b_{j}$

In this case:

First, you have to notice any differential operation over a constant vector is zero. Therefore, $(\nabla\times[\vec{a}\times(\vec{b}\times\vec{r})])_{s}=\varepsilon_{pqs}\partial_{p}[\vec{a}\times(\vec{b}\times\vec{r})]_{q}=\varepsilon_{pqs}\partial_{p}(\varepsilon_{ijq}a_{i}(\vec{b}\times\vec{r})_{j})=\varepsilon_{pqs}\partial_{p}(\varepsilon_{ijq}a_{i}\varepsilon_{nmj}b_{n}x_{m})=\varepsilon_{pqs}\partial_{p}(\varepsilon_{jqi}\varepsilon_{jnm}a_{i}b_{n}x_{m})=\varepsilon_{pqs}\partial_{p}([\delta_{qn}\delta_{im}-\delta_{qm}\delta_{in}]a_{i}b_{n}x_{m})=\varepsilon_{pqs}([\delta_{qn}\delta_{im}-\delta_{qm}\delta_{in}]a_{i}b_{n}\delta_{mp})=\varepsilon_{pqs}([\delta_{qn}\delta_{ip}-\delta_{qp}\delta_{in}]a_{i}b_{n})=\varepsilon_{pqs}(a_{p}b_{q}-0)=\vec{a}\times\vec{b}$