How to simplify $(\nabla \times \mathbf{A}) \cdot(\mathbf{B}\times \mathbf{C})?$

Question

We have $(\nabla \times \mathbf{A}) \cdot(\mathbf{B}\times \mathbf{C}).$ I have looked at a bunch of vectorial identites but none of them help me to simplify this one. Perhaps someone has more insight into this.

Answer 1

Just use the determinant notation, and compute the scalar product. It doesn't seem to exist a simplification for that.

Eventually:

$$(\partial_y a_z - \partial_z a_y)(b_yc_z - b_zc_y) + (\partial_z a_x - \partial_x a_z)(b_zc_x - b_xc_z) + (\partial_x a_y - \partial_y a_x)(b_xc_y - b_yc_x)$$

Answer 2

One needs to use Lagrange's Identity, thus we have: $$(\nabla \mathbf{B})(\mathbf{A\cdot C})-(\nabla \mathbf{C})(\mathbf{A\cdot B}).$$

Answer 3

You are looking at $$\varepsilon_{ijk}\partial_ja_k\varepsilon_{ilm} b_lc_m=\delta_{jl}\delta_{km}c_mb_l\partial_ja_k-\delta_{jm}\delta_{kl}c_mb_l\partial_ja_k=\\=c_kb_j\partial_ja_k-c_mb_l\partial_ma_l=((B\cdot\nabla) A)\cdot C-((C\cdot \nabla)A)\cdot B$$

Or, in more matrix-like form $$C^t(\nabla A)B- B^t(\nabla A)C$$