Divergence of Vector Cross Product

Definition
Let $R$ be a region of space embedded in Cartesian $3$ space $\R^3$.

Let $\mathbf A$ and $\mathbf B$ be vector fields over $R$.

Then:
 * $\map {\operatorname {div} } {\mathbf A \times \mathbf B} = \mathbf B \cdot \curl \mathbf A - \mathbf A \cdot \curl \mathbf B$

where:
 * $\operatorname {div}$ denotes the divergence operator
 * $\curl$ denotes the curl operator
 * $\times$ denotes vector cross product
 * $\cdot$ denotes dot product.

Proof
From Divergence Operator on Vector Space is Dot Product of Del Operator and Curl Operator on Vector Space is Cross Product of Del Operator:

where $\nabla$ denotes the del operator.

Hence we are to demonstrate that:
 * $\nabla \cdot \paren {\mathbf A \times \mathbf B} = \mathbf B \cdot \paren {\nabla \times \mathbf A} - \mathbf A \cdot \paren {\nabla \times \mathbf B}$

Let $\tuple {\mathbf i, \mathbf j, \mathbf k}$ be the standard ordered basis on $\R^3$.

Let $\mathbf A$ and $\mathbf B: \R^3 \to \R^3$ be expressed as vector-valued functions on $\R^3$:


 * $\mathbf A := \tuple {\map {A_x} {\mathbf r}, \map {A_y} {\mathbf r}, \map {A_z} {\mathbf r} }$


 * $\mathbf B := \tuple {\map {B_x} {\mathbf r}, \map {B_y} {\mathbf r}, \map {B_z} {\mathbf r} }$

where $\mathbf r = \tuple {x, y, z}$ is the position vector of an arbitrary point in $R$.

Then:

Also presented as
This result can also be presented as:


 * $\nabla \cdot \paren {\mathbf A \times \mathbf B} = \mathbf B \cdot \paren {\nabla \times \mathbf A} - \mathbf A \cdot \paren {\nabla \times \mathbf B}$

presupposing the implementations of $\operatorname {div}$ and $\curl$ as operations using the del operator.