Divergence of cross product of two vectors, intuition What is the geometrical/physical intuition behind this identity? (possibly in relation to the intuitions commonly given for curl, divergence and cross product)
$$\nabla\cdot\left ( \vec{A}\times\vec{B} \right )=\vec{B}\cdot\left ( \nabla \times\vec{A} \right )-\vec{A}\cdot\left ( \nabla \times\vec{B} \right )$$
 A: After hustling with this identity for a bit, this is what I was able to come up with. First thing to pay attention to is that $\nabla \cdot (\vec A\times \vec B )$ is the divergence of the cross product vector field.
The interpretation for the cross product vector field depends on the domain of the problem, but we can abstract that and just think that $\vec A\times \vec B$ generates a new vector field $\vec C$.
Hence, $\nabla \cdot \vec C$ can be understood using the common intuition for divergence, which is that it measures the amount of "fluid" being removed or added to the field (sink/source). For a better explanation on this, you can check out this videos by ThreeBrownOneBlue.
Now, to get a feel for what is going on in the identity, we will explore a very simple 3D case in which the vector field $\vec B$ varies only in the $\hat k$ axis, and $\vec A$ is equal to 1 in the $\hat i$ axis.
$$\vec A(x,y,z)= 1\hat i + 0 \hat j + 0\hat k$$
$$\vec B(x,y,z)= 0\hat i + (z+2) \hat j + 0\hat k$$

Below we draw a cube of 1x1x1, and show in a diagram the how the vector fields acts.
$$\\$$

Note that $\nabla \cdot \vec C$ is the difference between the flow leaving the top face and the flow entering the bottom face. From this we can start to understand how the curl of $\vec B$ affects the divergence of $\vec C$. The $\nabla \times \vec B$ is commonly viewed as the "rotation" of the field, but, such rotation implies that field $\vec B$ is changing in the perpendicular direction, which is exactly where $\vec A \times \vec B$ will be pointing. Hence, a curl in $\vec B$ means an acceleration in $\vec A \times \vec B$ and therefore, a variation in the divergence.
In this example, we can see that the divergence in the cross product field can be calculated in two ways. Either evaluating the change going inside one face and leaving other. Or by calculating the variation occurring inside the cube. This two ways of calculating are the equivalence shown in the identity.
Therefore, calculating the divergence using the surfaces we get
$$\nabla\cdot \vec C = +1\cdot A - 2 \cdot A = -1 \cdot A = -1$$
Where $A$ is the area of the face cube. And using the curl, we get;
$$\nabla\cdot \vec C = \vec B \cdot (\nabla \times \vec A) \cdot h - \vec A \cdot (\nabla \times \vec B) = 0 - 1\hat i \cdot (1\hat i) \cdot h = -1$$
Where $h$ is the height of the cube.
We calculated variation in the cube with 1x1x1, but the same could've been done for the infinitesimally small cube, which would eliminate $A$ and $h$ from the calculation.
Finally, the case illustrated is for very simple fields, but more complex fields can be thought in the same way, by just breaking the effect in each direction and applying the same logic. I also hope that it is clear that the same explanation shows how the curl of $\vec A$ appears in the equation. If $\vec A$ had the same curl as $\vec B$ $$\vec A(x,y,z) = (z+2)\hat i + 0 \hat j + 0 \hat k$$
Then, the curl of $\vec A$ would also "accelerate" the cross product field in the $\hat k$ axis, decreasing the divergence even more.
Hopefully this gives you some intuition on the identity. If something is not clear let me know, and I can try to do some more diagrams.
A: Due to bilinearity and some symmetries, this claim is equivalent to a special case of it, in which $A\parallel j$ and $B\parallel k$, so that $A\times B\parallel i$ and the left-hand side only gets a contribution from $\partial_x$. So the fact that everything in the original question is a $3$-dimensional vector (or $3$-dimensional vector operator) is really a distraction: that scalar functions $A(x,\,y,\,z),\,B(x,\,y,\,z)$ satisfy $(AB)_x=BA_x+AB_x$ is all that's really being claimed.
I realize these observations sound like I'm trying to prove the identity rather than give it an intuition, and if that were all I wanted to do, I'd be basically home by now: I'd just want to prove a product rule for $\partial_x$, in which the dependence on $y,\,z$ is also a distraction. But of course, I'm not here to rehearse a proof of the product rule. I'm here to note that if we want an intuition for all this, it's important to first note that, because it follows from the product rule and a couple of properties of vector calculus, the intuition we need concerns either whichever of those facts seems least intuitive, or how they collaborate. Whatever decision we make about that, we want that intuition to be geometric.
The usual geometric intuition for the product rule is actually in $2$ dimensions instead of $3$, noting that infinitesimally enlarging a rectangle adds two strips to it. I'd argue that's the crux here. The bilinearity & symmetry at work is really just the fact that multivariable calculus treats multiple variables as a single point roaming in a space of appropriate dimension. It's not just that we prove the identity with the product rule; it's that it doesn't really say anything other than the product rule.
