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In the identity: $\nabla\times(A\times B) = (B \cdot\nabla)A-(A \cdot\nabla)B+A(\nabla \cdot B)-B(\nabla \cdot A)$

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    Have you tried computing it with some exemplary $A$ and $B$? Or for what specific difference you are asking for? – 2017-02-22
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    I personally would be more confused about the difference between $(B\cdot\nabla)A$ and $B(\nabla\cdot A)$. In your case, $\nabla$ one time is applied to $A$ and the other time to $B$. Thats the difference. – 2017-02-22
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    It was in a book. I thought that dot product is commutative and is applied to $A$ in both cases. Now I got it thanks! – 2017-02-22
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    The dot product _is commutative_. Note that only one of the operations is a dot product. The other one is the application of an operator which yields another vector (dot product yields a scalar). – 2017-02-22

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Assuming we are in $\mathbb R^n$,

$$(A \cdot \nabla) = \sum_i A^i \frac{\partial}{\partial x^i}$$

for coordinates $\{x^i\}$ on $\mathbb R^n$. This denotes the directional derivative along $\vec A$ and moreover $A \cdot \nabla$ is simply a way of writing a vector field as a derivation. The Lie derivative of a scalar is precisely this:

$$\mathcal L_A \phi = \sum_i A^i \frac{\partial \phi}{\partial x^i}.$$

For the case of a vector, one has $\mathcal L_A B = [A,B]$ with $A$ and $B$ expressed as derivations as above. Now for $(\nabla \cdot A)$, this is simply a scalar and equal to the divergence of $\vec A$, namely,

$$\nabla \cdot A = \sum_i \frac{\partial A^i}{\partial x^i}.$$

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    But can I ask if you see any difference between $(B\cdot \nabla)A$ and $B(\nabla\cdot A)$? – 2017-02-22
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    So dot product between del operator and vector function isn't commutative! Thanks! – 2017-02-22