# Vector product

1. May 4, 2005

### Dinosaur

I looked up the formula for $$\nabla \times (\vec{A} \times \vec{B})$$:

$$\nabla \times (\vec{A} \times \vec{B}) = \vec{A}(\nabla \cdot \vec{B}) - \vec{B}(\nabla \cdot \vec{A}) + (\vec{B} \cdot \nabla)\vec{A} - (\vec{A} \cdot \nabla)\vec{B}$$

What does a vector followed by a del mean? Mathworld says that in the context of a unit vector it's the directional derivative. It's unclear to me how this works because then you have a vector times a vector for the last two terms. Can someone please clarify how this works?

2. May 4, 2005

### Data

Edit: IGNORE!!

$${\vec A} \cdot \nabla = \nabla \cdot {\vec A}.$$

The directional derivative is a scalar.

Last edited: May 5, 2005
3. May 4, 2005

### Dinosaur

Ah, ok. Dot products are commutative. Thanks.

4. May 5, 2005

### Galileo

The expression $(\vec A \cdot \vec \nabla )$ is an operator, like $\vec \nabla$. It has to work on a vector. You can see what it does by treating $\vec \nabla$ as a genuine vector:

$$\vec A \cdot \vec \nabla = A_x \frac{\partial}{\partial x}+A_y \frac{\partial}{\partial y}+ A_z\frac{\partial}{\partial z}$$

Ofcourse, you can always check it by writing out the components :P

Last edited: May 5, 2005
5. May 5, 2005

### Data

oops, yeah, I'm insane! Though I should note that it can actually also work on a scalar.

Last edited: May 5, 2005
6. May 5, 2005