Question in Curl of a cross product.

**yungman** · March 21st 2011, 06:00 PM

$\nabla X (\vec A X \vec B) \;=\; (\vec B \cdot \nabla)\vec A - \vec B(\nabla \cdot \vec A) -(\vec A \cdot \nabla)\vec B + \vec A ( \nabla \cdot \vec B)$ .

What is $\vec A \cdot \nabla$ ?

**Prove It** · March 21st 2011, 06:15 PM

Have they told you what $\displaystyle \mathbf{A}$ and $\displaystyle \mathbf{B}$ are equal to?

**yungman** · March 21st 2011, 06:17 PM

Originally Posted by Prove It

Have they told you what $\displaystyle \mathbf{A}$ and $\displaystyle \mathbf{B}$ are equal to?

Nop, this is just a general formula of the product of two vectors.

**Ackbeet** · March 22nd 2011, 02:02 AM

Originally Posted by yungman

$\nabla X (\vec A X \vec B) \;=\; (\vec B \cdot \nabla)\vec A - \vec B(\nabla \cdot \vec A) -(\vec A \cdot \nabla)\vec B + \vec A ( \nabla \cdot \vec B)$ .

What is $\vec A \cdot \nabla$ ?

It's the vector $\vec{A}$ dotted with the gradient operator. The result of this dot product is a scalar operator. So, for example, the expression

$\displaystyle(\vec{A}\cdot\nabla)\vec{B}=\left(\su m_{j=1}^{3}A_{j}\,\dfrac{\partial}{\partial x_{j}}\right)\vec{B}=\left\langle\sum_{j=1}^{3}A_{ j}\,\dfrac{\partial B_{1}}{\partial x_{j}},\sum_{j=1}^{3}A_{j}\,\dfrac{\partial B_{2}}{\partial x_{j}},\sum_{j=1}^{3}A_{j}\,\dfrac{\partial B_{3}}{\partial x_{j}}\right\rangle.$

Does that make sense?

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