# Vector cross product

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1. Sep 28, 2016

### Muthumanimaran

1. The problem statement, all variables and given/known data
This is not a homework problem, I am currently reading the Derivation of potential of a charged particle in Electric and Magnetic field from the book Mechanics by Symon (I attached the image of the page), I need to know how to expand the vector cross product
such as $$\vec{v}\times({\nabla}{\times}\vec{A})=?$$

2. Relevant equations
ax(bxc)=(a.c)b-(a.b)c (Lagrange's formula)

3. The attempt at a solution
I tried to expand the vector triple product using Lagrange's formula, but what i am getting is,
$$(\vec{v}.\vec{A})\nabla-(\nabla.\vec{A})\vec{v}$$
As far as I know $$(\nabla.\vec{A})\vec{v}\neq(\vec{v}.\nabla\vec{A})$$ Is there any mistake I did during my calculation or should I use some other formula? please help.

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2. Sep 28, 2016

### Orodruin

Staff Emeritus
You cannot just commute the differential operator with the fields in the way you have done. In order to apply the bac-cab rule to an expression with a $\nabla$ you must manipulate the end expression in such a way that the nabla acts on the same vectors before and after applying it.

3. Sep 28, 2016

### Ray Vickson

Write it out in detail. Using $D_u f$ to stand for $\partial f/\partial u$ for $u = x,y ,z$, the $x$-component of $\vec{v} \times (\nabla \times \vec{A})$ is
$$\left(\vec{v} \times (\nabla \times \vec{A}) \right)_x = v_y (\nabla \times \vec{A})_z - v_z (\nabla \times \vec{A})_y \\ = v_y (D_x A_y - D_y A_x) - v_z (D_z A_x -D_x A_z)$$.
Expand it out and see what you get, then do the same type of thing for the $y$- and $z$-components. In particular, note that the components of $\vec{v}$ are not differentiated, so cannot lie to the right of the $D$-sign, unless you close it out by putting it inside parentheses, like this: $(D_x A_y)$.

Last edited: Sep 28, 2016