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In the book Introduction to Plasma Physics by R. J. Goldston, in Appendix D (page 480), he writes the following vector identity

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

but I think the rhs is identically zero. In matrix notation the first term is

$$(\nabla\boldsymbol{B})\cdot\boldsymbol{A}=\left(\begin{aligned}&\frac{\partial B_x}{\partial x}&&\frac{\partial B_x}{\partial y}&&\frac{\partial B_x}{\partial z}\\ &\frac{\partial B_y}{\partial x}&&\frac{\partial B_y}{\partial y}&&\frac{\partial B_y}{\partial z}\\ &\frac{\partial B_z}{\partial x}&&\frac{\partial B_z}{\partial y}&&\frac{\partial B_z}{\partial z}\end{aligned}\right)\left(\begin{aligned}A_x\\A_y\\A_z\end{aligned}\right)=\left(\begin{aligned}\frac{\partial B_x}{\partial x}A_x+\frac{\partial B_x}{\partial y}A_y+\frac{\partial B_x}{\partial z}A_z\\ \frac{\partial B_y}{\partial x}A_x+\frac{\partial B_y}{\partial y}A_y+\frac{\partial B_y}{\partial z}A_z\\ \frac{\partial B_z}{\partial x}A_x+\frac{\partial B_z}{\partial y}A_y+\frac{\partial B_z}{\partial z}A_z\end{aligned}\right)$$ and the second term $$(\boldsymbol{A}\cdot\nabla)\boldsymbol{B}=\left(A_x\frac{\partial}{\partial x}+A_y\frac{\partial}{\partial y}+A_z\frac{\partial}{\partial z}\right)\left(\begin{aligned}B_x\\B_y\\B_z\end{aligned}\right)=\left(\begin{aligned}A_x\frac{\partial B_x}{\partial x}+A_y\frac{\partial B_x}{\partial y}+A_z\frac{\partial B_x}{\partial z}\\ A_x\frac{\partial B_y}{\partial x}+A_y\frac{\partial B_y}{\partial y}+A_z\frac{\partial B_y}{\partial z}\\ A_x\frac{\partial B_z}{\partial x}+A_y\frac{\partial B_z}{\partial y}+A_z\frac{\partial B_z}{\partial z}\end{aligned}\right)$$

For this reason, I think that this vector identity may be wrong. Also I have not found it in any other place.

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  • $\begingroup$ Becarefull there is no matrix multiplication in those term on the right hand side but only scalar products $\endgroup$ – Guy Fsone Oct 11 '17 at 18:30
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    $\begingroup$ Notation $\nabla \mathbf{B}$ is a bit confusing, but interpreting this formally as the result of matrix multiplication of the 'colum vector' $\nabla$ and the row vector $\mathbf{B}^{\mathsf{T}}$ then we obtain what velut luna gave. This also matches what caverac derived using Einstein summation. $\endgroup$ – Sangchul Lee Oct 11 '17 at 19:13
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Abusing a bit Einstein's notation

$$ F\times G = \epsilon_{ijk}F_iG_j\hat{e}_k $$

So that

\begin{eqnarray} {\bf A}\times(\nabla\times {\bf B})&=& \epsilon_{ijk}A_i(\nabla\times{\bf B})_j\hat{e}_k = \epsilon_{ijk}A_i(\epsilon_{abj}\partial_a B_b)\hat{e}_k\\ &=&-(\epsilon_{ikj}\epsilon_{abj})A_i\partial_aB_j\hat{e}_k\\ &=&-(\delta_{ia}\delta_{kb} - \delta_{ib}\delta_{ka})A_i\partial_aB_j\hat{e}_k \\ &=&-A_a\partial_a B_b \hat{e}_b + A_b \partial_a B_b\hat{e}_a \\ &=& -(A_a\partial_a)(B_b \hat{e}_b) + (\hat{e}_a\partial_a B_b)(\underbrace{A_b}_{A_c\delta_{bc}})\\ &=& -({\bf A}\cdot \nabla){\bf B} + (\hat{e}_a\partial_a B_b)(A_c\hat{e}_c\cdot \hat{e}_b) \\ &=& -({\bf A}\cdot \nabla){\bf B} + (\hat{e}_a\partial_a B_b\hat{e}_b)\cdot (A_c\hat{e}_c) \\ &=&-({\bf A}\cdot \nabla){\bf B} + (\nabla {\bf B})\cdot {\bf A} \end{eqnarray}

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$$(\nabla\boldsymbol{B})\cdot\boldsymbol{A}=\left(\begin{aligned}&\frac{\partial B_x}{\partial x}&&\frac{\partial B_y}{\partial x}&&\frac{\partial B_z}{\partial x}\\ &\frac{\partial B_x}{\partial y}&&\frac{\partial B_y}{\partial y}&&\frac{\partial B_z}{\partial y}\\ &\frac{\partial B_x}{\partial z}&&\frac{\partial B_y}{\partial z}&&\frac{\partial B_z}{\partial z}\end{aligned}\right)\left(\begin{aligned}A_x\\A_y\\A_z\end{aligned}\right)$$

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