Is this $\boldsymbol{A}\times(\nabla\times\boldsymbol{B})$ identity wrong? 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.
 A: 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}
A: $$(\nabla\boldsymbol{B})\cdot\boldsymbol{A}=\left(\begin{aligned}&\frac{\partial B_x}{\partial x}&&\frac{\partial B_y}{\partial x}&&\frac{\partial B_z}{\partial x}\\
&\frac{\partial B_x}{\partial y}&&\frac{\partial B_y}{\partial y}&&\frac{\partial B_z}{\partial y}\\
&\frac{\partial B_x}{\partial z}&&\frac{\partial B_y}{\partial z}&&\frac{\partial B_z}{\partial z}\end{aligned}\right)\left(\begin{aligned}A_x\\A_y\\A_z\end{aligned}\right)$$
