Let $\mathbf{A}$ be a constant vector and $\mathbf{r}=x\,\mathbf{i}+y\,\mathbf{j}+z\,\mathbf{k}$, verify the following identities:

(i) $\nabla\cdot[(\mathbf{r}\cdot\mathbf{r})\mathbf{A}]=2\mathbf{r}\cdot\mathbf{A}$

(ii) $\nabla\times[(\mathbf{r}\cdot\mathbf{r})\mathbf{A}]=2\mathbf{r}\times\mathbf{A}$

(iii) $\nabla\cdot(\mathbf{A}\times\mathbf{r})=0$

(iv) $(\mathbf{A}\times\nabla)\times\mathbf{r}=-2\mathbf{A}$.

in all these question when I use $\nabla$ with the constant vector $\mathbf{A}$ It should give me zero in all cases( div - grad - curl ) but when I come in question [iv] the answer is $2\mathbf{A}$ by using the triple cross product not calculating $\mathbf{A}\times\nabla = 0 $.

  • $\begingroup$ $\mathbf{r}$ is not a constant vector, and the differential operator $(\mathbf{A}\times\nabla)\times$ is acting on it. $\endgroup$ – user10354138 May 23 at 19:57
  • $\begingroup$ A is the constant vector not r $\endgroup$ – Omar Shawky May 23 at 19:58
  • 1
    $\begingroup$ No, $\nabla$ (or rather $(A\times\nabla)\times$) is acting on everything that comes after it, and in this case it is a nonconstant vector field $\mathbf{r}$. $\endgroup$ – user10354138 May 23 at 20:00
  • $\begingroup$ shouldn't i use the determent to calculate Ax∇ first due to the brackets ? $\endgroup$ – Omar Shawky May 23 at 20:01
  • $\begingroup$ Yes, and that give you a differential operator. $\endgroup$ – user10354138 May 23 at 20:01

You can't manipulate differential operators as if they are constant. In particular, you can't naively use the BAC-CAB identity because it switches the order around.

$\mathbf{A}\times\nabla$ is the vector-valued operator $$ \mathbf{i}\left(A_y\frac{\partial}{\partial z}-A_z\frac{\partial}{\partial y}\right) + \mathbf{j}\left(A_z\frac{\partial}{\partial x}-A_x\frac{\partial}{\partial z}\right) + \mathbf{k}\left(A_x\frac{\partial}{\partial y}-A_y\frac{\partial}{\partial x}\right) $$ where $\mathbf{A}=A_x\,\mathbf{i}+A_y\,\mathbf{j}+A_z\,\mathbf{k}$. You take the cross product again with $\mathbf{r}$: \begin{align*} (\mathbf{A}\times\nabla)\times\mathbf{r} &=\mathbf{i}\left[(\mathbf{A}\times\nabla)_y\mathbf{r}_z-(\mathbf{A}\times\nabla)_z\mathbf{r}_y\right]\\ &\quad+\mathbf{j}\left[(\mathbf{A}\times\nabla)_z\mathbf{r}_x-(\mathbf{A}\times\nabla)_x\mathbf{r}_z\right]\\ &\quad+\mathbf{k}\left[(\mathbf{A}\times\nabla)_x\mathbf{r}_y-(\mathbf{A}\times\nabla)_y\mathbf{r}_x\right]\\ &=\mathbf{i}\left[\left(A_z\frac{\partial}{\partial x}-A_x\frac{\partial}{\partial z}\right)z-\left(A_x\frac{\partial}{\partial y}-A_y\frac{\partial}{\partial x}\right)y\right] +\dots\\ &=-2A_x\,\mathbf{i}-2A_y\,\mathbf{j}-2A_z\,\mathbf{k}\\ &=-2\mathbf{A}. \end{align*}


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