Q:

Proving certain properties using curl, divergence, and gradient

Sorry about the confusing title; it was hard to phrase this.
For some context, I have an exam very soon and these are mock questions, my teacher has not provided any solutions to theses. So I am working my way through the questions.
I think I have solved part a) although this is probably relatively easy, although I know I have the correct idea in solving this I am unsure whether my working out is completely accurate.
For b) and c) I am confused of how to apply the cross product to two vectors in the form of $\bf{m}$ and $\bf{r}$. Whilst I am comfortable showing that vector fields are conservative and calculating the curl and divergence of a single vector, i.e. only $\bf{r}$, I have not came across how to apply this for two vectors.
Any help would be most welcome. Please note I am not asking for a whole answer, rather how to better apply my working out for part a), and the format or a starting point for b) and c).

If $\bf{m}$ is a constant vector and $\mathbf{r}=x\hspace{0.5mm}\mathbf{i}+y\hspace{0.5mm}\mathbf{j}+z\hspace{0.5mm}\mathbf{k}$, show that
$\hspace{4mm}$ a) $\nabla(\mathbf{m}\cdot\mathbf{r})=\mathbf{m}$
$\hspace{4mm}$ b) $\nabla\cdot(\mathbf{m}\times\mathbf{r})=0$
$\hspace{4mm}$ c) $\nabla\times(\mathbf{m}\times\mathbf{r})=2\mathbf{m}$

My work:
$\hspace{8mm}\nabla\,\times\implies$ Matrix
$\hspace{8mm}\nabla\,\cdot\implies$ Diff
$\mathbf{r}=x\hspace{0.5mm}\mathbf{i}+y\hspace{0.5mm}\mathbf{j}+z\hspace{0.5mm}\mathbf{k}$
$\mathbf{m}=a\hspace{0.5mm}\mathbf{i}+b\hspace{0.5mm}\mathbf{j}+c\hspace{0.5mm}\mathbf{k}$
$\text{a)}\:\:\nabla(\mathbf{m}\cdot\mathbf{r})=\dfrac{\partial}{\partial x}(ax)+\dfrac{\partial}{\partial y}(by)+\dfrac{\partial}{\partial z}(cz)=a+b+c=\mathbf{m}$
$\text{b)}\:\:(\mathbf{m}\times\mathbf{r})=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\a&b&c\\x&y&z\end{vmatrix}$
$\hspace{2.3cm}=(bz-cy)\hspace{0.5mm}\mathbf{i}+(cx-az)\hspace{0.5mm}\mathbf{j}+(ay-bx)\hspace{0.5mm}\mathbf{k}$
$\hspace{7mm}\nabla\cdot(\mathbf{m}\times\mathbf{r})=\dfrac{\partial}{\partial x}(bz-cy)\hspace{0.5mm}\mathbf{i}+\dfrac{\partial}{\partial y}(cx-az)\hspace{0.5mm}\mathbf{j}+\dfrac{\partial}{\partial z}(ay-bx)\hspace{0.5mm}\mathbf{k}$
Thanks if you read all this.
-Alexis.

A:

You're nearly there for part b).  Note, however, that $\nabla \cdot$ denotes the divergence, not the gradient.  So,
$$
\nabla \cdot (\mathbf m \times \mathbf r) = 
\frac{\partial }{\partial x}(bz - cy) +
\frac{\partial }{\partial y}(cx - az) + 
\frac{\partial }{\partial z}(ay - bx)
$$
That is, we should have no $\mathbf {i,j,k}$ in our answer.  Also, all of these partial derivatives are zero since, as you said in the comment, $a,b,c$ are constant.

Part c) is pretty tedious. You can save some effort if you avoid writing out cross products.  In particular, we can write
$$
\nabla \times (\mathbf m \times \mathbf r) = 
\mathbf i \times \frac{\partial }{\partial x}(\mathbf m \times \mathbf r) + 
\mathbf j \times \frac{\partial }{\partial y}(\mathbf m \times \mathbf r) + 
\mathbf k \times \frac{\partial }{\partial z}(\mathbf m \times \mathbf r)
\\ = 
\mathbf i \times \left(\mathbf m \times \frac{\partial \mathbf r}{\partial x}\right) + 
\mathbf j \times \left(\mathbf m \times \frac{\partial \mathbf r}{\partial y}\right) + 
\mathbf k \times \left(\mathbf m \times \frac{\partial \mathbf r}{\partial z}\right)
$$
and then use the BAC-CAB formula.

