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I'm having some trouble wrapping my mind around some vector problems. I'm hoping you could please help me to clarify it.

Let's say $\vec{r}=(x_{1},x_{2},…,x_{n})$, $\mathbf r = |\vec{r}|$. How should I correctly interpret $\mathbf r$?

In addition, I am looking for a way to compute $\nabla \cdot \frac{\vec{r}}{\mathbf r^k}$. Could you please walk me through how to do it?

Thank you so much in advance!

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    A single vector isn't a vector space, the span of a vector(s) may be a vector space (if it satisfies the requirements to be one). Your r looks like a position vector in $\mathbb{R}^n$ and so $|\vec{r}|$ may be interpreted as the length of the vector. – 2017-02-09

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First, $\bf r$, which is $|r|$, should be interpreted as the length of the vector $r$. So say we have a vector in $\Bbb R^{2}$. Let's pick $(1,2)$. Then $|(1,2)|$ means $\sqrt{(1 - 0)^{2} + (2-0)^{2}}$, which is the distance of the line with one endpoint the origin $(0,0)$ and the other endpoint $(1,2)$. For every point in $\Bbb R^{2}$, you can draw a line connecting $(0,0)$ and that point. The absolute value of the point means the length of this line. If instead of points you are considering $(1,2)$ as a vector, then $|(1,2)|$ can be interpreted as the "magnitude" or length of this vector (since vectors always start from the origin $(0,0)$).

Now, if I have a vector like $v = (2x_{1}, 3x_{2} - x_{2}^{2}, \sin(x_{3}) + 2x_{1})$, then $\nabla \cdot v$ means you apply the $\frac{\partial}{\partial x_{1}}$ operator to the first entry, PLUS the $\frac{\partial}{\partial x_{2}}$ operator applied to the second entry, PLUS the $\frac{\partial}{\partial x_{3}}$ applied to the third entry.

So I have \begin{split}\nabla \cdot v &= \frac{\partial}{\partial x_{1}}(2x_{1}) + \frac{\partial}{\partial x_{2}}(3x_{2} - x_{2}^{2}) + \frac{\partial}{\partial x_{3}}(\sin(x_{3}) + 2x_{1}) \\ &= 2 + 3 - 2x_{2} + \cos(x_{3}) \\ &= 5 - 2x_{2} + \cos(x_{3}). \end{split}

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    Yeah I got that part. But there is the $\mathbf r^k$, and I don't know how to deal with that when doing the calculation. – 2017-02-09
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    @f.w You do know! I added some more explanation about the absolute value. I will start typing a comment now to explain further. Hold tight. – 2017-02-09
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    @f.w So, since $r = (x_{1}, x_{2}, \dots, x_{n})$, and $|r| = \sqrt{x_{1}^{2} + x_{2}^{2} + \dots + x_{n}^{2}}$, which is just a number, then $|r|^{k} = (x_{1}^{2} + x_{2}^{2} + \dots + x_{n}^{2})^{\frac{k}{2}}$, right? So $\frac{1}{|r|^{k}}r = (\frac{x_{1}}{(x_{1}^{2} + x_{2}^{2} + \dots + x_{n}^{2})^{\frac{k}{2}}}, \frac{x_{2}}{(x_{1}^{2} + x_{2}^{2} + \dots + x_{n}^{2})^{\frac{k}{2}}}, \dots, \frac{x_{n}}{(x_{1}^{2} + x_{2}^{2} + \dots + x_{n}^{2})^{\frac{k}{2}}})$. Yes, it's ugly, but you *can* differentiate it using the method I talked about in my answer. Good luck! – 2017-02-09