It is given constant $a\in \mathbb R ^3$ and two vector fields:
$b(x)=x$ and $v(x)=\frac{x}{| | x | |^3 } $, $x\in \mathbb R^3$, $x\neq0$ Calculate: $div(a \times b)$ and $div $ $v$.
I know how to calculate $div$ and cross product generally, but I am not sure what here are excatly $a$,$b$ and $v$?
Can I for example write $a$ as $(a_1,a_2,a_3)$? What about $b$ and $v$?
On
$a$ is a constant 3-vector (see the first given) so yes, you can write it as $(a_1,a_2,a_3)$.
$\vec{b}(\vec{x})$ is a function of a 3-vector ( as indicated by the statement that $x\in\Bbb R^3$) that returns a 3-vector value: $\vec{b}((x_1,x_2,x_3)) = (x_1,x_2,x_3)$.
(I have written the vector sign to emphasize the nature of the three-component objects, but of course you could write $a$ rather than $\vec{a}$ and so forth.)
$\vec{v}(\vec{x})$ is a different function of a 3-vector that returns a 3-vector value: $$\vec{v}((x_1,x_2,x_3)) = \left(\frac{x_1}{(x_1^2+x_2^2+x_3^2)^{3/2}},\frac{x_2}{(x_1^2+x_2^2+x_3^2)^{3/2}},\frac{x_3}{(x_1^2+x_2^2+x_3^2)^{3/2}}\right)$$
So for example, $\vec{a}\cdot \vec{b}$ would be a scalar function of a vector argument such that $f(x_1,x_2,x_3) = a_1x_1+a_2x_2+a_3x_3$.
$\mathbf x = (x_1,x_2,x_3)\\ v(\mathbf x) = \frac {\mathbf x}{\|\mathbf x\|^3}\\ \nabla\cdot v(\mathbf x) = (\nabla\cdot\mathbf x)\|\mathbf x\|^{-3} + \mathbf x\cdot\nabla\|\mathbf x\|^{-3}\\ {3}\|\mathbf x\|^{-3} - 3(\mathbf x\cdot \mathbf x)\|\mathbf x\|^{-5}\\ 0$
$a\times b = a_2x_3 - a_3x_2, a_3x_1-a_1x_3, a_1x_2 - a_2 x_1\\ \nabla \cdot (a\times b) = \frac {\partial}{\partial x_1}(a_2x_3 - a_3x_2)+\frac {\partial}{\partial x_2}(a_3x_1-a_1x_3)+\frac {\partial}{\partial x_3}(a_1x_2 - a_2 x_1) = 0$