Divergence of vector multiplied to another vector? I'm working through an Electrodynamics book on my own, and can't understand the solution manual's logic on problem. Here's the problem:
Evaluate $(v_a \cdot \nabla)v_b        $
$v_a = x^2 \hat x  +3xz^2\hat y - 2xz\hat z    $
$v_b = xy\hat x + 2yz\hat y + 3zx\hat z$
I tried to to this by taking the divergence of $v_a$ and then multiplying it as a scalar to $v_b$. The solution's manual takes a  different approach, instead takes the partial derivative of the vector $v_b$'s components for each variable, multiplies each new vector to the corresponding component of $v_a$, and then simplifies. 
My question, what am I missing here? I guess I have several questions:
1. How can you do this problem the way the solutions manual does? I don't understand how to apply divergence that way.
2. Is the way I'm doing it wrong? Or are they equivalent?
3. If they are equivalent, why do it this way which seems more complex to me?
Thanks in advance. I've attached a picture of how the solutions manual works it out for reference. 
 A: No. They are not equivalent. $v_a \cdot \nabla$ is an operator while $\nabla \cdot v_a$ is a scalar. The solution manual is correct.
In general
$$v\cdot \nabla=v_x\frac{\partial}{\partial x}+v_y\frac{\partial}{\partial y}+v_z\frac{\partial}{\partial z}$$
which is not the same as
$$\nabla\cdot v=\frac{\partial v_x}{\partial x}+\frac{\partial v_y}{\partial y}+\frac{\partial v_z}{\partial z}$$
One can see that the order of terms matters. For example, in $v_x\frac{\partial}{\partial x}$, the differential operator does not act on $v_x$ but on what follows it. While in $\frac{\partial}{\partial x}v_x$ it acts on $v_x$ only.
Why does the solutions manual apply each partial derivative to every component of the vector instead of applying the $x$ partial to only the $x$ component, $y$ to only $y$, etc. the way the del operator is usually applied?
This is because in general a scalar operator $L$ acting on a vector $v$ is defined by
$$L(v)=(Lv_x)\hat{x}+(Lv_y)\hat{y}+(Lv_z)\hat{z}$$
A: The issue is that you are interpreting $v_a\cdot \nabla$ as $\nabla\cdot v_a$; the former is a differential operator, but the latter is a scalar. In particular,
$$
v_a\cdot\nabla = (x^2 \hat x  +3xz^2\hat y - 2xz\hat z)\cdot(\frac{\partial}{\partial x}\hat x+\frac{\partial}{\partial y}\hat y+\frac{\partial}{\partial z}\hat z) \\
=x^2\frac{\partial}{\partial x}+3xz^2\frac{\partial}{\partial y}-2xz\frac{\partial}{\partial z}.
$$
Once you've settled that, you just go on and apply the above operator to $v_b$.
