If $\vec{a}$ is a constant vector and $ϕ$ is a scalar field then what is $(\vec{a}\cdot\vec{∇}) ϕ$ equal to?

Question

I am confused about which solution of the following question is correct:

If $\vec{a}$ is a constant vector and $ϕ$ is a scalar field then $(\vec{a}\cdot\vec{∇}) ϕ$ is equal to: $0$ or $\vec{a}\cdot\vec{∇}ϕ$?

Consider $\vec{a} = a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\space$ and $\vec{∇} = \frac{\partial}{\partial x} \hat{i} + \frac{\partial}{\partial y} \hat{j} + \frac{\partial}{\partial z} \hat{k}$

Solution 1:

Since $\vec{a}$ is a constant vector; thus $\frac{\partial}{\partial x}a_1$ is $0$. So, $\vec{a}\cdot\vec{∇}$ = $\vec{∇}\cdot\vec{a} = 0$

Solution 2:

$\vec{a}\cdot\vec{∇} = a_1 \frac{\partial}{\partial x} + a_2 \frac{\partial}{\partial y} + a_3 \frac{\partial}{\partial z}$

$(\vec{a}\cdot\vec{∇}) ϕ = a_1 \frac{\partial}{\partial x} ϕ + a_2 \frac{\partial}{\partial y} ϕ + a_3 \frac{\partial}{\partial z} ϕ = \vec{a}\cdot\vec{∇}ϕ$

Which one is correct or wrong and why?

Thanks in advance...

Answer 1

The second solution is the good one.

You can also see it that way:

$(\vec{a}\cdot\vec{∇}) ϕ=\vec{a}\cdot(\vec{∇} ϕ)$

$\vec{∇}$ is not a vector, it is an operator that applies on a scalar field. It does not apply on a vector as well.

Answer 2

Solution 2 is correct. The operator should act on the latter $\phi$, not on the constant vector before it.