What does vector operator for angular momentum measure? Consider the vector operator for angular momentum $\hat L=\hat L_x \vec i +\hat L_y \vec j + \hat L_z \vec k$. 
Does this mean that if we want to measure the angular momentum of a particle in state $\psi$, we take $\hat L$ and let it act on $\psi$ to give us three possible eigenvalues of $\hat L_x$, $\hat L_y$ and $\hat L_z$, which will correspond to the $x,y,z$ components of the angular momentum? 
But since $\hat L_x$, $\hat L_y$ and $\hat L_z$ does not commute, this should not be the meaning of $\hat L$ because if we first make a measurement of $\hat L_x$, the state of the particle will be changed, and $\hat L_y$ should no longer act on the original state $\psi$. What then does the vector operator $\hat L$ gives us? More precisely, what is the measurement this operator $\hat L$ is trying to measure?
 A: You can only measure operators. A vector operator is a set of three operators, which we package into one piece of notation because they transform under rotations in a nice way. You can't measure it all in one go, any more than you can try to measure $\hat{x}$ and $\hat{p}$ at the same time. However, you can measure the components of $\hat{L}$, which are, as you would expect, the components of the angular momentum.
A: The fact that $\hat L_x$, $\hat L_y$ and $\hat L_z$ do not commute means that you cannot have a function which is eigenfunction of the three operators. Your function can only be an eigenfunction of one of the component operators, say $\hat L_z$ (it is the typically chosen one). However, the operator $\hat L^2$ does commute with each of the component operators. So if we have an eigenfunction of $\hat L_z$, it is also an eigenfunction of $\hat L^2$. (But not an eigenfunction of $\hat L_x$ and $\hat L_y$).
Therefore, the answer to your question is that the operator $\hat L$ is used to give us the magnitude of the angular momentum, by operating on the eigenfunction with its square: $\hat L^2$. The eigenvalue is then $l(l+1)ħ^2$, which is the magnitude of the angular momentum squared.
A: I like this question. What you describe is taking:
$${\bf \hat L}|l\rangle = \hat L_x|l\rangle {\bf \hat x} + \hat L_y \hat L_x|l\rangle {\bf \hat y} ... = {\bf \vec L}|l\rangle$$
which would lead to problems. I think an eigenvalue equation for a vector operator should be:
$${\bf \hat L}|l\rangle = \hat L_x|l\rangle {\bf \hat x} + \hat L_y|l\rangle {\bf \hat y} + \hat L_z|l\rangle {\bf \hat z} =  L_x|l\rangle {\bf \hat x} +  L_y|l\rangle {\bf \hat y} + L_z|l\rangle {\bf \hat z} = {\bf \vec L}|l>$$
where
$${\bf \hat L} = -i\hbar (y\frac{\partial}{\partial z} - z\frac{\partial}{\partial y}){\bf \hat x} + (permutations) $$
is the operator and 
$${\bf \vec L} = L_x {\bf \hat x}+ L_y {\bf \hat y} + L_z {\bf \hat z }$$ 
is an ordinary vector.
Now with:
$$ |l\rangle \equiv F_l(x, y, z)$$
the equation:
$$ {\bf \hat L}F_l(x, y, z) = {\bf \vec L}F_l(x, y, z)$$
doesn't have any non-trivial solutions: there are no eigenstates of completely known (non zero) angular momentum.
Regarding your question on "meaning": note that:
$$ L_z = -i\hbar\frac{\partial}{\partial\phi} $$
and eigenstates of that function look like:
$$ F(r, \theta, \phi) = R(r)\Theta(\theta)e^{\pm i m \phi} $$
which means that the function is invariant under rotations of $2\pi/m$.
For a wave-function to be an eigenstate of ${\bf \hat L}$ it would have to invariant under some non-trivial rotation about not just the $z$-axis, but the $x$ and $y$ axes also--which would mean it would have to be invariant under some rotation about any axis.
The only thing I can think of that can do that is a sphere, and this is, of course, the zero angular momentum eigenstate.
A: Basically you are asking does a state exist which is a simultaneous eigenstate of the three components. The answer is no. You can construct a state that is an eigenfunction of $L^2$ and any one component, usually the z-component is chosen. This is due to the commutation relations. If a state exists whose all three components are known then a rotation about any three axis will not change the state. But this would mean that the state would be spherically symmetric which is only possible if $L = 0$.
A: This is right. You cannot have an eigenstate and eigenvalue of this vectorial angular momentum operator $\hat{\mathbf{L}}$. In fact, one doesn't even need the commutation to see that this cannot work. The reasons are purely mathematical. Just consider that given an operator $\hat{O}$, its eigenvalue equation is
$$\hat{O} |\psi\rangle = O |\psi\rangle$$
where $O$ is the eigenvalue, which is generally a complex number, but should be a real number for operators representing physical parameters. Now consider what this should look like for vectorial operator $\hat{\mathbf{L}}$. We should, following the same logic, have it associated with an eigenvalue that is itself a vector, representing a specific angular momentum vector $\mathbf{L}$, giving
$$\hat{\mathbf{L}} |\psi\rangle = \mathbf{L} |\psi\rangle$$
But look at the right hand side. We have a vector multiplying a quantum state. That's not mathematically possible. You can only multiply a quantum state by a complex scalar, because that's the only other operation than addition and taking the inner product that is defined on the Hilbert space. So by that token alone, $\hat{\mathbf{L}}$ cannot have "eigenvalues" in this sense. And this makes sense: the output of applying your $\hat{\mathbf{L}}$ to a quantum state isn't even another quantum state, but instead the "vector of quantum states" (a "meta-vector", perhaps?)
$$\hat{\mathbf{L}} |\psi\rangle = (\hat{L}_x |\psi\rangle) \mathbf{i} + (\hat{L}_y |\psi\rangle) \mathbf{j} + (\hat{L}_z |\psi\rangle) \mathbf{k}$$
Thus even worse, technically your proposed "operator" isn't really a proper operator at all, but a map between two rather different vector spaces, and no such maps can have eigenvalues, only self-maps of the same vector space.
A: 
Does this mean that if we want to measure the angular momentum of a particle in state $ψ$, we take $\hat L$ and let it act on $ψ$ to give us three possible eigenvalues of $\hat L_x$, $\hat L_y$ and $\hat L_z$, which will correspond to the x,y,z components of the angular momentum?

As you know $L_x$, $L_y$ and $L_z$ are incompatible observables, but note that $L^2$ does commute with the three components of the angular momentum, so:
$$[L^2, Lx] = [L^2, Ly] =[L^2, Lz]=0$$
Which more compactly is:
$$[L^2, L] = 0$$
Based on it, it is feasible to find simultaneous eigenstates of $L^2$ and one of the three components of the angular momentum. Using as an example $L_z$ we would get:
$$L^2 f = \lambda f$$
$$L_z f = \lambda' f$$

What is the measurement this operator $\hat L$ is trying to measure?

Because of compatibility reasons, $L^2$ and the components of the angular momentum are worked out separately. 
Actually, to know $\hat L$ you should know simultaneously the three components of the angular momentum which is not the case. The uncertainty principle denies it.
