Euler-Lagrange equation in vector form If we have a Lagrangian which is a function of vectors $r$ and $r'$ where $r$ is the position with an $x$, $y$ and $z$ component and $r'$ is the time derivative of $r$ are we able to express the euler-lagrange equation as follows:
$$
\nabla_{r} \mathcal{L}-\frac{d}{dt}(\nabla_{r'} \mathcal{L})=0,
$$
where
$$\nabla_{r}=\left[\begin{array}{c} \dfrac{\partial }{\partial x}\\ \dfrac{\partial }{\partial y}\\\dfrac{\partial }{\partial z}\\ \end{array}\right].$$
I think this is equivalent to the euler-lagrange equation below, but I haven't seen it used anywhere, so dunno if im missing something stupid.
$$
\frac{\partial\mathcal{L}}{\partial r_{i}}-\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial r'_{i}}=0
$$
for $i$ belonging to $(x,y,z)$.
 A: These expressions are not just equivalent.
They are one and the same !
The very definition of $\nabla_r A $ is the vector of components ${\frac {\partial A} {\partial r_i}}$
A: You can write it, but in the "space of generalized coordinates", i.e. the configuration space, since Lagrange equation involves generalized coordinates and not (only) coordinates describing the physical space.
The difference can be subtle, but remember that in general generalized coordinates are not a full set of coordinates describing position in the physical space.
Remember where Lagrange equations come from, i.e. stationary-action principle, that involves variation of the action $S$,
$S = \displaystyle \int_{t_0}^{t_1} L(\dot{q}_k, q_k, t) \, dt $
and thus partial derivatives $\partial L /\partial q_k$,  $\partial L /\partial \dot{q}_k$, without the gradient of any physical quantity in the physical space.
If you really need a gradient, you can think at the generalized coordinates as the orthogonal coordinates of the configuration space (the space of generalized coordinates and their time derivatives), s.t. the gradient in this space becomes
$\nabla_{\mathbf{q}} \, f(q_k, \dot{q}_k,t) = \dfrac{\partial f}{\partial q_k} \mathbf{\hat{q}}_k + \dfrac{\partial f}{\partial \dot{q}_k} \mathbf{\hat{\dot{q}}}_k$,
but this is not the gradient you need/want.
A: The component form of Euler-Lagrange equation works for any cooordinate of any coordinate system while the vector form of Euler-Lagrange equation works only in Cartesian coordinate system.
The vector form should be interpreted as a shorthand for the three components of Euler-Lagrange equations coorresponding to three coordinates. If you interprete $\nabla_r$ as the gradient operator, for example, in the cylindrical coordinate system, you can not recover the component form of Euler-Lagrange equation.
