Geometric interpretation of $\vec v \cdot \operatorname{curl} \vec v = 0$ In this Math.SE question, I asked a question to which I was hoping to get a simple intuitive answer. Instead I received an otherwise perfectly correct but very mathematical one. Obviously, the words geometric and intuitive have very different meaning in that forum from what they mean to me... Can somebody help here?
There is a family of surfaces orthogonal to the vector field $\vec v \in \mathbb R^3$ iff $\vec v \cdot  \operatorname{curl} \vec v = 0 $. Now the necessity part is trivial, but the proof of sufficiency I have seen in physics textbooks, e.g. Kestin: Thermodynamics, is kind of mindless integration similar to the one usually offered to prove Poincare's lemma as in Flanders, or just asserted as in Born&Wolf: Optics. Is there an intuitive and geometric interpretation of this condition that would make it obvious why its sufficiency must be true?
 A: Edit: Now that I've read the original post more closely, I'm not sure that this answer actually addresses your question. Let me know if it doesn't and I can perhaps try again.
Remember: curl is only defined for vector fields and not for lone, individual vectors. With this in mind, curl is sometimes said to quantify the "vorticity" of a vector field (i.e. how swirly it is at a given point). Intuitively, if you imagine putting a little, tiny paddle-wheel into the vector field and think of the field as a flowing stream and the magnitudes of the nearby vectors as the strength of the current at that point, then you can tell whether the curl of the field is zero at the point described by the center of the paddle-wheel by figuring out whether or not the paddle-wheel is caused to turn by the current. Moreover, the faster the paddle-wheel turns the greater the vorticity and, thus, the greater the curl.
Now a spinning wheel lives in a plane, yes? That is to say, uniform rotation about a point is a fundamentally two-dimensional activity. Once we've observed this fact, we can easily see that the best way to define a plane is to give the components of its normal vector. Since the plane is flat and uniform out to infinity, the normal vector does not change and, more importantly, the normal vector is perpendicular to the plane at all points.
So now we see that the curl is giving us the information of a plane which we encode and compactify into a single vector. We do this again at every point in space and we get a new vector field that is the curl of the original vector field. This new vector vield has the interesting property that at any given point, the vector there is perpendicular to the vector at that point in the original field. 
And the dot product of perpendicular vectors is always zero. Intuitive enough for you?
So to summarize, the curl of a vector field at a point gives you a vector that is normal to the plane of rotation of the original vector field. The magnitude of the new vector tells you how curly it is, while the direction tells you which way the original field is spinning. By the way, don't forget the right-hand rule.
A: I) Consider a Riemannian 3-manifold $(M,g)$ with a non-vanishing vector field $V\in \Gamma(TM)$. Define a 2-dimensional distribution as
$$\tag{1} \Delta ~:=~ {\rm ran} (V)^{\perp}~\subseteq~ TM.$$
This definition (1) and the integrability condition
$$\tag{2} V \cdot \nabla\times V ~=~ 0$$ 
for $\Delta$ relies on a choice of dot product/metric tensor $g$. 
II) A more fundamental approach (that doesn't depend of a choice of metric) is
the following. Consider a 3-manifold $M$ with a non-vanishing$^1$ co-vector field/one-form $\vartheta\in \Gamma(T^{\ast}M)$. Define a 2-dimensional distribution as
$$\tag{3} \Delta ~:=~ {\rm ker}(\vartheta)~\subseteq~ TM.$$
The integrability condition for $\Delta$ reads
$$\tag{4} \vartheta \wedge \omega ~=~ 0 ,\qquad \omega~:=~\mathrm{d}\vartheta.$$
We will use eqs. (3)-(4) from now on [as opposed to eqs. (1)-(2)].
III) The integrability condition (4) is equivalent to
$$\tag{5} \exists \text{ locally defined gauge potential } A:~~ D\vartheta ~:=~ \mathrm{d}\vartheta -A \wedge \vartheta~=~0. $$
Note that eqs. (3)-(5) are covariant under a gauge transformation 
$$\tag{6} \vartheta~\longrightarrow ~ \vartheta^{\prime}~=~e^{f}\vartheta, \qquad A ~\longrightarrow ~A^{\prime}~=~A+ \mathrm{d} f, $$
where $f\in C^{\infty}(M)$ is a gauge function. Eq. (5) is also covariant under 
a shift symmetry
$$\tag{7} A~\longrightarrow ~A^{\prime}~=~A+ g\vartheta, $$
where $g\in C^{\infty}(M)$ is a function.
IV) One may show that the integrability condition (4) is equivalent to the involutive property of vector fields
$$\tag{8} \forall X,Y \text{ vector fields }\in~\Delta: ~~[X,Y]~\in~\Delta, $$
cf. Frobenius' theorem.
V) Moreover the integrability condition (4) implies that there is a 2-dimensional foliation of $M$, i.e. there exists an atlas of adapted local coordinate systems $(x,y,z)$ so that the third-component $\phi^3$ of a local coordinate transformation 
$$\tag{9}(x,y,z) ~\longrightarrow~ (x^{\prime},y^{\prime},z^{\prime})~=~(\phi^1(x,y,z),\phi^2(x,y,z),\phi^3(z)) $$
only depends on the third coordinate $z$. In these adapted coordinate systems, we have
$$\tag{10} \frac{\partial}{\partial x},  \frac{\partial}{\partial y} ~\in~\Delta,\qquad  \text{and} \qquad\vartheta~\propto~\mathrm{d}z.$$
VI) Further equivalent formulations of the integrability condition (4):
$$\tag{11}   \exists \text{ local integrating factor } e^f:~~ e^f\vartheta \text{ is closed}. $$
Equivalently, there exists a local gauge (6), where $\Delta={\rm ker}(\vartheta^{\prime})$, such that$^2$
$$\tag{12}  \vartheta^{\prime}\text{ is closed}. $$
By Poincare lemma, we have
$$\tag{13}  \exists \text{ local coordinate system } (x,y,z) :~~ \vartheta^{\prime}~=~\mathrm{d}z.$$
--
$^1$ If we instead of imposing that the 2-form $\omega:=\mathrm{d}\vartheta$ is non-vanishing (rather than the 1-form $\vartheta$ is non-vanishing), then $\omega$ must have constant rank 2, and we can use the Darboux' Theorem to conclude that  
$$\tag{14}  \exists \text{ local coordinate system } (x,y,z) :~~ \vartheta~=~y\mathrm{d}z.$$
$^2$ In particular, there exists one-forms $\vartheta$ in 3D which doesn't have an integrating factor.
