Interpretation of the curl of a vector field Let us assume the curl of a vector field is 
$$ P=(xy)(a_x)+ (y z) (a_y) +(z x) (a_z) $$
Where $ a_x, a_y, a_z $ are unit vectors along x y and z .
Then is the curl at a point in the field the "curlyness" of the field along a plane that is perpendicular to the vector P at that point?
 A: A nice definition, usually not shown for some reason (maybe because you need to shuffle some vector identites around, see Joos p.31), is $$\text{curl }\mathbf v:=\lim\frac{1}{\Delta\tau}\oint \mathrm d \mathbf S \times \mathbf v.$$
Also, because $\nabla\cdot\mathbf v:=\lim\frac{1}{\Delta\tau}\oint \mathrm d \mathbf S \cdot \mathbf v.$ and $\nabla u:=\lim\frac{1}{\Delta\tau}\oint \mathrm d \mathbf S u$, we can say $$\nabla:=\lim\frac{1}{\Delta\tau}\oint \mathrm d \mathbf S$$
which is a curious side-note. 
Now, a divergence measures how much a vector field 'sticks out of' an infinitesimal surface by integrating $\mathrm d \mathbf S\cdot\mathrm v$. If these contributions cancel each other out, nothing 'sticks out'. We can immediately visualize a field that 'sticks out' maximally - it has to be parallel to $\mathrm d\mathbf S$ on the surface of lets say a tiny sphere. This is achieved by any spherically symmetric source term! Divergence is simply infinitesimal flux. The gradient does the same thing but for scalar fields, and it's easy to visualize this as a directional derivative.
The curl similarly measures not how parallel the field is to the normals of the surface, but how perpendicular it is. To achieve maximal curl we want a field that is everywhere perpendicular to the surface in the same sense. If our surface is actually one dimensional, a circle, the circulation is maximal if the field lines also go in circles. Note that an analogous construction is prohibited on a two dimensional surface by the famous hairy ball theorem. 
Therefore, the curl indeed measures how closely the field lines follow circles, infinitesimally. 
A: Suppose $\bf F$ is a vector field and $C$ is some loop in space. Imagine the field is acting on a particle that is forced to remain on the loop $C$. At any given point $\bf r$ on the loop, with infinitessimal tangential displacement ${\rm d}{\bf r}$ along the curve, the force $\bf F$ at that point can be decomposed into two components: a parallel and a perpendicular component. Imagine the perpendicular component is somehow thrown out (since the particle is stuck in the contour, any force to move it out of the contour is cancelled out), and so what remains to act on the particle is the parallel component, which has magnitude ${\bf F}\cdot{\rm d}{\bf r}$. If we integrate this over the whole loop, we get the circulation:
$${\rm circulation}=\oint_C{\bf F}\cdot {\rm d}{\bf r}.$$
As we let $C$ be the nearly-circular boundary of a small, nearly-flat discus region of space and go about shrinking this cap, the circulation of $\bf F$ around $C$ shrinks too - it is proportional to the area enclosed by $C$. What is the constant of proportionality? If we consider circulation per area in the limit, we get the circulation density of $\bf F$ at the given point, 
$${\rm circulation~density}=\lim_{A\to0}\frac{1}{A}\oint_C {\bf F}\cdot{\rm d}{\bf r}.$$
One may derive an explicit formula for this quantity. As it happens, this quantity changes in a very interesting way relative to $\bf n=$ the normal vector of the discus region at the given point. It turns out to be a dot product ${\rm curl}\cdot{\bf n}$ for a vector field known as curl, given by
$$\begin{array}{ll} {\rm curl} & \displaystyle =\left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}\right){\bf i}+\left(\frac{\partial F_1}{\partial z}-\frac{\partial F_3}{\partial x}\right){\bf j}+\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right){\bf k} \\[5pt] & =``\det\begin{bmatrix}{\bf i} & {\bf j} & {\bf k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_1 & F_2 & F_3\end{bmatrix}" \\[5pt] & =``\nabla\times{\bf F}." \end{array}$$
A: The answer is no. The curl, in general, is not orthogonal to the vector field. In your example, the curl is $(-y,-z,-x)$ which is easy to see not to give a zero scalar product with $P$. However, for two dimensional vector fields it is true, since then $P=(f(x,y),g(x,y),0)$ and the curl is $(0,0,*)\bot P$. Here $*$ is marking a non-important element for the conclusion.
P.S. Look at this nice video for the geometrical interpretation of curls for 2D vector fields.
