The direction of the drag force From my reading, the drag force is opposite in direction to the velocity vector, but can't the pushing of the "air particles" in some direction be much tougher than in another, i.e, for some reason, the air is denser in some direction? Wouldn't then $\angle (\mathbf{F_d}, \mathbf{v})$ be not necessarily $\pi$?
 A: In everyday life the word drag tends to mean quadratic drag, and this is a complex process involving turbulent flow. The quadratic drag equation isn't a fundamental one but instead is an approximation. This makes it a poor example for your question.
It's easier to see this if you consider just the force when you make a fluid flow as this is nice and simple. For the fluid flow we use a strain rate $\dot\gamma$ rather then just a velocity, and if we make the fluid flow at this rate then we get a stress given by:
$$ \tau = \mu \dot\gamma $$
where $\mu$ is the viscosity of the liquid. For simple fluids the viscosity is just a number and the force is always opposite to the direction of motion as you describe. However for non-Newtonian fluids it is possible for the viscosity to be different in different directions. In that case we have to replace the viscosity by the viscous stress tensor.
This isn't as complicated as it sounds. A tensor is just something we can use to relate two vectors e.g.
$$ \mathbf a = \mathbf T \mathbf b $$
where $\mathbf a$ and $\mathbf b$ are vectors and $\mathbf T$ is the tensor. For the simple fluid $\mathbf T$ is just the viscosity, which is a rank $0$ tensor, and for more complicated fluids $\mathbf T$ would be a second rank tensor that is typically written as a matrix.
Anyhow the point of all this is that when the stress and strain are related by a second rank tensor the force and velocity are not necessarily colinear and the angle between them can be different from $\pi$. Nematic liquid crystals would be an example of a fluid like this.
