Air drag on a projectile A projectile moving through two spatial dimensions has a resolution of two velocities, $$\vec i(V_{0}\cos\theta)+\vec j(V_0\sin\theta -gt)$$ each of which should have their own air drag(?) But also  orthogonal to each other, just as the velocoties are.  Is it correct to sum the two air perpendicular air drags as:
$$\sum F_{drag}= \vec i(bV_0\cos\theta)+\vec j(bV_0\sin\theta-gtb)$$
where $b$ is supposed to represent the product of the drag coefficient, fluid density, etc. (using the Stokes model of proportionality)
seems like sensible reasoning (even if I do say so myself hehe)  but i still unsure. googling could not reveal :( so I beseech thee. 
 A: Your suggestion is correct in someways but incorrect in others. A better representation of the drag force would be
$$\vec{F}_{drag} = \alpha (V_x(t)\space\vec{i}\space +\space V_y(t)\space\vec{j})$$
Where $V_{x,y}$ are simply the time dependent velocities of the projectile. You will have to use this to set up differential equations in both the $x$ and $y$ directions in this case, and solve them subject to suitable boundary conditions (i.e. $V_x(0)=V_0)$ or something to that effect.
One of the issues with your formulation is that you have included the change in velocity due to gravity in your y-velocity, which makes it seems as though you believe your formula works at all times $t$. However, you have to remember that the drag force will alter the velocities as a function of time, so this doesn't work. Explicitly, what this means is that your initial two dimensional velocity decomposition is invalid.
It's also worth noting that the drag force is often proportional to velocity squared, instead of velocity, so maybe you should look at whether or not this is a more appropriate model for your system. In this case the formula above is the same, but you just change it to squared velocities, and the constant $\alpha$ will be different.
A: I recommend treating components as a way to calculate, not a way to think. The linear drag force is $-b\vec v$: proportional to, and opposite to, the velocity. If you expand that into components, you get what you wrote, except that you’re missing a negative sign. Thinking that there are “two drag forces” is a confusing approach.
A: The drag force $\vec{F}_d$ components are:
$$\vec{F}_d=-f(v)\,\vec{e}_v=-f(v)\,\frac{\vec{v}}{v}$$
where $f(v)$ is the drag law and v is the magnitude of the velocity vector $\vec v$
I) linear drag law : $f(v)=b\,v$
$$\vec{F}_d=-f(v)\frac{\vec{v}}{v}=-b\,\vec{v}=-b\,\begin{bmatrix}
  v_x \\
  v_y \\
\end{bmatrix}$$
II) quadratic  drag law : $f(v)=c\,v^2$
$$\vec{F}_d=-f(v)\frac{\vec{v}}{v}=-c\,v\,\vec{v}=-c\,\sqrt{v_x^2+v_y^2}\,\begin{bmatrix}
  v_x \\
  v_y \\
\end{bmatrix}$$
