How do I find a 3D vector when I'm given two of three direction angles? On an assignment, I've been given this task:

Find a vector that has direction angles $α = 75°$ and $β = 128°$.
Fully explain your method. Is there more than one possible answer?
Why?
What do they have in common?

I used to fact that $cos^2⁡α+cos^2⁡β+cos^2⁡γ=1$ to find that:
$γ=\cos^{-1}(±\sqrt{1-cos^2⁡ 75°-cos^2⁡ 128°})$
So I have the two possible exact values of the third direction angle, but it doesn't really help because I have no idea how to use this information to find a vector that creates these direction angles.
So I'm wondering what next, or possibly where to start from if what I did above wasn't necessary?
 A: I figured it out:
The answer lies in that the direction of a vector does not restrict its magnitude, in other words a vector with those angles can have any magnitude. Since I know the angles at this point, if I assign a magnitude, let's say 10, then I have everything I need to find the coordinates. This is true because we know that $$\cos\alpha=\frac{\mathbf{u}_{x}}{|\vec{u}|}$$
And I now know the value of both $|\vec u|$ and $\alpha$, meaning I can do this:
$$\cos75=\frac{x}{10}$$
$$x=10\cos75$$
Similarly, I can do the same with $\beta$ for $y$. With $\gamma$ I simply choose to use either 
$\cos^{-1}(\sqrt{1-cos^2⁡ 75°-cos^2⁡ 128°})$ 
or 
$\cos^{-1}(-\sqrt{1-cos^2⁡ 75°-cos^2⁡ 128°})$ 
as the angle for finding $z$, and do the same thing there. The corresponding values for $x, y, z$ then will be the coordinates for a vector with direction angles $α = 75°$ and $β = 128°$.
A: You're solving for another angle, but what you really desire are lengths (x, y, z coordinates).  Draw a rough picture of what your vector should look like using the two given angles.  Then draw some triangles in your 3 dimensional picture and use basic trigonometry to solve for side lengths that hopefully end up giving you the x, y, z components of your desired vector.  Basically, draw lots of pictures.
