Is it possible for a 3D vector to be drawn with the direction angles of $\alpha=45^\circ$ and $\beta=45^\circ$ ? if yes what is measure of $\gamma^\circ$?
I calculated $\cos^2(45^\circ )+\cos^2(45^\circ )+\cos^2(\gamma) = 1$ so $\gamma= \pi/2$. I am right?
The $\alpha$, $\beta$ and $\gamma$ are angles of the 3D vector against the x, y and z axis of the coordinate system. That does not mean that they are independent variables for defining a 3D vector in the space. For example, you cannot have a 3D vector that has 90 degree to all the 3 axes and you cannot have a 3D vector that has 0 degree to any 2 axes either. For any given angle (or angles), if you cannot find a solution for other angle(s), it simply means such a 3D vector does not exist.