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Short and sweet.

How does one calculate a directional vector in 3 dimensions by knowing the magnitude of the vector and the rotations about both the x and y axis?

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Check out spherical coordinates, specifically SS 3.1. –  Daryl Dec 15 '12 at 20:59
    
Ok so I read that (thanks for responding :D), and I am a bit confused as to what they are saying theta and alpha are in that section. Actually, a lot confused. Where do the rotations about the axis come into account anywhere in there? Additionally, do I even have the information I need to use those formulas? –  MrDoctorProfessorTyler Dec 15 '12 at 21:12
    
What apecifically do you mean by rotations? You would usually rotate a known vector by an angle about an axis? By rotation, do you mean angle? –  Daryl Dec 15 '12 at 21:44

1 Answer 1

Making the assumption of my comment, that your rotations are the angle between the $x-$ and $y-$axes, you can find the vector with the following hints.

Hints Let the vector being determined be $\vec{u}=[u_x,u_y,u_z]^T$.

  1. It follows that $\|\vec{u}\|^2=u_x^2+u_y^2+u_z^2$.
  2. You also have $\vec{u}\cdot\vec{i}=u_x=\|\vec{u}\|\cos\left(\theta_x\right)$.
  3. Furthermore, you have $\vec{u}\cdot\vec{j}=u_y=\|\vec{u}\|\cos\left(\theta_y\right)$.

You now have 3 equations with 3 unknowns, as $\|\vec{u}\|,\,\theta_x$ and $\theta_y$ are known.

Note: This vector will not be unique.

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