1
$\begingroup$

I have a tricky question and don't know if you can solve this one:

Given:

  • I have a vector $\vec{A}$.
  • I have the angle between vector $\vec{B}$ and the projection of $\vec{B}$ on the $xz$-plane.
  • I know that $\vec{B}$ is perpendicular to $\vec{A}$.
  • I know the length of $\vec{B}$ is $1$.

How to calculate the angle between the projection of $\vec{A}$ and the projection of $\vec{B}$ in the $xz$-plane, in function of the angle between $\vec{A}$ and the $xz$-plane?

I'm curious if someone can give a decent answer, succes !

$\endgroup$
  • $\begingroup$ looks like curiosity, smells like homework...What have you tried? $\endgroup$ – Paul Feb 28 '17 at 16:05
  • $\begingroup$ It's not really homework, it was for a bachelor thesis assignment :p . We have to program a virtual Drone in our own 3D testbed. $\endgroup$ – user6078387 May 8 '17 at 13:18
  • $\begingroup$ Anyway, we used this as a solution: dropbox.com/s/mqkfni4916fz9gr/… There's a set of possible solutions, but we had other information that made it possible to know which solution we needed. $\endgroup$ – user6078387 May 8 '17 at 13:27
1
$\begingroup$

I'm going to slightly change the problem: rather than have the angles & projections relative to the $xz$-plane, I'm going to do them relative to the $xy$-plane. This is really just a relabeling of the axes, but it allows us to use spherical polar coordinates for the vectors. In particular, we have $$ \vec{A} = |\vec{A}| \left[ (\sin \phi_A \cos \theta_A) \hat{x} + (\sin \phi_A \sin \theta_A) \hat{y} + (\cos \phi_A) \hat{z} \right] $$ $$ \vec{B} = (\sin \phi_B \cos \theta_B) \hat{x} + (\sin \phi_B \sin \theta_B) \hat{y} + (\cos \phi_B) \hat{z} $$ (note that $|\vec{B}| = 1$.) The angle between $\vec{A}$ and its projection on the $xy$-plane is then $\alpha_A = \frac{\pi}{2} - \phi_A$, and the angle between $\vec{B}$ and its projection is $\alpha_B = \frac{\pi}{2} - \phi_B$. WLOG we can choose the projection of $\vec{B}$ into the $xy$-plane to lie on the $x$-axis; this means that $\theta_B = 0$, and $\theta_A$ will be the angle between the vectors' projections into the $xy$-plane. Thus, our vectors are now $$ \vec{A} = |\vec{A}| \left[ (\cos \alpha_A \cos \theta_A) \hat{x} + (\cos \alpha_A \sin \theta_A) \hat{y} + (\sin \alpha_A) \hat{z} \right] $$ $$ \vec{B} = (\cos \alpha_B) \hat{x} + (\sin \alpha_B) \hat{z} $$

We also know that $\vec{A}$ and $\vec{B}$ are perpendicular, so we must have $$ 0 = \vec{A} \cdot \vec{B} = |\vec{A}| \left[ \cos \alpha_A \cos \theta_A \cos \alpha_B + \sin \alpha_A \sin \alpha_B \right] $$ which then implies that $$ \boxed{ \cos \theta_A = - \frac{ \sin \alpha_A \sin \alpha_B }{ \cos \alpha_A \cos \alpha_B} = - \tan \alpha_A \tan \alpha_B.} $$

Note that the right-hand side of this equation will not always be between -1 and 1 for certain values of $\alpha_A$ and $\alpha_B$. This is because the angles $\alpha_A$ and $\alpha_B$ cannot be freely specified and still satisfy all of above constraints, particularly the orthogonality constraint. For example, if $\alpha_A = \alpha_B = 75°$, then the angle between them cannot be greater than $30°$, and the orthogonality constraint cannot be satisfied.

$\endgroup$
  • $\begingroup$ Thanks for the answer, this was a very different method of solving the problem, but better than mine :) $\endgroup$ – user6078387 Feb 28 '17 at 17:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.