# Projecting an angle from one plane to another plane.

Consider I have two intersecting planes with an angle ($\theta$). I have two intersecting vectors ($\vec a$ and $\vec b$) on one of the planes that make an angle ($\gamma$). If I project these two vectors onto the other plane, what the projected angle ($\alpha$) between the projected vectors ($\vec a \prime$ and $\vec b \prime$) would be (as a function of the other angles)?

To make it simple, consider:

1. Vectors intersect on the hinge of the two planes.
2. Vectors are symmetric to the perpendicular line to the hinge.

Regards,

• A straightforward approach is to simply project the vectors onto the other plane and compute the angle between their images. Since the projection of a unit circle is an ellipse, you might be able to come up with a formula using the relationship between polar angle and parametric angle (that you can find here), but I expect it’ll be significantly more complicated than the first suggestion. – amd Mar 28 '17 at 22:09
• I made the question more clear and also simpler. – N.Madanian Mar 29 '17 at 8:35

This is how I understand the problem in a Cartesian coordinate system: And this is how I solved it (note that two angles are renamed for this via θ → α, α → β: The following equations are valid:

\begin{align} R_x = R_{xy} \cos \alpha = R_{xz} \cos \beta \\ R_y = R_{xy} \sin \alpha \\ R_z = R_{xz} \sin \beta = R \sin \gamma \\ R_{xy} = R \cos \gamma \end{align}

Putting them all together results in: \begin{align} \tan \gamma = \cos \alpha * \tan \beta \end{align}

So in the notation of the original question (with the angles restated as in the OP) the solution is: \begin{align} \tan \gamma = \cos \theta * \tan \alpha \end{align}

I had a similar problem and thought it might be useful for others, as the equation looks way less daunting and is easily solved for any of the 3 angles.

Here I found the answer. I'm not just good in explaining details but to give you a vision of how I solved it, I actually drew two new planes that each of which included one vector from the initial plane and its proper projection from the second plane. These two new planes intersect at the point where all four vectors and the two initial planes intersect. It's easy to notice that the angle between the new planes is equal to $\alpha$ and they both are perpendicular to the second plane. By applying some extra projections and doing some minor calculations, I finally reached to the answer which is shown below;

$\cos(\alpha/2)=\frac{\cos(\gamma/2)\cos (\theta)}{\sqrt {1-\cos^2(\gamma/2)\sin^2(\theta)}}$

or

$\cos(\gamma/2)=\frac{\cos(\alpha/2)}{\sqrt {\cos^2(\theta)+\cos^2(\alpha/2)\sin^2 (\theta)}}$

We know that for ($\theta \to 0$) $\Rightarrow$ ($\gamma \to \alpha$), since both planes overlap.

And for ($\theta \to \pi/2$) $\Rightarrow$ ($\gamma \to 0$), since both vectors $(\vec a$ and $\vec b$) overlap. This result actually stems from an additional assumption that I made, which was I assumed $\alpha$ to remain constant (since in my actual problem it's constant), though without this assumption the final result should be the same. to open a bit this last concept, note that when ($\theta \to \pi/2$) $\Rightarrow$ ($\alpha \to \pi$) which returns the following result;

$\cos(\gamma/2)=\frac{0}{0}$ $\Rightarrow$ $0\le \gamma=cte \le \pi$

which comes from the fact that $\gamma$ can possess any value independent of the actual angle between the initial plane and its projected plane ($\theta$).