# How to computer two Euler Angles between two vectors

I have looked on here for an answer to my question however the answers I found were how to find the actual angle and not broken down into theida and phi so here is my question:

If I have two vectors a and b how can I calculate the two Euler angles between them?

I do know the total angle is found by:

=arccos(Re(a⋅b)∥a∥∥b∥)


What I am looking for is this scenario:

where the red line is vector a and the purple line is vector b. The angle between them can be broke up onto the two planes. Those two angles are what I am trying to find.

• These are not Euler angles. And your theida is probably our theta. This said, drop one of the component (to achieve a projection) and compute the scalar product in 2D. – Yves Daoust Nov 7 '17 at 17:59
• @YvesDaoust I have posted an answer below. I think I follow the projection. Can you please let me know if you agree with what I posted please – Eric F Nov 7 '17 at 18:52
• I still disagree with your theida. – Yves Daoust Nov 7 '17 at 18:57
• @YvesDaoust What do you see wrong with my theida? It should just be the i and k components only right? I thought that is what I did – Eric F Nov 7 '17 at 19:01
• I tried to tell you but you did not understand. thinkbabynames.com/meaning/0/Theida – Yves Daoust Nov 7 '17 at 19:24

So given vector A = (2i,4j,5k) and vector B = (1i,3j,8k)

my theida (as shown in image above) would be:

cos(theida) = ((Ai,Ak) ⋅ (Bi,Bk)) / (||Ai,Ak|| * ||Bi,Bk||)


Therefore,

cos(theida) = (2*1) + (5*8) / (sqrt(2^2 + 5^2) * sqrt(1^2+8^2))

theida = acos(42/(sqrt(29)*sqrt(65)))

theida = 14.68 deg

and similarly:

cos(phi) = ((Ai,Aj) ⋅ (Bi,Bj)) / (||Ai,Aj|| * ||Bi,Bj||)


Therefore,

cos(phi) = (2*1) + (4*3) / (sqrt(2^2 + 4^2) * sqrt(1^2+3^2))

phi= acos(14/(sqrt(20)*sqrt(10)))

phi= 8.13 deg