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:


What I am looking for is this scenario:

enter image description here

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.

  • $\begingroup$ 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. $\endgroup$ – Yves Daoust Nov 7 '17 at 17:59
  • $\begingroup$ @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 $\endgroup$ – Eric F Nov 7 '17 at 18:52
  • $\begingroup$ I still disagree with your theida. $\endgroup$ – Yves Daoust Nov 7 '17 at 18:57
  • $\begingroup$ @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 $\endgroup$ – Eric F Nov 7 '17 at 19:01
  • $\begingroup$ I tried to tell you but you did not understand. thinkbabynames.com/meaning/0/Theida $\endgroup$ – 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||)


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||)


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


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.