Euler angles equivalence with a single rotation

[sentinel]

imagine we rotate a vector centered at the origin with euler angles alpha,beta,gamma.
now the question is, can we do this rotation by the means of defining a vector N(which its length is 1).and rotating the vector zeta radians counter clockwise around N?
I think it must be possible and I want to know if N is unique(we have only one N for the given euler angles) and I want to find N and zeta in terms of alpha beta and gamma.
thank you.(complete answers really apreciated!GOD bless those who give complete anwers!)
P.S: what I've done to find the answer:I used cartesian coordinates and found the rotatian matrix correspending to a rotation zeta over a given N.>>RA=A'
I found R.A is a column matrix (x,y,z) and A'(x',y',z').and R is a 3x3 matrix which I found in terms of N(nx,ny,nz) and zeta.

 

[mfb]

Do you want to transform a specific, known vector? If you want to express a general rotation with just N and zeta, those will depend on the rotated vector. You can look how this vector transforms under the euler rotations and compare this to a rotation around an arbitrary N. Solving the equations will give an appropriate N.

The transformation $\vec{N} \to -\vec{N}$ and $\zeta \to - \zeta +2 \pi k$ always works, so N is not unique. You can restrict ζ to [0,pi], and N will be unique apart from special cases (ζ=0 and ζ=pi).

 

[sentinel]

thank you sir.
of course we know the vector we want to rotate.its say;A(Ax,Ay,Az) but it doesn't matter I think.for what ever vector, we use we want to know N and zeta which must be functions of only alpha beta gamma(the euler angles correspending to rotation).
I mean imagine we have the rotation matrix (possibly NOT a function of A and only function of alpha beta gamma) and also we know the rotation matrix for the rotaion zeta around an N.(also NOT a function of A) so if we simply put this two matrix equal we should get N and zeta as a function of the euler angles.A is not involved in the answer.
now I want some angel! to write N and zeta as function of the euler angles.
it means around which axis and how much do we need to do a single rotation to get an euler rotation of alpha beta gamma.

 

[Muphrid]

Rotations are very easy to deal with in geometric algebra and/or with quaternions.

The rotation is entirely characterized by a rotor: $$R = e^{- i \theta /2}$$ where $i$ is some bivector, or a geometric object, that obeys $i^2 = -1$. There are three such objects in 3D space: we'll call them $e_{xy}, e_{yz}, e_{zx}$. They describe rotations in the xy, yz, and zx planes, respectively.

We can chain three such rotations together:

$$R = \exp(-e_{xy}\alpha/2) \exp(- e_{yz}\beta/2) \exp(- e_{zx} \gamma/2)$$

(You should already know that this is a perfectly valid means of parameterizing rotations; you can always go from fixed axes to rotated axes with Euler angles.)

When you multiply all three rotors out, you'll get something of the form

$$R = a + b e_{xy} + c e_{yz} + d e_{zx} = \cos \frac{\theta}{2} - u \sin \frac{\theta}{2}$$

where $u$ is a unit bivector. Taking $\cos^{-1} a = \theta/2$ gives the angle. Dividing out $\sin \theta/2$ gives the plane of rotation.

Example: consider a rotation composed of a $\pi/3$ rotation in the xy plane, $\pi/2$ in the yz plane, and $\pi/3$ in the zx plane. The rotor is then

$$R = \left(\frac{\sqrt{3}}{2} - e_{xy} \frac{1}{2} \right) \left(\frac{\sqrt{2}}{2} - e_{yz} \frac{\sqrt{2}}{2} \right) \left(\frac{\sqrt{3}}{2} - e_{zx} \frac{1}{2} \right)$$

Right away, just by multiplying all the first terms and all the last terms, we can find that $\cos \theta/2 = \sqrt{2}/4$.

The rest of the multiplication is best done by selecting for the other terms:

$$R = \frac{\sqrt{2}}{4} - e_{xy} \frac{\sqrt{6}}{4} - e_{yz} \frac{\sqrt{2}}{4} - e_{zx} \frac{\sqrt{6}}{4}$$

(May have done arithmetic wrong, but it should give the idea that this is not difficult--only tedious--to do.)

From here, the answer is within sight. The angle is $2\cos^{-1} (\sqrt{2}/4) \approx 2.42 = 139^\circ$. Written in terms of the unit bivector, the rotor is

$$R = \frac{\sqrt{2}}{4} - \frac{\sqrt{14}}{4} \left(\sqrt{\frac{3}{7}} e_{xy} +\sqrt{\frac{1}{7}} e_{yz} + \sqrt{\frac{3}{7}} e_{zx} \right)$$

Or, more plainly, we have a 139 degree rotation about $e_x + 3 e_y + 3 e_z$. And this was all possible through geometric algebra/quaternions.

 

[mfb]

If you want to express a general rotation with just N and zeta, those will depend on the rotated vector.

This is wrong - sorry, I was confused.
See Muphrid.