Why a finite angular displacement is not a vector?

[gracy]

One of my friend has answered this question in this way.

Angular displacement can't be a vector because addition is not commutative. Say we are looking at the Earth with North America facing us and the North Pole facing up:if we rotate the Earth so that we move 90 degrees north, now the NP is facing us and NA faces down. If we the rotate the Earth to move 90 degrees east, we end up with NA facing down and the NP facing to the left. But, if we reverse the order, we end up facing the NP while NA faces to the the left.

For very small displacements, however, the surface of the Earth is basically flat, so "rotating" the Earth is simply moving in a given direction without swapping any axis. So if we are facing some point A on the equator, either way we do the rotations we end up facing a point just NE of A, with the NP facing up.
But I am not getting this.Please help me out.

 

[Orodruin]

Your "very small" rotations give rise to differences that are also very small, but they are there unless you go to the limit of infinitesimal rotations. If you do that, you essentially get the tangent vector spaces of the sphere. But finite rotations of the sphere do not form a vector space.

 

[gleem]

Angular displacement can't be a vector because addition is not commutative. S

Vector addition is commutative. When you make several different rotations N- 90 and E 90 if reversing the order is E 90 and N 90 in that order that is not reversing the original rotations. You must us the opposite rotations in the reverse sequence i.e., west 90 and south 90 which will bring you to your initial orientation. Take a displacement on a plane use the sequence move x turn Rt move y turn Lt/ reversing the sequence will not bring you to your initial position without reversing the turn direction too. If B = A + ΔA then A ≠ B + ΔA but A = B - ΔA. - ΔA being the reverse of ΔA. Or have I missed the point?