4 ms·
Sorry, what you have stated is not very clear. If you use cos(theta) + sin(theta)*(ai + bj + ck) representation as you mentioned, what does happen to the orient
by tsarakoye_selo 2y ago
Sorry, what you have stated is not very clear. If you use cos(theta) + sin(theta)*(ai + bj + ck) representation as you mentioned, what does happen to the orientation information when sin(theta) becomes zero
- moefh 2y agoWhen theta=0, you have q = cos(0) + sin(0)*(ai+bj+ck) = 1 + 0 = 1 That means applying the "rotation" to the vector gives v' = q * v * q^-1 = 1 * v * 1^-1 = v So the output ("rotated") vector is the same as the input, as you would expect for a rotation by 0.