I am given this figure of a regular octahedron and I have to find the unit quaternion.
In the symmetry group of the regular octahedron what is the unit quaternion that represents a 90 degree rotation about the semi x-axis $\lambda i, \lambda >0$ . The rotation appears to be counter clockwise to an observer at point A.
At point A, I am just given the i vector and I know that I have a 90 degree rotation. So, $\theta = 90^{\circ}$. There is also $ \alpha = cos \theta$, but if I just plug in 90 degrees right now, I will just have 0.
If I have a 90 degree rotation. Then
$2\theta = 90^{\circ}$
$\theta = 45^{\circ}$
$ \alpha = 45^{\circ} = \frac{1}{\sqrt{2}}$
The norm of u is 1 and $u^2$ is 1 so I have $ u = \frac{1}{\sqrt{2}}+ \lambda i$ so $u^2 = \frac{1}{2} + (\lambda)^2(i)^2$ (that $i^2$ is not imaginary..it's a vector)
and then I have to solve for lambda
$1 = \frac{1}{2} + (\lambda)^2(i)^2$ using the ...alright I'm forgetting my terminology here but I know $\sqrt(1^2+0^2+0^2) = \sqrt(1) = 1$ because i is my only vector and I have (1,0,0)
$1 = \frac{1}{2} + (\lambda)^2$
$1-\frac{1}{2} = (\lambda)^2$
$\frac{1}{2} = (\lambda)^2$
$\frac{1}{\sqrt{2}}, - \frac{1}{\sqrt{2}} = \lambda$
$ u = \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} i$
Is that the unit quaternion? I'm kind of lost.