Rotation to a spherical coordinate vector

Question

To rotate a vector A in x-y plane through longitude $ \theta $ in the same x-y plane we multiply by $ e^{i \theta }$.

By what operation is A rotated through latitude $\phi$ out of the x-y plane?

Answer 1

A rotation in 3-D about an axis through the origin can be viewed as multiplication by $e^{\theta K}$, where $\theta$ is the rotation angle and $K$ is the “cross product matrix” of a unit vector $\mathbf v$ that gives the rotation axis.

For example, rotating the vector $(12,5,0)^T$ “up” to a latitude of $\phi$ corresponds to a rotation about the axis $\mathbf v = (5/13,-12/13,0)^T$, with corresponding “cross-product matrix” $$K=\pmatrix{0&0&-\frac{12}{13}\\0&0&-\frac5{13}\\\frac{12}{13}&\frac5{13}&0},$$ so we need to compute the matrix $\exp(\phi V)$. Rodrigues’ rotation formula gives a convenient way to compute this exponential: $$\exp(\phi K)=I+(\sin\phi)K+(1-\cos\phi)K^2.$$ For $\phi=\tan^{-1}(12/5)$, we have $\sin\phi=12/13$ and $\cos\phi=5/13$. Multiplying out $\exp(\phi K)(12,5,0)^T$ with these values yields $(60/13,25/13,12)^T\approx(4.62,1.92,12.0)^T$. Note that $K^2=\mathbf v^T\mathbf v-I$, which can sometimes be easier to compute than squaring $K$ directly.