I have to write an algorithm that can given a rotation matrix, find $k$ and $f_i$.
$R = \text{rotationMatrix}(k, f_i)$
I am given $R$ and need to find $k$ and $f_i$, but i don't know how to do this, and the only formula i know for converting $k$ and $f_i$ into a rotation matrix is
$$R=\left[\begin{array}{ccc}\cos \theta+u_{x}^{2}(1-\cos \theta) & u_{x} u_{y}(1-\cos \theta)-u_{z} \sin \theta & u_{x} u_{z}(1-\cos \theta)+u_{y} \sin \theta \\ u_{y} u_{x}(1-\cos \theta)+u_{z} \sin \theta & \cos \theta+u_{y}^{2}(1-\cos \theta) & u_{y} u_{z}(1-\cos \theta)-u_{x} \sin \theta \\ u_{z} u_{x}(1-\cos \theta)-u_{y} \sin \theta & u_{z} u_{y}(1-\cos \theta)+u_{x} \sin \theta & \cos \theta+u_{z}^{2}(1-\cos \theta)\end{array}\right]$$
Any ideas on how I can attack this problem? Maybe use another formula to figure it out from?
Edit:
Thank you for the help. This is the correct function
Reverse rotation
function [ k, fi ] = arot( R )
fi = acosd(0.5*(R(1,1)+R(2,2)+R(3,3)-1));
k = zeros(3,1);
k(1) = (R(3,2)-R(2,3))/(2*sind(fi));
k(2) = (R(1,3)-R(3,1))/(2*sind(fi));
k(3) = (R(2,1)-R(1,2))/(2*sind(fi));
end
The rotation matrix
function R = rot(k,fi)
% This is just to make it easyer to read!
x = k(1);
y = k(2);
z = k(3);
% Create a 3x3 zero matrix
R = zeros(3,3);
% We use the formual for rotationg matrix about a unit vector k
R(1,1) = cosd(fi)+x^2*(1-cosd(fi));
R(1,2) = x*y*(1-cosd(fi))-z*sind(fi);
R(1,3) = x*z*(1-cosd(fi))+y*sind(fi);
R(2,1) = y*x*(1-cosd(fi))+z*sind(fi);
R(2,2) = cosd(fi)+y^2*(1-cosd(fi));
R(2,3) = y*z*(1-cosd(fi))-x*sind(fi);
R(3,1) = z.*x.*(1-cosd(fi))-y.*sind(fi);
R(3,2) = z.*y.*(1-cosd(fi))+x.*sind(fi);
R(3,3) = cosd(fi)+z^2.*(1-cosd(fi));
end
Best Answer
The trace of the matrix will give a quantity related to the cosine of the angle of rotation. It should have one eigenvector with a real eigenvalue - that will be the axis of rotation (up to a sign).