Homogeneous transformation matrix (6)
The axis-angle representation (2).
As promised, we prove today that for every vector in the original coordinate system, the rotation matrix
indeed rotates around the unit direction by angle . Here
Without loss of generality, we assume is also a unit vector. After the rotation, we let denote the resulting unit vector
Our proof consists of two parts:
The angle between and equals the angle between and . That is, the dot products of these two pairs are equal.
Let denote the unit vector that is parallel with the cross product of and . Similarly, let be such a vector for and . Then the angle between and is , and their cross product is .
We prove (i) first by evaluating directly the dot product.
|(substitute using (2))|
|(substitute for using (3))|
|(use (2) again).|
Therefore, the angle between and and that between and are the same. We denote this angle by , so .
To prove (ii), we first note that
For any four vectors , an identity holds for the following scalar triple product:
The angle between and is revealed by their dot product, so we first derive
|(use the identity)|
|(plug in in (3))|
|(plug in in (2))|
|(simplify and do factorization)|
|(derive scalars below).|
In essence, this is the proof for for any vector and a skew-symmetric matrix . A matrix is skew-symmetric if it is the opposite of its transpose: .
For the second scalar , we have
for by (2). Hence
If and hence or , then vectors align, so the rotation angle simply does not matter. Otherwise if and hence nor , then and do not align, so
which means that the rotation is indeed .
Lastly, it is simple to show
because is perpendicular to both and and
In case where , we have
which means that the rotation axis is indeed .