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Mathematics-Online course: Linear Algebra - Analytic Geometry - Orthogonal Groups

Rotation Axis and Angle


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A rotation $ Q$ in $ \mathbb{R}^3$ has an axis of rotation, i.e. $ Q$ fixes an unit vector $ u$, and corresponds to a plane rotation by an angle $ \vartheta$ in the plane orthogonal to $ u$.

With respect to an orthonormal right-handed coordinate system $ u,v,w$, the matrix representation of $ Q$ is given by:

$\displaystyle \tilde Q =
\left(\begin{array}{ccc}
1&0&0 \\
0&\cos\vartheta & -\sin\vartheta \\
0&\sin\vartheta & \cos\vartheta \\
\end{array}\right)
\,.
$

For the angle of rotation

$\displaystyle \cos\vartheta = \frac{1}{2}\left(\operatorname{Spur} Q - 1\right)
$

holds.
(Authors: Höllig/Reble/Höfert)

(temporary unavailable)

With the rotation-matrix

$\displaystyle Q = \frac{1}{2} \left( \begin{array}{ccc} 1 & -\sqrt{2} & 1 \\ \s...
...& 0 & -\sqrt{2}
\\ 1 & \sqrt{2} & 1 \end{array} \right) \in \mathbb{R}^{3,3}
$

you get

$\displaystyle Q^{\operatorname t}Q = \frac{1}{4} \left( \begin{array}{ccc} 4 & ...
...\\ 0 & 0 & 4 \end{array} \right) =
E \Rightarrow Q^{\operatorname t}= Q^{-1}
$

and

$\displaystyle \operatorname{det} Q = \operatorname{det} \frac{1}{2} \left( \beg...
...\
\sqrt{2} & 0 & -\sqrt{2} \\ 1 & \sqrt{2} & 1 \end{array} \right) = + 1\,.
$

Compute the rotationaxis, the eigenvector $ v$ to the eigenvalue $ \lambda = 1$:

$\displaystyle 2(Q-1E) = \left( \begin{array}{ccc} -1 & -\sqrt{2} & 1 \\ \sqrt{2} & -2 & -\sqrt{2} \\
1 & \sqrt{2} & -1 \end{array} \right)
$

With 2nd row $ - \sqrt{2} \cdot$ 1st row and 3rd row $ +$ 1st row you get

$\displaystyle \left( \begin{array}{ccc} -1 & -\sqrt{2} & 1 \\ 0 & -4 & 0 \\ 0 & 0 & 0 \end{array} \right)
$

and so the eigenvector $ v = \frac{1}{\sqrt{2}} \left( \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \right)$.

To compute $ \varphi$ you compute the angle between an unit vector $ n$ with $ n \bot v$ and the unit vector $ x = Qn$.

$\displaystyle n = \left( \begin{array}{c} 0 \\ 1 \\ 0 \end{array} \right)\,,
x...
...{1}{2} \left( \begin{array}{c} -\sqrt{2} \\ 0 \\ \sqrt{2} \end{array} \right)
$

$\displaystyle cos \varphi = n \cdot x = 0 \Rightarrow \varphi = \pm \frac{\pi}{2}\,.
$


  automatically generated 4/21/2005