Past exam of the mathematical methods I course of the University of Oxford / 2013 resit / 10

Use active rotations on column vectors and positive angles counter-clockwise when viewed from the positive axis toward the origin. Rotation about $z$ takes ${\mathbf{e}}_x$ to $(\cos \alpha ,\sin \alpha ,0)$, ${\mathbf{e}}_y$ to $(-\sin \alpha ,\cos \alpha ,0)$, and leaves ${\mathbf{e}}_z$ fixed. These images are the columns of the rotation matrix:It satisfies $R_z^T{R}_z=I$ and $\det R_z=1$. Stating the viewing convention avoids confusing this active rotation with a passive change of axes.

With the same active rotation convention, the $x$ axis remains fixed, while ${\mathbf{e}}_y$ moves toward ${\mathbf{e}}_z$. ThereforeFor instance, a positive quarter-turn maps ${\mathbf{e}}_y$ to ${\mathbf{e}}_z$ and ${\mathbf{e}}_z$ to $-{\mathbf{e}}_y$, checking both signs. This is again an orthogonal matrix of determinant one.

The first operation acts nearest the column vector, so the composite rotation matrix is $M=R_z(\alpha )R_x(\beta )$:At $\alpha =\beta =\pi /2$ this reduces toSolving $M\mathbf{v}=\mathbf{v}$ gives $v_x=v_y=v_z$. Normalization yields the fixed-axis unit vectorsBoth signs describe the same unoriented axis. The operation is a cyclic permutation, and $M^3=I$; it is not generally obtained by reversing the multiplication order.

Choose the unit vector $\mathbf{p}=(1,-1,0)/\sqrt{2}$, which is perpendicular to $\mathbf{v}$. Then $M\mathbf{p}=(0,1,-1)/\sqrt{2}$. Their dot product is $-1/2$, and the rotation matrix preserves their norms, soThe same unsigned angle holds for every nonzero vector perpendicular to the axis: this is the rotation plane of a three-dimensional proper rotation. Alternatively, $\mathrm{tr}M=0=1+2\cos \chi$ gives the same rotation angle $\chi =2\pi /3$.