Question 9.21: (a) Write down the quaternion for a rotation about the x-axi......
(a) Write down the quaternion for a rotation about the x-axis through an angle θ. (b) Obtain the corresponding direction cosine matrix.
(a) According to Eqn (9.146),
q =\sin \cfrac{\theta}{2} \hat{ u } \quad q_4=\cos \cfrac{\theta}{2} (9.146)
{q =\sin (\theta / 2) \hat{ i } \quad q_4=\cos (\theta / 2)} (a)
so that
\boxed{\{\widehat{q}\}=\left\{\begin{array}{c}\sin (\theta / 2) \\0 \\0 \\\cos (\theta / 2)\end{array}\right\}} (b)
(b) Substituting q_1=\sin (\theta / 2), q_2=q_3=0, \text { and } q_4=\cos (\theta / 2) into Eqn (9.148) yields
[ Q ]=\left[\begin{array}{ccc}\sin ^2(\theta / 2)+\cos ^2(\theta / 2) & 0 & 0 \\0 & -\sin ^2(\theta / 2)+\cos ^2(\theta / 2) & 2 \sin (\theta / 2) \cos (\theta / 2) \\0 & -2 \sin (\theta / 2) \cos (\theta / 2) & -\sin ^2(\theta / 2)+\cos ^2(\theta / 2)\end{array}\right] (c)
From trigonometry, we recall that
\sin ^2 \cfrac{\theta}{2}+\cos ^2 \cfrac{\theta}{2}=1 \quad 2 \sin \cfrac{\theta}{2} \cos \cfrac{\theta}{2}=\sin \theta \quad \cos ^2 \cfrac{\theta}{2}-\sin ^2 \cfrac{\theta}{2}=\cos \thetaTherefore, Eqn (c) becomes
\boxed{[ Q ]=\left[\begin{array}{ccc}1 & 0 & 0 \\0 & \cos \theta & \sin \theta \\0 & -\sin \theta & \cos \theta\end{array}\right]} (d)
We recognize this as the direction cosine matrix \left[ R _1(\theta)\right] for a rotation θ around the x-axis.