Question 4.5 : If the DCM for the transformation from xyz to x′ y′ z ′ is......
If the DCM for the transformation from xyz to x^{′} y^{′} z^{′} is
[ Q] = \begin{bmatrix} 0.64050 & 0.75319 & -0.15038 \\ 0.76736 & – 0.63531 & -0.086824 \\ -0.030154 & -0.17101 & -0.98481 \end{bmatrix}
find the angles α, β, and γ of the classical Euler sequence.
Use Algorithm 4.3
Step 1:
\alpha=\tan^{-1}\left({\frac{Q_{31}}{-Q_{32}}}\right)=\tan^{-1}\left({\frac{-0.030154}{-[-0.17101]}}\right)
Since the numerator is negative and the denominator is positive, α must lie in the fourth quadrant. Thus
\tan^{-1}\left({\frac{-0.030154}{-1[-0.17101]}}\right)=\tan^{-1}(-0.1763)=-10^{\circ}\Rightarrow α = 350°
Step 2:
\beta=\cos^{-1}Q_{33}=\cos^{-1}(-0.98481)=170.0°
Step 3:
\gamma=\tan^{-1}\!{\frac{Q_{13}}{Q_{23}}}=\tan^{-1}\left({\frac{-0.15038}{0.086824}}\right)
The numerator is negative and the denominator is positive, so γ lies in the fourth quadrant,
\tan^{-1}\left({\frac{-0.15038}{0.086824}}\right)=\tan^{-1}(-0.17320)=-60^{\circ}\Rightarrow γ = 300°