Question 4.4: In Figure 4.10, the x′-axis is defined by the line segment O......
In Figure 4.10, the x′-axis is defined by the line segment O′P. The x′y′ plane is defined by the intersecting line segments O′P and O′Q. The z′-axis is normal to the plane of O′P and O′Q and obtained by rotating O′P toward O′Q and using the right-hand rule.
(a) Find the transformation matrix [Q].
(b) If \{ v \}=\left[\begin{array}{lll}2 & 4 & 6\end{array}\right]^T, \text { find }\{ v ^{\prime}\}.
(c) If \left\{ v ^{\prime}\right\}=\left[\begin{array}{lll}2 & 4 & 6\end{array}\right]^{T}, \text { find }\{ v \}.
(a) Resolve the directed line segments \overrightarrow{O^{\prime} P} \text { and } \overrightarrow{O^{\prime} Q} into components along the unprimed system:
Taking the cross product of \overrightarrow{O^{\prime} P} \text { into } \overrightarrow{O^{\prime} Q} yields a vector Z^{\prime}, which lies in the direction of the desired positive z^{\prime}-axis:
Z ^{\prime}=\overrightarrow{O^{\prime} P} \times \overrightarrow{O^{\prime} Q}=8 \hat{ i }+6 \hat{ j }+20 \hat{ k }Taking the cross product of Z^{\prime} \text { into } \overrightarrow{O^{\prime} P} then yields a vector Y^{\prime}, which points in the positive y^{\prime} direction:
Y ^{\prime}= Z ^{\prime} \times \overrightarrow{O^{\prime} P}=-68 \hat{ i }-176 \hat{ j }+80 \hat{ k }Normalizing the vectors \overrightarrow{O^{\prime} P}, Y ^{\prime} \text { and } \bar{Z}^{\prime} produces the \hat{ i }^{\prime}, \hat{ j }^{\prime} \text {, and } \widehat{ k }^{\prime} unit vectors, respectively. Thus
\begin{aligned}& \hat{ i }^{\prime}=\cfrac{\overrightarrow{O^{\prime} P}}{\left\|\overrightarrow{O^{\prime} P}\right\|}=-0.8729 \hat{ i }+0.4364 \hat{ j }+0.2182 \hat{ k } \\\\& \hat{ j }^{\prime}=\cfrac{ Y ^{\prime}}{\left\| Y ^{\prime}\right\|}=-0.3318 \hat{ i }-0.8588 \hat{ j }+0.3904 \hat{ k }\end{aligned}and
\hat{ k }^{\prime}=\cfrac{ Z ^{\prime}}{\left\| Z ^{\prime}\right\|}=0.3578 \hat{ i }+0.2683 \hat{ j }+0.8944 \hat{ k }\hat{ k }^{\prime}=\cfrac{ Z ^{\prime}}{\left\| Z ^{\prime}\right\|}=0.3578 \hat{ i }+0.2683 \hat{ j }+0.8944 \hat{ k }
The components of \hat{ i }^{\prime}, \hat{ j }^{\prime} \text {, and } \hat{ k }^{\prime} are the rows of the DCM [Q]. Thus,
[ Q ]=\left[\begin{array}{ccc}-0.8729 & 0.4364 & 0.2182 \\-0.3318 & -0.8588 & 0.3904 \\0.3578 & 0.2683 & 0.8944\end{array}\right](b)
\left\{ v ^{\prime}\right\}=[ Q ]\{ v \}=\left[\begin{array}{ccc}-0.8729 & 0.4364 & 0.2182 \\-0.3318 & -0.8588 & 0.3904 \\0.3578 &0.2683 & 0.8944\end{array}\right]\left\{\begin{array}{l}2 \\4 \\6\end{array}\right\}=\boxed{\left\{\begin{array}{c}1.309 \\-1.756 \\7.155\end{array}\right\}}(c)
\{ v \}=[ Q ]^T\left\{ v ^{\prime}\right\}=\left[\begin{array}{ccc}-0.8729 & -0.3318 & 0.3578 \\0.4364 & -0.8588 & 0.2683 \\0.2182 & 0.3904 & 0.8944\end{array}\right]\left\{\begin{array}{l}2 \\4 \\6\end{array}\right\}=\boxed{\left\{\begin{array}{c}-0.9263 \\-0.9523 \\7.364\end{array}\right\}}