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 towards O'Q and using the right-hand rule. (a) Find the transformation matrix [Q].

Question

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 towards O′Q and using the right-hand rule. (a) Find the transformation matrix [Q]. (b) If [latex]\{ v \}=\left\lfloor\begin{array}{lll}2 & 4 & 6\end{array}\right\rfloor^{T}[/latex], find {v′}. (c) If [latex]\left\{ v ^{\prime}\right\}=\left\lfloor\begin{array}{lll}2 & 4 & 0\end{array}\right\rfloor^{T}[/latex], find {v}.

Accepted Answer

(a) Resolve the directed line segments [latex]\overrightarrow{O^{\prime} P} ext { and } \overrightarrow{O^{\prime} Q}[/latex] into components along the unprimed system:

[latex]\overrightarrow{O^{\prime} P}=(-5-3) \hat{ i }+(5-1) \hat{ j }+(4-2) \hat{ k }=-8 \hat{ i }+4 \hat{ j }+2 \hat{ k }[/latex]
[latex]\overrightarrow{O^{\prime} Q}=(-6-3) \hat{ i }+(3-1) \hat{ j }+(5-2) \hat{ k }=-9 \hat{ i }+2 \hat{ j }+3 \hat{ k }[/latex]

Taking the cross product of [latex]\overrightarrow{O^{\prime} P} ext { into } \overrightarrow{O^{\prime} Q}[/latex] yields a vector Z' which lies in the direction of the desired positive z' axis:

[latex]Z ^{\prime}=\overrightarrow{O^{\prime} P} imes \overrightarrow{O^{\prime} Q}=8 \hat{ i }+6 \hat{ j }+20 \hat{ k }[/latex]

Taking the cross product of Z' into [latex]\overrightarrow{O^{\prime} P}[/latex] then yields a vector Y' which points in the positive y' direction:

[latex]Y ^{\prime}= Z imes \overrightarrow{O^{\prime} P}=-68 \hat{ i }-176 \hat{ j }+80 \hat{ k }[/latex]

Normalizing the vectors [latex] \overrightarrow{O^{\prime} P}[/latex],Y' and Z' produces the [latex]\hat{ i }^{\prime}, \hat{ j }^{\prime} ext { and } \hat{ k }^{\prime}[/latex] unit vectors, respectively. Thus

[latex]\hat{ i }^{\prime}=\frac{\overrightarrow{O^{\prime} P}}{\| \overrightarrow{O^{\prime} P \|}}=-0.8729 \hat{ i }+0.4364 \hat{ j }+0.2182 \hat{ k }[/latex]
[latex]\hat{ j }^{\prime}=\frac{ Y ^{\prime}}{\left\| Y ^{\prime} ight\|}=-0.3318 \hat{ i }-0.8588 \hat{ j }+0.3904 \hat{ k }[/latex]

and

[latex]\hat{ k }^{\prime}=\frac{ Z ^{\prime}}{\left\| Z ^{\prime} ight\|}=0.3578 \hat{ i }+0.2683 \hat{ j }+0.8944 \hat{ k }[/latex]

The components of [latex]\hat{ i }^{\prime}, \hat{ j }^{\prime} ext { and } \hat{ k }^{\prime}[/latex] are the rows of the orthogonal transformation matrix [Q]. Thus,

[latex][ Q ]=\left[\begin{array}{rrr}-0.8729 & 0.4364 & 0.2182 \-0.3318 & -0.8588 & 0.3904 \0.3578 & 0.2683 & 0.8944\end{array} ight][/latex]

(b)

[latex]\left\{ v ^{\prime} ight\}=[ Q ]\{ v \}=\left[\begin{array}{rrr}-0.8729 & 0.4364 & 0.2182 \-0.3318 & -0.8588 & 0.3904 \0.3578 & 0.2683 & 0.8944\end{array} ight]\left\{\begin{array}{l}2 \4 \6\end{array} ight\}=\underline{\left\{\begin{array}{r}1.309 \-1.756 \7.155\end{array} ight\}} [/latex]

(c)

[latex]\{ v \}=[ Q ]^{T}\left\{ v ^{\prime} ight\}=\left[\begin{array}{rrr}-0.8729 & -0.3318 & 0.3578 \0.4364 & -0.8588 & 0.2683 \0.2182 & 0.3904 & 0.8944\end{array} ight]\left\{\begin{array}{l}2 \4 \0\end{array} ight\}=\underline{\left\{\begin{array}{r}-3.073 \-2.562 \1.998\end{array} ight\}} [/latex]

Question 4.4: In Figure 4.10, the x' axis is defined by the line segment O...

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