Consider the vector v = v j. Using the quaternion and corresponding direction cosine matrix in Example 11.22, carry out the following operations and interpret the results geometrically: (i) v' = q ⊗ v ⊗ q* (ii) {v'} = [Q]{v} where v = {v j/0} q = {sin(θ/2) i/cos(θ/2)}

Question

Consider the vector [latex]v =v \hat{ j }[/latex]. Using the quaternion and corresponding direction cosine matrix in Example 11.22, carry out the following operations and interpret the results geometrically:

(i) [latex]\widehat{ v }^{\prime}=\widehat{ q } \otimes \widehat{ v } \otimes \widehat{ q }^*[/latex]

(ii) {v'} = [Q] {v}

where

[latex]\widehat{ v }=\left\{\begin{array}{c}v \hat{ j } \\hdashline 0\end{array} ight\} \quad \widehat{ q }=\left\{\begin{array}{c}\sin ( heta / 2) \hat{ i } \\hdashline cos ( heta / 2)\end{array} ight\} \quad[ Q ]=\left[\begin{array}{ccc}1 & 0 & 0 \0 & \cos heta & \sin heta \0 & -\sin heta & \cos heta\end{array} ight][/latex]

Accepted Answer

(i) We do the two quaternion products one after the other using Eq. (11.146). The first product is

[latex]\widehat{ q } \otimes \widehat{ v }=\left\{\frac{\cos ( heta / 2)(v \hat{ j })+0 \cdot \sin ( heta / 2) \hat{ i }+\sin ( heta / 2) \hat{ i } imes v \hat{ j }}{\cos ( heta / 2) \cdot 0-\sin ( heta / 2) \hat{ i } \cdot v \hat{ j }} ight\}[/latex]

[latex]=\left\{\frac{v \cos ( heta / 2) \hat{ j }+v \sin ( heta / 2) \hat{ k }}{0} ight\}[/latex]

[latex]\widehat{ p } \otimes \widehat{ q }=\left\{\frac{p_4 q +q_4 p + p imes q }{p_4 q_4- p \cdot q } ight\}[/latex] (11.146)

Following this by the second product, we get

[latex]\widehat{ q } \otimes \widehat{ v } \otimes \widehat{ q }^*=\left\{\frac{ν \cos ( heta / 2) \hat{ j }+ν \sin ( heta / 2) \hat{ k }}{0} ight\} \otimes\left\{\frac{- \sin ( heta/2) \hat{i}}{\cos ( heta/2)} ight\}[/latex]

[latex]=\left\{\frac{0 \cdot[-\sin ( heta / 2) \hat{ i }]+\cos ( heta / 2) \cdot[ν \cos ( heta / 2) \hat{ j }+ν \sin ( heta / 2) \hat{ k }]+[ν\cos ( heta / 2) \hat{ j }+ν \sin ( heta / 2) \hat{ k }] imes[-\sin ( heta / 2) \hat{ i }]} {0 \cdot[\cos ( heta / 2)]-[ν \cos ( heta / 2) \hat{ j }+ν \sin ( heta / 2) \hat{ k }] \cdot[-\sin ( heta / 2) \hat{ i }]} ight\}[/latex]

[latex]=\{\frac{\overbrace{\left[ν \cos ^2( heta / 2)-ν \sin ^2( heta / 2) ight]}^{\cos heta} \hat{ j }+\overbrace{[2 ν \sin ( heta / 2) \cos ( heta / 2)]}^{\sin heta} \hat{ k }}{0}\}=\left\{\frac{ν \cos heta \hat{ j }+ν \sin heta \hat{ k } }{0} ight\}[/latex]

Finally, therefore,

[latex]v ^{\prime}=ν \cos heta \hat{ j }+ν \sin heta \hat{ k }[/latex] (a)

(ii)

[latex]\left\{ v ^{\prime} ight\}=\left[\begin{array}{ccc}1 & 0 & 0 \0 & \cos heta & \sin heta \0 & -\sin heta & \cos heta\end{array} ight]\left\{\begin{array}{l}0 \ν \0\end{array} ight\}=\left\{\begin{array}{c}0 \ν \cos heta \-ν \sin heta\end{array} ight\}[/latex]

or

[latex]v ^{\prime}=ν \cos heta \hat{ j }-ν \sin heta \hat{ k }[/latex] (b)

These two results are illustrated in Fig. 11.29.

Question 11.24: Consider the vector v = v j. Using the quaternion and corres...

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