Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/82

as will be evident on considering separately in the expression $\Phi .\rho$ the components perpendicular and parallel to $\theta ,$ or on substituting in

$ii+\cos q(jj+kk)+\sin q(kj-jk)$

for $\cos q$ and $\sin q$ their values in terms of $\tan {\tfrac {1}{2}}q.$

If we set, in either of these equations,

$\theta =ai+bj+ck,$

we obtain, on reduction, the formula

$\Phi ={\frac {\begin{Bmatrix}(1+a^{2}-b^{2}-c^{2})ii+(2ab-2c)ij+(2ac+2b)ik\\+(2ab+2c)ji+(1-a^{2}+b^{2}-c^{2})jj+(2bc-2a)jk\\+(2ac-2b)ki+(2bc+2a)kj+(1-a^{2}-b^{2}+c^{2})kk\end{Bmatrix}}{1+a^{2}+b^{2}+c^{2}}},$

in which the versor is expressed in terms of the rectangular components of the vector semitangent of version.

146. If $\alpha ,\beta ,\gamma$ are unit vectors, expressions of the form

$2\alpha \alpha -{\text{I}}$⁠$2\beta \beta -{\text{I}},$⁠$2\gamma \gamma -{\text{I}},$

are biquadrantal versors. A product like

$\{2\beta \beta -{\text{I}}\}.\{2\alpha \alpha -{\text{I}}\}$

is a versor of which the axis is perpendicular to $\alpha$ and $\beta ,$ and the amount of rotation twice that which would carry $\alpha$ to $\beta .$ It is evident that any versor may be thus expressed, and that either $\alpha$ or $\beta$ may be given any direction perpendicular to the axis of rotation. If

$\Phi =\{2\beta \beta -{\text{I}}\}.\{2\alpha \alpha -{\text{I}}\},$⁠and⁠$\Psi =\{2\gamma \gamma -{\text{I}}\}.\{2\beta \beta -{\text{I}}\},$

we have for the resultant of the successive rotations

$\Psi .\Phi =\{2\gamma \gamma -{\text{I}}\}.\{2\alpha \alpha -{\text{I}}\}.$

This may be applied to the composition of any two successive rotations, $\beta$ being taken perpendicular to the two axes of rotation, and affords the means of determining the resultant rotation by construction on the surface of a sphere. It also furnishes a simple method of finding the relations of the vector semitangents of version for the versors $\Phi ,\Psi ,$ and $\Psi .\Phi .$ Let

$\theta _{1}={\frac {-\Phi _{\text{X}}}{1+\Phi _{\text{S}}}},$⁠$\theta _{2}={\frac {-\Psi _{\text{X}}}{1+\Psi _{\text{S}}}},$⁠$\theta _{3}={\frac {-\{\Psi .\Phi \}_{\text{X}}}{1+\{\Psi .\Phi \}_{\text{S}}}}$

Then, since

${\begin{aligned}\Phi &=4\alpha .\beta \beta \alpha -2\alpha \alpha -2\beta \beta +{\text{I}},\\\theta _{1}&={\frac {\alpha \times \beta }{\alpha .\beta }},\end{aligned}}$

which is moreover geometrically evident. In like manner,

$\theta _{2}={\frac {\beta \times \gamma }{\beta .\gamma }}$⁠$\theta _{3}={\frac {\alpha \times \gamma }{\alpha .\gamma }}\cdot$