式展開: 6月 2020

四元数の行列表現において，可逆元とその逆元を左右から作用させる

$$ \begin{eqnarray} \mathbf{q}=w_q+x_qi+y_qj+z_qk &\leftrightarrow& \href{https://shikitenkai.blogspot.com/2020/06/blog-post_99.html}{ \begin{bmatrix} w_q+x_qi&y_q+z_qi\\ -(y_q-z_qi)&w_q-x_qi\\ \end{bmatrix} } \\ \mathbf{p}=w_p+x_pi+y_pj+z_pk &\leftrightarrow& \begin{bmatrix} w_p+x_pi&y_p+z_pi\\ -(y_p-z_pi)&w_p-x_pi\\ \end{bmatrix} \end{eqnarray} $$ $$ \begin{eqnarray} \mathbf{q}^{-1}=\frac{\bar{\mathbf{q}}}{|\mathbf{q}|^2} &\leftrightarrow& \frac{1}{w_q^2+x_q^2+y_q^2+z_q^2} \begin{bmatrix} w_q-x_qi&-y_q-z_qi\\ -(-y_q+z_qi)&w_q+x_qi\\ \end{bmatrix} \end{eqnarray} $$ $$ \begin{eqnarray} \mathbf{q}\mathbf{p}\mathbf{q}^{-1} &\leftrightarrow& \frac{1}{w_q^2+x_q^2+y_q^2+z_q^2} \begin{bmatrix} w_q+x_qi&y_q+z_qi\\ -(y_q-z_qi)&w_q-x_qi\\ \end{bmatrix} \begin{bmatrix} w_p+x_pi&y_p+z_pi\\ -(y_p-z_pi)&w_p-x_pi\\ \end{bmatrix} \begin{bmatrix} w_q-x_qi&-y_q-z_qi\\ -(-y_q+z_qi)&w_q+x_qi\\ \end{bmatrix} \\&=& \frac{1}{w_q^2+x_q^2+y_q^2+z_q^2} \begin{bmatrix} (w_q^2+x_q^2+y_q^2+z_q^2) w_p +( (w_q^2+x_q^2-y_q^2-z_q^2) x_p +2(x_qy_q-w_qz_q) y_p +2(w_qy_q+x_qz_q) z_p )i & 2(w_qz_q+x_qy_q)x_p+(w_q^2-x_q^2+y_q^2-z_q^2)y_p+2(y_qz_q-w_qx_q)z_p +( 2(x_qz_q-w_qy_q)x_p+2(w_qx_q+y_qz_q)y_p+(w_q^2-x_q^2-y_q^2+z_q^2)z_p )i \\ -( 2(w_qz_q+x_qy_q)x_p+(w_q^2-x_q^2+y_q^2-z_q^2)y_p+2(y_qz_q-w_qx_q)z_p -( 2(x_qz_q-w_qy_q)x_p+2(w_qx_q+y_qz_q)y_p+(w_q^2-x_q^2-y_q^2+z_q^2)z_p )i ) & (w_q^2+x_q^2+y_q^2+z_q^2)w_p -( (w_p^2+x_q^2-y_q^2-z_q^2)x_p+2(x_qy_q-w_qz_q)y_p+2(w_qy_q+x_qz_q)z_p )i \\ \end{bmatrix} \\&&\;\cdots\;式展開は最後に添付 \end{eqnarray} $$ $$ \begin{eqnarray} w_{qpq^{-1}}&=&\frac{(w_q^2+x_q^2+y_q^2+z_q^2) w_p}{w_q^2+x_q^2+y_q^2+z_q^2}=w_p \\x_{qpq^{-1}}&=&\frac{(w_q^2+x_q^2-y_q^2-z_q^2)x_p+2(x_qy_q-w_qz_q)y_p+2(w_qy_q+x_qz_q)z_p}{w_q^2+x_q^2+y_q^2+z_q^2} \\y_{qpq^{-1}}&=&\frac{2(w_qz_q+x_qy_q)x_p+(w_q^2-x_q^2+y_q^2-z_q^2)y_p+2(y_qz_q-w_qx_q)z_p}{w_q^2+x_q^2+y_q^2+z_q^2} \\z_{qpq^{-1}}&=&\frac{2(x_qz_q-w_qy_q)x_p+2(w_qx_q+y_qz_q)y_p+(w_q^2-x_q^2-y_q^2+z_q^2)z_p}{w_q^2+x_q^2+y_q^2+z_q^2} \end{eqnarray} $$

特殊なp, qを考える

$ w_q=\cos{\left(\frac{\theta}{2}\right)},x_q^2+y_q^2+z_q^2=\sin^2{\left(\frac{\theta}{2}\right)} $となる $\mathbf{q}$及び $ w_p=0 $となる$\mathbf{p}$を考える．
また，この時 $v_x=\frac{x_q}{\sin{\left(\frac{\theta}{2}\right)}}, \;v_y=\frac{y_q}{\sin{\left(\frac{\theta}{2}\right)}}, \;v_z=\frac{z_q}{\sin{\left(\frac{\theta}{2}\right)}}, \;v_x^2+v_y^2+v_z^2=1$となるような$v_x, v_y, v_z$を用意する． $$ \begin{eqnarray} w_q^2+x_q^2+y_q^2+z_q^2 &=& \left( \cos^2{ \left( \frac{\theta}{2} \right) } +\left( \sin{ \left( \frac{\theta}{2} \right) } v_x \right)^2 +\left( \sin{ \left( \frac{\theta}{2} \right) } v_y \right)^2 +\left( \sin{ \left( \frac{\theta}{2} \right) } v_z \right)^2 \right) \\&=&\left( \cos^2{ \left( \frac{\theta}{2} \right) } +\sin^2{ \left( \frac{\theta}{2} \right) } \left( v_x^2+v_y^2+v_z^2 \right) \right) \\&=&\left( \cos^2{ \left( \frac{\theta}{2} \right) } +\sin^2{ \left( \frac{\theta}{2} \right) } \right) \\&=&1 \\w_{qpq^{-1}}&=&w_p \\&=&0 \\x_{qpq^{-1}}&=&(w_q^2+x_q^2-y_q^2-z_q^2)x_p+2(x_qy_q-w_qz_q)y_p+2(w_qy_q+x_qz_q)z_p \\&=& \left( \cos^2{ \left( \frac{\theta}{2} \right) } +\left( \sin{ \left( \frac{\theta}{2} \right) } v_x \right)^2 -\left( \sin{ \left( \frac{\theta}{2} \right) } v_y \right)^2 -\left( \sin{ \left( \frac{\theta}{2} \right) } v_z \right)^2 {\color{red} -2\left( \sin{ \left( \frac{\theta}{2} \right) } v_x \right)^2 +2\left( \sin{ \left( \frac{\theta}{2} \right) } v_x \right)^2 } \right)x_p +2\left( \sin{ \left( \frac{\theta}{2} \right) } v_x \sin{ \left( \frac{\theta}{2} \right) } v_y - \cos{ \left( \frac{\theta}{2} \right) } \sin{ \left( \frac{\theta}{2} \right) } v_z \right)y_p +2\left( \cos{ \left( \frac{\theta}{2} \right) } \sin{ \left( \frac{\theta}{2} \right) } v_y + \sin{ \left( \frac{\theta}{2} \right) } v_x \sin{ \left( \frac{\theta}{2} \right) } v_z \right)z_p \\&=& \left( \cos^2{ \left( \frac{\theta}{2} \right) } - \sin^2{ \left( \frac{\theta}{2} \right) } \left( v_x^2 +v_y^2 +v_z^2 \right) +2\sin^2{ \left( \frac{\theta}{2} \right) } v_x^2 \right)x_p +2 \sin^2{ \left( \frac{\theta}{2} \right) } v_xv_yy_p -2 \cos{ \left( \frac{\theta}{2} \right) } \sin{ \left( \frac{\theta}{2} \right) } v_zy_p +2 \cos{ \left( \frac{\theta}{2} \right) } \sin{ \left( \frac{\theta}{2} \right) } v_yz_p +2 \sin^2{ \left( \frac{\theta}{2} \right) } v_xv_zz_p \\&=& \left( \cos^2{ \left( \frac{\theta}{2} \right) } - \sin^2{ \left( \frac{\theta}{2} \right) }\cdot1 \right)x_p +2\sin^2{ \left( \frac{\theta}{2} \right) } v_x^2x_p +2 \sin^2{ \left( \frac{\theta}{2} \right) } \left(v_xv_yy_p + v_xv_zz_p\right) +2 \cos{ \left( \frac{\theta}{2} \right) } \sin{ \left( \frac{\theta}{2} \right) } \left(v_yz_p - v_zy_p\right) \\&=& \left( \cos^2{ \left( \frac{\theta}{2} \right) } - \sin^2{ \left( \frac{\theta}{2} \right) } \right)x_p +2 \sin^2{ \left( \frac{\theta}{2} \right) } \left(v_x^2x_p + v_xv_yy_p + v_xv_zz_p\right) +2 \cos{ \left( \frac{\theta}{2} \right) } \sin{ \left( \frac{\theta}{2} \right) } \left(v_yz_p - v_zy_p\right) \\&=& \cos{\left(\theta\right)}x_p +\left(1-\cos{\left(\theta\right)}\right)\left(\left(v_xx_p + v_yy_p + v_zz_p\right)v_x\right) +\sin{\left(\theta\right)}\left(v_yz_p - v_zy_p\right) \\&&\;\cdots\;\cos^2{\left(\frac{\theta}{2}\right)}-\sin^2{\left(\frac{\theta}{2}\right)} =\cos{\left(\frac{\theta}{2}+\frac{\theta}{2}\right)} =\cos{\left(\theta\right)} \\&&\;\cdots\;2\sin^2{\left(\frac{\theta}{2}\right)} =\sin^2{\left(\frac{\theta}{2}\right)}+\sin^2{\left(\frac{\theta}{2}\right)} =\left(1-\cos^2{\left(\frac{\theta}{2}\right)}\right)+\sin^2{\left(\frac{\theta}{2}\right)} =1-\left(\cos^2{\left(\frac{\theta}{2}\right)}-\sin^2{\left(\frac{\theta}{2}\right)}\right) =1-\cos{\left(\theta\right)} \\&&\;\cdots\;2\cos{\left(\frac{\theta}{2}\right)}\sin{\left(\frac{\theta}{2}\right)} =\sin{\left(\frac{\theta}{2}+\frac{\theta}{2}\right)} =\sin{\left(\theta\right)} \\&=& \left(v_xx_p + v_yy_p + v_zz_p\right)v_x +\cos{\left(\theta\right)}\left(x_p - \left(v_xx_p + v_yy_p + v_zz_p\right)v_x\right) +\sin{\left(\theta\right)}\left(v_yz_p - v_zy_p\right) \\&=&\left(\mathbf{V}\cdot \mathbf{V}_p\right)v_x +\cos{\left(\theta\right)}\left(x_p - \left(\mathbf{V}\cdot \mathbf{V}_p\right)v_x\right) +\sin{\left(\theta\right)}\left(v_yz_p - v_zy_p\right) \;\cdots\;\mathbf{V}=(v_x, v_y, v_z) , \mathbf{V}_p=(x_p, y_p, z_p) \\y_{qpq^{-1}}&=&2(w_qz_q+x_qy_q)x_p+(w_q^2-x_q^2+y_q^2-z_q^2)y_p+2(y_qz_q-w_qx_q)z_p \\&=& \left(v_xx_p + v_yy_p + v_zz_p\right)v_y +\cos{\left(\theta\right)}\left(y_p - \left(v_xx_p + v_yy_p + v_zz_p\right)v_y\right) +\sin{\left(\theta\right)}\left(v_zx_p - v_xz_p\right) \\&=& \left(\mathbf{V}\cdot \mathbf{V}_p\right)v_y +\cos{\left(\theta\right)}\left(y_p - \left(\mathbf{V}\cdot \mathbf{V}_p\right)v_y\right) +\sin{\left(\theta\right)}\left(v_zx_p - v_xz_p\right) \\z_{qpq^{-1}}&=&2(x_qz_q-w_qy_q)x_p+2(w_qx_q+y_qz_q)y_p+(w_q^2-x_q^2-y_q^2+z_q^2)z_p \\&=& \left(v_xx_p + v_yy_p + v_zz_p\right)v_z +\cos{\left(\theta\right)}\left(z_p - \left(v_xx_p + v_yy_p + v_zz_p\right)v_z\right) +\sin{\left(\theta\right)}\left(v_xy_p - v_yx_p\right) \\&=& \left(\mathbf{V}\cdot \mathbf{V}_p\right)v_z +\cos{\left(\theta\right)}\left(z_p - \left(\mathbf{V}\cdot \mathbf{V}_p\right)v_z\right) +\sin{\left(\theta\right)}\left(v_xy_p - v_yx_p\right) \end{eqnarray} $$ $$ \begin{eqnarray} \mathbf{P}_{qpq^{-1}} &=&w_{qpq^{-1}}+x_{qpq^{-1}}i+y_{qpq^{-1}}j+z_{qpq^{-1}}k \\&=&0 \\&&+\left( \left(\mathbf{V}\cdot \mathbf{V}_p\right)v_x +\cos{\left(\theta\right)}\left(x_p - \left(\mathbf{V}\cdot \mathbf{V}_p\right)v_x\right) +\sin{\left(\theta\right)}\left(v_yz_p - v_zy_p\right) \right)i \\&&+\left( \left(\mathbf{V}\cdot \mathbf{V}_p\right)v_y +\cos{\left(\theta\right)}\left(y_p - \left(\mathbf{V}\cdot \mathbf{V}_p\right)v_y\right) +\sin{\left(\theta\right)}\left(v_zx_p - v_xz_p\right) \right)j \\&&+\left( \left(\mathbf{V}\cdot \mathbf{V}_p\right)v_z +\cos{\left(\theta\right)}\left(z_p - \left(\mathbf{V}\cdot \mathbf{V}_p\right)v_z\right) +\sin{\left(\theta\right)}\left(v_xy_p - v_yx_p\right) \right)k \\&=& \left(\mathbf{V}\cdot \mathbf{V}_p\right)\left(v_xi+v_yj+v_zk\right) +\cos{\left(\theta\right)}\left(x_pi+y_pj+z_pk\right) -\cos{\left(\theta\right)}\left(\mathbf{V}\cdot \mathbf{V}_p\right)\left(v_xi+v_yj+v_zk\right) +\sin{\left(\theta\right)}\left(\left(v_yz_p - v_zy_p\right)i+\left(v_zx_p - v_xz_p\right)j\left(v_xy_p - v_yx_p\right)k\right) \\&=& \left(\mathbf{V}\cdot \mathbf{V}_p\right)\mathbf{V} +\cos{\left(\theta\right)}\mathbf{V}_p -\cos{\left(\theta\right)}\left(\mathbf{V}\cdot \mathbf{V}_p\right)\mathbf{V} +\sin{\left(\theta\right)}\left(\mathbf{V}\times\mathbf{V}_p\right) \\&=& \left(\mathbf{V}\cdot \mathbf{V}_p\right)\mathbf{V} +\cos{\left(\theta\right)}\left( \mathbf{V}_p-\left(\mathbf{V}\cdot \mathbf{V}_p\right) \right)\mathbf{V} +\sin{\left(\theta\right)}\left(\mathbf{V}\times\mathbf{V}_p\right) \end{eqnarray} $$ これはベクトル$\mathbf{V}_p=(x_p, y_p, z_p)$を，回転軸$\mathbf{V}=(v_x, v_y, v_z)$周りに$\theta$だけ回す変換となる．

$$ \begin{eqnarray} \mathbf{q}\mathbf{p}\mathbf{q}^{-1} &\leftrightarrow& \frac{1}{w_q^2+x_q^2+y_q^2+z_q^2} \begin{bmatrix} w_q+x_qi&y_q+z_qi\\ -(y_q-z_qi)&w_q-x_qi\\ \end{bmatrix} \begin{bmatrix} w_p+x_pi&y_p+z_pi\\ -(y_p-z_pi)&w_p-x_pi\\ \end{bmatrix} \begin{bmatrix} w_q-x_qi&-y_q-z_qi\\ -(-y_q+z_qi)&w_q+x_qi\\ \end{bmatrix} \\&=& \frac{1}{w_q^2+x_q^2+y_q^2+z_q^2} \begin{bmatrix} (w_q+x_qi)(w_p+x_pi) +(y_q+z_qi)(-(y_p-z_pi)) & (w_q+x_qi)(y_p+z_pi) +(y_q+z_qi)(w_p-x_pi) \\ (-(y_q-z_qi))(w_p+x_pi) +(w_q-x_qi)(-(y_p-z_pi)) & (-(y_q-z_qi))(y_p+z_pi) +(w_q-x_qi)(w_p-x_pi) \\ \end{bmatrix} \begin{bmatrix} w_q-x_qi&-y_q-z_qi\\ -(-y_q+z_qi)&w_q+x_qi\\ \end{bmatrix} \\&=& \frac{1}{w_q^2+x_q^2+y_q^2+z_q^2} \begin{bmatrix} w_qw_p+w_qx_pi+x_qiw_p+x_qix_pi + y_q(-y_p)+y_qz_pi+z_qi(-y_p)+z_qiz_pi & w_qy_p+w_qz_pi+x_qiy_p+x_qiz_pi + y_qw_p+y_q(-x_pi)+z_qiw_p+z_qi(-x_pi) \\ (-y_q)w_p+(-y_q)x_pi+z_qiw_p+z_qix_pi + w_q(-y_p)+w_qz_pi+(-x_qi)(-y_p)+(-x_qi)z_pi & (-y_q)y_p+(-y_q)z_pi+z_qiy_p+z_qiz_pi + w_qw_p+w_q(-x_pi)+(-x_qi)w_p+(-x_qi)(-x_pi) \\ \end{bmatrix} \begin{bmatrix} w_q-x_qi&-y_q-z_qi\\ -(-y_q+z_qi)&w_q+x_qi\\ \end{bmatrix} \\&=& \frac{1}{w_q^2+x_q^2+y_q^2+z_q^2} \begin{bmatrix} w_qw_p -x_qx_p -y_qy_p -z_qz_p +(w_qx_p +x_qw_p +y_qz_p -z_qy_p)i & w_qy_p +y_qw_p +z_qx_p -x_qz_p +(w_qz_p +z_qw_p +x_qy_p -y_qx_p)i \\ -(w_qy_p -x_qz_p +y_qw_p +z_qx_p -(w_qz_p +x_qy_p -y_qx_p +z_qw_p)i & w_qw_p -x_qx_p -y_qy_p -z_qz_p -(w_qx_p +x_qw_p +y_qz_p -z_qy_p)i \\ \end{bmatrix} \begin{bmatrix} w_q-x_qi&-y_q-z_qi\\ -(-y_q+z_qi)&w_q+x_qi\\ \end{bmatrix} \\&=& \frac{1}{w_q^2+x_q^2+y_q^2+z_q^2} \begin{bmatrix} (w_qw_p -x_qx_p -y_qy_p -z_qz_p +(w_qx_p +x_qw_p +y_qz_p -z_qy_p)i)(w_q-x_qi) +(w_qy_p +y_qw_p +z_qx_p -x_qz_p +(w_qz_p +z_qw_p +x_qy_p -y_qx_p)i)(-(-y_q+z_qi)) & (w_qw_p -x_qx_p -y_qy_p -z_qz_p +(w_qx_p +x_qw_p +y_qz_p -z_qy_p)i)(-y_q-z_qi) +(w_qy_p +y_qw_p +z_qx_p -x_qz_p +(w_qz_p +z_qw_p +x_qy_p -y_qx_p)i)(w_q+x_qi)\\ (-(w_qy_p -x_qz_p +y_qw_p +z_qx_p -(w_qz_p +x_qy_p -y_qx_p +z_qw_p)i)(w_q-x_qi) +(w_qw_p -x_qx_p -y_qy_p -z_qz_p -(w_qx_p +x_qw_p +y_qz_p -z_qy_p)i)(-(-y_q+z_qi)) &(-(w_qy_p -x_qz_p +y_qw_p +z_qx_p -(w_qz_p +x_qy_p -y_qx_p +z_qw_p)i)(-y_q-z_qi) +(w_qw_p -x_qx_p -y_qy_p -z_qz_p -(w_qx_p +x_qw_p +y_qz_p -z_qy_p)i)(w_q+x_qi)\\ \end{bmatrix} \\&=& \frac{1}{w_q^2+x_q^2+y_q^2+z_q^2} \begin{bmatrix} (w_qw_p -x_qx_p -y_qy_p -z_qz_p +(w_qx_p +x_qw_p +y_qz_p -z_qy_p)i)(w_q-x_qi) +(w_qy_p +y_qw_p +z_qx_p -x_qz_p +(w_qz_p +z_qw_p +x_qy_p -y_qx_p)i)(y_q-z_qi)) & (w_qw_p -x_qx_p -y_qy_p -z_qz_p +(w_qx_p +x_qw_p +y_qz_p -z_qy_p)i)(-y_q-z_qi) +(w_qy_p +y_qw_p +z_qx_p -x_qz_p +(w_qz_p +z_qw_p +x_qy_p -y_qx_p)i)(w_q+x_qi)\\ (-w_qy_p +x_qz_p -y_qw_p -z_qx_p +(w_qz_p +x_qy_p -y_qx_p +z_qw_p)i)(w_q-x_qi) +(w_qw_p -x_qx_p -y_qy_p -z_qz_p -(w_qx_p +x_qw_p +y_qz_p -z_qy_p)i)(y_q-z_qi) &(-w_qy_p +x_qz_p -y_qw_p -z_qx_p +(w_qz_p +x_qy_p -y_qx_p +z_qw_p)i)(-y_q-z_qi) +(w_qw_p -x_qx_p -y_qy_p -z_qz_p -(w_qx_p +x_qw_p +y_qz_p -z_qy_p)i)(w_q+x_qi)\\ \end{bmatrix} \\&=& \frac{1}{w_q^2+x_q^2+y_q^2+z_q^2} \begin{bmatrix} (w_q^2+x_q^2+y_q^2+z_q^2) w_p +( (w_q^2+x_q^2-y_q^2-z_q^2) x_p +2(x_qy_q-w_qz_q) y_p +2(w_qy_q+x_qz_q) z_p )i & 2(w_qz_q+x_qy_q)x_p+(w_q^2-x_q^2+y_q^2-z_q^2)y_p+2(y_qz_q-w_qx_q)z_p +( 2(x_qz_q-w_qy_q)x_p+2(w_qx_q+y_qz_q)y_p+(w_q^2-x_q^2-y_q^2+z_q^2)z_p )i \\ -( 2(w_qz_q+x_qy_q)x_p+(w_q^2-x_q^2+y_q^2-z_q^2)y_p+2(y_qz_q-w_qx_q)z_p -( 2(x_qz_q-w_qy_q)x_p+2(w_qx_q+y_qz_q)y_p+(w_q^2-x_q^2-y_q^2+z_q^2)z_p )i ) & (w_q^2+x_q^2+y_q^2+z_q^2)w_p -( (w_p^2+x_q^2-y_q^2-z_q^2)x_p+2(x_qy_q-w_qz_q)y_p+2(w_qy_q+x_qz_q)z_p )i \\ \end{bmatrix} \\&&\;\cdots\;各要素の式展開は以降に添付 \end{eqnarray} $$

(1,1)要素

(1,1)要素の第一項
$$ \begin{eqnarray} (w_qw_p -x_qx_p -y_qy_p -z_qz_p + (w_qx_p +x_qw_p +y_qz_p -z_qy_p)i)(w_q-x_qi) &=& (w_qw_p -x_qx_p -y_qy_p -z_qz_p +(w_qx_p +x_qw_p +y_qz_p -z_qy_p)i)(w_q) + (w_qw_p -x_qx_p -y_qy_p -z_qz_p +(w_qx_p +x_qw_p +y_qz_p -z_qy_p)i)(-x_qi) \\&=& w_pw_qw_q -w_qx_px_q -w_qy_py_q -w_qz_pz_q + w_qx_px_q +w_px_qx_q +x_qy_qz_p -x_qy_pz_q +( w_qw_qx_p +w_pw_qx_q +w_qy_qz_p -w_qy_pz_q -w_pw_qx_q +x_px_qx_q +x_qy_py_q +x_qz_pz_q )i \end{eqnarray} $$ (1,1)要素の第二項
$$ \begin{eqnarray} (w_qy_p +y_qw_p +z_qx_p -x_qz_p + (w_qz_p +z_qw_p +x_qy_p -y_qx_p)i)(-(-y_q+z_qi)) &=& (w_qy_p +y_qw_p +z_qx_p -x_qz_p +(w_qz_p +z_qw_p +x_qy_p -y_qx_p)i)(y_q) + (w_qy_p +y_qw_p +z_qx_p -x_qz_p +(w_qz_p +z_qw_p +x_qy_p -y_qx_p)i)(-z_qi) \\&=& w_qy_py_q +w_py_qy_q +x_py_qz_q -x_qy_qz_p + w_qz_pz_q +w_pz_qz_q +x_qy_pz_q -x_py_qz_q +( w_qy_qz_p +w_py_qz_q +x_qy_py_q -x_py_qy_q -w_qy_pz_q -w_py_qz_q -x_pz_qz_q +x_qz_pz_q )i \end{eqnarray} $$ (1,1)要素の第一，二項の和
$$ \begin{eqnarray} (w_qw_p -x_qx_p -y_qy_p -z_qz_p + (w_qx_p +x_qw_p +y_qz_p -z_qy_p)i)(w_q-x_qi)\\ +(w_qy_p +y_qw_p +z_qx_p -x_qz_p + (w_qz_p +z_qw_p +x_qy_p -y_qx_p)i)(-(-y_q+z_qi)) &=& w_pw_qw_q -w_qx_px_q -w_qy_py_q -w_qz_pz_q + w_qx_px_q +w_px_qx_q +x_qy_qz_p -x_qy_pz_q + w_qy_py_q +w_py_qy_q +x_py_qz_q -x_qy_qz_p + w_qz_pz_q +w_pz_qz_q +x_qy_pz_q -x_py_qz_q +( w_qw_qx_p +w_pw_qx_q +w_qy_qz_p -w_qy_pz_q -w_pw_qx_q +x_px_qx_q +x_qy_py_q +x_qz_pz_q +w_qy_qz_p +w_py_qz_q +x_qy_py_q -x_py_qy_q -w_qy_pz_q -w_py_qz_q -x_pz_qz_q +x_qz_pz_q )i \\ &=& (w_q^2+x_q^2+y_q^2+z_q^2) w_p +( (w_q^2+x_q^2-y_q^2-z_q^2) x_p +2(x_qy_q-w_qz_q) y_p +2(w_qy_q+x_qz_q) z_p )i \end{eqnarray} $$

(1,2)要素

(1,2)要素の第一項
$$ \begin{eqnarray} (w_qw_p -x_qx_p -y_qy_p -z_qz_p +(w_qx_p +x_qw_p +y_qz_p -z_qy_p)i)(-y_q-z_qi) &=& (w_qw_p -x_qx_p -y_qy_p -z_qz_p +(w_qx_p +x_qw_p +y_qz_p -z_qy_p)i)(-y_q) +(w_qw_p -x_qx_p -y_qy_p -z_qz_p +(w_qx_p +x_qw_p +y_qz_p -z_qy_p)i)(-z_qi) \\&=& -w_pw_qy_q +x_px_qy_q +y_py_qy_q +z_pz_qy_q +w_qx_pz_q +w_px_qz_q +y_qz_pz_q -y_pz_qz_q +(-w_qx_py_q -w_px_qy_q -y_qy_qz_p +y_py_qz_q -w_pw_qz_q +x_px_qz_q +y_py_qz_q +z_pz_qz_q)i \end{eqnarray} $$ (1,2)要素の第二項
$$ \begin{eqnarray} ( w_qy_p +y_qw_p +z_qx_p -x_qz_p +(w_qz_p +z_qw_p +x_qy_p -y_qx_p)i)(w_q+x_qi) &=& ( w_qy_p +y_qw_p +z_qx_p -x_qz_p +(w_qz_p +z_qw_p +x_qy_p -y_qx_p)i)w_q +( w_qy_p +y_qw_p +z_qx_p -x_qz_p +(w_qz_p +z_qw_p +x_qy_p -y_qx_p)i)x_qi \\&=& w_qw_qy_p +w_pw_qy_q +w_qx_pz_q -w_qx_qz_p -w_qx_qz_p -w_px_qz_q -x_qx_qy_p +x_px_qy_q +( w_qw_qz_p +w_pw_qz_q +w_qx_qy_p -w_qx_py_q +w_qx_qy_p +w_px_qy_q +x_px_qz_q -x_qx_qz_p )i \end{eqnarray} $$ (1,2)要素の第一，二項の和
$$ \begin{eqnarray} (w_qw_p-x_qx_p -y_qy_p -z_qz_p+(w_qx_p+x_qw_p +y_qz_p -z_qy_p)i)(-y_q-z_qi)\\ + (w_qy_p +y_qw_p +z_qx_p -x_qz_p+(w_qz_p +z_qw_p +x_qy_p -y_qx_p)i)(w_q+x_qi) &=& -w_pw_qy_q +x_px_qy_q +y_py_qy_q +z_pz_qy_q +w_qx_pz_q +w_px_qz_q +y_qz_pz_q -y_pz_qz_q +w_qw_qy_p +w_pw_qy_q +w_qx_pz_q -w_qx_qz_p -w_qx_qz_p -w_px_qz_q -x_qx_qy_p +x_px_qy_q +( -w_qx_py_q -w_px_qy_q -y_qy_qz_p +y_py_qz_q -w_pw_qz_q +x_px_qz_q +y_py_qz_q +z_pz_qz_q +w_qw_qz_p +w_pw_qz_q +w_qx_qy_p -w_qx_py_q +w_qx_qy_p +w_px_qy_q +x_px_qz_q -x_qx_qz_p )i \\&=& 2(w_qz_q+x_qy_q)x_p+(w_q^2-x_q^2+y_q^2-z_q^2)y_p+2(y_qz_q-w_qx_q)z_p +( 2(x_qz_q-w_qy_q)x_p+2(w_qx_q+y_qz_q)y_p+(w_q^2-x_q^2-y_q^2+z_q^2)z_p )i \end{eqnarray} $$

(2,1)要素

(2,1)要素の第一項
$$ \begin{eqnarray} (-w_qy_p +x_qz_p -y_qw_p -z_qx_p +(w_qz_p +x_qy_p -y_qx_p +z_qw_p)i)(w_q-x_qi) &=& -w_qy_p +x_qz_p -y_qw_p -z_qx_p +( w_qz_p +x_qy_p -y_qx_p +z_qw_p )i )(w_q) -w_qy_p +x_qz_p -y_qw_p -z_qx_p +( w_qz_p +x_qy_p -y_qx_p +z_qw_p )i )(-x_qi) \\&=& -w_qy_pw_q +x_qz_pw_q -y_qw_pw_q -z_qx_pw_q +w_qz_px_q +x_qy_px_q -y_qx_px_q +z_qw_px_q +( w_qz_pw_q +x_qy_pw_q -y_qx_pw_q +z_qw_pw_q +w_qy_px_q -x_qz_px_q +y_qw_px_q +z_qx_px_q )i \end{eqnarray} $$ (2,1)要素の第二項
$$ \begin{eqnarray} ( w_qw_p -x_qx_p -y_qy_p -z_qz_p -( w_qx_p +x_qw_p +y_qz_p -z_qy_p )i )(y_q-z_qi) &=& ( w_qw_p -x_qx_p -y_qy_p -z_qz_p -( w_qx_p +x_qw_p +y_qz_p -z_qy_p )i )(y_q) +( w_qw_p -x_qx_p -y_qy_p -z_qz_p -( w_qx_p +x_qw_p +y_qz_p -z_qy_p )i )(-z_qi) \\&=& w_qw_py_q -x_qx_py_q -y_qy_py_q -z_qz_py_q -w_qx_pz_q -x_qw_pz_q -y_qz_pz_q +z_qy_pz_q +( -w_qx_py_q -x_qw_py_q -y_qz_py_q +z_qy_py_q -w_qw_pz_q +x_qx_pz_q +y_qy_pz_q +z_qz_pz_q )i \end{eqnarray} $$ (2,1)要素の第一，二項の和
$$ \begin{eqnarray} ( -w_qy_p +x_qz_p -y_qw_p -z_qx_p +( w_qz_p +x_qy_p -y_qx_p +z_qw_p )i )(w_q-x_qi) \\ +( w_qw_p -x_qx_p -y_qy_p -z_qz_p -( w_qx_p +x_qw_p +y_qz_p -z_qy_p )i )(y_q-z_qi) &=& -w_qy_pw_q +x_qz_pw_q -y_qw_pw_q -z_qx_pw_q +w_qz_px_q +x_qy_px_q -y_qx_px_q +z_qw_px_q +( w_qz_pw_q +x_qy_pw_q -y_qx_pw_q +z_qw_pw_q +w_qy_px_q -x_qz_px_q +y_qw_px_q +z_qx_px_q )i + w_qw_py_q -x_qx_py_q -y_qy_py_q -z_qz_py_q -w_qx_pz_q -x_qw_pz_q -y_qz_pz_q +z_qy_pz_q +( -w_qx_py_q -x_qw_py_q -y_qz_py_q +z_qy_py_q -w_qw_pz_q +x_qx_pz_q +y_qy_pz_q +z_qz_pz_q )i \\&=& -w_qy_pw_q +x_qz_pw_q -y_qw_pw_q -z_qx_pw_q +w_qz_px_q +x_qy_px_q -y_qx_px_q +z_qw_px_q +w_qw_py_q -x_qx_py_q -y_qy_py_q -z_qz_py_q -w_qx_pz_q -x_qw_pz_q -y_qz_pz_q +z_qy_pz_q +( w_qz_pw_q +x_qy_pw_q -y_qx_pw_q +z_qw_pw_q +w_qy_px_q -x_qz_px_q +y_qw_px_q +z_qx_px_q -w_qx_py_q -x_qw_py_q -y_qz_py_q +z_qy_py_q -w_qw_pz_q +x_qx_pz_q +y_qy_pz_q +z_qz_pz_q )i \\&=& -2(w_qz_q+x_qy_q)x_p(-w_q^2+x_q^2-y_q^2+z_q^2)y_p+2(w_qx_q-y_qz_q)z_p +( (w_q^2-x_q^2-y_q^2+z_q^2)z_p+2(w_qx_q+y_qz_q)y_p+2(x_qz_q-w_qy_q)x_p )i \\&=& -( 2(w_qz_q+x_qy_q)x_p+(w_q^2-x_q^2+y_q^2-z_q^2)y_p+2(y_qz_q-w_qx_q)z_p -( 2(x_qz_q-w_qy_q)x_p+2(w_qx_q+y_qz_q)y_p+(w_q^2-x_q^2-y_q^2+z_q^2)z_p )i ) \end{eqnarray} $$

(2,2)要素

(2,2)要素の第一項
$$ \begin{eqnarray} (-w_qy_p +x_qz_p -y_qw_p -z_qx_p +(w_qz_p +x_qy_p -y_qx_p +z_qw_p)i)(-y_q-z_qi) &=& (-w_qy_p +x_qz_p -y_qw_p -z_qx_p +(w_qz_p +x_qy_p -y_qx_p +z_qw_p)i)(-y_q) +(-w_qy_p +x_qz_p -y_qw_p -z_qx_p +(w_qz_p +x_qy_p -y_qx_p +z_qw_p)i)(-z_qi) \\&=& w_qy_py_q -x_qz_py_q +y_qw_py_q +z_qx_py_q +w_qz_pz_q +x_qy_pz_q -y_qx_pz_q +z_qw_pz_q +( -w_qz_py_q -x_qy_py_q +y_qx_py_q -z_qw_py_q +w_qy_pz_q -x_qz_pz_q +y_qw_pz_q +z_qx_pz_q )i \end{eqnarray} $$ (2,2)要素の第二項
$$ \begin{eqnarray} (w_qw_p -x_qx_p -y_qy_p -z_qz_p -(w_qx_p +x_qw_p +y_qz_p -z_qy_p)i)(w_q+x_qi) &=& (w_qw_p -x_qx_p -y_qy_p -z_qz_p -(w_qx_p +x_qw_p +y_qz_p -z_qy_p)i)(w_q) +(w_qw_p -x_qx_p -y_qy_p -z_qz_p -(w_qx_p +x_qw_p +y_qz_p -z_qy_p)i)(x_qi) \\&=& w_qw_pw_q -x_qx_pw_q -y_qy_pw_q -z_qz_pw_q +w_qx_px_q +x_qw_px_q +y_qz_px_q -z_qy_px_q +( -w_qx_pw_q -x_qw_pw_q -y_qz_pw_q +z_qy_pw_q +w_qw_px_q -x_qx_px_q -y_qy_px_q -z_qz_px_q )i \end{eqnarray} $$ (2,1)要素の第一，二項の和
$$ \begin{eqnarray} (-w_qy_p +x_qz_p -y_qw_p -z_qx_p +(w_qz_p +x_qy_p -y_qx_p +z_qw_p)i)(-y_q-z_qi) \\+(w_qw_p -x_qx_p -y_qy_p -z_qz_p -(w_qx_p +x_qw_p +y_qz_p -z_qy_p)i)(w_q+x_qi) &=& (w_qy_py_q -x_qz_py_q +y_qw_py_q +z_qx_py_q +w_qz_pz_q +x_qy_pz_q -y_qx_pz_q +z_qw_pz_q +( -w_qz_py_q -x_qy_py_q +y_qx_py_q -z_qw_py_q +w_qy_pz_q -x_qz_pz_q +y_qw_pz_q +z_qx_pz_q )i) + (w_qw_pw_q -x_qx_pw_q -y_qy_pw_q -z_qz_pw_q +w_qx_px_q +x_qw_px_q +y_qz_px_q -z_qy_px_q +( -w_qx_pw_q -x_qw_pw_q -y_qz_pw_q +z_qy_pw_q +w_qw_px_q -x_qx_px_q -y_qy_px_q -z_qz_px_q )i) \\&=& w_qy_py_q -x_qz_py_q +y_qw_py_q +z_qx_py_q +w_qz_pz_q +x_qy_pz_q -y_qx_pz_q +z_qw_pz_q +w_qw_pw_q -x_qx_pw_q -y_qy_pw_q -z_qz_pw_q +w_qx_px_q +x_qw_px_q +y_qz_px_q -z_qy_px_q +( -w_qz_py_q -x_qy_py_q +y_qx_py_q -z_qw_py_q +w_qy_pz_q -x_qz_pz_q +y_qw_pz_q +z_qx_pz_q -w_qx_pw_q -x_qw_pw_q -y_qz_pw_q +z_qy_pw_q +w_qw_px_q -x_qx_px_q -y_qy_px_q -z_qz_px_q )i) \\&=& (w_q^2+x_q^2+y_q^2+z_q^2)w_p +( (-w_p^2-x_q^2+y_q^2+z_q^2)x_p+2(w_qz_q-x_qy_q)y_p-2(w_qy_q+x_qz_q)z_p )i \\&=& (w_q^2+x_q^2+y_q^2+z_q^2)w_p -( (w_p^2+x_q^2-y_q^2-z_q^2)x_p+2(x_qy_q-w_qz_q)y_p+2(w_qy_q+x_qz_q)z_p )i \end{eqnarray} $$

四元数の行列表現での逆元

$$ \begin{eqnarray} \mathbf{q}=w+xi+yj+zk &\leftrightarrow& \href{https://shikitenkai.blogspot.com/2020/06/blog-post_99.html}{ \begin{bmatrix} w+xi&y+zi\\ -(y-zi)&w-xi\\ \end{bmatrix} } \end{eqnarray} $$ $$ \begin{eqnarray} |\mathbf{q}|^2=|w+xi+yj+zk|^2 &\leftrightarrow& \begin{vmatrix} w+xi&y+zi\\ -(y-zi)&w-xi\\ \end{vmatrix} \\&=&(w+xi)(w-xi)-(y+zi)(-(y-zi)) \\&=&(w+xi)(w-xi)+(y+zi)(y-zi) \\&=&w^2+x^2+y^2+z^2 \end{eqnarray} $$ $$ \begin{eqnarray} \bar{\mathbf{q}}=w-xi-yj-zk &\leftrightarrow& \begin{bmatrix} w+(-x)i&(-y)+(-z)i\\ -((-y)-(-z)i)&w-(-x)i\\ \end{bmatrix} \\&=& \begin{bmatrix} w-xi&-y-zi\\ -(-y+zi)&w+xi\\ \end{bmatrix} \end{eqnarray} $$ $$ \begin{eqnarray} \mathbf{q}\bar{\mathbf{q}} &=&(w+xi+yj+zk)(w-xi-yj-zk) \\&\leftrightarrow& \begin{bmatrix} w+xi&y+zi\\ -(y-zi)&w-xi\\ \end{bmatrix} \begin{bmatrix} w-xi&-y-zi\\ -(-y+zi)&w+xi\\ \end{bmatrix} \\&=& \begin{bmatrix} (w+xi)(w-xi)+(y+zi)(-(-y+zi)) &(w+xi)(-y-zi)+(y+zi)(w+xi)\\ (-(y-zi))(w-xi)+(w-xi)(-(-y+zi)) &(-(y-zi))(-y-zi)+(w-xi)(w+xi)\\ \end{bmatrix} \\&=& \begin{bmatrix} (w+xi)(w-xi)+(y+zi)(y-zi) &(w+xi)(-y-zi)+(y+zi)(w+xi)\\ (-y+zi)(w-xi)+(w-xi)(y-zi) &(-y+zi)(-y-zi)+(w-xi)(w+xi)\\ \end{bmatrix} \\&=& \begin{bmatrix} w^2-wxi+wxi+x^2 +y^2-yzi+yzi+z^2 &-wy-wzi-xyi+xz +wy+xyi+wzi-xz\\ -wy+xyi+wzi+xz +wy-wzi-xyi-xz &y^2+yzi-yzi+z^2 +w^2+wxi-wxi+x^2\\ \end{bmatrix} \\&=& \begin{bmatrix} w^2+x^2+y^2+z^2 &0\\ 0 &w^2+x^2+y^2+z^2\\ \end{bmatrix} \end{eqnarray} $$ $$ \begin{eqnarray} \bar{\mathbf{q}}\mathbf{q} &=&(w-xi-yj-zk)(w+xi+yj+zk) \\&\leftrightarrow& \begin{bmatrix} w-xi&-y-zi\\ -(-y+zi)&w+xi\\ \end{bmatrix} \begin{bmatrix} w+xi&y+zi\\ -(y-zi)&w-xi\\ \end{bmatrix} \\&=& \begin{bmatrix} (w-xi) (w+xi) +(-y-zi) (-(y-zi)) & (w-xi) (y+zi) +(-y-zi) (w-xi)\\ (-(-y+zi)) (w+xi) +(w+xi) (-(y-zi)) & (-(-y+zi)) (y+zi) +(w+xi) (w-xi)\\ \end{bmatrix} \\&=& \begin{bmatrix} (w-xi) (w+xi) +(-y-zi) (-y+zi) & (w-xi) (y+zi) +(-y-zi) (w-xi)\\ (y-zi) (w+xi) +(w+xi) (-y+zi) & (y-zi) (y+zi) +(w+xi) (w-xi)\\ \end{bmatrix} \\&=& \begin{bmatrix} w^2 +wxi -wxi +x^2 +y^2 +yzi -yzi +z^2 &wy +wzi -xyi +xz -wy +xyi -wzi -xz\\ wy +xyi -wzi +xz -wy +wzi -xyi -xz &y^2 +yzi -yzi +z^2 +w^2 -wxi +wxi +x^2\\ \end{bmatrix} \\&=& \begin{bmatrix} w^2+x^2+y^2+z^2 &0\\ 0 &w^2+x^2+y^2+z^2\\ \end{bmatrix} \;\cdots\;行列の積は一般に非可換だが，この形の行列では可換である \end{eqnarray} $$ $$ \begin{eqnarray} \frac{\mathbf{q}\bar{\mathbf{q}}}{|\mathbf{q}|^2} &=&\frac{\bar{\mathbf{q}}\mathbf{q}}{|\mathbf{q}|^2} =\mathbf{q}\frac{\bar{\mathbf{q}}}{|\mathbf{q}|^2} =\frac{\bar{\mathbf{q}}}{|\mathbf{q}|^2}\mathbf{q} =\frac{1}{|\mathbf{q}|^2}\mathbf{q}\bar{\mathbf{q}} =\frac{1}{|\mathbf{q}|^2}\bar{\mathbf{q}}\mathbf{q} \\&\leftrightarrow& \frac{1}{w^2+x^2+y^2+z^2} \begin{bmatrix} w^2+x^2+y^2+z^2 &0\\ 0 &w^2+x^2+y^2+z^2\\ \end{bmatrix} \\&=& \begin{bmatrix} 1&0\\ 0&1\\ \end{bmatrix} \\&=&\mathbf{E} \end{eqnarray} $$ $$ \begin{eqnarray} \mathbf{q}^{-1} &=&\frac{\bar{\mathbf{q}}}{|\mathbf{q}|^2} \\&\leftrightarrow& \frac{1}{w^2+x^2+y^2+z^2} \begin{bmatrix} w-xi&-y-zi\\ -(-y+zi)&w+xi\\ \end{bmatrix} \end{eqnarray} $$

四元数の行列表現での積

$$ \begin{eqnarray} w+xi+yj+zk &\leftrightarrow& \href{https://shikitenkai.blogspot.com/2020/06/blog-post_99.html}{ \begin{bmatrix} w+xi&y+zi\\ -(y-zi)&w-xi\\ \end{bmatrix} } \\ \\ (w_1+x_1i+y_1j+z_1k)(w_2+x_2i+y_2j+z_2k) &\leftrightarrow& \begin{bmatrix} w_1+x_1i&y_1+z_1i\\ -(y_1-z_1i)&w_1-x_1i\\ \end{bmatrix} \begin{bmatrix} w_2+x_2i&y_2+z_2i\\ -(y_2-z_2i)&w_2-x_2i\\ \end{bmatrix} \\&=& \begin{bmatrix} (w_1+x_1i)(w_2+x_2i)+(y_1+z_1i)(-(y_2-z_2i)) & (w_1+x_1i)(y_2+z_2i)+(y_1+z_1i)(w_2-x_2i) \\(-(y_1-z_1i))(w_2+x_2i)+(w_1-x_1i)(-(y_2-z_2i)) & (-(y_1-z_1i))(y_2+z_2i)+(w_1-x_1i)(w_2-x_2i)\\ \end{bmatrix} \\&=& \begin{bmatrix} w_1w_2 +w_1x_2i +x_1iw_2 +x_1ix_2i +y_1(-y_2) +y_1z_2i +z_1i(-y_2) +z_1iz_2i & w_1y_2 +w_1z_2i +x_1iy_2 +x_1iz_2i +y_1w_2 +y_1(-x_2i) +z_1iw_2 +z_1i(-x_2i) \\(-y_1)w_2 +(-y_1)x_2i +z_1iw_2 +z_1ix_2i +w_1(-y_2) +w_1z_2i +(-x_1i)(-y_2) +(-x_1i)z_2i & (-y_1)y_2 +(-y_1)z_2i +z_1iy_2 +z_1iz_2i +w_1w_2 +w_1(-x_2i) +(-x_1i)w_2 +(-x_1i)(-x_2i) \end{bmatrix} \\&=& \begin{bmatrix} ( w_1w_2 -x_1x_2 -y_1y_2 -z_1z_2) +( w_1x_2 +w_2x_1 +y_1z_2 -z_1y_2)i & ( w_1y_2 +w_2y_1 +z_1x_2 -x_1z_2) +( w_1z_2 +w_2z_1 +x_1y_2 -y_1x_2)i \\-((w_1y_2 +w_2y_1 +z_1x_2 -x_1z_2) -(w_1z_2 +w_2z_1 +x_1y_2 -y_1x_2)i) & ( w_1w_2 -x_1x_2 -y_1y_2 -z_1z_2) -( w_1x_2 +w_2x_1 +y_1z_2 -z_1y_2)i \end{bmatrix} \\&\leftrightarrow& ( w_1w_2 -x_1x_2 -y_1y_2 -z_1z_2) \\&&+( w_1x_2 +w_2x_1 +y_1z_2 -z_1y_2)i \\&&+( w_1y_2 +w_2y_1 +z_1x_2 -x_1z_2)j \\&&+( w_1z_2 +w_2z_1 +x_1y_2 -y_1x_2)k \end{eqnarray} $$ $$ \begin{eqnarray} (w_2+x_2i+y_2j+z_2k)(w_1+x_1i+y_1j+z_1k) &\leftrightarrow& \begin{bmatrix} w_2+x_2i&y_2+z_2i\\ -(y_2-z_2i)&w_2-x_2i\\ \end{bmatrix} \begin{bmatrix} w_1+x_1i&y_1+z_1i\\ -(y_1-z_1i)&w_1-x_1i\\ \end{bmatrix} \\&=& \begin{bmatrix} (w_2+x_2i)(w_1+x_1i)+(y_2+z_2i)(-(y_1-z_1i)) & (w_2+x_2i)(y_1+z_1i)+(y_2+z_2i)(w_1-x_1i) \\(-(y_2-z_2i))(w_1+x_1i)+(w_2-x_2i)(-(y_1-z_1i)) & (-(y_2-z_2i))(y_1+z_1i)+(w_2-x_2i)(w_1-x_1i)\\ \end{bmatrix} \\&=& \begin{bmatrix} w_2w_1 +w_1x_2i +x_1iw_2 +x_1ix_2i +y_2(-y_1) +y_2z_1i +z_2i(-y_1) +z_2iz_1i & w_2y_1 +w_2z_1i +x_2iy_1 +x_2iz_1i +y_2w_1 +y_2(-x_1i) +z_2iw_1 +z_2i(-x_1i) \\(-y_2)w_1 +(-y_2)x_1i +z_2iw_1 +z_2ix_1i +w_2(-y_1) +w_2z_1i +(-x_2i)(-y_1) +(-x_2i)z_1i & (-y_2)y_1 +(-y_2)z_1i +z_2iy_1 +z_2iz_1i +w_2w_1 +w_2(-x_1i) +(-x_2i)w_1 +(-x_2i)(-x_1i) \end{bmatrix} \\&=& \begin{bmatrix} ( w_1w_2 -x_1x_2 -y_1y_2 -z_1z_2) +( w_1x_2 +w_2x_1 +y_1z_2 -z_1y_2)i & ( w_1y_2 +w_2y_1 +z_1x_2 -x_1z_2) +( w_1z_2 +w_2z_1 +x_1y_2 -y_1x_2)i \\-((w_1y_2 +w_2y_1 +z_1x_2 -x_1z_2) -(w_1z_2 +w_2z_1 +x_1y_2 -y_1x_2)i) & ( w_1w_2 -x_1x_2 -y_1y_2 -z_1z_2) -( w_1x_2 +w_2x_1 +y_1z_2 -z_1y_2)i \end{bmatrix} \\&&\;\cdots\;行列の積は一般に非可換だが，この形の行列では可換である \\&\leftrightarrow& ( w_1w_2 -x_1x_2 -y_1y_2 -z_1z_2) \\&&+( w_1x_2 +w_2x_1 +y_1z_2 -z_1y_2)i \\&&+( w_1y_2 +w_2y_1 +z_1x_2 -x_1z_2)j \\&&+( w_1z_2 +w_2z_1 +x_1y_2 -y_1x_2)k \end{eqnarray} $$
四元数の積としては以下のようにまとまる． $$ \begin{eqnarray} w&=&( w_1w_2 -x_1x_2 -y_1y_2 -z_1z_2) \\x&=&( w_1x_2 +w_2x_1 +y_1z_2 -z_1y_2) \\y&=&( w_1y_2 +w_2y_1 +z_1x_2 -x_1z_2) \\z&=&( w_1z_2 +w_2z_1 +x_1y_2 -y_1x_2) \end{eqnarray} $$ これは以前の計算結果と一致する．

四元数の行列表現

$$ \begin{eqnarray} w+xi+yj+zk &\leftrightarrow& \begin{bmatrix} w+xi&y+zi\\ -(y-zi)&w-xi\\ \end{bmatrix} \\&=& w\begin{bmatrix} 1&0\\ 0&1\\ \end{bmatrix} +x\begin{bmatrix} i&0\\ 0&-i\\ \end{bmatrix} +y\begin{bmatrix} 0&1\\ -1&0\\ \end{bmatrix} +z\begin{bmatrix} 0&i\\ i&0\\ \end{bmatrix} \\&=&w\mathbf{E}+x\mathbf{I}+y\mathbf{J}+z\mathbf{K} \end{eqnarray} $$

四元数の単位同士の積の確認

$$ \begin{eqnarray} i\cdot i \leftrightarrow \mathbf{I}\cdot\mathbf{I} &=& \begin{bmatrix} i&0\\ 0&-i\\ \end{bmatrix} \cdot \begin{bmatrix} i&0\\ 0&-i\\ \end{bmatrix} \\&=&\begin{bmatrix} i\cdot i+0\cdot0&i\cdot0+0\cdot (-i)\\ 0\cdot i+(-i)\cdot0&0\cdot 0+(-i)\cdot(-i)\\ \end{bmatrix} \\&=& \begin{bmatrix} -1&0\\ 0&-1\\ \end{bmatrix} \\&=&-\mathbf{E} \\j\cdot j \leftrightarrow \mathbf{J}\cdot\mathbf{J} &=& \begin{bmatrix} 0&1\\ -1&0\\ \end{bmatrix} \cdot \begin{bmatrix} 0&1\\ -1&0\\ \end{bmatrix} \\&=&\begin{bmatrix} 0\cdot0+1\cdot(-1)&0\cdot1+1\cdot0\\ (-1)\cdot0+0\cdot(-1)&(-1)\cdot1+0\cdot0\\ \end{bmatrix} \\&=& \begin{bmatrix} -1&0\\ 0&-1\\ \end{bmatrix} \\&=&-\mathbf{E} \\k\cdot k \leftrightarrow \mathbf{K}\cdot\mathbf{K} &=& \begin{bmatrix} 0&i\\ i&0\\ \end{bmatrix} \cdot \begin{bmatrix} 0&i\\ i&0\\ \end{bmatrix} \\&=&\begin{bmatrix} 0\cdot0+i\cdot i&0\cdot i+i\cdot0\\ i\cdot0+0\cdot i&i\cdot i+0\cdot0\\ \end{bmatrix} \\&=& \begin{bmatrix} -1&0\\ 0&-1\\ \end{bmatrix} \\&=&-\mathbf{E} \\i\cdot j \leftrightarrow \mathbf{I}\cdot\mathbf{J} &=& \begin{bmatrix} i&0\\ 0&-i\\ \end{bmatrix} \cdot \begin{bmatrix} 0&1\\ -1&0\\ \end{bmatrix} \\&=&\begin{bmatrix} i\cdot0+0 \cdot (-1)&i\cdot 1+0 \cdot0\\ 0\cdot0+(-i)\cdot (-1)&0\cdot 1+(-i)\cdot0\\ \end{bmatrix} \\&=& \begin{bmatrix} 0&i\\ i&0\\ \end{bmatrix} \\&=&\mathbf{K} \\j\cdot i \leftrightarrow \mathbf{J}\cdot\mathbf{I} &=& \begin{bmatrix} 0&1\\ -1&0\\ \end{bmatrix} \cdot \begin{bmatrix} i&0\\ 0&-i\\ \end{bmatrix} \\&=&\begin{bmatrix} 0 \cdot i+1 \cdot 0&0 \cdot 0+1\cdot(-i)\\ (-1)\cdot i+0 \cdot 0&(-1)\cdot 0+0\cdot(-i)\\ \end{bmatrix} \\&=& \begin{bmatrix} 0&-i\\ -i&0\\ \end{bmatrix} \\&=&-\mathbf{K} \\i\cdot k \leftrightarrow \mathbf{I}\cdot\mathbf{K} &=& \begin{bmatrix} i&0\\ 0&-i\\ \end{bmatrix} \cdot \begin{bmatrix} 0&i\\ i&0\\ \end{bmatrix} \\&=&\begin{bmatrix} i\cdot0+0 \cdot i&i\cdot i+0 \cdot0\\ 0\cdot0+(-i)\cdot i&0\cdot i+(-i)\cdot0\\ \end{bmatrix} \\&=& \begin{bmatrix} 0&-1\\ 1&0\\ \end{bmatrix} \\&=&-\mathbf{J} \\k\cdot i \leftrightarrow \mathbf{K}\cdot\mathbf{I} &=& \begin{bmatrix} 0&i\\ i&0\\ \end{bmatrix} \cdot \begin{bmatrix} i&0\\ 0&-i\\ \end{bmatrix} \\&=&\begin{bmatrix} 0\cdot i+i\cdot0&0\cdot0+i\cdot(-i)\\ i\cdot i+0\cdot0&i\cdot0+0\cdot(-i)\\ \end{bmatrix} \\&=& \begin{bmatrix} 0&1\\ -1&0\\ \end{bmatrix} \\&=&\mathbf{J} \\j\cdot k \leftrightarrow \mathbf{J}\cdot\mathbf{K} &=& \begin{bmatrix} 0 & 1\\ -1 & 0\\ \end{bmatrix} \cdot \begin{bmatrix} 0&i\\ i&0\\ \end{bmatrix} \\&=&\begin{bmatrix} 0 \cdot 0+1\cdot i&0 \cdot i+1\cdot 0\\ (-1)\cdot 0+0\cdot i&(-1)\cdot i+0\cdot 0\\ \end{bmatrix} \\&=& \begin{bmatrix} i&0\\ 0&-i\\ \end{bmatrix} \\&=&\mathbf{I} \\k\cdot j \leftrightarrow \mathbf{K}\cdot\mathbf{J} &=& \begin{bmatrix} 0&i\\ i&0\\ \end{bmatrix} \cdot \begin{bmatrix} 0 & 1\\ -1 & 0\\ \end{bmatrix} \\&=&\begin{bmatrix} 0 \cdot 0+i\cdot (-1)&0 \cdot 1+i\cdot 0\\ i \cdot 0+0\cdot (-1)&i \cdot 1+0\cdot 0\\ \end{bmatrix} \\&=& \begin{bmatrix} -i&0\\ 0&i\\ \end{bmatrix} \\&=&-\mathbf{I} \\ \end{eqnarray} $$

複素数の行列表現

$$ \begin{eqnarray} a+bi &\leftrightarrow& \begin{bmatrix} a&-b\\ b&a\\ \end{bmatrix} \\ \\積 \\ (a+bi)\cdot(c+di) &=&ac+adi+bic+bidi \\&=&ac-bd+adi+bci \\&=&(ac-bd)+(ad+bc)i \\ \begin{bmatrix} a&-b\\ b&a\\ \end{bmatrix} \cdot \begin{bmatrix} c&-d\\ d&c\\ \end{bmatrix} &=& \begin{bmatrix} ac-bd&-(ad+bc)\\ ad+bc&ac-bd\\ \end{bmatrix} \\ \\積(交換) \\ (c+di)\cdot(a+bi) &=&(ac-bd)+(ad+bc)i \\ \begin{bmatrix} c&-d\\ d&c\\ \end{bmatrix} \cdot \begin{bmatrix} a&-b\\ b&a\\ \end{bmatrix} &=& \begin{bmatrix} ac-bd&-(ad+bc)\\ ad+bc&ac-bd\\ \end{bmatrix} \;\cdots\;行列の積は一般に非可換だが，この形の行列では可換である \\ \\和 \\ (a+ib)+(c+id) &=&(a+c)+(b+d)i \\ \begin{bmatrix} a&-b\\ b&a\\ \end{bmatrix} + \begin{bmatrix} c&-d\\ d&c\\ \end{bmatrix} &=& \begin{bmatrix} a+c&-(b+d)\\ b+d&a+c\\ \end{bmatrix} \\ \\差 \\ (a+bi)-(c+di) &=&(a-c)+(b-d)i \\ \begin{bmatrix} a&-b\\ b&a\\ \end{bmatrix} - \begin{bmatrix} c&-d\\ d&c\\ \end{bmatrix} &=& \begin{bmatrix} a-c&-(b-d)\\ b-d&a-c\\ \end{bmatrix} \\ \\逆数 \\ \frac{1}{c+di} &=&\frac{1}{c+di}\frac{c-di}{c-di} \\&=&\frac{c-di}{c^2+d^2} \\ \frac{1}{ \begin{bmatrix} c&-d\\ d&c\\ \end{bmatrix} } &=& \begin{bmatrix} c&-d\\ d&c\\ \end{bmatrix}^{-1} \;\cdots\;\frac{1}{A}=A^{-1} \\&=& \frac{1}{ \begin{vmatrix} c&d\\ -d&c\\ \end{vmatrix} } \begin{bmatrix} c&d\\ -d&c\\ \end{bmatrix} \;\cdots\; \begin{bmatrix} a&b\\ c&d\\ \end{bmatrix}^{-1} = \frac{1}{ \begin{vmatrix} a&b\\ c&d\\ \end{vmatrix} } \begin{bmatrix} d&-b\\ -c&a\\ \end{bmatrix} \\&=& \frac{1}{c^2+d^2} \begin{bmatrix} c&d\\ -d&c\\ \end{bmatrix} \\ \\商 \\ \frac{a+bi}{c+di} &=&\frac{(a+bi)(c-di)}{(c+di)(c-di)} \\&=&\frac{(ac+bd)+(bc-ad)i}{c^2+d^2} \\ \frac{ \begin{bmatrix} a&-b\\ b&a\\ \end{bmatrix} }{ \begin{bmatrix} c&-d\\ d&c\\ \end{bmatrix} } &=& \begin{bmatrix} a&-b\\ b&a\\ \end{bmatrix} \cdot \frac{1}{ \begin{bmatrix} c&-d\\ d&c\\ \end{bmatrix} } \\&=& \begin{bmatrix} a&-b\\ b&a\\ \end{bmatrix} \cdot \left( \frac{1}{c^2+d^2} \begin{bmatrix} c&d\\ -d&c\\ \end{bmatrix} \right) \;\cdots\; \frac{1}{ \begin{bmatrix} c&-d\\ d&c\\ \end{bmatrix} } = \frac{1}{c^2+d^2} \begin{bmatrix} c&d\\ -d&c\\ \end{bmatrix} \\&=& \frac{1}{c^2+d^2} \begin{bmatrix} a&-b\\ b&a\\ \end{bmatrix} \cdot \begin{bmatrix} c&d\\ -d&c\\ \end{bmatrix} \\&=& \frac{1}{c^2+d^2} \begin{bmatrix} ac+bd&-(bc-ad)\\ bc-ad&ac+bd\\ \end{bmatrix} \end{eqnarray} $$

外積に外積を組み合わされた式　( ベクトル三重積 )

外積に外積を組み合わされた式

$$ \begin{eqnarray} \left(\mathbf{A}\times\mathbf{B}\right)\times\mathbf{C} &=& \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ x_a&y_a&z_a\\ x_b&y_b&z_b\\ \end{vmatrix} \cdot \left(x_c\mathbf{i}+y_c\mathbf{j}+z_c\mathbf{k}\right) \\&=& \left( \left(y_az_b-y_bz_a\right)\mathbf{i} +\left(x_bz_a-x_az_b\right)\mathbf{j} +\left(x_ay_b-x_by_a\right)\mathbf{k} \right) \times \left( x_c\mathbf{i} +y_c\mathbf{j} +z_c\mathbf{k} \right) \\&=& \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ \left(y_az_b-y_bz_a\right)\mathbf{i} &\left(x_bz_a-x_az_b\right)\mathbf{j} &\left(x_ay_b-x_by_a\right)\mathbf{k}\\ x_c&y_c&z_c\\ \end{vmatrix} \\&=& \left\{ \left(x_bz_a-x_az_b\right)z_c-\left(x_ay_b-x_by_a\right)y_c \right\}\mathbf{i} \\&&+\left\{ \left(x_ay_b-x_by_a\right)x_c-\left(y_az_b-y_bz_a\right)z_c \right\}\mathbf{j} \\&&+\left\{ \left(y_az_b-y_bz_a\right)y_c-\left(x_bz_a-x_az_b\right)x_c \right\}\mathbf{k} \\&=& \left\{ \left(x_b\;z_az_c-x_a\;z_bz_c\right)-\left(x_a\;y_by_c-x_b\;y_ay_c\right) \right\}\mathbf{i} \\&&+\left\{ \left(y_b\;x_ax_c-y_a\;x_bx_c\right)-\left(y_a\;z_bz_c-y_b\;z_az_c\right) \right\}\mathbf{j} \\&&+\left\{ \left(z_b\;y_ay_c-z_a\;y_by_c\right)-\left(z_a\;x_bx_c-z_b\;x_ax_c\right) \right\}\mathbf{k} \\&=& \left\{\left(x_b\;z_az_c-x_a\;z_bz_c\right)-\left(x_a\;y_by_c-x_b\;y_ay_c\right) {\color{red}+x_b\;x_ax_c-x_a\;x_bx_c} \right\}\mathbf{i} \\&&+\left\{\left(y_b\;x_ax_c-y_a\;x_bx_c\right)-\left(y_a\;z_bz_c-y_b\;z_az_c\right) {\color{red}+y_b\;y_ay_c-y_a\;y_by_c} \right\}\mathbf{j} \\&&+\left\{\left(z_b\;y_ay_c-z_a\;y_by_c\right)-\left(z_a\;x_bx_c-z_b\;x_ax_c\right) {\color{red}+z_b\;z_az_c-z_a\;z_bz_c} \right\}\mathbf{k} \\&=& \left\{ \left(x_b\;x_ax_c+x_b\;y_ay_c+x_b\;z_az_c\right)\mathbf{i} +\left(y_b\;x_ax_c+y_b\;y_ay_c+y_b\;z_az_c\right)\mathbf{j} +\left(z_b\;x_ax_c+z_b\;y_ay_c+z_b\;z_az_c\right)\mathbf{k} \right\} \\&&-\left\{ \left(x_a\;x_bx_c+x_a\;y_by_c+x_a\;z_bz_c\right)\mathbf{i} +\left(y_a\;x_bx_c+y_a\;y_by_c+y_a\;z_bz_c\right)\mathbf{j} +\left(z_a\;x_bx_c+z_a\;y_by_c+z_a\;z_bz_c\right)\mathbf{k} \right\} \\&=& \left\{ \left(x_ax_c+y_ay_c+z_az_c\right)x_b\;\mathbf{i} +\left(x_ax_c+y_ay_c+z_az_c\right)y_b\;\mathbf{j} +\left(x_ax_c+y_ay_c+z_az_c\right)z_b\;\mathbf{k} \right\} \\&&-\left\{ \left(x_bx_c+y_by_c+z_bz_c\right)x_a\;\mathbf{i} +\left(x_bx_c+y_by_c+z_bz_c\right)y_a\;\mathbf{j} +\left(x_bx_c+y_by_c+z_bz_c\right)z_a\;\mathbf{k} \right\} \\&=& \left(x_ax_c+y_ay_c+z_az_c\right)\left(x_b\;\mathbf{i}+y_b\;\mathbf{j}+z_b\;\mathbf{k}\right) -\left(x_bx_c+y_by_c+z_bz_c\right)\left(x_a\;\mathbf{i}+y_a\;\mathbf{j}+z_a\;\mathbf{k}\right) \\&=&\left(\mathbf{A}\cdot\mathbf{C}\right)\mathbf{B}-\left(\mathbf{B}\cdot\mathbf{C}\right)\mathbf{A} \end{eqnarray} $$ $$ \begin{eqnarray} \mathbf{A}\times\left(\mathbf{B}\times\mathbf{C}\right) &=& \left(x_a\mathbf{i}+y_a\mathbf{j}+z_a\mathbf{k}\right) \times \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ x_b&y_b&z_b\\ x_c&y_c&z_c\\ \end{vmatrix} \\&=& \left( x_a\mathbf{i} +y_a\mathbf{j} +z_a\mathbf{k} \right) \times \left( \left(y_bz_c-y_cz_b\right)\mathbf{i} +\left(x_cz_b-x_bz_c\right)\mathbf{j} +\left(x_by_c-x_cy_b\right)\mathbf{k} \right) \\&=& \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ x_a&y_a&z_a\\ \left(y_bz_c-y_cz_b\right)&\left(x_cz_b-x_bz_c\right)&\left(x_by_c-x_cy_b\right)\\ \end{vmatrix} \\&=& \left\{ y_a\left(x_by_c-x_cy_b\right)-z_a\left(x_cz_b-x_bz_c\right) \right\} \mathbf{i} \\&&+ \left\{ z_a\left(y_bz_c-y_cz_b\right)-x_a\left(x_by_c-x_cy_b\right) \right\} \mathbf{j} \\&&+ \left\{ x_a\left(x_cz_b-x_bz_c\right)-y_a\left(y_bz_c-y_cz_b\right) \right\} \mathbf{k} \\&=& \left\{ \left(x_b\;y_ay_c-x_c\;y_ay_b\right)-\left(x_c\;z_az_b-x_b\;z_az_c\right) \right\} \mathbf{i} \\&&+ \left\{ \left(y_b\;z_az_c-y_c\;z_az_b\right)-\left(y_c\;x_ax_b-y_b\;x_ax_c\right) \right\} \mathbf{j} \\&&+ \left\{ \left(z_b\;x_ax_c-z_c\;x_ax_b\right)-\left(z_c\;y_ay_b-z_b\;y_ay_c\right) \right\} \mathbf{k} \\&=& \left\{ \left(x_b\;y_ay_c-x_c\;y_ay_b\right)-\left(x_c\;z_az_b-x_b\;z_az_c\right) {\color{red}+x_b\;x_ax_c-x_c\;x_ax_b} \right\} \mathbf{i} \\&&+ \left\{ \left(y_b\;z_az_c-y_c\;z_az_b\right)-\left(y_c\;x_ax_b-y_b\;x_ax_c\right) {\color{red}+y_b\;y_ay_c-y_c\;y_ay_b} \right\} \mathbf{j} \\&&+ \left\{ \left(z_b\;x_ax_c-z_c\;x_ax_b\right)-\left(z_c\;y_ay_b-z_b\;y_ay_c\right) {\color{red}+z_b\;z_az_c-z_c\;z_az_b} \right\} \mathbf{k} \\&=& \left\{ \left( x_b\;x_ax_c+x_b\;y_ay_c+x_b\;z_az_c \right) \mathbf{i} + \left( y_b\;x_ax_c+y_b\;y_ay_c+y_b\;z_az_c \right) \mathbf{j} + \left( z_b\;x_ax_c+z_b\;y_ay_c+z_b\;z_az_c \right) \mathbf{k} \right\} \\&&-\left\{ \left( x_c\;x_ax_b+x_c\;y_ay_b+x_c\;z_az_b \right) \mathbf{i} + \left( y_c\;x_ax_b+y_c\;y_ay_b+y_c\;z_az_b \right) \mathbf{j} + \left( z_c\;x_ax_b+z_c\;y_ay_b+z_c\;z_az_b \right) \mathbf{k} \right\} \\&=& \left\{ \left( x_ax_c+y_ay_c+z_az_c \right) x_b\;\mathbf{i} + \left( x_ax_c+y_ay_c+z_az_c \right) y_b\;\mathbf{j} + \left( x_ax_c+y_ay_c+z_az_c \right) z_b\;\mathbf{k} \right\} \\&&-\left\{ \left( x_ax_b+y_ay_b+z_az_b \right) x_c\;\mathbf{i} + \left( x_ax_b+y_ay_b+z_az_b \right) y_c\;\mathbf{j} + \left( x_ax_b+y_ay_b+z_az_b \right) z_c\;\mathbf{k} \right\} \\&=& \left( x_ax_c+y_ay_c+z_az_c \right) \left(x_b\;\mathbf{i}+y_b\;\mathbf{j}+z_b\;\mathbf{k}\right) - \left( x_ax_b+y_ay_b+z_az_b \right) \left(x_c\;\mathbf{i}+y_c\;\mathbf{j}+z_c\;\mathbf{k}\right) \\&=&\left(\mathbf{A}\cdot\mathbf{C}\right)\mathbf{B}-\left(\mathbf{A}\cdot\mathbf{B}\right)\mathbf{C} \\ \\ \left(\mathbf{A}\times\mathbf{B}\right)\times\mathbf{C} &=&-\mathbf{C}\times\left(\mathbf{A}\times\mathbf{B}\right) \;\cdots\;\mathbf{A}\times\mathbf{B}=-\mathbf{B}\times\mathbf{A} \\&=&\left((-\mathbf{C})\cdot\mathbf{B}\right)\mathbf{A}-\left((-\mathbf{C})\cdot\mathbf{A}\right)\mathbf{B} \;\cdots\;\mathbf{A}\times\left(\mathbf{B}\times\mathbf{C}\right)=\left(\mathbf{A}\cdot\mathbf{C}\right)\mathbf{B}-\left(\mathbf{A}\cdot\mathbf{B}\right)\mathbf{C} \\&=&\left(\mathbf{A}\cdot\mathbf{C}\right)\mathbf{B}-\left(\mathbf{B}\cdot\mathbf{C}\right)\mathbf{A} \;\cdots\;\mathbf{A}\cdot\mathbf{B}=\mathbf{B}\cdot\mathbf{A},\;\left(\left(-\mathbf{A}\right)\cdot\mathbf{B}\right)=-\left(\mathbf{A}\cdot\mathbf{B}\right) \end{eqnarray} $$

内積と外積が組み合わされた式 ( スカラー三重積 )

内積と外積が組み合わされた式

外積したものと内積をとった場合 $$ \begin{eqnarray} \mathbf{A}\cdot\left(\mathbf{B}\times\mathbf{C}\right) &=&\left(x_a\mathbf{i}+y_a\mathbf{j}+z_a\mathbf{k}\right) \cdot \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ x_b&y_b&z_b\\ x_c&y_c&z_c\\ \end{vmatrix} \\&=& \left( x_a\mathbf{i} +y_a\mathbf{j} +z_a\mathbf{k} \right) \cdot \left( \left(y_bz_c-y_cz_b\right)\mathbf{i} +\left(x_cz_b-x_bz_c\right)\mathbf{j} +\left(x_by_c-x_cy_b\right)\mathbf{k} \right) \\&=& x_a\left(y_bz_c-y_cz_b\right) +y_a\left(x_cz_b-x_bz_c\right) +z_a\left(x_by_c-x_cy_b\right) \\&=& x_ay_bz_c +x_by_cz_a +x_cy_az_b -x_ay_cz_b -x_by_az_c -x_cy_bz_a \end{eqnarray} $$ $$ \begin{eqnarray} \left(\mathbf{A}\times\mathbf{B}\right)\cdot\mathbf{C} &=& \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ x_a&y_a&z_a\\ x_b&y_b&z_b\\ \end{vmatrix} \cdot \left(x_c\mathbf{i}+y_c\mathbf{j}+z_c\mathbf{k}\right) \\&=& \left( \left(y_az_b-y_bz_a\right)\mathbf{i} +\left(x_bz_a-x_az_b\right)\mathbf{j} +\left(x_ay_b-x_by_a\right)\mathbf{k} \right) \cdot \left( x_c\mathbf{i} +y_c\mathbf{j} +z_c\mathbf{k} \right) \\&=& \left(y_az_b-y_bz_a\right)x_c +\left(x_bz_a-x_az_b\right)y_c +\left(x_ay_b-x_by_a\right)z_c \\&=& x_cy_az_b-x_cy_bz_a +x_by_cz_a-x_ay_cz_b +x_ay_bz_c-x_by_az_c \\&=& x_ay_bz_c +x_by_cz_a +x_cy_az_b -x_ay_cz_b -x_by_az_c -x_cy_bz_a \end{eqnarray} $$ 以上より内積と外積の位置が入れ替わった上記2式が等しいことがわかった．この計算をスカラー三重積と呼ぶ． $$ \begin{eqnarray} \mathbf{A}\cdot\left(\mathbf{B}\times\mathbf{C}\right) &=&\left(\mathbf{A}\times\mathbf{B}\right)\cdot\mathbf{C} \end{eqnarray} $$

スカラー三重積と循環シフト

前述の等式の各ベクトルの表す文字を循環シフトさせて文字を入れ替え，更に2つの等式を用意する． $$ \begin{eqnarray} \mathbf{A}\cdot\left(\mathbf{B}\times\mathbf{C}\right) &=&\left(\mathbf{A}\times\mathbf{B}\right)\cdot\mathbf{C}\;\cdots\;前述の等式 \\ \mathbf{B}\cdot\left(\mathbf{C}\times\mathbf{A}\right) &=&\left(\mathbf{B}\times\mathbf{C}\right)\cdot\mathbf{A} =\mathbf{A}\cdot\left(\mathbf{B}\times\mathbf{C}\right) \;\cdots\;\mathbf{A}\cdot\mathbf{B}=\mathbf{B}\cdot\mathbf{A} \\ \mathbf{C}\cdot\left(\mathbf{A}\times\mathbf{B}\right) &=&\left(\mathbf{C}\times\mathbf{A}\right)\cdot\mathbf{B} =\mathbf{B}\cdot\left(\mathbf{C}\times\mathbf{A}\right) \;\cdots\;\mathbf{A}\cdot\mathbf{B}=\mathbf{B}\cdot\mathbf{A} \end{eqnarray} $$ 以上より6つの“内積と外積が組み合わされた式”は等しいことがわかった． $$ \begin{eqnarray} \mathbf{A}\cdot\left(\mathbf{B}\times\mathbf{C}\right) &=&\mathbf{B}\cdot\left(\mathbf{C}\times\mathbf{A}\right) \\&=&\mathbf{C}\cdot\left(\mathbf{A}\times\mathbf{B}\right) \\&=&\left(\mathbf{A}\times\mathbf{B}\right)\cdot\mathbf{C} \\&=&\left(\mathbf{B}\times\mathbf{C}\right)\cdot\mathbf{A} \\&=&\left(\mathbf{C}\times\mathbf{A}\right)\cdot\mathbf{B} \end{eqnarray} $$
(これらは$\mathbf{A}$,$\mathbf{B}$,$\mathbf{C}$を循環シフトした並びであり，$\mathbf{A}$,$\mathbf{C}$,$\mathbf{B}$のような入替えを含めた並びにはならない点に注意)

点と直線の距離 (直角三角形の面積を経由して求める)

直角三角形を作成し面積の公式より求める方法

$点P(x_p, y_p)$から$x軸$と水平に$直線ax+by+c=0$に向かって線分を引き$交点A(x_a, y_a=y_p)$と，$点P$から$x軸$と垂直に$直線ax+by+c=0$に向かって線分を引き$交点B(x_b=x_p, y_b)$を用意する．
$\triangle ABP$は$\angle P$を直角とした直角三角形のため，$\triangle ABP$の面積は$\frac{1}{2}\cdot\overline{AP}\cdot\overline{BP}$となる．一方で，$\overline{AB}$を底辺として考えると$点P$への$高さd$を用いることで，$\triangle ABP$の面積は$\frac{1}{2}\cdot\overline{AB}\cdot d$で表せる．この2式が同一の面積を示すので等式にして，式変形で$d$を求める($d$は$点P(x_p, y_p)$と$直線ax+by+c=0$との距離となる)．

$$ \begin{eqnarray} \frac{1}{2}\cdot\overline{AP}\cdot\overline{BP}&=&\frac{1}{2}\cdot\overline{AB}\cdot d \\\overline{AP}\cdot\overline{BP}&=&\overline{AB}\cdot d \\d&=&\frac{\overline{AP}\cdot\overline{BP}}{\overline{AB}} \end{eqnarray} $$ $\overline{AP}$を求める． $$ \begin{eqnarray} ax_a+by_p+c&=&0\;\cdots\;y_a=y_pなのでy=y_pとしてx_aを求める． \\x_a&=&-\frac{b}{a}y_p-\frac{c}{a} \\\overline{AP}&=&\left|x_a-x_p\right| \\&=&\left|-\frac{b}{a}y_p-\frac{c}{a}-x_p\right| \\&=&\left|-\frac{1}{a}\left(ax_p+by_p+c\right)\right| \\&=&\left|\frac{1}{a}\left(ax_p+by_p+c\right)\right| \end{eqnarray} $$ $\overline{BP}$を求める． $$ \begin{eqnarray} ax_p+by_b+c&=&0\;\cdots\;x_b=x_pなのでx=x_pとしてy_bを求める． \\y_b&=&-\frac{a}{b}x_p-\frac{c}{b} \\\overline{BP}&=&\left|y_b-y_p\right| \\&=&\left|-\frac{a}{b}x_p-\frac{c}{b}-y_p\right| \\&=&\left|-\frac{1}{b}\left(ax_p+by_p+c\right)\right| \\&=&\left|\frac{1}{b}\left(ax_p+by_p+c\right)\right| \end{eqnarray} $$ $\overline{AB}$を求める． $$ \begin{eqnarray} \overline{AB}&=&\sqrt{\overline{AP}^2+\overline{BP}^2} \\&=&\sqrt{\left|\frac{1}{a}\left(ax_p+by_p+c\right)\right|^2+\left|\frac{1}{b}\left(ax_p+by_p+c\right)\right|^2} \\&=&\sqrt{\frac{1}{a^2}\left|ax_p+by_p+c\right|^2+\frac{1}{b^2}\left|ax_p+by_p+c\right|^2} \\&=&\sqrt{\left(\frac{1}{a^2}+\frac{1}{b^2}\right) \left|ax_p+by_p+c\right|^2} \\&=&\sqrt{\left(\frac{a^2+b^2}{a^2b^2}\right) \left|ax_p+by_p+c\right|^2} \\&=&\frac{\sqrt{a^2+b^2}}{\left|ab\right|}\left|ax_p+by_p+c\right| \end{eqnarray} $$ $d$を求める． $$ \begin{eqnarray} \\d&=&\frac{\overline{AP}\cdot\overline{BP}}{\overline{AB}} \\&=&\frac{\overline{AP}\cdot\overline{BP}}{\sqrt{\overline{AP}^2+\overline{BP}^2}} \\&=&\frac{\left|\frac{1}{a}\left(ax_p+by_p+c\right)\right|\cdot\left|\frac{1}{b}\left(ax_p+by_p+c\right)\right|}{\frac{\sqrt{a^2+b^2}}{\left|ab\right|}\left|ax_p+by_p+c\right|} \\&=&\frac{\frac{1}{\left|ab\right|}\left|ax_p+by_p+c\right|^2}{\frac{\sqrt{a^2+b^2}}{\left|ab\right|}\left|ax_p+by_p+c\right|} \\&=&\frac{\left|ax_p+by_p+c\right|}{\sqrt{a^2+b^2}} \end{eqnarray} $$

点と直線の距離 (直交する直線の式を経由して求める)

直交する直線の式より交点を求め，距離を求める方法

$点P(x_p, y_p)$と$直線ax+by+c=0$の距離の公式を導く． $$ \begin{eqnarray} ax+by+c&=&0 \\y&=&-\frac{a}{b}x-\frac{c}{b} \end{eqnarray} $$ 傾き$\alpha$の直線に直交する直線の傾きは$-\frac{1}{\alpha}$となるので上記の直線の場合，直交する直線の傾きは以下のようになる． $$ \begin{eqnarray} -\frac{1}{\alpha}&=&-\frac{1}{-\frac{a}{b}} \\&=&\frac{b}{a} \end{eqnarray} $$ 上記傾きの直線が$点P(x_p, y_p)$を通るのでこの直線の式は以下のようになる． $$ \begin{eqnarray} y-y_p&=&\frac{b}{a}\left(x-x_p\right) \\y&=&\frac{b}{a}\left(x-x_p\right)+y_p \\&=&\frac{b}{a}x-\frac{b}{a}x_p+y_p \end{eqnarray} $$ もとの$直線 y=-\frac{a}{b}x-\frac{c}{b}$と上記の直交する$直線 y=\frac{b}{a}x-\frac{b}{a}x_p+y_p$との$交点Q(x_q, y_q)$を求める． $$ \begin{eqnarray} \\-\frac{a}{b}x_q-\frac{c}{b}&=&\frac{b}{a}x_q-\frac{b}{a}x_p+y_p \\-\frac{a}{b}x_q-\frac{b}{a}x_q&=&-\frac{b}{a}x_p+y_p+\frac{c}{b} \\\left(-\frac{a}{b}-\frac{b}{a}\right)x_q&=&-\frac{b}{a}x_p+y_p+\frac{c}{b} \\-\frac{a^2+b^2}{ab}x_q&=&-\frac{b}{a}x_p+y_p+\frac{c}{b} \\x_q&=&\frac{-ab}{a^2+b^2}\left(-\frac{b}{a}x_p+y_p+\frac{c}{b}\right) \\&=&\frac{1}{a^2+b^2}\left(b^2x_p-aby_p-ac\right) \\ \\y_q&=&-\frac{a}{b}x_q-\frac{c}{b} \\&=&-\frac{a}{b}\left\{\frac{1}{a^2+b^2}\left(b^2x_p-aby_p-ac\right)\right\}-\frac{c}{b} \\&=&\frac{1}{a^2+b^2}\left(-abx_p+a^2y_p+\frac{a^2}{b}c\right)-\frac{c}{b} \\&=&\frac{1}{a^2+b^2}\left\{-abx_p+a^2y_p+\frac{a^2}{b}c-\left(a^2+b^2\right)\frac{c}{b}\right\} \\&=&\frac{1}{a^2+b^2}\left\{-abx_p+a^2y_p+\frac{c}{b}\left(a^2-a^2-b^2\right)\right\} \\&=&\frac{1}{a^2+b^2}\left(-abx_p+a^2y_p-\frac{b^2c}{b}\right) \\&=&\frac{1}{a^2+b^2}\left(-abx_p+a^2y_p-bc\right) \end{eqnarray} $$ 以上より$点P(x_p, y_p)$と$点Q(x_q, y_q)$の距離を求める． $$ \begin{eqnarray} \\x_q-x_p&=&\frac{1}{a^2+b^2}\left(b^2x_p-aby_p-ac\right)-x_p \\&=&\frac{1}{a^2+b^2}\left\{b^2x_p-aby_p-ac-\left(a^2+b^2\right)x_p\right\} \\&=&\frac{1}{a^2+b^2}\left(b^2x_p-aby_p-ac-a^2x_p-b^2x_p\right) \\&=&\frac{1}{a^2+b^2}\left(-a^2x_p-aby_p-ac\right) \\&=&\frac{-a}{a^2+b^2}\left(ax_p+by_p+c\right) \\ \\\left(x_q-x_p\right)^2&=&\left\{\frac{-a}{a^2+b^2}\left(ax_p+by_p+c\right)\right\}^2 \\&=&\left(\frac{-a}{a^2+b^2}\right)^2\left(ax_p+by_p+c\right)^2 \\&=&\frac{ \left(-a\right)^2 }{ \left(a^2+b^2\right)^2 }\left(ax_p+by_p+c\right)^2 \\&=&\frac{ a^2 }{ \left(a^2+b^2\right)^2 }\left(ax_p+by_p+c\right)^2 \\ \\y_q-y_p&=&\frac{1}{a^2+b^2}\left(-abx_p+a^2y_p-bc\right)-y_p \\&=&\frac{1}{a^2+b^2}\left\{-abx_p+a^2y_p-bc-\left(a^2+b^2\right)y_p\right\} \\&=&\frac{1}{a^2+b^2}\left(-abx_p+a^2y_p-bc-a^2y_p-b^2y_p\right) \\&=&\frac{1}{a^2+b^2}\left(-abx_p-b^2y_p-bc\right) \\&=&\frac{-b}{a^2+b^2}\left(ax_p+by_p+c\right) \\ \\\left(y_q-y_p\right)^2&=&\left\{\frac{-b}{a^2+b^2}\left(ax_p+by_p+c\right)\right\}^2 \\&=&\left(\frac{-b}{a^2+b^2}\right)^2\left(ax_p+by_p+c\right)^2 \\&=&\frac{ \left(-b\right)^2 }{ \left(a^2+b^2\right)^2 }\left(ax_p+by_p+c\right)^2 \\&=&\frac{ b^2 }{ \left(a^2+b^2\right)^2 }\left(ax_p+by_p+c\right)^2 \\ \\\left(x_q-x_p\right)^2+\left(y_q-y_p\right)^2 &=&\frac{ a^2 }{ \left(a^2+b^2\right)^2 }\left(ax_p+by_p+c\right)^2 + \frac{ b^2 }{ \left(a^2+b^2\right)^2 }\left(ax_p+by_p+c\right)^2 \\&=&\left\{ \frac{ a^2 }{ \left(a^2+b^2\right)^2 } + \frac{ b^2 }{ \left(a^2+b^2\right)^2 }\right\} \left(ax_p+by_p+c\right)^2 \\&=&\frac{a^2+b^2}{\left(a^2+b^2\right)^2}\left(ax_p+by_p+c\right)^2 \\&=&\frac{1}{a^2+b^2}\left(ax_p+by_p+c\right)^2 \\&=&\frac{\left(ax_p+by_p+c\right)^2}{a^2+b^2} \\ \\\sqrt{\left(x_q-x_p\right)^2+\left(y_q-y_p\right)^2} &=&\sqrt{\frac{\left(ax_p+by_p+c\right)^2}{a^2+b^2}} \\&=&\frac{\left|ax_p+by_p+c\right|}{\sqrt{a^2+b^2}} \end{eqnarray} $$ 与えられる$直線$が$直線y=ax+b$の形の場合，上記公式の$b=1$で$c$を$b$と表すことになる．よって公式は以下のように変わることになる． $$ \begin{eqnarray} \sqrt{\left(x_q-x_p\right)^2+\left(y_q-y_p\right)^2} &=&\frac{\left|ax_p+y_p+b\right|}{\sqrt{a^2+1^2}} \end{eqnarray} $$

2点を補間した位置への回転操作のための四元数

位置ベクトル$\mathbf{p}$への四元数$\mathbf{q_1}$, $\mathbf{q_2}$による回転操作を行った結果を$\mathbf{p_1}$，$\mathbf{p_2}$とする．また$\mathbf{p_1}$と$\mathbf{p_2}$との間を補間した位置を$\mathbf{p_t}$とした時，$\mathbf{p}$に対する回転操作で$\mathbf{p_t}$となる四元数$\mathbf{q_{t}}$を考える． $$ \begin{eqnarray} \mathbf{q}&=&w+x\mathbf{i}+y\mathbf{j}+z\mathbf{k} \\&=&(w, x\mathbf{i}+y\mathbf{j}+z\mathbf{k}) \;\cdots\;\left(実部(スカラー),ijk部(ベクトル)\right) \\&=&\left(w,\mathbf{V}\right) \;\cdots\;\mathbf{V}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k} \\\mathbf{\overline{q}} &=&w-x\mathbf{i}-y\mathbf{j}-z\mathbf{k} \\&=&(w, -x\mathbf{i}-y\mathbf{j}-z\mathbf{k}) \\&=&\left(w,-\mathbf{V}\right) \\\mathbf{q_{回転}}&=&\left(\cos{\left(\frac{\theta}{2}\right)}, \sin{\left(\frac{\theta}{2}\right)}\mathbf{V}\right) \\&&\;\cdots\;特に位置ベクトルを回転軸\mathbf{V}(ただし\mathbf{V}は単位ベクトル.\;\left|\mathbf{V}\right|=1),回転角\thetaで回す際の四元数 \\&&\href{https://shikitenkai.blogspot.com/2020/05/blog-post_41.html}{\mathbf{q_{回転}}と\mathbf{q_{回転}^{-1}}で位置ベクトル\mathbf{p}へ左右から作用させ\mathbf{q_{回転}}\mathbf{p}\mathbf{q_{回転}^{-1}}として用いる.} \\\mathbf{q_1} &=&\left(\cos{\left(\frac{\theta_1}{2}\right)}, \sin{\left(\frac{\theta_1}{2}\right)}\mathbf{V_1}\right) \\&&\;\cdots\;\left|q_1\right|=1, \left|\mathbf{V_1}\right|=1,\; 位置ベクトルpをp_1に向けて回転させる際の軸 \\&=&\left(C_1, S_1\mathbf{V_1}\right) \\\mathbf{q_2} &=&\left(\cos{\left(\frac{\theta_2}{2}\right)}, \sin{\left(\frac{\theta_2}{2}\right)}\mathbf{V_2}\right) \\&&\;\cdots\;\left|q_2\right|=1, \left|\mathbf{V_2}\right|=1位置ベクトルpをp_2に向けて回転させる際の軸 \\&=&\left(C_2, S_2\mathbf{V_2}\right) \\\mathbf{q_{1 \rightarrow2}} &=&\left(\cos{\left(\frac{\theta_{1 \rightarrow2}}{2}\right)}, \sin{\left(\frac{\theta_{1 \rightarrow2}}{2}\right)}\mathbf{V_{1 \rightarrow2}}\right) \\&&\;\cdots\;\left|q_{1 \rightarrow2}\right|=1, \left|\mathbf{V_{1 \rightarrow2}}\right|=1,\; 位置ベクトルpをp_1からp_2に向けて回転させる際の軸 \\&=&\left(C_{1 \rightarrow2}, S_{1 \rightarrow2}\mathbf{V_{1 \rightarrow2}}\right) \\\mathbf{q_{1 \rightarrow t}} &=&\left(\cos{\left(t\frac{\theta_{1 \rightarrow2}}{2}\right)}, \sin{\left(t\frac{\theta_{1 \rightarrow2}}{2}\right)}\mathbf{V_{1 \rightarrow2}}\right) \\&&\;\cdots\;0\leq t \leq 1,\; 位置ベクトルpをp_1からp_2に向けて回転させる途中のp_tまでの際の軸なので\mathbf{V_{1 \rightarrow2}}で変わらない.\;回転量だけtで調整される. \\&=&\left(C_{1 \rightarrow t}, S_{1 \rightarrow t}\mathbf{V_{1 \rightarrow2}}\right) \\\mathbf{q_{t}} &=&\left(\cos{\left(\frac{\theta_{t}}{2}\right)}, \sin{\left(\frac{\theta_{t}}{2}\right)}\mathbf{V_{t}}\right) \\&&\;\cdots\;\left|q_{t}\right|=1, \left|\mathbf{V_{t}}\right|=1,\; 位置ベクトルpをp_tに向けて回転させる際の軸 \\&=&\left(C_{t}, S_{t}\mathbf{V_{t}}\right) \end{eqnarray} $$

$\mathbf{p}$，$\mathbf{p_1}$，$\mathbf{p_2}$，$\mathbf{q_1}$，$\mathbf{q_2}$の関係性から$\mathbf{q_{1 \rightarrow2}}=\left(C_{1 \rightarrow2}, S_{1 \rightarrow2}\mathbf{V_{1 \rightarrow2}}\right)$を求める． $$ \begin{eqnarray} \mathbf{p_1}&=&\mathbf{q_1}\mathbf{p}\mathbf{q_1^{-1}} \\ \\\mathbf{p_2}&=&\mathbf{q_2}\mathbf{p}\mathbf{q_2^{-1}} \\ \\\mathbf{p_2}&=&\mathbf{q_{1\rightarrow2}}\mathbf{p_1}\mathbf{q_{1\rightarrow2}^{-1}} \\&=&\mathbf{q_{1 \rightarrow2}}\left(\mathbf{q_1}\mathbf{p}\mathbf{q_1^{-1}}\right)\mathbf{q_{1 \rightarrow2}^{-1}} \;\cdots\;\mathbf{p_1}=\mathbf{q_1}\mathbf{p}\mathbf{q_1^{-1}} \\&=&\left(\mathbf{q_{1 \rightarrow2}}\mathbf{q_1}\right)\mathbf{p}\left(\mathbf{q_1^{-1}}\mathbf{q_{1 \rightarrow2}^{-1}}\right) \\ \\\mathbf{q_2}&=&\mathbf{q_{1 \rightarrow2}}\mathbf{q_1} \\ \\\mathbf{q_{1 \rightarrow2}}&=&\mathbf{q_2}\mathbf{q_1^{-1}} \\&=&\mathbf{q_2}\frac{\mathbf{\overline{q_1}}}{|\mathbf{q_1}|^2} \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/05/blog-post_58.html}{\mathbf{q^{-1}}=\frac{\mathbf{\overline{q}}}{|\mathbf{q}|^2}} \\&=&\mathbf{q_2}\mathbf{\overline{q_1}} \;\cdots\;|\mathbf{q_1}|^2=1^2=1 \\&=&\left(C_2, S_2\mathbf{V_2}\right)\left(C_1, -S_1\mathbf{V_1}\right) \\&=&\left( C_1C_2-\{-S_1\}S_2\mathbf{V_2}\cdot\mathbf{V_1} ,\; -C_2S_1\mathbf{V_1}+C_1S_2\mathbf{V_2}+\{-S_1\}S_2\mathbf{V_2}\times\mathbf{V_1} \right) \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/05/quaternion.html}{\mathbf{q}_1\mathbf{q}_2=\left(w_1,\mathbf{V_1}\right)\left(w_2,\mathbf{V_2}\right)=\left(w_1w_2-\mathbf{V}_1\cdot\mathbf{V}_2,\;w_1\mathbf{V}_2+w_2\mathbf{V}_1+\mathbf{V}_1\times\mathbf{V}_2\right)} \\&=&\left( C_1C_2+S_1S_2\mathbf{V_1}\cdot\mathbf{V_2} ,\; -C_2S_1\mathbf{V_1}+C_1S_2\mathbf{V_2}+S_1S_2\mathbf{V_1}\times\mathbf{V_2} \right) \;\cdots\;\mathbf{A}\cdot\mathbf{B}=\mathbf{B}\cdot\mathbf{A},\;\mathbf{A}\times\mathbf{B}=-\mathbf{B}\times\mathbf{A} \\ \\C_{1 \rightarrow2}&=&C_1C_2+S_1S_2\mathbf{V_1}\cdot\mathbf{V_2} \\S_{1 \rightarrow2}\mathbf{V_{1 \rightarrow2}}&=&-C_2S_1\mathbf{V_1}+C_1S_2\mathbf{V_2}+S_1S_2\mathbf{V_1}\times\mathbf{V_2} \end{eqnarray} $$

$\mathbf{p}$，$\mathbf{p_1}$，$\mathbf{p_t}$，$\mathbf{q_1}$，$\mathbf{q_{1 \rightarrow t}}$の関係性から$\mathbf{q_t}=\left(C_t, S_t\mathbf{V_t}\right)$を求める． $$ \begin{eqnarray} \mathbf{p_t}&=&\mathbf{q_{t}}\mathbf{p}\mathbf{q_{t}^{-1}} \\ \\\mathbf{p_t}&=&\mathbf{q_{1\rightarrow t}}\mathbf{p_1}\mathbf{q_{1\rightarrow t}^{-1}} \\&=&\mathbf{q_{1\rightarrow t}}\left(\mathbf{q_1}\mathbf{p}\mathbf{q_1^{-1}}\right)\mathbf{q_{1\rightarrow t}^{-1}} \;\cdots\;\mathbf{p_1}=\mathbf{q_1}\mathbf{p}\mathbf{q_1^{-1}} \\&=&\left(\mathbf{q_{1\rightarrow t}}\mathbf{q_1}\right)\mathbf{p}\left(\mathbf{q_1^{-1}}\mathbf{q_{1\rightarrow t}^{-1}}\right) \\ \\\mathbf{q_{t}}&=&\mathbf{q_{1 \rightarrow t}}\mathbf{q_1} \\&=&\left(C_{1 \rightarrow t}, S_{1 \rightarrow t}\mathbf{V_{1 \rightarrow 2}}\right)\left(C_1, S_1\mathbf{V_1}\right) \\&=&\left( C_1C_{1 \rightarrow t}-S_1S_{1 \rightarrow t}\mathbf{V_{1 \rightarrow 2}}\cdot\mathbf{V_1} ,\; C_{1 \rightarrow t}S_{1}\mathbf{V_1}+C_1S_{1 \rightarrow t}\mathbf{V_{1 \rightarrow 2}}+S_1S_{1 \rightarrow t}\mathbf{V_{1 \rightarrow 2}}\times\mathbf{V_1} \right) \\&=&\left( C_1C_{1 \rightarrow t}-S_1S_{1 \rightarrow t}\mathbf{V_1}\cdot\mathbf{V_{1 \rightarrow 2}} ,\; C_{1 \rightarrow t}S_{1}\mathbf{V_1}+C_1S_{1 \rightarrow t}\mathbf{V_{1 \rightarrow 2}}+S_1S_{1 \rightarrow t}\mathbf{V_{1 \rightarrow 2}}\times\mathbf{V_1} \right) \;\cdots\;\mathbf{A}\cdot\mathbf{B}=\mathbf{B}\cdot\mathbf{A} \\ \\C_{t}&=&C_1C_{1 \rightarrow t}-S_1S_{1 \rightarrow t}\mathbf{V_1}\cdot\mathbf{V_{1 \rightarrow 2}} \\S_{t}\mathbf{V_{t}}&=&C_{1 \rightarrow t}S_{1}\mathbf{V_1}+C_1S_{1 \rightarrow t}\mathbf{V_{1 \rightarrow 2}}+S_1S_{1 \rightarrow t}\mathbf{V_{1 \rightarrow 2}}\times\mathbf{V_1} \end{eqnarray} $$

上記で求めた$S_{1 \rightarrow2}\mathbf{V_{1 \rightarrow2}}$を用いて，$\mathbf{V_1}\cdot\mathbf{V_{1 \rightarrow2}}$を求めておく． $$ \begin{eqnarray} S_{1 \rightarrow2}\mathbf{V_{1 \rightarrow2}}&=&-C_2S_1\mathbf{V_1}+C_1S_2\mathbf{V_2}+S_1S_2\mathbf{V_1}\times\mathbf{V_2} \\ \\\mathbf{V_{1 \rightarrow2}} &=&\frac{1}{S_{1 \rightarrow2}}\left\{-C_2S_1\mathbf{V_1}+C_1S_2\mathbf{V_2}+S_1S_2\mathbf{V_1}\times\mathbf{V_2}\right\} \\ \\\mathbf{V_1}\cdot\mathbf{V_{1 \rightarrow2}} &=&\mathbf{V_1}\cdot\left[ \frac{1}{S_{1 \rightarrow2}}\left\{ -C_2S_1\mathbf{V_1} +C_1S_2\mathbf{V_2} +S_1S_2\left(\mathbf{V_1}\times\mathbf{V_2}\right) \right\} \right] \\&=&\frac{1}{S_{1 \rightarrow2}}\left\{ -C_2S_1\mathbf{V_1}\cdot\mathbf{V_1} +C_1S_2\mathbf{V_1}\cdot\mathbf{V_2} +S_1S_2\mathbf{V_1}\cdot\left(\mathbf{V_1}\times\mathbf{V_2}\right) \right\} \\&=&\frac{1}{S_{1 \rightarrow2}}\left\{ -C_2S_1 +C_1S_2\mathbf{V_1}\cdot\mathbf{V_2} +S_1S_2\mathbf{V_2}\cdot\left(\mathbf{V_1}\times\mathbf{V_1}\right)\right\} \;\cdots\;\mathbf{V_1}\cdot\mathbf{V_1}=\left|V_1\right|^2=1^2=1 ,\;\href{https://shikitenkai.blogspot.com/2020/06/blog-post_42.html}{\mathbf{A}\cdot\left(\mathbf{B}\times\mathbf{C}\right)=\mathbf{C}\cdot\left(\mathbf{A}\times\mathbf{B}\right)(スカラー三重積)} \\&=&\frac{1}{S_{1 \rightarrow2}}\left\{ -C_2S_1 +C_1S_2\mathbf{V_1}\cdot\mathbf{V_2} \right\} \;\cdots\;\mathbf{A}\times\mathbf{A}=0 \end{eqnarray} $$

上記で求めた$\mathbf{V_1}\cdot\mathbf{V_{1 \rightarrow 2}}$を用いて，$C_{t}$を変形する． $$ \begin{eqnarray} C_{t} &=&C_1C_{1 \rightarrow t}-S_1S_{1 \rightarrow t}\mathbf{V_1}\cdot\mathbf{V_{1 \rightarrow 2}} \\&=&C_1C_{1 \rightarrow t}-S_1S_{1 \rightarrow t}\left[\frac{1}{S_{1 \rightarrow2}}\left\{-C_2S_1+C_1S_2\mathbf{V_1}\cdot\mathbf{V_2}\right\}\right] \\&&\;\cdots\;\mathbf{V_1}\cdot\mathbf{V_{1 \rightarrow 2}}=\frac{1}{S_{1 \rightarrow2}}\left\{-C_2S_1+C_1S_2\mathbf{V_1}\cdot\mathbf{V_2}\right\} \\&=&C_1C_{1 \rightarrow t}-\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}\left\{-C_2S_1^2+C_1S_1S_2\mathbf{V_1}\cdot\mathbf{V_2}\right\} \\&=&C_1C_{1 \rightarrow t}-\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}\left\{-C_2\left(1-C_1^2\right)+C_1S_1S_2\mathbf{V_1}\cdot\mathbf{V_2}\right\} \;\cdots\;\sin^2{\left(\theta\right)}=1-\cos^2{\left(\theta\right)} \\&=&C_1C_{1 \rightarrow t}-\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}\left\{-C_2+C_1^2C_2+C_1S_1S_2\mathbf{V_1}\cdot\mathbf{V_2}\right\} \\&=&C_1C_{1 \rightarrow t}-\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}\left\{-C_2+C_1\left(C_1C_2+S_1S_2\mathbf{V_1}\cdot\mathbf{V_2}\right)\right\} \\&=&C_1C_{1 \rightarrow t}-\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}\left\{-C_2+C_1C_{1 \rightarrow 2}\right\} \;\cdots\;C_{1 \rightarrow 2}=C_1C_2+ S_1S_2\mathbf{V_1}\cdot\mathbf{V_2} \\&=&C_1C_{1 \rightarrow t} +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}C_2 -\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}C_1C_{1 \rightarrow 2} \\&=&\frac{S_{1 \rightarrow 2}C_{1 \rightarrow t}-C_{1 \rightarrow 2}S_{1 \rightarrow t}} {S_{1 \rightarrow2}}C_1 +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}C_2 \\&=&\frac{ \sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}\cos{\left(t\frac{\theta_{1 \rightarrow 2}}{2}\right)} -\cos{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}\sin{\left(t\frac{\theta_{1 \rightarrow 2}}{2}\right)} }{S_{1 \rightarrow2}}C_1 +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}C_2 \\&=&\frac{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}-t\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{S_{1 \rightarrow2}}C_1 +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}C_2 \;\cdots\;\sin{\left(\alpha\right)}\cos{\left(\beta\right)}-\cos{\left(\alpha\right)}\sin{\left(\beta\right)}=\sin{\left(\alpha-\beta\right)} \\&=& \frac{\sin{\left(\left(1-t\right)\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{S_{1 \rightarrow2}} C_1 +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}} C_2 \\&=& \frac{\sin{\left(\left(1-t\right)\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} C_1 +\frac{ \sin{\left( t\frac{\theta_{1 \rightarrow 2}}{2} \right)} } { \sin{\left( \frac{\theta_{1 \rightarrow 2}}{2} \right)} } C_2 \end{eqnarray} $$

上記で求めた$\mathbf{V_{1 \rightarrow 2}}$を用いて，$S_{t}\mathbf{V_{t}}$を変形する． $$ \begin{eqnarray} S_{t}\mathbf{V_{t}} &=&C_{1 \rightarrow t}S_{1}\mathbf{V_1} +C_1S_{1 \rightarrow t}\mathbf{V_{1 \rightarrow 2}} +S_1S_{1 \rightarrow t}\mathbf{V_{1 \rightarrow 2}}\times\mathbf{V_1} \\&=&C_{1 \rightarrow t}S_{1}\mathbf{V_1} +C_1S_{1 \rightarrow t}\left[\frac{1}{S_{1 \rightarrow2}}\left\{-C_2S_1\mathbf{V_1}+C_1S_2\mathbf{V_2}+S_1S_2\mathbf{V_1}\times\mathbf{V_2}\right\}\right] +S_1S_{1 \rightarrow t}\left[\frac{1}{S_{1 \rightarrow2}}\left\{-C_2S_1\mathbf{V_1}+C_1S_2\mathbf{V_2}+S_1S_2\mathbf{V_1}\times\mathbf{V_2}\right\}\right]\times\mathbf{V_1} \\&&\;\cdots\;\mathbf{V_{1 \rightarrow 2}}=\frac{1}{S_{1 \rightarrow2}}\left\{-C_2S_1\mathbf{V_1}+C_1S_2\mathbf{V_2}+S_1S_2\mathbf{V_1}\times\mathbf{V_2}\right\} \\&=&C_{1 \rightarrow t}S_{1}\mathbf{V_1} +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}\left\{-C_1C_2S_1\mathbf{V_1}+C_1^2S_2\mathbf{V_2}+C_1S_1S_2\mathbf{V_1}\times\mathbf{V_2}\right\} +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}\left\{-C_2S_1^2\mathbf{V_1}\times\mathbf{V_1}+C_1S_1S_2\mathbf{V_2}\times\mathbf{V_1}+S_1^2S_2\mathbf{V_1}\times\mathbf{V_2}\times\mathbf{V_1}\right\} \\&=&C_{1 \rightarrow t}S_{1}\mathbf{V_1} +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}\left\{ -C_1C_2S_1\mathbf{V_1} +C_1^2S_2\mathbf{V_2} +C_1S_1S_2\mathbf{V_1}\times\mathbf{V_2} -C_2S_1^2\mathbf{V_1}\times\mathbf{V_1} +C_1S_1S_2\mathbf{V_1}\times\mathbf{V_2} +S_1^2S_2\mathbf{V_1}\times\mathbf{V_2}\times\mathbf{V_1} \right\} \\&=&C_{1 \rightarrow t}S_{1}\mathbf{V_1} +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}\left[ -C_1C_2S_1\mathbf{V_1} +C_1^2S_2\mathbf{V_2} +S_1^2S_2\left\{ \left(\mathbf{V_1}\cdot\mathbf{V_1}\right)\mathbf{V_2}-\left(\mathbf{V_2}\cdot\mathbf{V_1}\right)\mathbf{V_1} \right\} \right] \;\cdots\;\mathbf{A}\times\mathbf{A}=0,\; \href{https://shikitenkai.blogspot.com/2020/06/blog-post_35.html}{\mathbf{A}\times\mathbf{B}\times\mathbf{C}=\left(\mathbf{A}\cdot\mathbf{C}\right)\mathbf{B}-\left(\mathbf{B}\cdot\mathbf{C}\right)\mathbf{A}} \\&=&C_{1 \rightarrow t}S_{1}\mathbf{V_1} +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}\left[ -C_1C_2S_1\mathbf{V_1} +C_1^2S_2\mathbf{V_2} +S_1^2S_2\left\{ \left|\mathbf{V_1}\right|^2\mathbf{V_2}-\left|\mathbf{V_2}\right|\left|\mathbf{V_1}\right|\cos{\left(\theta_{1 \rightarrow2}\right)}\mathbf{V_1} \right\} \right] \;\cdots\;\mathbf{A}\cdot\mathbf{A}=\left|\mathbf{A}\right|^2 ,\;\mathbf{A}\cdot\mathbf{B}=\left|\mathbf{A}\right|\left|\mathbf{B}\right|\cos{\left(\theta\right)} \\&=&C_{1 \rightarrow t}S_{1}\mathbf{V_1} +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}\left\{ -C_1C_2S_1\mathbf{V_1} +C_1^2S_2\mathbf{V_2} +S_1^2S_2\mathbf{V_2}-S_1^2S_2\cos{\left(\theta_{1 \rightarrow2}\right)}\mathbf{V_1} \right\} \;\cdots\;\left|\mathbf{V_1}\right|=1 ,\;\left|\mathbf{V_2}\right|=1 \\&=&C_{1 \rightarrow t}S_{1}\mathbf{V_1} +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}\left[ \left\{-C_1C_2S_1-S_1^2S_2\cos{\left(\theta_{1 \rightarrow2}\right)}\right\} \mathbf{V_1} +\left\{C_1^2S_2+S_1^2S_2\right\} \mathbf{V_2} \right] \\&=&C_{1 \rightarrow t}S_{1}\mathbf{V_1} +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}\left[ -S_1\left\{C_1C_2+S_1S_2\cos{\left(\theta_{1 \rightarrow2}\right)}\right\} \mathbf{V_1} +S_2\left\{C_1^2+S_1^2\right\} \mathbf{V_2} \right] \\&=&C_{1 \rightarrow t}S_{1}\mathbf{V_1} +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}\left\{ -S_1C_{1 \rightarrow2}\mathbf{V_1} +S_2\mathbf{V_2} \right\} \;\cdots\;C_{1 \rightarrow2}=C_1C_2+S_1S_2\cos{\left(\theta_{1 \rightarrow2}\right)} ,\;\cos^2{\left(\theta\right)}+\sin^2{\left(\theta\right)}=1 \\&=& \frac{S_{1 \rightarrow2}C_{1 \rightarrow t}-C_{1 \rightarrow2}S_{1 \rightarrow t}}{S_{1 \rightarrow2}} S_{1}\mathbf{V_1} +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}S_2 \mathbf{V_2} \\&=& \frac{ \sin{ \left(\frac{\theta_{1 \rightarrow 2}}{2}\right) } \cos{ \left(t\frac{\theta_{1 \rightarrow 2}}{2}\right) } -\cos{ \left(\frac{\theta_{1 \rightarrow 2}}{2}\right) } \sin{ \left(t\frac{\theta_{1 \rightarrow 2}}{2}\right) } } { S_{1 \rightarrow2} } S_{1}\mathbf{V_1} +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}} S_2\mathbf{V_2} \\&=& \frac{ \sin{ \left(\frac{\theta_{1 \rightarrow 2}}{2}-t\frac{\theta_{1 \rightarrow 2}}{2}\right) } } { S_{1 \rightarrow2} } S_{1}\mathbf{V_1} +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}} S_2\mathbf{V_2} \;\cdots\;\sin{\left(\alpha\right)}\cos{\left(\beta\right)}-\cos{\left(\alpha\right)}\sin{\left(\beta\right)}=\sin{\left(\alpha-\beta\right)} \\&=& \frac{ \sin{ \left(\left(1-t\right)\frac{\theta_{1 \rightarrow 2}}{2}\right) } } { S_{1 \rightarrow2} } S_{1}\mathbf{V_1} +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}} S_2\mathbf{V_2} \\&=& \frac{\sin{\left(\left(1-t\right)\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} S_{1}\mathbf{V_1} +\frac{\sin{\left(t\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} S_2\mathbf{V_2} \end{eqnarray} $$

以上により$\mathbf{p_{t}}$へ$\mathbf{p}$を回転させる四元数$\mathbf{q_{t}}$が$\mathbf{q_{1}}$，$\mathbf{q_{2}}$(及び$\theta_{1 \rightarrow 2}$と$t$)によって表すことができた． $$ \begin{eqnarray} \mathbf{q_{t}}&=&\left(C_{t},\;S_{t}\mathbf{V_{t}}\right) \\&=&\left( \frac{\sin{\left(\left(1-t\right)\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} C_1 +\frac{\sin{\left(t\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} C_2 ,\; \frac{\sin{\left(\left(1-t\right)\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} S_{1}\mathbf{V_1} +\frac{\sin{\left(t\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} S_2\mathbf{V_2} \right) \\&=&\frac{\sin{\left(\left(1-t\right)\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} C_1 +\frac{\sin{\left(t\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} C_2 +\frac{\sin{\left(\left(1-t\right)\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} S_{1}\mathbf{V_1} +\frac{\sin{\left(t\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} S_2\mathbf{V_2} \\&=&\frac{\sin{\left(\left(1-t\right)\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} C_1 +\frac{\sin{\left(\left(1-t\right)\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} S_{1}\mathbf{V_1} +\frac{\sin{\left(t\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} C_2 +\frac{\sin{\left(t\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} S_2\mathbf{V_2} \\&=&\frac{\sin{\left(\left(1-t\right)\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} \left(C_1,S_{1}\mathbf{V_1}\right) +\frac{\sin{\left(t\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} \left(C_2,S_{2}\mathbf{V_2}\right) \\&=&\frac{\sin{\left(\left(1-t\right)\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} \mathbf{q_1} +\frac{\sin{\left(t\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} \mathbf{q_2} \end{eqnarray} $$ これは“ 2点を補間した位置への回転操作のための三角凾数”と同様の式の形をしている．

2点を補間した位置への回転操作のための三角凾数

$原点Oを中心とした等距離R上の点Pの点P_1から点P_2への移動(半径Rの円周上)を考える.\overrightarrow{OP_1}と\overrightarrow{OP_2}のなす角を\thetaとする.$ $$ \begin{eqnarray} \left|\overrightarrow{OP}\right|&=&\left|\overrightarrow{OP_1}\right|=\left|\overrightarrow{OP_2}\right|=R \\\overrightarrow{OP_1}\cdot\overrightarrow{OP_2}&=&\left|\overrightarrow{OP_1}\right|\left|\overrightarrow{OP_2}\right|\cos{\left(\theta\right)} \end{eqnarray} $$ パラメータ$t$(ただし$t$は$0\leq t \leq 1$の実数)を用いて$\overrightarrow{OP}$との関係(内積)を求めると以下のようになる $$ \begin{eqnarray} \\\overrightarrow{OP}\cdot\overrightarrow{OP_1}&=&\left|\overrightarrow{OP}\right|\left|\overrightarrow{OP_1}\right|\cos{\left(t\theta\right)} \\&=&R^2\cos{\left(t\theta\right)}\;\cdot\;ただしtは0\leq t \leq 1の実数 \\\overrightarrow{OP}\cdot\overrightarrow{OP_2}&=&\left|\overrightarrow{OP}\right|\left|\overrightarrow{OP_2}\right|\cos{\left(\left(1-t\right)\theta\right)} \\&=&R^2\cos{\left(\left(1-t\right)\theta\right)} \end{eqnarray} $$ $\overrightarrow{OP}=\alpha \overrightarrow{OP_1}+ \beta \overrightarrow{OP_2}$で表すと仮定する． $$ \begin{eqnarray} \\\overrightarrow{OP}\cdot\overrightarrow{OP_1} &=&(\alpha \overrightarrow{OP_1}+\beta \overrightarrow{OP_2})\cdot\overrightarrow{OP_1} \\&=&\alpha \overrightarrow{OP_1}\cdot\overrightarrow{OP_1} +\beta \overrightarrow{OP_2}\cdot\overrightarrow{OP_1} \\&=&\alpha \left|\overrightarrow{OP_1}\right|\left|\overrightarrow{OP_1}\right|\cos{\left(0\right)} +\beta \left|\overrightarrow{OP_1}\right|\left|\overrightarrow{OP_2}\right|\cos{\left(\theta\right)} \\&=&\alpha R^2\cdot1+\beta R^2\cos{\left(\theta\right)} \\&=&R^2\left(\alpha +\beta \cos{\left(\theta\right)} \right) \\\overrightarrow{OP}\cdot\overrightarrow{OP_2} &=&(\alpha \overrightarrow{OP_1}+ \beta \overrightarrow{OP_2})\cdot\overrightarrow{OP_2} \\&=&\alpha \overrightarrow{OP_1}\cdot\overrightarrow{OP_2} +\beta \overrightarrow{OP_2}\cdot\overrightarrow{OP_2} \\&=&\alpha \left|\overrightarrow{OP_1}\right|\left|\overrightarrow{OP_2}\right|\cos{\left(\theta\right)} +\beta \left|\overrightarrow{OP_2}\right|\left|\overrightarrow{OP_2}\right|\cos{\left(0\right)} \\&=&\alpha R^2\cos{\left(\theta\right)}+\beta R^2\cdot1 \\&=&R^2\left(\alpha \cos{\left(\theta\right)} +\beta \right) \end{eqnarray} $$ 以上より以下の連立方程式が作れる． $$ \left\{\begin{align} \alpha +\beta \cos{\left(\theta\right)}&=&\cos{\left(t\theta\right)}\\ \alpha \cos{\left(\theta\right)} +\beta&=&\cos{\left(\left(1-t\right)\theta\right)}\\ \end{align}\right.\\ $$ 連立方程式を解く． $$ \begin{eqnarray} \left (\alpha+\beta \cos{\left(\theta\right)}\right)\cos{\left(\theta\right)}&=&\cos{\left(t\theta\right)}\cos{\left(\theta\right)} \;\cdots\;連立方程式上側の式の両辺に\cos{\left(\theta\right)}を掛ける． \\\alpha\cos{\left(\theta\right)}+\beta \cos^2{\left(\theta\right)}&=&\cos{\left(t\theta\right)}\cos{\left(\theta\right)}\\ \left(\alpha\cos{\left(\theta\right)}+\beta \cos^2{\left(\theta\right)}\right) -\left(\alpha \cos{\left(\theta\right)} +\beta\right) &=&\left(\cos{\left(t\theta\right)}\cos{\left(\theta\right)}\right) -\left(\cos{\left(\left(1-t\right)\theta\right)}\right) \;\cdots\;連立方程式上側の式の変形から連立方程式下側の式を引く． \\\beta \cos^2{\left(\theta\right)}-\beta &=&\cos{\left(t\theta\right)}\cos{\left(\theta\right)} -\left(\cos{\left(\theta-t\theta\right)}\right) \\\beta \left(\cos^2{\left(\theta\right)}-1\right) &=&\cos{\left(t\theta\right)}\cos{\left(\theta\right)} -\left( \cos{\left(\theta\right)}\cos{\left(t\theta\right)} +\sin{\left(\theta\right)}\sin{\left(t\theta\right)} \right) \;\cdots\;\cos{\left(\alpha-\beta\right)}=\cos{\left(\alpha\right)}\cos{\left(\beta\right)}+\sin{\left(\alpha\right)}\sin{\left(\beta\right)} \\\beta \left(\cos^2{\left(\theta\right)}-1\right) &=&-\sin{\left(\theta\right)}\sin{\left(t\theta\right)} \\\beta \left(1-\cos^2{\left(\theta\right)}\right) &=&\sin{\left(\theta\right)}\sin{\left(t\theta\right)} \\\beta \sin^2{\left(\theta\right)} &=&\sin{\left(\theta\right)}\sin{\left(t\theta\right)} \\\beta &=&\frac{\sin{\left(\theta\right)}\sin{\left(t\theta\right)}}{\sin^2{\left(\theta\right)}} \\\beta &=&\frac{\sin{\left(t\theta\right)}}{\sin{\left(\theta\right)}} \\\alpha +\left(\frac{\sin{\left(t\theta\right)}}{\sin{\left(\theta\right)}}\right)\cos{\left(\theta\right)}&=&\cos{\left(t\theta\right)} \;\cdots\;連立方程式上側の式に求めた\betaを代入する．\\\alpha&=&\cos{\left(t\theta\right)}-\left(\frac{\sin{\left(t\theta\right)}}{\sin{\left(\theta\right)}}\right)\cos{\left(\theta\right)} \\\alpha&=&\frac{ \sin{\left(\theta\right)} \cos{\left(t\theta\right)} -\cos{\left(\theta\right)} \sin{\left(t\theta\right)} }{ \sin{\left(\theta\right)} } \\\alpha&=&\frac{ \sin{\left(\theta-t\theta\right)} }{ \sin{\left(\theta\right)} } \;\cdots\;\sin{\left(\alpha\right)}\cos{\left(\beta\right)}-\cos{\left(\alpha\right)}\sin{\left(\beta\right)}=\sin{\left(\alpha-\beta\right)} \\&=&\frac{ \sin{\left(\left(1-t\right)\theta\right)} }{ \sin{\left(\theta\right)} } \\\alpha,\;\beta&=&\frac{\sin{\left(\left(1-t\right)\theta\right)}}{\sin{\left(\theta\right)}} ,\;\frac{\sin{\left(t\theta\right)}}{\sin{\left(\theta\right)}} \end{eqnarray} $$ 解より点$P_1$から点$P_2$への移動の間の点$P$を変数$t$を用いて以下のように表現できることが示せた． $$ \begin{eqnarray} \overrightarrow{OP}&=&\alpha \overrightarrow{OP_1}+ \beta \overrightarrow{OP_2} \\&=&\frac{\sin{\left(\left(1-t\right)\theta\right)}}{\sin{\left(\theta\right)}}\overrightarrow{OP_1} +\frac{\sin{\left(t\theta\right)}}{\sin{\left(\theta\right)}}\overrightarrow{OP_2} \end{eqnarray} $$ 三次元でも四元数を用いると同様の形の式となることを示した記事が“ 2点を補間した位置への回転操作のための四元数”です．

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