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Exercise 8.1

(1)​

x=(x1,x2,x3),y=(y1,y2,y3)\textbf{x} = (x_1, x_2, x_3), \textbf{y} = (y_1, y_2, y_3) に対し、

xΓ—y=(x2y3βˆ’x3y2,x3y1βˆ’x1y3,x1y2βˆ’x2y1)=(y3x2βˆ’y2x3,y1x3βˆ’y3x1,y2x1βˆ’y1x2)=βˆ’(y2x3βˆ’y3x2,y3x1βˆ’y1x3,y1x2βˆ’y2x1)=βˆ’yΓ—x\begin{aligned} \textbf{x} \times \textbf{y} &= (x_2y_3 - x_3y_2, x_3y_1 - x_1y_3, x_1y_2 - x_2y_1) \\ &= (y_3x_2 - y_2x_3, y_1x_3 - y_3x_1, y_2x_1 - y_1x_2) \\ &= - (y_2x_3 - y_3x_2, y_3x_1 - y_1x_3, y_1x_2 - y_2x_1) \\ &= - \textbf{y} \times \textbf{x} \end{aligned}

とγͺる。 γ—γŸγŒγ£γ¦γ€ xΓ—y=βˆ’yΓ—x\textbf{x} \times \textbf{y} = - \textbf{y} \times \textbf{x} を得る。

(2)​

x=(x1,x2,x3),y=(y1,y2,y3)\textbf{x} = (x_1, x_2, x_3), \textbf{y} = (y_1, y_2, y_3) に対し、

xβ‹…(xΓ—y)=xβ‹…(x2y3βˆ’x3y2,x3y1βˆ’x1y3,x1y2βˆ’x2y1)=x1x2y3~βˆ’x1x3y2β€Ύ+x2x3y1undefinedβˆ’x2x1y3~+x3x1y2β€Ύβˆ’x3x2y1undefined=0\begin{aligned} \textbf{x} \cdot (\textbf{x} \times \textbf{y}) &= \textbf{x} \cdot (x_2y_3 - x_3y_2, x_3y_1 - x_1y_3, x_1y_2 - x_2y_1)\\ &= \utilde{x_1x_2y_3} - \underline{x_1x_3y_2} + \underlinesegment{x_2x_3y_1} - \utilde{x_2x_1y_3} + \underline{x_3x_1y_2} - \underlinesegment{x_3x_2y_1} \\ &= 0 \\ \end{aligned}
yβ‹…(xΓ—y)=yβ‹…(x2y3βˆ’x3y2,x3y1βˆ’x1y3,x1y2βˆ’x2y1)=y1x2y3~βˆ’y1x3y2β€Ύ+y2x3y1β€Ύβˆ’y2x1y3undefined+y3x1y2undefinedβˆ’y3x2y1~=0\begin{aligned} \textbf{y} \cdot (\textbf{x} \times \textbf{y}) &= \textbf{y} \cdot (x_2y_3 - x_3y_2, x_3y_1 - x_1y_3, x_1y_2 - x_2y_1)\\ &= \utilde{y_1x_2y_3} - \underline{y_1x_3y_2} + \underline{y_2x_3y_1} - \underlinesegment{y_2x_1y_3} + \underlinesegment{y_3x_1y_2} - \utilde{y_3x_2y_1} \\ &= 0 \\ \end{aligned}

とγͺる。 γ—γŸγŒγ£γ¦γ€ xβ‹…(xΓ—y)=yβ‹…(xΓ—y)=0\textbf{x} \cdot (\textbf{x} \times \textbf{y}) = \textbf{y} \cdot (\textbf{x} \times \textbf{y}) = 0 を得る。