内積・外積についての公式 1 の証明

スカラー3重積の公式 $3$\displaystyle{ \displaystyle{A}\cdot $\displaystyle{B}\times \displaystyle{C}$ =\displaystyle{B}\cdot $\displaystyle{C}\times \displaystyle{A}$ =\displaystyle{C}\cdot $\displaystyle{A}\times \displaystyle{B}$ }$ $3$\displaystyle{ =\det ${ \displaystyle{A} }^{T}{ \displaystyle{B} }^{T}{ \displaystyle{C} }^{T}$ }$ の証明

[証明]

ベクトル $3$\displaystyle{ \displaystyle{A}= $\hspace{1}{A}_{x}\text{\hspace{1}},\text{\hspace{1}}\hspace{1}{A}_{y}\text{\hspace{1}},\text{\hspace{1}}\hspace{1}{A}_{z}$ }$ ， $3$\displaystyle{ \displaystyle{B}= $\hspace{1}{B}_{x}\text{\hspace{1}},\text{\hspace{1}}\hspace{1}{B}_{y}\text{\hspace{1}},\text{\hspace{1}}\hspace{1}{B}_{z}$ }$ ， $3$\displaystyle{ \displaystyle{C}= $\hspace{1}{C}_{x}\text{\hspace{1}},\text{\hspace{1}}\hspace{1}{C}_{y}\text{\hspace{1}},\text{\hspace{1}}\hspace{1}{C}_{z}$ }$ に対して，

$3$\displaystyle{ \displaystyle{A}\times \displaystyle{B} }$ $3$\displaystyle{ = $\hspace{1}{A}_{y}\hspace{1}{B}_{z}- \hspace{1}{A}_{z}\hspace{1}{B}_{y}\text{\hspace{1}},\text{\hspace{1}}\hspace{1}{A}_{z}\hspace{1}{B}_{x}- \hspace{1}{A}_{x}\hspace{1}{B}_{z}\text{\hspace{1}},\text{\hspace{1}}\hspace{1}{A}_{x}\hspace{1}{B}_{y}- \hspace{1}{A}_{y}\hspace{1}{B}_{x}$ }$

$3$\displaystyle{ \displaystyle{B}\times \displaystyle{C} }$ $3$\displaystyle{ = $\hspace{1}{B}_{y}\hspace{1}{C}_{z}- \hspace{1}{B}_{z}\hspace{1}{C}_{y}\text{\hspace{1}},\text{\hspace{1}}\hspace{1}{B}_{z}\hspace{1}{C}_{x}- \hspace{1}{B}_{x}\hspace{1}{C}_{z}\text{\hspace{1}},\text{\hspace{1}}\hspace{1}{B}_{x}\hspace{1}{C}_{y}- \hspace{1}{B}_{y}\hspace{1}{C}_{x}$ }$

$3$\displaystyle{ \displaystyle{C}\times \displaystyle{A} }$ $3$\displaystyle{ = $\hspace{1}{C}_{y}\hspace{1}{A}_{z}- \hspace{1}{C}_{z}\hspace{1}{A}_{y}\text{\hspace{1}},\text{\hspace{1}}\hspace{1}{C}_{z}\hspace{1}{A}_{x}- \hspace{1}{C}_{x}\hspace{1}{A}_{z}\text{\hspace{1}},\text{\hspace{1}}\hspace{1}{C}_{x}\hspace{1}{A}_{y}- \hspace{1}{C}_{y}\hspace{1}{A}_{x}$ }$

より，

$3$\displaystyle{ \displaystyle{A}\cdot $\displaystyle{B}\times \displaystyle{C}$ }$ $3$\displaystyle{ =\hspace{1}{A}_{x} $\hspace{1}{B}_{y}\hspace{1}{C}_{z}- \hspace{1}{B}_{z}\hspace{1}{C}_{y}$ }$ $3$\displaystyle{ +\hspace{1}{A}_{y} $\hspace{1}{B}_{z}\hspace{1}{C}_{x}- \hspace{1}{B}_{x}\hspace{1}{C}_{z}$ }$ $3$\displaystyle{ +\hspace{1}{A}_{z} $\hspace{1}{B}_{x}\hspace{1}{C}_{y}- \hspace{1}{B}_{y}\hspace{1}{C}_{x}$ }$

$3$\displaystyle{ =\hspace{1}{A}_{x}\hspace{1}{B}_{y}\hspace{1}{C}_{z}- \hspace{1}{A}_{x}\hspace{1}{B}_{z}\hspace{1}{C}_{y} }$ $3$\displaystyle{ +\hspace{1}{A}_{y}\hspace{1}{B}_{z}\hspace{1}{C}_{x}- \hspace{1}{A}_{y}\hspace{1}{B}_{x}\hspace{1}{C}_{z} }$ $3$\displaystyle{ +\hspace{1}{A}_{z}\hspace{1}{B}_{x}\hspace{1}{C}_{y}- \hspace{1}{A}_{z}\hspace{1}{B}_{y}\hspace{1}{C}_{x} }$

$3$\displaystyle{ \displaystyle{B}\cdot $\displaystyle{C}\times \displaystyle{A}$ }$ $3$\displaystyle{ =\hspace{1}{B}_{x} $\hspace{1}{C}_{y}\hspace{1}{A}_{z}- \hspace{1}{C}_{z}\hspace{1}{A}_{y}$ }$ $3$\displaystyle{ +\hspace{1}{B}_{y} $\hspace{1}{C}_{z}\hspace{1}{A}_{x}- \hspace{1}{C}_{x}\hspace{1}{A}_{z}$ }$ $3$\displaystyle{ +\hspace{1}{B}_{z} $\hspace{1}{C}_{x}\hspace{1}{A}_{y}- \hspace{1}{C}_{y}\hspace{1}{A}_{x}$ }$

$3$\displaystyle{ =\hspace{1}{B}_{x}\hspace{1}{C}_{y}\hspace{1}{A}_{z}- \hspace{1}{B}_{x}\hspace{1}{C}_{z}\hspace{1}{A}_{y} }$ $3$\displaystyle{ +\hspace{1}{B}_{y}\hspace{1}{C}_{z}\hspace{1}{A}_{x}- \hspace{1}{B}_{y}\hspace{1}{C}_{x}\hspace{1}{A}_{z} }$ $3$\displaystyle{ +\hspace{1}{B}_{z}\hspace{1}{C}_{x}\hspace{1}{A}_{y}- \hspace{1}{B}_{z}\hspace{1}{C}_{y}\hspace{1}{A}_{x} }$

$3$\displaystyle{ \displaystyle{C}\cdot $\displaystyle{A}\times \displaystyle{B}$ }$ $3$\displaystyle{ =\hspace{1}{C}_{x} $\hspace{1}{A}_{y}\hspace{1}{B}_{z}- \hspace{1}{A}_{z}\hspace{1}{B}_{y}$ }$ $3$\displaystyle{ +\hspace{1}{C}_{y} $\hspace{1}{A}_{z}\hspace{1}{B}_{x}- \hspace{1}{A}_{x}\hspace{1}{B}_{z}$ }$ $3$\displaystyle{ +\hspace{1}{C}_{z} $\hspace{1}{A}_{x}\hspace{1}{B}_{y}- \hspace{1}{A}_{y}\hspace{1}{B}_{x}$ }$

$3$\displaystyle{ =\hspace{1}{C}_{x}\hspace{1}{A}_{y}\hspace{1}{B}_{z}- \hspace{1}{C}_{x}\hspace{1}{A}_{z}\hspace{1}{B}_{y} }$ $3$\displaystyle{ +\hspace{1}{C}_{y}\hspace{1}{A}_{z}\hspace{1}{B}_{x}- \hspace{1}{C}_{y}\hspace{1}{A}_{x}\hspace{1}{B}_{z} }$ $3$\displaystyle{ +\hspace{1}{C}_{z}\hspace{1}{A}_{x}\hspace{1}{B}_{y}- \hspace{1}{C}_{z}\hspace{1}{A}_{y}\hspace{1}{B}_{x} }$

となるので $3$\displaystyle{ \displaystyle{A}\cdot $\displaystyle{B}\times \displaystyle{C}$ =\displaystyle{B}\cdot $\displaystyle{C}\times \displaystyle{A}$ =\displaystyle{C}\cdot $\displaystyle{A}\times \displaystyle{B}$ }$ が成り立つ．また，

$3$\displaystyle{ \det ${ \displaystyle{A} }^{T}{ \displaystyle{B} }^{T}{ \displaystyle{C} }^{T}$ = \| \begin{array} \hspace{1}{A}_{x} & \hspace{1}{B}_{x} & \hspace{1}{C}_{x} & \vspace{6}\\ \hspace{1}{A}_{y} & \hspace{1}{B}_{y} & \hspace{1}{C}_{y} & \vspace{6}\\ \hspace{1}{A}_{z} & \hspace{1}{B}_{z} & \hspace{1}{C}_{z} & \vspace{6}\\ \end{array} \| }$ $3$\displaystyle{ =\hspace{1}{A}_{x} \| \begin{array} \hspace{1}{B}_{y} & \hspace{1}{C}_{y} & \vspace{6}\\ \hspace{1}{B}_{z} & \hspace{1}{C}_{z} & \vspace{6}\\ \end{array} \| }$ $3$\displaystyle{ - \hspace{1}{A}_{y} \| \begin{array} \hspace{1}{B}_{x} & \hspace{1}{C}_{x} & \vspace{6}\\ \hspace{1}{B}_{z} & \hspace{1}{C}_{z} & \vspace{6}\\ \end{array} \| }$ $3$\displaystyle{ +\hspace{1}{A}_{z} \| \begin{array} \hspace{1}{B}_{x} & \hspace{1}{C}_{x} & \vspace{6}\\ \hspace{1}{B}_{y} & \hspace{1}{C}_{y} & \vspace{6}\\ \end{array} \| }$

$3$\displaystyle{ =\hspace{1}{A}_{x} $\hspace{1}{B}_{y}\hspace{1}{C}_{z}- \hspace{1}{B}_{z}\hspace{1}{C}_{y}$ }$ $3$\displaystyle{ - \hspace{1}{A}_{y} $\hspace{1}{B}_{x}\hspace{1}{C}_{z}- \hspace{1}{B}_{z}\hspace{1}{C}_{x}$ }$ $3$\displaystyle{ +\hspace{1}{A}_{z} $\hspace{1}{B}_{x}\hspace{1}{C}_{y}- \hspace{1}{B}_{y}\hspace{1}{C}_{x}$ }$

より， $3$\displaystyle{ \det ${ \displaystyle{A} }^{T}{ \displaystyle{B} }^{T}{ \displaystyle{C} }^{T}$ =\displaystyle{A}\cdot $\displaystyle{B}\times \displaystyle{C}$ }$ が成り立つ．（証明終）

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最終更新日2026年3月10日