内積・外積についての公式 2 の証明

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

なので，

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

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

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

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

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

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

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

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

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

$3$\displaystyle{ = \{ $\displaystyle{A}\cdot \displaystyle{C}$ \hspace{1}{B}_{x}- $\displaystyle{A}\cdot \displaystyle{B}$ \hspace{1}{C}_{x}\} \displaystyle{i} }$

$3$\displaystyle{ + \{ $\displaystyle{A}\cdot \displaystyle{C}$ \hspace{1}{B}_{y}- $\displaystyle{A}\cdot \displaystyle{B}$ \hspace{1}{C}_{y}\} \displaystyle{j} }$

$3$\displaystyle{ + \{ $\displaystyle{A}\cdot \displaystyle{C}$ \hspace{1}{B}_{z}- $\displaystyle{A}\cdot \displaystyle{B}$ \hspace{1}{C}_{z}\} \displaystyle{k} }$

$3$\displaystyle{ = $\displaystyle{A}\cdot \displaystyle{C}$ \displaystyle{B}- $\displaystyle{A}\cdot \displaystyle{B}$ \displaystyle{C} }$

となる．（証明終）

ちなみに，

$3$\displaystyle{ \displaystyle{B}\times $\displaystyle{C}\times \displaystyle{A}$ }$ $3$\displaystyle{ = $\displaystyle{B}\cdot \displaystyle{A}$ \displaystyle{C}- $\displaystyle{B}\cdot \displaystyle{C}$ \displaystyle{A} }$
$3$\displaystyle{ \displaystyle{C}\times $\displaystyle{A}\times \displaystyle{B}$ }$ $3$\displaystyle{ = $\displaystyle{C}\cdot \displaystyle{B}$ \displaystyle{A}- $\displaystyle{C}\cdot \displaystyle{A}$ \displaystyle{B} }$

であり，

$3$\displaystyle{ \displaystyle{A}\times $\displaystyle{B}\times \displaystyle{C}$ +\displaystyle{B}\times $\displaystyle{C}\times \displaystyle{A}$ +\displaystyle{C}\times $\displaystyle{A}\times \displaystyle{B}$ =0 }$

が成り立つので，ヤコビ恒等式を満たす．

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最終更新日2023年2月20日