# 内積・外積についての公式 2 の証明 (proof of formula 2 for inner product and cross product)

[証明]

ベクトル  $A=\left({A}_{x}\text{\hspace{0.17em}},\text{\hspace{0.17em}}{A}_{y}\text{\hspace{0.17em}},\text{\hspace{0.17em}}{A}_{z}\right)$$B=\left({B}_{x}\text{\hspace{0.17em}},\text{\hspace{0.17em}}{B}_{y}\text{\hspace{0.17em}},\text{\hspace{0.17em}}{B}_{z}\right)$$C=\left({C}_{x}\text{\hspace{0.17em}},\text{\hspace{0.17em}}{C}_{y}\text{\hspace{0.17em}},\text{\hspace{0.17em}}{C}_{z}\right)$  に対して，

$B×C$$=\left({B}_{y}{C}_{z}-{B}_{z}{C}_{y}\text{\hspace{0.17em}},\text{\hspace{0.17em}}{B}_{z}{C}_{x}-{B}_{x}{C}_{z}\text{\hspace{0.17em}},\text{\hspace{0.17em}}{B}_{x}{C}_{y}-{B}_{y}{C}_{x}\right)$
$=\left({B}_{y}{C}_{z}-{B}_{z}{C}_{y}\right)i+\left({B}_{z}{C}_{x}-{B}_{x}{C}_{z}\right)j+\left({B}_{x}{C}_{y}-{B}_{y}{C}_{x}\right)k$

なので，

$A×\left(B×C\right)$$=\left\{{A}_{y}\left({B}_{x}{C}_{y}-{B}_{y}{C}_{x}\right)-{A}_{z}\left({B}_{z}{C}_{x}-{B}_{x}{C}_{z}\right)\right\}i$

$+\left\{{A}_{z}\left({B}_{y}{C}_{z}-{B}_{z}{C}_{y}\right)-{A}_{x}\left({B}_{x}{C}_{y}-{B}_{y}{C}_{x}\right)\right\}j$

$+\left\{{A}_{x}\left({B}_{z}{C}_{x}-{B}_{x}{C}_{z}\right)-{A}_{y}\left({B}_{y}{C}_{z}-{B}_{z}{C}_{y}\right)\right\}k$

$=\left\{\left({A}_{y}{C}_{y}+{A}_{z}{C}_{z}\right){B}_{x}-\left({A}_{y}{B}_{y}+{A}_{z}{B}_{z}\right){C}_{x}\right\}i$

$+\left\{\left({A}_{x}{C}_{x}+{A}_{z}{C}_{z}\right){B}_{y}-\left({A}_{x}{B}_{x}+{A}_{z}{B}_{z}\right){C}_{y}\right\}j$

$+\left\{\left({A}_{x}{C}_{x}+{A}_{y}{C}_{y}\right){B}_{z}-\left({A}_{x}{B}_{x}+{A}_{y}{B}_{y}\right){C}_{z}\right\}k$

$=\left\{\left({A}_{x}{C}_{x}+{A}_{y}{C}_{y}+{A}_{z}{C}_{z}\right){B}_{x}-\left({A}_{x}{B}_{x}+{A}_{y}{B}_{y}+{A}_{z}{B}_{z}\right){C}_{x}\right\}i$

$+\left\{\left({A}_{x}{C}_{x}+{A}_{y}{C}_{y}+{A}_{z}{C}_{z}\right){B}_{y}-\left({A}_{x}{B}_{x}+{A}_{y}{B}_{y}+{A}_{z}{B}_{z}\right){C}_{y}\right\}j$

$+\left\{\left({A}_{x}{C}_{x}+{A}_{y}{C}_{y}+{A}_{z}{C}_{z}\right){B}_{z}-\left({A}_{x}{B}_{x}+{A}_{y}{B}_{y}+{A}_{z}{B}_{z}\right){C}_{z}\right\}k$

$=\left\{\left(A\cdot C\right){B}_{x}-\left(A\cdot B\right){C}_{x}\right\}i$

$+\left\{\left(A\cdot C\right){B}_{y}-\left(A\cdot B\right){C}_{y}\right\}j$

$+\left\{\left(A\cdot C\right){B}_{z}-\left(A\cdot B\right){C}_{z}\right\}k$

$=\left(A\cdot C\right)B-\left(A\cdot B\right)C$

となる．（証明終）

ちなみに，

$B×\left(C×A\right)$$=\left(B\cdot A\right)C-\left(B\cdot C\right)A$
$C×\left(A×B\right)$$=\left(C\cdot B\right)A-\left(C\cdot A\right)B$

であり，

$A×\left(B×C\right)+B×\left(C×A\right)+C×\left(A×B\right)=0$

が成り立つので，ヤコビ恒等式を満たす．

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