# 内積・外積についての公式 3 の証明 (proof of formula 3 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)$$D=\left({D}_{x}\text{\hspace{0.17em}},\text{\hspace{0.17em}}{D}_{y}\text{\hspace{0.17em}},\text{\hspace{0.17em}}{D}_{z}\right)$  に対して，

$A×B$$=\left({A}_{y}{B}_{z}-{A}_{z}{B}_{y}\text{\hspace{0.17em}},\text{\hspace{0.17em}}{A}_{z}{B}_{x}-{A}_{x}{B}_{z}\text{\hspace{0.17em}},\text{\hspace{0.17em}}{A}_{x}{B}_{y}-{A}_{y}{B}_{x}\right)$

$C×D$$=\left({C}_{y}{D}_{z}-{C}_{z}{D}_{y}\text{\hspace{0.17em}},\text{\hspace{0.17em}}{C}_{z}{D}_{x}-{C}_{x}{D}_{z}\text{\hspace{0.17em}},\text{\hspace{0.17em}}{C}_{x}{D}_{y}-{C}_{y}{D}_{x}\right)$

より，

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

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

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

$={A}_{y}{B}_{z}{C}_{y}{D}_{z}+{A}_{z}{B}_{y}{C}_{z}{D}_{y}-{A}_{y}{B}_{z}{C}_{z}{D}_{y}-{A}_{z}{B}_{y}{C}_{y}{D}_{z}$

$+{A}_{z}{B}_{x}{C}_{z}{D}_{x}+{A}_{x}{B}_{z}{C}_{x}{D}_{z}-{A}_{z}{B}_{x}{C}_{x}{D}_{z}-{A}_{x}{B}_{z}{C}_{z}{D}_{x}$

$+{A}_{x}{B}_{y}{C}_{x}{D}_{y}+{A}_{y}{B}_{x}{C}_{y}{D}_{x}-{A}_{x}{B}_{y}{C}_{y}{D}_{x}-{A}_{y}{B}_{x}{C}_{x}{D}_{y}$

となる．ここで，右辺に

$0=$ ${A}_{x}{B}_{x}{C}_{x}{D}_{x}+{A}_{y}{B}_{y}{C}_{y}{D}_{y}+{A}_{z}{B}_{z}{C}_{z}{D}_{z}$ $-{A}_{x}{B}_{x}{C}_{x}{D}_{x}-{A}_{y}{B}_{y}{C}_{y}{D}_{y}-{A}_{z}{B}_{z}{C}_{z}{D}_{z}$

を加えると，

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

$+{A}_{y}{B}_{x}{C}_{y}{D}_{x}+{A}_{y}{B}_{y}{C}_{y}{D}_{y}+{A}_{y}{B}_{z}{C}_{y}{D}_{z}$ $-{A}_{y}{B}_{x}{C}_{x}{D}_{y}-{A}_{y}{B}_{y}{C}_{y}{D}_{y}-{A}_{y}{B}_{z}{C}_{z}{D}_{y}$

$+{A}_{z}{B}_{x}{C}_{z}{D}_{x}+{A}_{z}{B}_{y}{C}_{z}{D}_{y}+{A}_{z}{B}_{z}{C}_{z}{D}_{z}$ $-{A}_{z}{B}_{x}{C}_{x}{D}_{z}-{A}_{z}{B}_{y}{C}_{y}{D}_{z}-{A}_{z}{B}_{z}{C}_{z}{D}_{z}$

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

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

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

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

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

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

が成り立つ．（証明終）

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