内積・外積についての公式 1 の証明 (proof of formula 1 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，

$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)$

$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)$

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

より，

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

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

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

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

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

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

となるので $A\cdot \left(B×C\right)=B\cdot \left(C×A\right)=C\cdot \left(A×B\right)$ が成り立つ．また，

$\mathrm{det}\left({A}^{T}{B}^{T}{C}^{T}\right)=|\begin{array}{ccc}{A}_{x}& {B}_{x}& {C}_{x}\\ {A}_{y}& {B}_{y}& {C}_{y}\\ {A}_{z}& {B}_{z}& {C}_{z}\end{array}|$ $={A}_{x}|\begin{array}{cc}{B}_{y}& {C}_{y}\\ {B}_{z}& {C}_{z}\end{array}|$ $-{A}_{y}|\begin{array}{cc}{B}_{x}& {C}_{x}\\ {B}_{z}& {C}_{z}\end{array}|$ $+{A}_{z}|\begin{array}{cc}{B}_{x}& {C}_{x}\\ {B}_{y}& {C}_{y}\end{array}|$

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

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

より， $\mathrm{det}\left({A}^{T}{B}^{T}{C}^{T}\right)=A\cdot \left(B×C\right)$ が成り立つ．（証明終）

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