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)

公式 $(A \times B) \cdot (C \times D)$ $= (A \cdot C) (B \cdot D) - (A \cdot D) (B \cdot C)$ の証明

[証明]

ベクトル $A = (A_{x}, A_{y}, A_{z})$ ， $B = (B_{x}, B_{y}, B_{z})$ ， $C = (C_{x}, C_{y}, C_{z})$ ， $D = (D_{x}, D_{y}, D_{z})$ に対して，

$A \times B$ $= (A_{y} B_{z} - A_{z} B_{y}, A_{z} B_{x} - A_{x} B_{z}, A_{x} B_{y} - A_{y} B_{x})$

$C \times D$ $= (C_{y} D_{z} - C_{z} D_{y}, C_{z} D_{x} - C_{x} D_{z}, C_{x} D_{y} - C_{y} D_{x})$

より，

$(A \times B) \cdot (C \times D)$ $= (A_{y} B_{z} - A_{z} B_{y}) (C_{y} D_{z} - C_{z} D_{y})$

$+ (A_{z} B_{x} - A_{x} B_{z}) (C_{z} D_{x} - C_{x} D_{z})$

$+ (A_{x} B_{y} - A_{y} B_{x}) (C_{x} D_{y} - C_{y} D_{x})$

$= A_{y} B_{z} C_{y} D_{z} + A_{z} B_{y} C_{z} D_{y} - A_{y} B_{z} C_{z} D_{y} - A_{z} B_{y} C_{y} D_{z}$

$+ A_{z} B_{x} C_{z} D_{x} + A_{x} B_{z} C_{x} D_{z} - A_{z} B_{x} C_{x} D_{z} - A_{x} B_{z} C_{z} D_{x}$

$+ A_{x} B_{y} C_{x} D_{y} + A_{y} B_{x} C_{y} D_{x} - A_{x} B_{y} C_{y} D_{x} - A_{y} B_{x} C_{x} D_{y}$

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，

$(A \times B) \cdot (C \times D)$ $= A_{x} B_{x} C_{x} D_{x} + A_{x} B_{y} C_{x} D_{y} + A_{x} B_{z} C_{x} D_{z}$ $- A_{x} B_{x} C_{x} D_{x} - A_{x} B_{y} C_{y} D_{x} - A_{x} B_{z} C_{z} D_{x}$

$+ A_{y} B_{x} C_{y} D_{x} + A_{y} B_{y} C_{y} D_{y} + A_{y} B_{z} C_{y} D_{z}$ $- A_{y} B_{x} C_{x} D_{y} - A_{y} B_{y} C_{y} D_{y} - A_{y} B_{z} C_{z} D_{y}$

$+ A_{z} B_{x} C_{z} D_{x} + A_{z} B_{y} C_{z} D_{y} + A_{z} B_{z} C_{z} D_{z}$ $- A_{z} B_{x} C_{x} D_{z} - A_{z} B_{y} C_{y} D_{z} - A_{z} B_{z} C_{z} D_{z}$

$= A_{x} C_{x} (B_{x} D_{x} + B_{y} D_{y} + B_{z} D_{z})$ $- A_{x} D_{x} (B_{x} C_{x} + B_{y} C_{y} + B_{z} C_{z})$

$+ A_{y} C_{y} (B_{x} D_{x} + B_{y} D_{y} + B_{z} D_{z})$ $- A_{y} D_{y} (B_{x} C_{x} + B_{y} C_{y} + B_{z} C_{z})$

$+ A_{z} C_{z} (B_{x} D_{x} + B_{y} D_{y} + B_{z} D_{z})$ $- A_{z} D_{z} (B_{x} C_{x} + B_{y} C_{y} + B_{z} C_{z})$

$= (A_{x} C_{x} + A_{y} C_{y} + A_{z} C_{z}) (B_{x} D_{x} + B_{y} D_{y} + B_{z} D_{z})$

$- (A_{x} D_{x} + A_{y} D_{y} + A_{z} D_{z}) (B_{x} C_{x} + B_{y} C_{y} + B_{z} C_{z})$

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

が成り立つ．（証明終）

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最終更新日2023年2月20日