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)

スカラー3重積の公式   ${\displaystyle A\cdot (B\times C)=B\cdot (C\times A)=C\cdot (A\times B)}$ ${\displaystyle =\det \left(A^T{B}^T{C}^T\right)}$   の証明

[証明]

ベクトル  ${\displaystyle A=(A_x,A_y,A_z)}$ ， ${\displaystyle B=(B_x,B_y,B_z)}$ ， ${\displaystyle C=(C_x,C_y,C_z)}$  に対して，

${\displaystyle A\times B}$${\displaystyle =(A_y{B}_z-A_z{B}_y,A_z{B}_x-A_x{B}_z,A_x{B}_y-A_y{B}_x)}$

${\displaystyle B\times C}$${\displaystyle =(B_y{C}_z-B_z{C}_y,B_z{C}_x-B_x{C}_z,B_x{C}_y-B_y{C}_x)}$

${\displaystyle C\times A}$${\displaystyle =(C_y{A}_z-C_z{A}_y,C_z{A}_x-C_x{A}_z,C_x{A}_y-C_y{A}_x)}$

より，

${\displaystyle A\cdot (B\times C)}$ ${\displaystyle =A_x(B_y{C}_z-B_z{C}_y)}$ ${\displaystyle +A_y(B_z{C}_x-B_x{C}_z)}$ ${\displaystyle +A_z(B_x{C}_y-B_y{C}_x)}$

${\displaystyle =A_x{B}_y{C}_z-A_x{B}_z{C}_y}$ ${\displaystyle +A_y{B}_z{C}_x-A_y{B}_x{C}_z}$ ${\displaystyle +A_z{B}_x{C}_y-A_z{B}_y{C}_x}$

${\displaystyle B\cdot (C\times A)}$ ${\displaystyle =B_x(C_y{A}_z-C_z{A}_y)}$ ${\displaystyle +B_y(C_z{A}_x-C_x{A}_z)}$ ${\displaystyle +B_z(C_x{A}_y-C_y{A}_x)}$

${\displaystyle =B_x{C}_y{A}_z-B_x{C}_z{A}_y}$ ${\displaystyle +B_y{C}_z{A}_x-B_y{C}_x{A}_z}$ ${\displaystyle +B_z{C}_x{A}_y-B_z{C}_y{A}_x}$

${\displaystyle C\cdot (A\times B)}$ ${\displaystyle =C_x(A_y{B}_z-A_z{B}_y)}$ ${\displaystyle +C_y(A_z{B}_x-A_x{B}_z)}$ ${\displaystyle +C_z(A_x{B}_y-A_y{B}_x)}$

${\displaystyle =C_x{A}_y{B}_z-C_x{A}_z{B}_y}$ ${\displaystyle +C_y{A}_z{B}_x-C_y{A}_x{B}_z}$ ${\displaystyle +C_z{A}_x{B}_y-C_z{A}_y{B}_x}$

となるので ${\displaystyle A\cdot (B\times C)=B\cdot (C\times A)=C\cdot (A\times B)}$ が成り立つ．また，

${\displaystyle \det \left(A^T{B}^T{C}^T\right)=\left|\begin{array}{ccc}{A}_x& B_x& C_x\\ A_y& B_y& C_y\\ A_z& B_z& C_z\end{array}\right|}$ ${\displaystyle =A_x\left|\begin{array}{cc}{B}_y& C_y\\ B_z& C_z\end{array}\right|}$ ${\displaystyle -A_y\left|\begin{array}{cc}{B}_x& C_x\\ B_z& C_z\end{array}\right|}$ ${\displaystyle +A_z\left|\begin{array}{cc}{B}_x& C_x\\ B_y& C_y\end{array}\right|}$

${\displaystyle =A_x(B_y{C}_z-B_z{C}_y)}$ ${\displaystyle -A_y(B_x{C}_z-B_z{C}_x)}$ ${\displaystyle +A_z(B_x{C}_y-B_y{C}_x)}$

${\displaystyle =A_x{B}_y{C}_z-A_x{B}_z{C}_y}$ ${\displaystyle +A_y{B}_z{C}_x-A_y{B}_x{C}_z}$ ${\displaystyle +A_z{B}_x{C}_y-A_z{B}_y{C}_x}$

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

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最終更新日2026年3月10日