Summarize Timeline Top Qs Fact Check
n = 3,
K
=
R
{\displaystyle \mathbb {K} =\mathbb {R} }
のとき
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3
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1
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2
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2
{\displaystyle {\begin{aligned}\left(\textstyle \sum \limits _{i=1}^{n}a_{i}c_{i}\right)\left(\textstyle \sum \limits _{j=1}^{n}b_{j}d_{j}\right)-\left(\textstyle \sum \limits _{i=1}^{n}a_{i}d_{i}\right)\left(\textstyle \sum \limits _{j=1}^{n}b_{j}c_{j}\right)&=(a_{1}b_{2}-a_{2}b_{1})(c_{1}d_{2}-c_{2}d_{1})\\&+(a_{2}b_{3}-a_{3}b_{2})(c_{2}d_{3}-c_{3}d_{2})\\&+(a_{1}b_{3}-a_{3}b_{1})(c_{1}d_{3}-c_{3}d_{1})\\({\boldsymbol {a}}\cdot {\boldsymbol {c}})({\boldsymbol {b}}\cdot {\boldsymbol {d}})-({\boldsymbol {a}}\cdot {\boldsymbol {d}})({\boldsymbol {b}}\cdot {\boldsymbol {c}})&=({\boldsymbol {a}}\times {\boldsymbol {b}})_{3}({\boldsymbol {c}}\times {\boldsymbol {d}})_{3}\\&+({\boldsymbol {a}}\times {\boldsymbol {b}})_{1}({\boldsymbol {c}}\times {\boldsymbol {d}})_{1}\\&+({\boldsymbol {a}}\times {\boldsymbol {b}})_{2}({\boldsymbol {c}}\times {\boldsymbol {d}})_{2}\end{aligned}}}
すなわち、クロス積 のスカラー四重積 の公式
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a
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∈
R
3
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{\displaystyle ({\boldsymbol {a}}\times {\boldsymbol {b}})\cdot ({\boldsymbol {c}}\times {\boldsymbol {d}})=({\boldsymbol {a}}\cdot {\boldsymbol {c}})({\boldsymbol {b}}\cdot {\boldsymbol {d}})-({\boldsymbol {a}}\cdot {\boldsymbol {d}})({\boldsymbol {b}}\cdot {\boldsymbol {c}})\quad ({\boldsymbol {a}},{\boldsymbol {b}},{\boldsymbol {c}},{\boldsymbol {d}}\in \mathbb {R} ^{3})}
が得られる。(この式をビネ・コーシーの恒等式ということもある。)
この式をスカラー三重積 の性質を使って変形すれば
{
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∴
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{\displaystyle {\begin{aligned}\{({\boldsymbol {a}}\times {\boldsymbol {b}})\times {\boldsymbol {c}}\}\cdot {\boldsymbol {d}}&=\left\{({\boldsymbol {a}}\cdot {\boldsymbol {c}}){\boldsymbol {b}}-({\boldsymbol {b}}\cdot {\boldsymbol {c}}){\boldsymbol {a}}\right\}\cdot {\boldsymbol {d}}\\\therefore ~({\boldsymbol {a}}\times {\boldsymbol {b}})\times {\boldsymbol {c}}&=({\boldsymbol {a}}\cdot {\boldsymbol {c}}){\boldsymbol {b}}-({\boldsymbol {b}}\cdot {\boldsymbol {c}}){\boldsymbol {a}}\end{aligned}}}
とベクトル三重積 の公式が得られる。
また、c = a , d = b とおくと、
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2
(
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∈
R
3
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{\displaystyle \|{\boldsymbol {a}}\times {\boldsymbol {b}}\|^{2}=\|{\boldsymbol {a}}\|^{2}\|{\boldsymbol {b}}\|^{2}-({\boldsymbol {a}}\cdot {\boldsymbol {b}})^{2}\quad ({\boldsymbol {a}},{\boldsymbol {b}}\in \mathbb {R} ^{3})}
と、ベクトル解析におけるラグランジュの恒等式が得られる。