Re: [線代] Back-Cab rule 的證明

看板Math作者ntust661 (XDeutesh)時間12年前 (2011/09/27 01:43)推噓3(3推 0噓 1→)

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※ 引述《obelisk0114 (追風箏的孩子)》之銘言： : Back-Cab rule : : → → → → → → → → → : A ×(B ×C) = B (A‧C) - C (A‧B) : 如何證明? 預備已知 A ×(B ×C) 的向量落在 B 與 C 的展開的平面。 → → → → → 故 A ×(B ×C) = αB + βC → → → → → → → → A ‧{ A ×(B ×C) } = α(A‧B) + β(C‧A) → → → → → → 而因為 A ×B ×C 的向量既垂直 A 也垂直 B ×C → → → → → → → → A ‧{ A ×(B ×C) } = α(A‧B) + β(C‧A) = 0 → → α = γ(A‧C) → → β = -γ(A‧B) → → → → → → → → → 故 A ×(B ×C) = γ{(A‧C) B -(A‧B) C } → → → → → → 考慮 B ×(B ×C) = γ1{ (B‧C) B - (B‧B) C } 考慮大小 {B ×(B ×C)}‧C 2 = (C ×B)‧(B ×C) = γ1{ (B‧C) -(B‧B)(C‧C) } 2 2 2 2 2 2 2 2 = │B││C│sin φ = γ1 ( │B││C│ cos φ - │B││C│ ) γ1 = -1 故繼續證 B‧(A ×(B ×C)) = -B‧((B ×C) ×A) = -(B ×(B ×C))‧A → → → → → → → = -( (B‧C) B - (B‧B) C )‧A = - (B‧C)(B‧A) + (B‧B)(C‧A) 比照 = γ{(A‧C)(B‧B) - (A‧B)(C‧B) } γ = 1 故得證 -- -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 218.161.120.11

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obelisk0114

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obelisk0114

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yan12125

09/28 21:31, , 3^F

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