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平行六面體ABCD-A'B'C'D'中,AB=5,AD=3,AA'=7,角BAD=60度,角BAA'=角DAA'=45度,求AC'的長。 題示:結合空間向量來解題~ 要過程~

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Q: 平行六面體 ABCD ~ A'B'C'D'中, AB = 5, AD = 3, AA' = 7, ∠BAD = 60o, ∠BAA' = ∠DAA' = 45o, 求 AC' 的長 Sol: ∵ 平行六面體 ∴ BC = AD = 3, CC’ = AA’ = 7, ∠CAA' = ∠BAA' = 45o ∠ABC = 180o - ∠BAD = 180o - 60o = 120o 首先先看 △ABC, AB = 5, BC = 3, ∠B = 120o 根據餘弦定理 AC2 = AB2 + BC2 - 2 * AB * BC * cos∠B AC2 = 52 + 32 - 2 * 5 * 3 * ( - 1/2 ) AC2 = 49 AC = 7 ∠ACC’ = 180o - ∠CAA’ = 180o - 45o = 135o 再看 △ACC’, AC = 7, CC’ = 7, ∠ACC’ = 135o AC’2 = AC^2 + CC’2 - 2 * AC * CC’ * cos∠C AC'2 = 72 + 72 - 2 * 7 * 7 * ( - √2/2 ) AC'2 = 98 + 49√2 AC' = √( 98 + 49√2 ) Ans: AC' = √( 98 + 49√2 ) P.S 抱歉, 因為我已經回答 10 個問題了, 所以要過 12 點才能回答你的問題. 2008-01-07 14:55:42 補充: sorry, 答案更正 ( 因為我覺得 AC 與 CC’ 的夾角並非 135o )so, forget the answer above. 2008-01-07 14:56:32 補充: correction:AC' = AB + AD + AA'| AC' |2 = | AB + AD + AA' |2AC'2 = | AB |2 + | AD |2 + | AA' |2 + 2 ( AB.AD + AB.AA' + AD.AA' )= 25 + 9 + 49 + 2 ( 5 * 3 * cos60o + 5 * 7 cos45o + 3 * 7 cos45o )= 98 + 56√2 2008-01-07 14:56:49 補充: AC’ = √( 98 + 56√2 )Ans: AC’ = √( 98 + 56√2 )

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