[解題] 三角形OAB的面積
年級:高二
科目:數學
章節:三維空間向量
請問三角形OAB面積等於向量OA和向量OB外積向量的二分之一長度,原因為何?
以下我的想法:
假設 O=(0,0,0) A=(a,b,c) B=(p,q,r)
∣OA∣^2=a^2+b^2+c^2
∣OB∣^2=p^2+q^2+r^2
∣AB∣^2=(p-a)^2+(q-b)^2+(r-c)^2
三角形OAB面積
=(1/2)|OA|∣OB∣sin∠AOB
=(1/2)|OA|∣OB∣√(1-cos^2∠AOB)
=(1/2)|OA|∣OB∣√[1-(∣OA∣^2+∣OB∣^2-∣AB∣^2)^2/(2|OA|∣OB∣)^2]
=(1/4)√[4|OA|^2∣OB∣^2-(∣OA∣^2+∣OB∣^2-∣AB∣^2)^2]
接著把∣OA∣^2 ∣OB∣^2 ∣AB∣^2 分別代入上式
經過繁複的計算整理
=(1/2)√[(br-cq)^2+(cp-ar)^2+(aq-bp)^2]
除此之外,不曉得各位在教學的時候有無較簡易的方法或利用什麼觀念來解釋此公式?
以上
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※ 編輯: jackal594 來自: 220.138.16.71 (08/29 12:16)
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