Re: [中學] 空間向量
: 想請問這題應該要如何解呢
: 希望大大給個提示
: 感恩
AP = (1/3)AB + 2AD + aAE
BD = AD - AB
BG = AE + AD
BD x BG = -(|AD||AE|/|AB|)AB - (|AB||AE|/|AD|)AD - (|AB||AD|/|AE|)AE
=> AD * BD x BG =/= 0
AD無法以xBD + yBG表達出來
AP = AB + BP = (1/3)AB + 2AD + aAE
=> BP = (-2/3)AB + 2AD + aAE
= (-2/3)[AD - BD] + 2AD + a[BG - AD]
= (2/3)BD + (2 - 2/3 - a)AD + aBG
=> 2 - 2/3 - a = 0
=> a = 4/3
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