Re: [閒聊]
一樣騙P幣
※ 引述《maxcsh (ㄚ翰)》之銘言：
※ 引述《tobe6104 (拖拖拖拖比)》之銘言：
: 空間中給A(0,0,0),B(1,1,√2),C(2,0,0)三點座標
: 如ABCD為一正四面體，求D點座標。
難得沒事來騙騙P幣好了，
設D(x,y,z)，由四邊等距：
x^2 + y^2 + z^2 = (x-1)^2 + (y-1)^2 + (z-√2)^2 = (x-2)^2 + y^2 + z^2 = 4
=> 2x-1 + 2y-1 + 2√2z-2 = 0
4x-4 = 0
x^2 + y^2 + z^2 = 4
=> x = 1
=> 2y + 2√2z = 2
y^2 + z^2 = 3
=> z = －(y - 1)/√2 代入得 3y^2 - 2y - 5 = 0
=> y = -1 or 5/3 => z = √2 or -√2/3
=> D( 1 , -1 , √2 ) or D( 1 , 5/3 ,-√2/3 )
以上沒拿筆純靠爛爛的心算，算錯概不負責=.=
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