→ crazystan :謝謝~~ 06/02 17:36
※ 引述《crazystan (米漢堡)》之銘言:
: 設空間中兩直線L1:x/2 = (y+1)/3 = z+3
: L2:(x+1)/4 = (y-4)/-2 = z+2/-1
: 若直線L過P(1,2,-1)且與L1,L2交於A,B兩點
: 求線段AB長?
: 想不出來 麻煩版上大大了
: 謝謝
L1, L2為歪斜線
設A( 2s, 3s-1, s-3)
B(4t-1,-2t+4,-t-2)
向量PA=(2s-1, 3s-3, s-2)
向量PB=(4t-2,-2t+2,-t-1)
2s-1 3s-3 s-2
∵APB三點共線 ∴─── = ─── = ───
4t-2 -2t+2 -t-1
(1) -4st+4s+2t-2 = 12st-6s-12t+6 16st-10s-14t+8 = 0
(2) -3st- s+3t+3 = -2st+2s+ 4t-4 st+ 3s+ t-7 = 0
(3) -2st-2s+ t+1 = 4st-2s- 8t+4 6st - 9t+3 = 0
16(7-3s-t)-10s-14t+8 = 0 58s+30t=120
6(7-3s-t) - 9t+3 = 0 18s+15t= 45
解出s,t之後再用距離公式求解
我計算能力比較差,中間有算錯的話還請包涵,算法應該沒錯
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