Re: [多變] 求參數方程式~
※ 引述《chat543 (小ㄐ)》之銘言:
: 2 2 2 2 2
: x +2y +2z = 20與 x + y + z =4 之交點(0,1,3)之公切線參數方程式為?
2 2 2
(i)令 F(x,y,z) = x + 2y + 2z - 20
在其曲面上任一點(x,y,z)的切平面法向量為
▽F(x,y,z) = 2xi + 4yj + 4zk
則在(0,1,3)的切平面法向量為
▽F(0,1,3) = 4j + 12k = 4(j+3k)
取 u = j + 3k
2 2
(ii)令G(x,y,z) = x + y + z - 4
在其曲面上任一點(x,y,z)的切平面法向量為
▽G(x,y,z) = 2xi + 2yj + k
則在(0,1,3)的切平面法向量為
▽G(0,1,3) = 2j + k
取 v = 2j + k
則所求在交點(0,1,3)的公切線,其方向向量為
u x v = (j+3k)x(2j+k) = -5i
取 p = i + 0j + 0k
則在交點(0,1,3)的公切線為 (x-0)i+(y-1)j+(z-3)k = tp = ti + 0j + 0k
根據向量獨立性 得在交點(0,1,3)的公切線參數式為
x = t , y = 1 , z = 3 for all t in R
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※ 編輯: bighead0720 來自: 219.91.94.199 (06/15 12:54)
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