Question

高数微分题求空间曲面x*2+2y*2+3z*2=6在点（1,1,1）处的切平面及法线方程

Answer 1

设 F(x,y,z) = x*2+2y*2+3z*2 - 6

Fx' = 2x,Fy'= 4y,Fz' = 6z

法向量 n = {Fx',Fy',Fz'} |(1,1,1) = {2,4,6} // {1,2,3}

切平面方程 (x-1) + 2(y-1) + 3(z-1) =0

法线方程 (x-1)/1 = (y-1)/2 + (z-1)/3