x^3+y^3+z^3+xyz-6=0在(1,2,-1)处的切平面方程

Question

x^3+y^3+z^3+xyz-6=0在(1,2,-1)处的切平面方程

Answer 1

设:F(x,y,z) = x^3+y^3+z^3+xyz-6 .
则:Fx(x,y,z,)=3x^2+yz, Fy(x,y,z,)=3y^2+xz, Fz(x,y,z,)=3z^2+xy,
Fx(1,2.-1) =3+(-2)=1, Fy(1,2,-1) =12-1=11, Fz(1,2,-1)= 3+2=5.
故: 切平面方程为:(x-1) +11(y-2) +5(z+1)=0