題目 6. (10 Points) Find the equation of the tangent plane and the normal line to the surface S:x2y+exyz−2cos(xz)=0S:x^2y+e^{xyz}-2\cos(xz)=0S:x2y+exyz−2cos(xz)=0 at (1,1,0)(1,1,0)(1,1,0). 解答 解法一 令左側為 F(x,y,z)F(x,y,z)F(x,y,z)。在 (1,1,0)(1,1,0)(1,1,0)， ∇F=(2,1,1).\nabla F=(2,1,1).∇F=(2,1,1). 故切平面和法線分別為 2(x−1)+(y−1)+z=0,(x,y,z)=(1,1,0)+t(2,1,1).\boxed{2(x-1)+(y-1)+z=0}, \qquad \boxed{(x,y,z)=(1,1,0)+t(2,1,1)}.2(x−1)+(y−1)+z=0,(x,y,z)=(1,1,0)+t(2,1,1).
