问题标题:
【曲面x2+2y2+3z2=21在点(1,-2,-2)的法线方程为x−11=y+2−4=z+2−6x−11=y+2−4=z+2−6.】
问题描述:

曲面x2+2y2+3z2=21在点(1,-2,-2)的法线方程为x−11=y+2−4=z+2−6

x−11=y+2−4=z+2−6

谭新莲回答:
  令F(x,y,z)=x2+2y2+3z2-21,则得:   F′x(1,-2,-2)=2x|(1,-2,-2)=2   F′y(1,-2,-2)=4y|(1,-2,-2)=-8   F′z(1,-2,-2)=6z|(1,-2,-2)=-12   ∴法线方程为:   x−11=y+2−4=z+2−6
查看更多
数学推荐
热门数学推荐