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
.
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
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