已知空间曲面的方程怎样设在任意点的切平面方程?
曲面上一点(x,y,z)处的法向量为n=(x/2,2y,2z/9)
把点p带入得到n=(1,
-2,
2/3)
可以取n0=(3,
-6,
2)
所以切平面为3(x-2)-6(y+1)+2(z-3)=0
整理后3x-6y+2z=18
法线为(x-2)/3=(y+1)/(-6)=(z-3)/2
满意请采纳,谢谢支持。
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