曲面在点P（1,2,0)处的切平面方程为（).

A . 2x+4y+z=11 B . -2x-4y+z=-1 C . 2x-4y-z=-15 D . 2x-4y+z=-5

A .https://assets.asklib.com/psource/2015102817185777015.jpg B .https://assets.asklib.com/psource/201510281719144880.jpg C .https://assets.asklib.com/psource/2015102817192862954.jpg D .https://assets.asklib.com/psource/2015102817195036093.jpg

A . （x-1）／1=（y-1）／1=（z-1）／-1 B . （x-1）／1=（y-1）／1=（z-1）／-2 C . （x-1）／1=（y-1）／-1=（z-1）／-2 D . x+y-z=1

A . 2χ+4y+z=11 B . -2χ-4y+z=-1 C . 2χ-4y-z=-15 D . 2χ-4y+z=-5

A . （x-1）+2（y-2）-（z-3）=0 B . （x+1）+2（y+2）+3（z+3）=0 C . （x-1）+2（y-2）+3（z-3）=0 D . （x+1）+2（y+2）-（z+3）=0

A .https://assets.asklib.com/psource/2015110315421942037.png B .https://assets.asklib.com/psource/2015110315422529820.png C .https://assets.asklib.com/psource/2015110315422933663.png D .https://assets.asklib.com/psource/201511031542321565.png

A .https://assets.asklib.com/psource/2015102915073495353.jpg B .https://assets.asklib.com/psource/2015102915074686160.jpg C .https://assets.asklib.com/psource/2015102915075983857.jpg D .https://assets.asklib.com/psource/2015102915081374089.jpg

A . 2x-y+2=0 B . 2x+y+1=0 C . 2x+y-3=0 D . 2x-y+3=0

A．x+2y+3z+6=0 B．x+2y+3z-6=0 C．x-2y+3z+6=0 D．x-2y+3z-6=0

<img src='https://img2.soutiyun.com/ask/2021-01-10/979120559151028.png' />

曲面z=2x²+4y²在点(1，1，6)处的切平面方程为______，法线方程为______．

曲面x2+2y2+3t2=6在点(-1，1，-1)处的切平面方程为()。 A．x+2y+3z-6=0 B．x+2y+3z+6=0 C．2x-y-3=0 D．x-2y+3z-6=0

求曲线x²-z=0，3x+2y+1=0在点(1，-2，1)处的法平面与直线<img src='https://img2.soutiyun.com/latex/latex.action' />间的夹角．

设z=f(x，y)由方程x-yz+cosxyz=2确定，求曲面z=f(x，y)在P₀(1，1，0)处的切平面方程与法线方程

证明:若两曲面F₁(x,y,z)=0,F₂(x,y,z)=0在点P(x₀,y₀,z₀)正交(两曲面在点P的法线垂直),则在点P(x₀,y₀,z₀)有 <img src='https://img2.soutiyun.com/ask/2020-11-13/974138679474776.png' /> 并验证两曲面3x²+2y²=2x+1,x²+y²+z²-4y-2z+2=0在点(1,1,2)正交.

设曲面<img src='https://img2.soutiyun.com/ask/2020-12-07/976208897532342.jpg' />，平面π：2x+2y+z+5=0。 (1)求曲面S上与π平行的切平面； (2)求曲面S与平面π之间的最短距离。