Find equation for tangent plane

**joepinsley** · March 16th 2010, 04:44 PM

Find the equation for the tangent plane to the surface x^2+3xyz+y^2=-5z^2 at the point (1,1,1).

I found the answer to be x+y+3z=5 but my solution manual says -5x-5y+7z=-3.

**redsoxfan325** · March 16th 2010, 10:50 PM

Originally Posted by joepinsley

Find the equation for the tangent plane to the surface x^2+3xyz+y^2=-5z^2 at the point (1,1,1).

I found the answer to be x+y+3z=5 but my solution manual says -5x-5y+7z=-3.

It should be $x^2+3xyz+y^2=5z^2$ (no negative sign before $5z^2$ ). Then

$f(x,y,z)=x^2+3xyz+y^2-5z^2=0$

$\nabla f = \langle 2x+3yz,3xz+2y,3xy-10z\rangle$

Plug in $(1,1,1)$ to get $\langle 5,5,-7\rangle$

So it's $5(x-1)+5(y-1)-7(z-1)=0\implies 5x+5y-7z=3$ (or $-5x-5y+7z=-3$ if you prefer).

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