I would like to find the normal vector to the surface $$xz + z^2 - xy^2 = 5$$

at the point $(1,1,2)$, and further deduce the equation of the tangent plane at this point.

By letting the surface be some $f$, I have found $\nabla f$ to be the vector $i -2 j + 5k$, but I have no idea whether this vector is normal or the tangential one, since ordinarily when differentiating we get a tangential line. How should I further proceed?


1 Answer 1


$\nabla f$ is a vector normal to the surface.

You use $\nabla f$ as your normal to the tangent plain at $(1,1,2)$.

The equation of your tangent plain is

$$ 1(x-1)-2(y-1)+5(z-2)=0 $$ You may solve for $z$ if you wish.

  • $\begingroup$ Why is it equals to zero and not 5? $\endgroup$ Mar 5, 2018 at 1:38
  • $\begingroup$ Because you want to pass through the point (1,1,2) $\endgroup$ Mar 5, 2018 at 1:43

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