# Find the tangent plane to an arbitrary implicit surface?

Being given an implicit surface of the form f(x,y,z)=0

And assuming we know the gradient G=(xi,yi,zi) at a known point P (x0,y0,z0) on the surface, how do you compute the tangent plane to that surface at P ?

You don't know f, only it's gradient.

To my understanding the gradient is perpendicular to the tangent plane at point P, so it suffices to say that (P-X).G=0 where X is an arbitrary point which should lead to an implicit linear equation of the form:

ax+by+cz+d = 0

and then if one wished to express this as a function of 2 variables the final solution would be for example:

(-a/c)x+(-b/c)y-d/c = z

However it seems this is wrong, and I don't understand why

• What happens if $c=0$? – amd Oct 22 '17 at 4:00
• Then you can't have a function of the form z=f(x,y), so that's not really an issue :p – Makogan Oct 22 '17 at 4:58
• Well, you asked what’s wrong with your reasoning, and I’ve pointed out one of the flaws. Trying to solve for $z$ fails for tangent planes that are parallel to the $z$-axis, which are perfectly reasonable things to have for the level surface of a function $f:\mathbb R^3\to\mathbb R$. – amd Oct 22 '17 at 5:11