# Formula for the tangent plane to a functional surface

I'm having a hard time grasping what they're saying in my textbook for this chapter.

If $$z=f(x,y)$$ defines a surface, then the tangent plane to the surface at (a,b) may be obtained as $$(-f_x(a,b),-f_y(a,b),1) * (x-a,y-b,z-f(a,b))=0$$ This is the result of applying the gradient-of-a-surface formula to the surface $$z-f(x,y)=0$$ at the point $$(a,b,f(a,b))$$.

How is the formula comparable to the formula for finding tangent planes to a surface which is $$(f_x(a,b,c),f_y(a,b,c),f_z(a,b,c))*(x-a,y-b,z-c)=0$$ and why are the $$f_x$$ and $$f_y$$ components negative in the first formula?

• Have you at least tried to compute the gradient of $z-f(x,y)$? – amd Jan 21 '19 at 23:18
• Essentially a duplicate of math.stackexchange.com/q/3073917/265466. – amd Jan 21 '19 at 23:20

The two formulations are the same. We are taking $$F(x,y,z)=z-f(x,y)$$ and setting $$F=0$$. Taking partial derivatives of $$F$$ gives $$F_x(a,b,c)=-f_x(a,b)$$ $$F_y(a,b,c)=-f_y(a,b)$$ $$F_z(a,b,c)=1$$
The partial derivatives with respect to $$x$$ and $$y$$ have negative coefficients because we are taking the gradient of $$z-f(x,y)$$. They would have positive coefficients if we defined $$F(x,y,z)=f(x,y)-z$$ but then $$F_z(a,b,c)$$ would be $$-1$$ instead of $$1$$.