Equation of a Tangent plane an the normal line to a given point

rubecuber

1. The problem statement, all variables and given/known data

xy +yz + zx = 3 (1,1,1)

2. Relevant equations
equation of tangent plane is z-z0 = fx(x0,y0)(x-x0) +fy(x0,y0)(y-y0)


3. The attempt at a solution

Right, I've been a few of these exercises, however, this is the first one I've seen that equals a number and not "z". So, when I take the partial derivatives with respect to x and y I get, respectively, y+z, and x+z. Now, I'm sort of guessing here, but I'm taking fz and I'll get y+x. Then I plug the values into the equation and get fx = 2, fy = 2 and fz = 2? The thing is that the book doesn't say anything about an fz. What now?

lanedance

the direction you have found (2,2,2) is the gradient of the function f(x,y,z) = xy +yz + zx at (1,1,1), and will be normal to the tangent plane (why?)

use it to find the equation of your tangent plane (the dot product of any vetcor in the plane and the normal must be zero...)

HallsofIvy

In general, if you have a surface writtten as f(x,y,z)= constant, then the gradient vector,
[tex]\nabla f= f_x\vec{i}+ f_y\vec{j}+ f_z\vec{k}[/itex]
is normal to the surface. It is the "normal vector" and gives you the tangent plane.

I consider this situation easier than "z= f(x,y)". In fact, to find a normal vector and tangent plane to z= f(x,y), I would immediately write it as G(x,y,z)= z- f(x,y)= 0.