# Homework Help: Tangent plane to a surface, no need for cross product?

1. Oct 10, 2011

### d86

For tangent plane equation
z-z0 = $f{x}$(x0,y0)(x-x0) + $f{y}$(x0,y0)(y-y0)

how come there is no cross product of the partial derivatives $f{x}$ X $f{y}$ to give the normal vector for the plane?

2. Oct 10, 2011

### LCKurtz

The cross product is already built in. If you parametrize the surface as x = x, y = y, z = f(x,y) you have the parameterization

R(x,y) = < x, y, f(x,y)>

Then Rx = < 1, 0, fx>
Ry = <0, 1, fy>

n = Rx X Ry= <-fx, -fy, 1> or you can change the signs <fx, fy, -1>.

Using that normal and the point (x0,y0,z0> gives you the same equation you have.

3. Oct 10, 2011

### d86

i see
so as long as the slopes of two intersecting lines are known, the plane they are on can be defined that way w/o going through the cross product?

4. Oct 10, 2011

### LCKurtz

In 3D we don't talk about "slopes", only directions. But I don't see what you are getting at with this comment. You would get the normal by crossing the two direction vectors.

5. Oct 10, 2011

### d86

i just got confused by why the equation looked different from the regular plane equation. i see that it's the same thing now. the book does refer to $f_{x}$ and $f_{y}$ as slopes tho.

6. Oct 10, 2011

### LCKurtz

Yes. The partials represent the slope of a two variable function where you hold the other variable constant. They are sometimes called "the slope in the x or y direction". Still, for a general straight line in 3-space, you wouldn't use the term slope.