Tangent plane to a surface, no need for cross product?

[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?

 

[LCKurtz]

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?

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 R_x = < 1, 0, f_x>
R_y = <0, 1, f_y>

n = R_x X R_y= <-f_x, -f_y, 1> or you can change the signs <f_x, f_y, -1>.

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

 

[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?

 

[LCKurtz]

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?

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.

 

[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.

 

[LCKurtz]

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.

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.