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

  1. Oct 10, 2011 #1

    d86

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    For tangent plane equation
    z-z0 = [itex]f{x}[/itex](x0,y0)(x-x0) + [itex]f{y}[/itex](x0,y0)(y-y0)

    how come there is no cross product of the partial derivatives [itex]f{x}[/itex] X [itex]f{y}[/itex] to give the normal vector for the plane?
     
  2. jcsd
  3. Oct 10, 2011 #2

    LCKurtz

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    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.
     
  4. Oct 10, 2011 #3

    d86

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    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?
     
  5. Oct 10, 2011 #4

    LCKurtz

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    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.
     
  6. Oct 10, 2011 #5

    d86

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    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 [itex]f_{x}[/itex] and [itex]f_{y}[/itex] as slopes tho.
     
  7. Oct 10, 2011 #6

    LCKurtz

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    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.
     
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