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Tangent Plane And Normal Vector.

  1. Jun 6, 2004 #1

    dcl

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    I'm having trouble working out the tangent plane of an equation at a specified point (4,1,-2)
    The equation being [tex]9x^2 - 4y^2 - 25z^2 = 40[/tex]

    now
    [tex]\nabla f = (18x, -8y, -50z)[/tex] yeh?
    Just reading off this should give us the normal vector shouldn't it? (18,-8,-50)
    and from that we can work out the equation of the plane.
    18(x-4) - 8(y-1) -50(z-(-2)) = 0
    Is this corrent or am I using a horribly flawed method?
     
  2. jcsd
  3. Jun 6, 2004 #2

    dcl

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    Think I've worked it out for myself.
    Method was sorta wrong.
    Once I have Grad F, all I need to do is sub in the values of the point and It will give me the normal vector and from that I can work out the equation.
    I think thats right.
     
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