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Calculus 3, dealing with tangent planes and surfaces.

  1. May 3, 2009 #1

    okay i came up with doing the gradient of the ellipsoid. Then set that equal to the vector, <4,-4,6>. I solved and got x,y,z = 1,-2,1

    I looked at the answer key and it said (1/3) (1,-2,1)

    Does anyone know where the 1/3 came from?
    Last edited: May 3, 2009
  2. jcsd
  3. May 4, 2009 #2


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    Hi yopy! :smile:
    You'll only get a proportion out of that.

    2*12 +(-2)2 +3*1 = 9,

    so you have to divide everything by 3. :wink:
  4. May 4, 2009 #3


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    One obvious point should be that (1, -2, 1) is not even on the ellipse! [itex]2(1)^2+ (-2)^2+ 3(1)^2= 9[/itex] not 1!

    In order that two planes be parallel, their normal vectors must be parallel, but not necessarily equal. As tiny-tim said, you should have [itex]<4x, 2y, 6z>= \lambda<4, -4, 6>[/itex] for some number [itex]\lambda[/itex]. That together with [itex]2x^2+ y^2+ 3z^2= 1[/itex] gives x, y, and z.
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