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Gradient and line

  1. Feb 25, 2007 #1
    Sorry to ask yet another question, but it's very hard to contact my teacher.

    1. The problem statement, all variables and given/known data

    [tex]f(x,y,z) = x^2-yz+z^2[/tex]

    a = (0,1,1), b = (1,3,2). Find a point c on the line joining a and b such that;

    [tex]f(b)-f(a)=\nabla f(c)\bullet(b-a)[/tex]

    2. The attempt at a solution

    f(b) = 1-3*2+2^2=-1
    f(a) = 0-1*1+1 = 0

    [tex]\nabla f(c) = (2x)i - (z)j + (2z-y)k[/tex]

    (b - a) = i + 2j + k

    [tex]-1 = [(2x)i - (z)j + (2z-y)k]\bullet[i + 2j + k][/tex]

    [tex]-1 = 2x -2z + 2z-y[/tex]

    [tex]0 = 2x -y +1[/tex]

    Let L be the line between a and b

    [tex]L = ti+(1+2t)+(1+t)k[/tex]

    By substitution to find the point on the line that is also on the plane;

    [tex]0 = 2t -1-2t +1[/tex]

    [tex]0 = [/tex]

    So, it seems to be impossible
     
  2. jcsd
  3. Feb 25, 2007 #2
    Your result is 0=0 so EVERY point on the line would work (unless there is a mistake somewhere).
     
  4. Feb 25, 2007 #3
    Exactly. But the question seem to imply there's only 1 point (find THE point).
     
  5. Feb 25, 2007 #4
    A quick check: Mathematica or Maple graphs.
     
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