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Help with Rolle's theorem

  1. Sep 22, 2010 #1
    Given is a one-parameter family of planes, through

    [tex]x \cdot n(u) + p(u) = 0[/tex]

    with normal vector n and base point p, both depending on the parameter u.

    Two planes with parameters [tex]u_1[/tex] and [tex]u_2[/tex], with [tex]u_1 < u_2[/tex], intersect in a line (planes are assumed to be non-parallel). This line also lies in the plane

    [tex]x \cdot (n(u_1) - n(u_2)) + p(u_1) - p(u_2) = 0[/tex]

    Now, the book I am reading claims that, "by Rolle's theorem, we get:"

    [tex]x_1 n_1'(v_1) + x_2 n_2'(v_2) + x_3 n_3'(v_3) + p'(v_4) = 0[/tex] with [tex]u_1 \leq v_i \leq u_2[/tex].

    However, I don't see how the theorem applies here... for starters, I don't see anything of the form [tex]f(a) = f(b)[/tex], as required by the theorem.
  2. jcsd
  3. Sep 24, 2010 #2


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    I'm trying to understand your notation. If n and p are vectors, I supose x is a vector dotted into n, which gives a scalar?? How do you add a scalar to a vector? Is the 0 on the right side a scalar or vector? And you say you have a one parameter family of planes through

    [tex]x \cdot n(u) + p(u) = 0[/tex]

    What do you mean by that?
    Last edited: Sep 24, 2010
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