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Homework Help: Equation of path

  1. Oct 12, 2006 #1
    The equation of a path is of the form: [itex]\vec r = \vec r_0 + \vec A t[/itex]
    If 't' represents time, show that the time of closest approach is:
    [tex]t = -\frac{\vec r_0 \cdot \vec A}{|\vec A|^2}[/tex]

    I am not really sure on how to proceed about this, but I made a crude approach by assuming [itex]\vec r[/itex] and [itex]\vec r_0[/itex] to be perpendicular. I took the dot product with r0 on both sides of given equation.
    [tex]-\vec r_0^2 = \vec A \cdot \vec r_0 t[/tex]

    I don't think this a right way to solve, please give some suggestions.
     
    Last edited: Oct 12, 2006
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  3. Oct 12, 2006 #2

    HallsofIvy

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    Closest approach to what?
     
  4. Oct 15, 2006 #3
    Oh sorry, I forget to add that. Find the distance of closest approach to the origin i. e. the distance from the origin to the line.
     
  5. Oct 15, 2006 #4

    HallsofIvy

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    It's not [itex]\vec r_0[/itex] that's perpendicular to [itex]\vec r[/itex], it's [itex]\vec A[/itex].
    At the point of closest approach, the vector in the direction of the line, that is, [itex]\vec A[/itex], is perpendicular to the position vector, [itex]\vec r_0+ \vec At[/itex] itself. That is [itex]\vec A \cdot (\vec r_0+ \vec At= 0[/itex]. That is, [itex]\vec A \cdot \vec r_0+ \vec A \cdot \vec A t= 0[/itex]. Can you solve that for t?
     
  6. Oct 16, 2006 #5
    Wow, thanks! That makes sense. :smile:
    [tex]-\vec A \cdot \vec r_0 = |\vec A|^2t[/tex]

    [tex]t = -\frac{\vec r_0 \cdot \vec A}{|\vec A|^2}[/tex]
     
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