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Simple Work On Line Question.

  1. May 17, 2009 #1
    Simple question.

    So in an example question I'm lookin at finding the work between 2 points on a line.

    The points are A(1, 0, 0) and B(0, 1, (pi^3)/8)

    The vector of the line joining these points is then given as:

    r = i + t(-i + j + [(pi^3)/8]k )

    Is that correct? I can't see how that equation arises? It could just be a typo, or I'm just having a stupid day!!

  2. jcsd
  3. May 17, 2009 #2


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    Staff: Mentor

    No, that's the parametric equation of the (straight) line connecting those points. Taking different values of t in the range [0,1] gives you different points along that line.
  4. May 17, 2009 #3
    Yes, but is it correct? Shud there be a -i in there?

    Is it r = a + tb ????
  5. May 18, 2009 #4


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    Staff: Mentor

    It looks OK to me. Substitute t = 0 and you get point A. Substitute t = 1 and you get point B. In unit vector notation,

    [tex]\vec A = 1 \hat i + 0 \hat j + 0 \hat k[/tex]

    [tex]\vec B = 0 \hat i + 1 \hat j + \frac{\pi^3}{8} \hat k[/tex]
  6. May 18, 2009 #5
    I guess I'm confused as to that vector. How do you make that position vector from those 2 points?
  7. May 18, 2009 #6
    The vector from B to A is simply
    [tex]\vec{B} - \vec{A}[/tex]
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