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Line equation

  1. Mar 4, 2006 #1
    How do you find the equation of a line between two points in three dimensional space? I sorta forgot. =\
  2. jcsd
  3. Mar 4, 2006 #2


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    c(t) = (P-Q)t+P
  4. Mar 4, 2006 #3
    Shouldn't that be c(t) = (P-Q)t+Q

    because at t = 0 you should get Q, and at t = 1 you should get P which isn't what happens in your equation.
  5. Mar 4, 2006 #4


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    It really doesn't matter how you parametrize it so long as it is the same line. c(t) = .3455(P-Q)t+(P+Q)/2 is equally valid.
  6. Mar 5, 2006 #5


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    Orthodontist's form, c= (P- Q)t+ P, gives P when t=0 and Q when t= -1. That perfectly valid.

    By the way, Tandoorichicken, since a line is one-dimensional, in a three dimensional space you need more than one equation. Orthodontist gave a "vector" equation where P and Q are the position vectors of two points and t is a parameter. Writing that in component form gives three parametric equations. Given the single variable t, you can calculate x, y, and z.
    Last edited by a moderator: Mar 5, 2006
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