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Homework Help: Parametrization of a line formed by 3 points

  1. Jun 13, 2012 #1
    1. The problem statement, all variables and given/known data

    Find a parametrization of the equation of the line formed by the points A, B, and P.

    A(2,-1,3) B(4,3,1) P(3,1,2)

    2. Relevant equations


    3. The attempt at a solution

    Alright, so, I've already determined that P is equidistant from the points A and B, and therefore the center of a sphere. I'm having a hard time finding a parametrization since we're dealing with 3 points, as oppose to 2. If, for example, it asked for the parametrization of the equation of the line formed by A and P, I would find the vector AP, and then simply use A as my (x_0, y_0, z_0) and plug in the v-values as well. Since it's along the line formed by all three points, however, I'm a bit confused.

    Finding the vector AP suggests a line parallel to the line I'm setting up parameters for, but what about a third point? The point BP would be parallel to said line as well, but can I assume it's along the same line as AP, or is the fact that it's parallel sufficient? Even so, how do I include B in this?

    Hope this makes sense.

    Any help would be greatly appreciated! Thank you!

    EDIT// So I've just realized that PA and PB are anti-parallel. If that's the case, can I use just one or the other, and would either parametrization be correct?
    Last edited: Jun 13, 2012
  2. jcsd
  3. Jun 13, 2012 #2

    Simon Bridge

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    Circle... with A and B on the circumference. But |AB| != |PA|? Of course all three could be on the circumference of a conic section?

    Have you checked that A B and P are not co-linear?

    What is the definition of a straight line in 3 dimensions? (i.e. why would you need three points - normally you'd just use two points that are on the line right?)
  4. Jun 13, 2012 #3
    A, B, and P are co-linear, right? And I'm saying that based on the fact that AP and BP are anti-parallel vectors.

    And since they all lie along the same line, I wouldn't need three points, and so a parametrization using AP is just as valid as one using BP, even though they appear different?

    I'm not completely comfortable with the concept of parametrizations, but it seems to me, these equations could describe the same line using any point along the line, and so it'd make sense that there would be various forms, right?

    Hopefully I'm on the right track...thanks for the help! (:
  5. Jun 13, 2012 #4

    Simon Bridge

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    If A B and P are co-linear, then you just need two of them to construct a vector that points along the line. Any other point on the line will be some scale-factor along that vector from any other point on the line.

    so construct a vector v from A to B (say) it will have components (vx, vy, vz) ... if A has components (u,v,w) then the line parametrizes to the set of points Q=(x,y,z) such that:

    (x,y,z)=(u,v,w)+k(vx,vy,vz): k is an arbitrary scalar.

    I'm not saying that A B and P are co-linear here - you have to check.
    Usually three points are used to define a plane.
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