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How to find which of three points are on a line?

  1. Sep 13, 2013 #1
    1. The problem statement, all variables and given/known data
    Which of these three points falls on the line?

    l: r(t)=(i+2j)+t(6i+j-5k)

    P(1,2,0); Q(-5,1,5); R(-4,2,5)

    I have the answer, but I don't understand why P and Q fall on the line but R does not. Is it because the i and j magnitudes are different for R?


    2. Relevant equations


    3. The attempt at a solution

    P:
    (1+6t) = 1
    t=0

    2=2+t
    t=0

    t=0

    Q:
    -5=1+6t
    t=-1

    1=2+t
    t=-1

    t=0

    R:
    -4=1+6t
    t=-5/6

    2=t+2
    t=0

    t=0
     
  2. jcsd
  3. Sep 13, 2013 #2
    First, you are right that P and Q fall on the line and R does not. And as you see, R is not on the line because using those points you get 3 incompatible equations in t.

    Whether a point falls on a line depends on all 3 magnitudes, i, j and k, and the trivial answer to your question is that most combinations will not work i.e. most 3 dimensional points are not on any given line.

    However, working backwards from the equation we can get some insight. For example r(0) = i + 2j. So the point that fits there is P = (1,2,0). When you look at r(-1) = -5i + j - 5k, you see that Q = (-5,1,-5) is a point on the line.

    But your incompatible equation for R tells you that there is no possible t for which r(t) = R. It's not really deeper than that.
     
  4. Sep 13, 2013 #3
    Thanks! I think I understand, but what about the t=0 in R? Does this value not count since there is already an incompatible value in its components?
     
  5. Sep 14, 2013 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    The line has parametric equations x = 1 + 6t, y = 2 + t, z = -5t. A point (a,b,c) lies on the line if the equations a = 1+6t, b = 2+t, c = -5t are compatible; that is, we must get the same t from all three equations.
     
  6. Sep 14, 2013 #5
    Thanks a bunch! That's what I needed.
     
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