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Vector equation for a line segment in 3 dimensions

  1. Dec 21, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the vector equation for a line segment from (2,-1,4) to (4,6,1).


    2. Relevant equations
    Arithmetic :p


    3. The attempt at a solution
    What I did was that I simply solved for the distance between the two points and used it as my vector, I then said

    <4-2t, 6-7t, 1+3t> when 0≤t≤1

    The book didn't do anything like this at all. They threw in some weird formula that seems like it complicated everything for no reason at all. Is my way of doing it wrong?
     
  2. jcsd
  3. Dec 21, 2013 #2
    What is the formula the book threw in ?
    Your solution appears to be in the vector form of a line.

    Namely:
    $$<x_0, y_0, z_0>+t < a, b, c>$$
     
    Last edited: Dec 21, 2013
  4. Dec 21, 2013 #3
    Yeah, well isn't that exactly what they were looking for?
    They thew in some (1-t)<vector> + <vector2>. Whatever. I'm not interested in their way as I found a much more intuitive way of doing it on sladder. I was just curious if my answer is OK. It gives the exact same result as their does but it's in a different format.
     
  5. Dec 21, 2013 #4

    Ray Vickson

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    The segment from ##\vec{a}## to ##\vec{b}## has the form
    [tex] \vec{v} = \vec{v}(t) \equiv (1-t) \vec{a} + t \vec{b}, \; 0 \leq t \leq 1.[/tex]
    When ##t = 0## we have ##\vec{v} =\vec{a}## and when ##t = 1## we have ##\vec{v} = \vec{b}##.

    Note that just saying ##\vec{v} = \vec{a} + t \vec{b}## won't work (why not?).
     
  6. Dec 21, 2013 #5

    SammyS

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    For one thing your solution goes from (4,6,1) to (2,-1,4) as t goes from 0 to 1 .

    ... if that makes any difference.
     
  7. Dec 21, 2013 #6
    Okay, does it matter? It's still the same line, just drawn in a different direction.
     
  8. Dec 22, 2013 #7

    HallsofIvy

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    Tha'ts up to you. You said "Find the vector equation for a line segment from (2,-1,4) to (4,6,1)." Was the "from", "to" important?
     
  9. Dec 22, 2013 #8
    I guess it matters then. Didn't really think about that. Thanks.
     
  10. Dec 22, 2013 #9

    SammyS

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    Even if direction is important, your initial solution is good if t goes from 1 to 0 .
     
  11. Dec 23, 2013 #10
    Easy fix! I like it, thanks! :)
     
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