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Homework Help: Proving A,B and C are Collinear

  1. Apr 1, 2006 #1

    danago

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    I merged the duplicate thread. This post cannot be deleted!


    Integral
     
    Last edited by a moderator: Apr 1, 2006
  2. jcsd
  3. Apr 1, 2006 #2

    Fermat

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    If A, B and C all lie on the same line, then what can you say about the direction of the lines, AB,AC ?
     
  4. Apr 1, 2006 #3

    danago

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    Hey. Heres the question:

    Ive drawn a diagram:
    http://img2.freeimagehosting.net/uploads/b74251caf2.gif [Broken]

    a=3i-j
    b=-i+15j
    c=9i-25j

    So pretty much, i think i need to prove that [tex]\overrightarrow{BA}=h\overrightarrow{BC}[/tex]

    Ive found that
    [tex]\overrightarrow{BA}=\overrightarrow{BO}+\overrightarrow{OA}[/tex]
    [tex]=-b+a[/tex]
    [tex]=4i-16j[/tex]

    [tex]\overrightarrow{BC}=\overrightarrow{BO}+\overrightarrow{OC}[/tex]
    [tex]=-b+c[/tex]
    [tex]=10i-40j[/tex]

    From that, i can see that the i and j components have a set ratio. ie. i:j = 1:4.

    For this question, what would i write as my final proof that the three points are collinear? I would use the answers page in my textbook, but it doesnt give answers to questions that are more than 1 line :devil:

    Thanks in advance,
    Dan.
     
    Last edited by a moderator: May 2, 2017
  5. Apr 1, 2006 #4

    Fermat

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    You solution is correct up to the same ratio bit.
    Thereafter, sinmply say that, because of the same ratio,
    |BC| is a multiple of |AB|
    hence BC and AB are parallel.
    Since they share a common point, B, then they are collinear.


    BTW, your sketch looks like it has OA at (3i + j) rather than (3i - j)
     
  6. Apr 1, 2006 #5

    danago

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    ok thanks very much for that.

    And yea, i made a mistake in my sketch.
     
  7. Apr 1, 2006 #6

    NateTG

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    Hmm...
    Is there an example in the text somewhere?
    Since 'use vectors to' is pretty vauge, you can do this a bunch of ways.

    For example, you could use the dot product
    [tex]\frac{(\vec{b}-\vec{c}) \cdot (\vec{a}-\vec{b})}{| (\vec{b}-\vec{c})| |(\vec{a}-\vec{b})|}=\pm 1[/tex]
    or the cross product
    [tex](\vec{b}-\vec{c}) \times (\vec{a}-\vec{b}) = \vec{0}[/tex]

    Or you could show that all the vectors are on the line
    [tex]y=-4x+11[/tex]

    Or your drawing works
     
    Last edited: Apr 1, 2006
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