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Help! Vectors

  1. Mar 26, 2009 #1
    Help! Vectors!!!

    1. The problem statement, all variables and given/known data

    In the triangle ABC, D divides AB in the ratio 3:2, E divides DC in the ratio 1:5 and F divides AC in the ratio 1:2. Show (using position vectors and the section formula) that B, E and F are collinear and find BE:EF


    2. Relevant equations

    Section formula = [tex]\frac{m \textbf{a} = n\textbf{b}}{m+n}[/tex]

    3. The attempt at a solution
    Other than drawing this out, and spotting that they look like they're kind of in a line.
    This isn't due for homework or anything, I'm just revising, so it would be really useful if someone cold explain this to me.

    Any ideas?
     
  2. jcsd
  3. Mar 26, 2009 #2

    tiny-tim

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    Hi andrew.c! :smile:

    Use the section formula to find the vectors D E and F

    for example, F = … ? :smile:
     
  4. Mar 26, 2009 #3
    Re: Help! Vectors!!!

    F = [tex]\frac{a+2c}{3}[/tex] ?
     
  5. Mar 26, 2009 #4

    tiny-tim

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    Almost … but it's closer to A,

    so f = (2a + c)/3 :wink:

    Next, what are D and E? :smile:
     
  6. Mar 26, 2009 #5
    Re: Help! Vectors!!!

    AD/DB = 3/2

    sec. formula...

    [tex]\frac{3a+2b}{5}[/tex]

    ---

    DE/EC = 1/5

    sec. formula...

    [tex]\frac{d+5c}{6}[/tex]
    ---
    I still dont really understand which value in the ratio is m and which is n, I am just reading it as AD/DB = m/n. Is this right?
     
  7. Mar 26, 2009 #6

    tiny-tim

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    Forget the formula

    I never remember which way round it is …

    I just ask myself each time "which one is it nearer?"

    so if DE:EC = 1:5, then it's nearer D, so … ? :smile:
     
  8. Mar 26, 2009 #7
    Re: Help! Vectors!!!

    ok, i get that logic :)

    so it would be...

    [tex]D = \frac{2a+3b}{5}[/tex]

    and

    [tex]E = \frac{5d+6}{c}[/tex]
     
  9. Mar 26, 2009 #8

    tiny-tim

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    Yup! :biggrin:

    (except for the obvious mistake … quick! edit it before anyone else notices! :wink:)

    ok, now you have the vectors for B E and F …

    how can you prove they're collinear? :smile:
     
  10. Mar 30, 2009 #9
    Re: Help! Vectors!!!

    OK, so I have...

    [tex]D = \frac{2a+3b}{5}[/tex] [tex]E=\frac{5d+c}{6}[/tex] and [tex]F = \frac{2a+C}{3}[/tex]

    For collinearity, I need to prove that BE is parallel to EF, with a common point at E?

    I tried this, but not sure if its right - i got the bottom lines to be the same, but my notes indicate that they should be a scalar multiple of each other!

    ------

    Sub. D into E to get...

    [tex] E = \frac{2a+3b+c}{6}[/tex]

    [tex]BE = e-b
    =\frac{2a+3b+c}{6} - b
    =\frac{2a+3b+c}{6}-\frac{6b}{6}
    =\frac{2a-3b+c}{6}[/tex]

    [tex]EF = f-e
    =\frac{2a+c}{3}-\frac{2a+3b+c}{6}
    =\frac{4a-2c}{6} -\frac{2a+3b+c}{6}
    =\frac{2a-3b+c}{6}[/tex]

    Since they are equal, and E was a common point, they are collinear?

    ----------------------

    Is BE:EF just 1:1 ?

    Thanks for your help btw!
     
  11. Mar 30, 2009 #10

    tiny-tim

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    Yes, that's very good

    (except you typed a minus for a plus in the last line :rolleyes:)

    the vectors BE and EF are exactly the same, in other words BE = EF, so yes, obviously it's 1:1 :smile:
     
  12. Mar 30, 2009 #11
    Re: Help! Vectors!!!

    Ta muchly for your help!

    Now i need to tackle vectors intercepting planes. Oh joy of joys!
     
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