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Vector problem

  1. Jan 12, 2004 #1
    Hmm...having a problem with a proof here:

    [​IMG]

    the information given is in the picture

    I have to prove that BE:EF = 1:2, and also that CD:DF = 1:1

    sorry if it is drawn badly, I did it quickly..but you get the idea.

    This needs to be a vector proof. So i'm assuming it has to have something to do with linear independence, thought I don't exactly know how to do it. I'd appreciate if anyone could help me out.
     
  2. jcsd
  3. Jan 13, 2004 #2
    It wont help if i solve this problem for u

    So i give u some hints which will start ur quest

    Consider E as your origin With A and C as Position vectors(PV) a and b {bold represents vectors} write PV for B using section formula PV for D will be -a/2

    Assume the ratio DF/CD=m and similarly for FE/EB=n

    Now again using section formula find PV in terms of m And then in terms of n

    Equate the two PV thus obtained And i hope u ur Q. will be concluded
     
  4. Jan 13, 2004 #3
    I still can't get it :(

    I really want to know how to do this...Can someone please just give me the full answer. I really think i'm missing something big here.
     
    Last edited: Jan 13, 2004
  5. Jan 13, 2004 #4
    *Bump*

    This is getting kind of urgent...I need to know this for a test tomorrow and I have no idea how to do it! =(
     
  6. Jan 13, 2004 #5
    Sorry about all this bumping...but I am in dire need. Please, I'm begging someone to finish this >_<
     
  7. Jan 14, 2004 #6
    Ok have u tried The hints i have given
     
  8. Jan 14, 2004 #7
    Consider E as origin And PV of A(a) and C(c) where a and c are PV



    now from section formula
    PV of B(b)

    [tex] \mbox{b} = \frac{\mbox{a}+\mbox{c}}{2} [/tex] and note the direction as well

    Also PV for D(d) = - a/2 note down the direction

    now consider the ratio CD:DF=1:m

    Again from section formula PV for D is given by
    [tex]=\frac{\mbox{f}+m\mbox{b}}{m+1} = - \frac{\mbox{a}}{2}[/tex]

    After rearrangement u get
    f= -(m+1)a/2 - mb ...............1

    Similarly consider BE:EF=1:n

    for which PV of F
    [tex](\mbox{f}) = n\mbox{b}=n \frac{\mbox{a}+\mbox{c}}{2} [/tex] .........2

    Equatin 1 and 2
    u get

    [tex] \frac{n\mbox{a]}{2}+\frac{n\mbox{b}}{2}=-\frac{(m+1)\mbox{a}}{2} - m\mbox{b}[/tex]
    Now from the properties of vectors or say uniqueness

    We have
    n=-(m+1) ............3
    and n=-2m ...........4

    from 3&4
    m=1 i.e CD:DF=1:1
    n=2 i.e BE:EF=1:2
     
    Last edited: Jan 14, 2004
  9. Jan 14, 2004 #8
    Here is the attachment for fig
     

    Attached Files:

    Last edited: Jan 14, 2004
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