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Geometrical proof using vectors

  1. Sep 11, 2008 #1
    Hello again, another vector question.

    The diagram (not included) shows a parallelogram OABC with OA=a and OC=c; AC and OB intersect at point D such that OD=hOB and AD=kAC.
    How do I show what h and k equal?
    I managed to achieve:
    a+c=2DB+AD-DC=2OD+DC-AD=OB
    and
    c-a=2DC+OD-DB=2AD+DB-OD=AC
    Furthermore, I know that AD+DB=OD+DC and OD-AD=DB-DC
    I can 'see' that h and k = 1/2; but how do I show it?
    As always, the help is appreciated. I love you people :).
     
  2. jcsd
  3. Sep 11, 2008 #2

    tiny-tim

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    Hi nobahar! :smile:

    Hint: what is the vector equation for a general point on AC? :wink:
     
  4. Sep 12, 2008 #3
    The equation of the line AC could be: r=zAD+c; r=zAD+a; r=zAD+xOB? (xOB being OD.)
    I'm guessing any of these are suitable since it satisfies the need for a parallel vector and a position vector?
    Then, any point that meets the line AC should =r, with different values for z. Where do I go from here? (If anywhere!)
     
  5. Sep 12, 2008 #4

    tiny-tim

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    Hi nobahar! :smile:
    hmm … sort-of the right idea …

    but AD is unknown …

    look for something similar that only involves a and c (which are given) and numbers. :smile:
     
  6. Sep 13, 2008 #5
    Hello again!
    Right, my maths tutor offered me this solution by e-mail:
    "OD = hOB = h(a+c) = ha +hc
    OD = OA + AD = a + kAC = a + k(c - a)

    Hence ha + hc= a + kc -ka
    Equating coefficients of a: h = 1 - k
    Equating coefficients of c: h = k
    Hence k=1-k
    2k = 1
    k=0.5 = h"

    From this (and other comments from her e-mail) I established the 'relationships' I first offered served no purpose; and (thanks to Tiny Tims input) neither did the vector equations of a line I proposed.
    However, I still have some further questions; since I like to understand as much as possible and as clearly as possible before moving on.
    I'm guessing the above isn't a vector equation of a line at any point, since it doesn't contain a parallel vector?
    On that basis, presumably there is going to be an alternative method?
    As always, thanks in advance.
     
  7. Sep 13, 2008 #6

    tiny-tim

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    Hello nobahar! :smile:

    The vector equation for the line AC (ie for a general point P on AC) is OP = OA + kAC for any number k,

    (of course, it's the same as OP = OC + k'AC, with k' = k + 1)

    because you go to A first, and then as far as you like along the direction of AC.

    (This is the equation your teacher used, with D = P. :wink:)

    But I don't understand what you mean by "it doesn't contain a parallel vector". :confused:
     
  8. Sep 13, 2008 #7
    For the parallel vector I was just going by the textbook, which proposes that: "...a line is defined if we know:
    i) a vector that is parallel to the line,
    and ii) the position vector of a point on the line.
    (Sadler, A.J., Thorning, D.W.S (2007, pg.65). Understanding Pure Mathematics, Glasgow: Oxford Universty Press).
    I now realise that a parallel vector is unnecessary if the direction of the vector in question is already identified (?).
    One final question :tongue:
    why k'=k+1 in OP=OC+k'AC?
    OP=OC-AC+kAC or OP=OC-(AC-kAC)
    OP=c-(c-a)+k(c-a)
    OP=c-c+a+kc-ka
    OP=a+kc-ka= OA+kAC
     
  9. Sep 13, 2008 #8

    tiny-tim

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    don't forget … c - a isn't the vector AC, it's a vector from O parallel to AC. :wink:

    so c - a was "a vector that is parallel to the line" :smile:
    hmm … perhaps that should have been k' = k - 1 :redface::

    OP = c - (c-a) + k(c-a)
    OP = c + (k - 1)(c-a) :smile:
     
  10. Sep 14, 2008 #9
    That is excellent!
    Thankyou for your help; as always Mr. Tiny Tim, it's much appreciated. :smile:
     
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