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Cartesian vector question

  1. Apr 20, 2010 #1
    1. The problem statement, all variables and given/known data

    Use Cartesian vectors in two-space to prove that the line segments joining midpoints of the consecutive sides of a quadrilateral form a parallelogram.


    2. Relevant equations

    We could say vectors a and b

    a = [a1, a2]
    b = [b1, b2]

    3. The attempt at a solution

    Am really confused by this question..I would really appreciate it if anyone could offer tips to help me get started on this problem.

    Thanks!
     
  2. jcsd
  3. Apr 20, 2010 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi spoc21! :smile:
    So far so good! :smile:

    Next would be

    c = [c1, c2]
    d = [d1, d2]

    Now what are the midpoints of ab, bc, cd, and da ? :wink:
     
  4. Apr 20, 2010 #3
    Hi tiny-tim,

    so if we use the midpoint formula, we would get the following:

    midpoint of ab = (a1+b1/2, a2 + b2/2)

    midpoint of bc = (b1 + c1/2, b2c2/2)

    midpoint of cd = ( c1+ d1/2, c2 + d2/2)

    midpoint of da = (d1 + a1/2, d2 + a2/2)

    Now I'm stuck, how would I show that joining these points would produce a parallelogram.

    Thank you very much for your help!
     
  5. Apr 20, 2010 #4

    Mark44

    Staff: Mentor

    You should identify the midpoints - give them names, such as by saying the E is the midpoint between A and B, F is the midpoint between B and C, G is the midpoint between C and D, H is the midpoint between C and D, and I is the midpoint between D and A.

    Now if the segments EF and GH are parallel, and the segments FG and HI are parallel, then EFGH is a parallelogram.

    One nit: In your midpoint formulas, add parentheses around the sums. E.g., ((a1 + b1)/2, (a2 + b2)/2)
     
  6. Apr 20, 2010 #5
    ok, so I get the following results

    EF = [(c1 - a1)/2, (c2 - a2)/2]
    GH = [(a1 - c1)/2, a2 - c2)/2]

    I'm a little lost here..(don't know how to proceed)

    Also, you said "G is the midpoint between C and D, H is the midpoint between C and D, and I is the midpoint between D and A."

    Are there two midpoints G, as well as H between C, and D?

    Thanks for your help
     
  7. Apr 20, 2010 #6

    Mark44

    Staff: Mentor

    I lost track of my midpoints. There should only be four of them E, F, G, and H - no I needed. I meant "H is the midpoint between D and A."

    You can check that your vectors EF and GH are parallel by showing that their slopes are the same. The slope of a vector v = <a, b> is b/a.
     
  8. Apr 21, 2010 #7
    ok thanks Mark..also, I was wondering if there is a way to solve this using the properties of cross products?
     
  9. Apr 21, 2010 #8

    Mark44

    Staff: Mentor

    |u X v| = |u||v|sin(theta), where theta is the angle between u and v. If u and v are parallel, the angle between them will be 0 or pi, so the magnitude of the cross product will be zero.
     
  10. Apr 25, 2010 #9
    So I have attached my working in pdf format, and would appreciate if you could just take a quick look at it, as I am unsure if this is correct...Thanks!!
     

    Attached Files:

  11. Apr 26, 2010 #10

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi spoc21! :smile:

    ok, but a little long-winded.

    (and why are you using square brackets instead of the usual round ones? has your professor told you to? :confused:)

    It would have saved time and space to choose an easier notation

    let A = (xA,yA) etc

    then the midpoints are MAB = 1/2 (xB + xA, yB + yA) etc

    and the vectors joining consecutive midpoints are VAC = 1/2 (xC - xA, yC - yA) etc

    so VAC = -VCA and VBD = -VDB

    in other words, opposite sides of the quadrilateral are parallel, and so it must be a parallelogram ("equal in magnitude" is a bonus, but not necessary for the proof) :wink:

    (note incidentally that there would have been no need to use Cartesian coordinates if the question hadn't required them … you can just use vector symbols, eg mAB = 1/2 (b + a) etc :smile:)

    EDIT: just noticed another way, that throws slightly more light on the position of the parallelogram …

    use the fact that if the midpoints of the two diagonals of a quadrilateral coincide, then it's a parallelogram …

    and those midpoints, for the new quadrilateral, are (both) 1/4 (a + b + c + d) …

    so it's a parallelogram, and "centred" in the same place as the original quadrilateral :wink:
     
    Last edited: Apr 26, 2010
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