1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Prove that ABCD is a parallelogram

  1. Sep 21, 2010 #1
    1. The problem statement, all variables and given/known data
    Let [tex]z_1[/tex], [tex]z_2[/tex], [tex]z_3[/tex] and [tex]z_4[/tex] be te position vectors of the vertices of quadrilateral AMCD. Prove that ABCD is a parallelogram if and only if [tex]z_1-z_2-z_3+z_4=0[/tex].

    2. Relevant equations

    3. The attempt at a solution
    The solution obviously uses the fact that collinear vectors of equal magnitude are equal, but I get [tex]z_1-z_2+z_3-z_4=0[/tex]. Am I missing something obvious or is it just notations issue with the vertices of the quadrilateral.
     
  2. jcsd
  3. Sep 21, 2010 #2

    HallsofIvy

    User Avatar
    Science Advisor

    Re: Parallelogram

    [itex]z_1- z_2[/itex] is the side from "[itex]z_2[/itex]" to "[itex]z_1[/itex]" and [itex]z_3- z_4[/itex] is the side from "[itex]z_4[/itex]" to "[itex]z_3[/itex]". Since this is a parallelogram those sides are parallel and equal in length. In other words, the vectors are equal: [itex]z_1- z_2= z_3- z_4[/itex] whence [itex]z_1- z_2- z_3+ z_4= 0[/itex]
     
  4. Sep 21, 2010 #3
    Re: Parallelogram

    That is true only if you haven't numbered the vertics clockwise, but in the form of a 'z' (if you get what I mean). In my picture, [tex]z_1[/tex] is adjacent to [tex]z_2[/tex] and [tex]z_4[/tex], and [tex]z_2[/tex] is adjacent to [tex]z_1[/tex] and [tex]z_3[/tex], so in my case the vector [tex]z_1-z_2[/tex] is opposite to the vector [tex]z_3-z_4[/tex].
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook