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Unit vector and distance

  1. Oct 7, 2013 #1
    Given a trapezium ABCD with AB parallel to CD. Show that the line joining the midpoints of the diagonals is parallel to AB & CD (done). Show also that its length is the difference of AB and CD.


    I called the midpoint of AC, E and the midpoint of BD, F

    Then found that ## \displaystyle \vec{EF} = \frac{1}{2}b + \frac{1}{2}(d-c) ## now I wasn't sure where to go from here so took a look at the solutions,

    they change ## \displaystyle \vec{EF} = \frac{1}{2}b - \frac{1}{2}(c-d) ## which I understand why, however the next step baffles me

    they go from:
    ## \displaystyle \vec{EF} = \frac{1}{2}b - \frac{1}{2}(c-d) ##
    ## \displaystyle |\vec{EF}|\hat{u} = \frac{1}{2} |\vec{AB}|\hat{u} - \frac{1}{2}|\vec{DC}|\hat{u} ##

    then go on to say that the length EF is ## \frac{1}{2}(|\vec{AB}| - |\vec{DC}|) ##

    what I don't understand is how the unit vector "u" got introduced, and why was it introduced? I'm not very good at maths, so please explain this in as most basic terms as possible.

    I apologise for not being able to bold "a" "b" "c" "u" etc, I have no idea how to do it via latex, but they are vectors (supposed to be).
  2. jcsd
  3. Oct 7, 2013 #2


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    What are ##b,~d,~c##?
  4. Oct 7, 2013 #3
    The position vector of point B,D,C I'm assuming
  5. Oct 7, 2013 #4
    position vector of B D C
  6. Oct 7, 2013 #5


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    Given that ##a,b,c,d## are position vectors to ##A,B,C,D##, that is not the correct expression for ##\vec{EF}##. You should get$$
    \vec{EF} =\frac 1 2(b-a) + \frac 1 2(d-c)$$From this it follows immediately that the length of EF is half the difference of the lengths of AB and CD, which leads me to believe the result of the problem is also stated incorrectly. You don't need to mess around with unit vectors at all.
    Last edited: Oct 7, 2013
  7. Oct 8, 2013 #6
  8. Oct 8, 2013 #7
    I took A to be the centre (as did the solutions)
  9. Oct 8, 2013 #8


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    You "took A to be the centre" of what? The problem says that A is one vertex of the trapezium.

    The last part, "its length is the difference of AB and CD" simply isn't true. The distance between the midpoints is half the difference between the lengths of AB and CD.

    For the entire problem you can do this (not using vectors): set up a Cartesian coordinate system with the origin at A and positive x-axis along side AB. Then A is (0, 0) and B is (a, 0) for some positive number, a. We can take D to be (b, c) and C to be (d, c) with b< d. (Since CD is parallel to AB, it is parallel to the x-axis so C and D have the same y coordinate.)

    The midpoint of diagonal AC is (d/2, c/2) and the midpoint of diagonal BD is ((a+ b)/2, c/2). Since the two midpoints have the same y coordinate the line through them is parallel to the x-axis and so parallel to the two parallel sides of the trapezium.

    The distance between the two midpoints is (a+b)/2- d/2= (a+b- d)/2. The length of AB is a and the length of CD is d- b. The difference in lengths is a- (d- b)= a+ b- d.
    Last edited by a moderator: Oct 8, 2013
  10. Oct 10, 2013 #9
    OK perhaps I should have started this off with the diagram I drew (it is attached).

    Perhaps I typed up the question wrong (I don't have it right now), but the diagram goes with the working above, if you could please take a look at it.

    Attached Files:

  11. Oct 10, 2013 #10


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    That is the diagram I used for post #5, to which you responded in post #6 "nevermind", whatever that is supposed to mean. You never did address the issue of whether you understood the problem is stated incorrectly or that the proof is apparent from what I posted.
  12. Oct 10, 2013 #11
    I understand the problem is stated incorrectly, but I'm still confused as why they used a unit vector. I don't think you took A to be the origin, else you would not have gotten (b-a) in your expression.
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