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Homework Help: Help visualising this triangle!

  1. Apr 14, 2012 #1
    1. The problem statement, all variables and given/known data

    I think this must be really easy but im not getting a visual for this triangle description.

    Let ABC be a triangle with [itex]\UparrowOA[/itex]=a and [itex]/UparrowOB[/itex]=b and [itex]/UparrowOC[/itex] = c
    Where O is the origin .

    2. Relevant equations

    3. The attempt at a solution

    How can I have a triangle ABC where all A,B,C are sides coming from O

    The way i'm looking at this is I have a point O, where the lines A,B,C start from ie, A,B,C only have one available endpoint.

    So how can these lines make a triangle without introducing a new line??

    Clearly, I am imagining things wrong here, I tried drawing it but it still didn't work. Help!!
  2. jcsd
  3. Apr 14, 2012 #2


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    See attachment.


    Attached Files:

  4. Apr 14, 2012 #3
    thank you sir!!
  5. Apr 14, 2012 #4
    Hmm I have a question related to this triangle and it asks

    i) If M is the midpoint of the line segment AB and P is the midpoint of the line segment CB express the vectors [itex]\vec OM[/itex] and [itex]\vec OP[/itex] in terms of a,b, and c.

    ii) Show that [itex]\vec MP[/itex] is parallel to [itex]\vec AC[/itex] and has half its length.

    Ok for part i) the answer is 1/2(b-a), 1/2(b-c), but to me it should be 1/2(b+a), 1/2(b+c).... as if you were to minus a or c does that not imply that you 'attach a/c to the end of vector b or in this case 1/2b and 1/2 (c,a)...
    I think the difficulty im having is understanding the direction in the vector, how do I know which way it is 'pointing' so to speak??

    in this situation what would the addition of the vectors look like?

    Last edited: Apr 14, 2012
  6. Apr 14, 2012 #5


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    If your answer is that [itex]\vec{OM} = \frac{1}{2}[/itex](a + b) and that [itex]\vec{OP} = \frac{1}{2}[/itex](b + c),

    then you are, in fact, right (and the book is wrong). a, b and c are position vectors - which means they define the positions of points A, B and C respectively with respect to a common origin O. By convention, they *always* point outward from the origin O toward the terminal point. So a = [itex]\vec{OA}[/itex], etc.
  7. Apr 14, 2012 #6
    OK just wanted to clear that up yeah in the book the have 1/2(b-a),1/2(b-c)... thanks for the quick reply! Keep it ℝeal!
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