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Thread: Triangle Vector Problem

  1. #1
    Member jacs's Avatar
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    Triangle Vector Problem

    Greetings,
    Have just started with vectors and struggling with a triangle problem. Seem to be going around in circles ... have a concept (more or less) how this is supposed to work, but just not seeing how to pull it together.
    Any help would be greatly appreciated. Have screen capped the question and attempted working thus far.

    Thanks, appreciate any help
    Triangle Vector Problem-vector.jpgTriangle Vector Problem-attemptedanswer.jpg

    PS Wasn't really sure which forum to post this in as vectors seemed to land in multiple places when i was looking through search results.
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  2. #2
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    Re: Triangle Vector Problem

    Quote Originally Posted by jacs View Post
    Greetings,
    Have just started with vectors and struggling with a triangle problem. Seem to be going around in circles ... have a concept (more or less) how this is supposed to work, but just not seeing how to pull it together.
    Any help would be greatly appreciated. Have screen capped the question and attempted working thus far.

    Thanks, appreciate any help
    Click image for larger version. 

Name:	vector.jpg 
Views:	19 
Size:	9.8 KB 
ID:	38510Click image for larger version. 

Name:	attemptedanswer.JPG 
Views:	17 
Size:	51.9 KB 
ID:	38509

    PS Wasn't really sure which forum to post this in as vectors seemed to land in multiple places when i was looking through search results.
    @jacs, I wish I could read the diagrams in your post. But it is all a jumble as far as I can see.
    I seems to me that someone is mixing directions and lengths. Again, I cannot read the diagrams.
    Is it the case that $\overrightarrow {OA} = 2\overrightarrow {AN}~\bf{?~?~}$ If so the in terms of distance, the lengths are :
    $\|\overrightarrow {OA}\|=a$, so that $\|\overrightarrow {AN}\|=2a$ AND $\|\overrightarrow {ON}\|=3a$

    Is it the case that $\overrightarrow {MN} = \overrightarrow {MO} + \overrightarrow {ON}$ So that $\|\overrightarrow {MN}\| = \|\overrightarrow {MO}\| + \|\overrightarrow {ON}\|$ or $\|\overrightarrow {MN}\| =0.5(b)+3a$

    Again, If I read this correctly you can use this: $\overrightarrow {OA} + \overrightarrow {AP} = \overrightarrow {OM} + \overrightarrow {MP}$.

    If you understand my points can you finish?
    Thanks from jacs
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  3. #3
    Member jacs's Avatar
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    Re: Triangle Vector Problem

    Sorry about the diagrams, that is how the teacher told us to do it...
    thanks for the jump start on this, i understand better now the directions vs magnitude issue
    and can see how $\overrightarrow {OA} + \overrightarrow {AP} = \overrightarrow {OM} + \overrightarrow {MP}$ works, which was a connection i didn't previously make
    I don't know if it is a forest for the trees issue, or i have been looking at this way too long, i feel that it should be pretty obvious, just can't see it. i keep getting myself in circles with k = 1 and that can't be right.
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  4. #4
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    Re: Triangle Vector Problem

    I don't see how to use "vector" geometry to solve this, but here's an algebraic solution :

    Thanks from jacs
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  5. #5
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    Re: Triangle Vector Problem

    vector equations

    MP=MO+OA+AP=-b/2+a+k(-a+b)=(1-k)a+(k-1/2)b

    and

    NP=NA+AP=-2a+k(-a+b)=(-k-2)a+k b

    now MP is parallel to NP therefore

    $\frac{1-k}{-k-2}=\frac{k-1/2}{k}$

    $k=2/5$
    Thanks from jacs
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  6. #6
    Member jacs's Avatar
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    Re: Triangle Vector Problem

    Thanks everyone, really appreciate the help here.
    The teacher is struggling to teach this and I suspect is learning and teaching it on the fly, so we are getting a bit of a jumbled mess.

    This question is way beyond the level we have been working at, and probably shouldn't have been given to us yet, but the different approaches that have been shown here have actually gone a very long way to helping understand the links between the bits and pieces we have been taught.

    thanks again for your time and patience. it is greatly appreciated.
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