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Area of triangle given 3 vectors pointing to vertices

  1. Oct 8, 2013 #1
    1. The problem statement, all variables and given/known data

    Three vectors A, B, C point from the origin O to the three corners of a triangle. Show that the area of the triangle is given by

    area = [itex]\frac{1}{2}[/itex]|(B[itex]\times[/itex]C) + (C[itex]\times[/itex]A) + (A[itex]\times[/itex]C)|

    2. Relevant equations

    area of triangle with sides a, b, c = [itex]\frac{1}{2}|[/itex]a[itex]\times[/itex]c|


    3. The attempt at a solution

    I can't figure out how to go about it. I know I need to find the lengths between the vertices and use that equation, but I don't know how to make it look like the expected result. I'm totally lost.
     
  2. jcsd
  3. Oct 8, 2013 #2

    SteamKing

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    You won't get anywhere with this problem if you don't try something.
     
  4. Oct 8, 2013 #3

    haruspex

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    What is the area of the triangle between two vectors a, b? Can you subdivide the triangle formed by the endpoints of three vectors into smaller triangles in a useful way?
     
  5. Oct 8, 2013 #4
    So, the area of the triangle between two vectors (let's say A and B) is 0.5|Axb| right? I still don't see how I can use that to solve this. I can find the area of every triangle but the one I need.

    EDIT: Alright, I was just being a dummy. I redrew my picture so that each of the vectors point away from eachother (into three different quadrants) and I realized the if I added up all of the triangles (each of which being contained in the big triangle) I'd get the area of the big one. So the area is the sum of the smaller areas. Thanks!
     
  6. Oct 8, 2013 #5

    UltrafastPED

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    Your area equation must have a typo: the vectors products CxA + AxC sum to zero!

    To solve this problem reference the vectors to the vertices; for example (B-A) and (C-A). This gives you vectors with the lengths of two of the sides ... which you already know how to do.
     
  7. Oct 8, 2013 #6
    Yep, you're right. One of those was supposed to be an A cross B. Thanks.
     
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