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Math Help - Finding the area of a triangle in two-space.

  1. #1
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    Finding the area of a triangle in two-space.

    I know how to find the area of a triangle in three-space, say ABC: First I would find \vec{AB} and then \vec{AC} and then use the formula for the area of a parallelogram:
    A = {\vec{AB} \times \vec{AC}
    Finding the magnitude of that and dividing it by two (since the area of a triangle is half that of the area of a parallelogram), I would find the area of the triangle ABC.

    But, how would I find the area of a triangle that does not have a Z component (is in two-space)? For example: Find the area of the triangle with vertices of (1, -2), (-1, 3), (2, 4). Since this is in two-space, I can't take the cross product of any two of these vectors to find the area. According to my book, the area of this triangle is \frac{17}{2} and I have no idea how they computed that. Any help? I would appreciate learning a systematic approach to solving the areas of a triangle in two-space. Thanks in advance.
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  2. #2
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    Re: Finding the area of a triangle in two-space.

    Why can't you add z=0 to all of them?
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  3. #3
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    Re: Finding the area of a triangle in two-space.

    Quote Originally Posted by TKHunny View Post
    Why can't you add z=0 to all of them?
    I did just that:
    A(1, -2, 0), B(-1, 3, 0), and C(2, 4, 0)
    \vec{AB} = (-1, 3, 0) - (1, -2, 0) = (-2, 5, 0)
    \vec{AC} = (2, 4, 0) - (1, -2, 0) = (1, 6, 0)
    The cross product of these two direction vectors:
    \vec{AB} \times \vec{AC} = (0, 0, -18)
    Now to find the magnitude of the normal:
    |\vec{AB} \times \vec{AC}| = \sqrt{(0)^{2}+(0)^{2}+(-18)^{2}
    |\vec{AB} \times \vec{AC}| = 18
    Since this is only the area of the parallelogram and the area of the triangle is half of that:
    A = \frac{18}{2} = 9 square units.

    But in my textbook, the area of the triangle is \frac{17}{2} or 8.5. So, did I compute the area incorrectly?
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  4. #4
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    Re: Finding the area of a triangle in two-space.

    Check the calculations. How did you get -18?

    -2*6 - 1*5 = ?
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  5. #5
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    Re: Finding the area of a triangle in two-space.

    Quote Originally Posted by sa-ri-ga-ma View Post
    Check the calculations. How did you get -18?

    -2*6 - 1*5 = ?
    Minor mistake of mine, thanks for catching it. I usually use the dot product to confirm I computed the cross product correctly, but since z = 0 for all z, I foolishly assumed it was correct even though I was off.

    Many thanks to TKHunny for giving me an invaluable tool to calculating problems like these.
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