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Homework Help: Find the area of a triangle with vector sides

  1. Jan 17, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the area of the triangle with vertices at the points with coordinates (1,2,3), (4,-3,2) and (8,1,1)

    2. Relevant equations

    3. The attempt at a solution

    well area of a triangle is 0.5 x base x height . We can get the length of each side by taking the modulus of each vector, but the problem we have is finding the height. We need to find a the point where a line between 2 vectors cuts the other vector at 90° (the base)

    How do we find this line?

    Last edited: Jan 17, 2010
  2. jcsd
  3. Jan 17, 2010 #2


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    Hi Thomas! :smile:

    Hint: area = absinθ = a x b :wink:
  4. Jan 17, 2010 #3
    you mean it's simply the "cross product" or "vector product". Of course
    axb = |a||b|sinx e

    (where e is direction +/-)

    do the modulus make any difference?

    I would simply the determinant rule for finding it

    i j k
    1 2 3
    8 1 1

    -i + 23j - 15k

    then take the modulus



  5. Jan 17, 2010 #4


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    ooh yes, I left out the | | symbols! :redface:

    yes, that's the right method, but you've only found the area of the triangle formed by points 1 and 3 and the origin, haven't you? :wink:
  6. Jan 17, 2010 #5

    Well, this may not be what your exercise is all about, but you can try "[URL [Broken] formula[/URL].
    Last edited by a moderator: May 4, 2017
  7. Jan 18, 2010 #6
    Tim, havn't I got to half it to get that area :) Right? How does that help though. :cry:

    I can remember from my A levels that there are position vectors, which are relative to the centre, direction vectors and I think they have a parameter in them telling you how far "along" the vector to go. I'm getting myself confused over the basics. :sigh:

    Anyway thinking about it again it's the area = 0.5 |AB x AC|

    a = (1,2,3), b = (4,-3,2) and c = (8,1,1)

    AB = b - a
    = 3,-5,-1
    AC = c - a
    = 4, 4, -1

    but what does |AB x AC| mean? |AB| x |AC|?

    Am I getting there?

    @ Altabeh - I saw this mentioned on the WIKI page for triangles, though useful as it looks I want I'm having conflicting thoughts and confusing myself with the fundamentals which I want to sort out.

  8. Jan 18, 2010 #7


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    oooh, i'm not doing very well on this :redface: … yes, half! :rolleyes:
    Yes, that's right …

    the formula for a general point on the line AB is a + k(b - a), for any number k.

    (though of course you don't need that in this case)
    That's right :smile: (except the last one is BC :wink:, but it doesn't matter anyway).
    No, it means the modulus of the vector AB x AC (or AB x BC or AC x BC).

    So just do (3,-5,-1) x (4, 4, -1), and take half the modulus. :wink:
  9. Jan 18, 2010 #8
    And at the end of your calculation, check your result to see if it matches the one I've prepared through Herron's formula here:

    [tex](1/2)*sqrt(2\sqrt(14)+2\sqrt(66)+2\sqrt(29)) \approx 2.936908289[/tex].

  10. Jan 19, 2010 #9
    So (3,-5,-1) x (4, 4, -1),

    = (12, -20, 1)

    = sqrt(144 + 400 + 1)

    = sqrt(545)

    = 5sqrt(109)

    That's not what Altabeh got. Guessing im the one who is wrong :eek:
  11. Jan 19, 2010 #10


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    Nooo :redface:, you need the cross product (you did it in your post #3, except you used OA x OB instead of AB x BC)
  12. Jan 19, 2010 #11
    i j k
    3 - 5 -1
    4 4 -1

    (5 -- 4)i - (-3 - -4)j + (12 - - 20)k

    9i -j + 32k

    sqrt(81 + 1 +1024) = sqrt(1106) ~= 33.3

    i take it I'm wrong, as Altabeh's answer is right? :rolleyes:

  13. Jan 19, 2010 #12
    and take half of that

    = 0.5sqrt(1106)

    = 16.6
  14. Jan 19, 2010 #13


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    The cross product method is the most straightforward, but you could also use the dot product to find the angle between two sides and use it to calculate the height of the triangle.

    [tex]\cos\theta = \frac{(3, -5, -1)\cdot(4, 4, -1)}{|(3, -5, -1)| |(4, 4, -1)|} = \frac{3\times4+(-5)\times4+(-1)\times(-1)}{\sqrt{3^2+(-5)^2+(-1)^2}\sqrt{4^2+4^2+(-1)^2}}=\frac{-7}{\sqrt{33}\sqrt{35}}[/tex]


    [tex]\sin\theta = \sqrt{1-\cos^2\theta}=\sqrt{1-\frac{(-7)^2}{33\times35}}=\frac{\sqrt{1106}}{\sqrt{33}\sqrt{35}}[/tex]

    You can arbitrarily choose one side to be the base of the triangle. The height will then be the length of the other side multiplied by the sine of the angle between them, so you get

    [tex]A=\frac{1}{2}bh = \frac{1}{2}\sqrt{33}\sqrt{35}\sin\theta = 0.5\sqrt{1106} = 16.6[/tex]
  15. Jan 19, 2010 #14


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    Looks like it's 16.6. :smile:
  16. Jan 19, 2010 #15
    No! I was wrong! The correct calculation is:

    [tex]sqrt(s*(s-\sqrt{33})*(s-\sqrt{35})*(s-\sqrt{54}) \approx 16.62828916[/tex],

    where [tex] s=(\sqrt{54}+\sqrt{35}+\sqrt{33})/2[/tex].

    Last edited: Jan 19, 2010
  17. Jan 20, 2010 #16


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    Thomas, you'd better use either my method or vela's!! :biggrin:
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