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Math Help - Area of the Triangle (in 3D)

  1. #1
    Tau
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    Area of the Triangle (in 3D)

    Hi everyone, so here's the problem:

    We got a triangle in a 3D room defined by the coordinates:

    A = (1, 2, -1)
    B = (3, 1, 4)
    C = (0, 2, 3)

    What's the area of the triangle?

    ----

    If it were just defined in a 2D plane that'd would have been easier.
    So maybe it would be good to redefine the triangle into a plane?
    Last edited by Isomorphism; February 5th 2009 at 07:36 AM.
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    Quote Originally Posted by Tau View Post
    Hi everyone, so here's the problem:

    We got a triangle in a 3D room defined by the coordinates:

    A = (1, 2, -1)
    B = (3, 1, 4)
    C = (0, 2, 3)

    What's the area of the triangle?

    ----

    If it were just defined in a 2D plane that'd would have been easier.
    So maybe it would be good to redefine the triangle into a plane?
    The area of a triangle is half the base times the height.

    Pick any two points to be the base and find the distance between them. To find the height, take the other point, and find the distance between that point and the midpoint of the base. This will be the height.

    I trust you know how to find the distance between two 3D points, and also how to find the midpoint between two points?
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    Quote Originally Posted by Tau View Post
    We got a triangle in a 3D room defined by the coordinates:

    A = (1, 2, -1)
    B = (3, 1, 4)
    C = (0, 2, 3)

    What's the area of the triangle?
    If you know about cross products of vectors there is a neat way to do this. Given vectors u and v, the length of the vector uv is |u||v|sin(θ), where θ is the angle between u and v. But if u and v represent two sides of a triangle then the area of the triangle is |u||v|sin(θ). So given a triangle with vertices at A, B and C, its area is |(B–A)(C–A)|.

    Quote Originally Posted by Mush View Post
    Pick any two points to be the base and find the distance between them. To find the height, take the other point, and find the distance between that point and the midpoint of the base. This will be the height.
    No, the height is not in general the distance from a vertex to the midpoint of the opposite side.
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    Quote Originally Posted by Tau View Post
    Hi everyone, so here's the problem:

    We got a triangle in a 3D room defined by the coordinates:

    A = (1, 2, -1)
    B = (3, 1, 4)
    C = (0, 2, 3)

    What's the area of the triangle?

    ----

    If it were just defined in a 2D plane that'd would have been easier.
    So maybe it would be good to redefine the triangle into a plane?
    You can also use Heron's formula.
    Just compute the length of the three sides using the distance formula and then apply this formula for area.
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    Hello, Tau!

    As Opalg pointed out, if we're allowed to use vectors, it's quite simple.


    We have a triangle defined by the points: . A(1, 2, -1),\;B(3, 1, 4),\;C(0, 2, 3)

    Find the area of the triangle.

    The area of the parallelogram determined by two vectors, \vec u\text{ and }\vec v,

    . . is given by: . |\vec u <br />
\times \vec v| . . . the magnitude of their cross-product.


    We have: . \begin{array}{ccccccc}\vec u &=& \overrightarrow{AB} &=& \langle 2,\text{-}1,5\rangle \\ \vec v &=& \overrightarrow{AC} &=& \langle \text{-}1,0,4\rangle \end{array}

    Then: . \vec u \times \vec v \;=\;\left|\begin{array}{ccc}i & j & k \\ 2 &\text{-}1 & 5 \\ \text{-}1 & 0 & 4 \end{array}\right| \;=\;i(-4-0) -j(8+5) + k(0-1) \;=\;\langle\text{-}4,\text{-}13,\text{-}1\rangle


    The area of the parallelogram is: . \sqrt{(\text{-}4)^2 + (\text{-}13)^2 + (\text{-}1)^2} \;=\;\sqrt{186}

    Therefore, the area of the triangle is: . \frac{1}{2}\sqrt{186}

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