Finding Triangle Area using Cross Product

  1. Apr 5, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the area of a triangle `PQR`, where `P=(0,4,4)`, `Q=(2,-6,-5)`, and `R=(-3,-5,6)`

    3. The attempt at a solution

    The vector PQ = (2, -10, -9)
    The vector PR = (-3, -9, 2)

    Using matrixes I set up something that looks like this...

    I J K
    2 -10 -9
    -3 -9 2

    Then using the matrix methods I get.

    I(-20 - 81) - J(4 - 27) + K(-18 - 30)

    I(-101) - J(-23) + K(48)

    I take the square root of the squares and get.

    109.4349122

    Answer = 57.0832725061

    What's the problem?
     
  2. jcsd
  3. Apr 5, 2008 #2
    %was too lazy to do computations, so here's the matlab code (might help you)
    u =

    -3 -9 2

    >> v

    v =

    2 -10 -9

    >> cross(v,u)

    ans =

    -101 23 -48

    ans =

    114.1665

    >> ans/2

    ans =

    57.0833

    why divided by 2?
    because axb = |a|.|b|.sin theta
    and area is 1/2 of that
     
  4. Apr 6, 2008 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The area of a parallelogram formed by two vectors, [itex]\vec{u}[/itex] and [itex]\vec{v}[/itex], is [itex]|\vec{u}\times\vec{v}|[/itex]. Since a triangle is half a parallelogram, the the area of a triangle having sides [itex]\vec{u}[/itex] and [itex]\vec{v}[/itex] is half that.
     
  5. Apr 6, 2008 #4
    RootX, the square root of what you have square doesn't match up with the answer you have.
     
  6. Apr 6, 2008 #5
    If you take the magnitude the negative numbers will become positive

    [tex]
    \sqrt{(-101)^2 + (23)^2 + (-48)^2}= 114.167 ~ /2 = 57.0833

    [/tex]
     
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