Finding Triangle Area using Cross Product

  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. %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. HallsofIvy

    HallsofIvy 40,549
    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. RootX, the square root of what you have square doesn't match up with the answer you have.
     
  6. 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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