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Area of a triangle by vectors - getting different answers

  1. Jan 24, 2010 #1
    1. The problem statement, all variables and given/known data
    A = (7,6,8)
    B = (2,3,-1)
    C = (2, -7, 9)

    Find the area of the triangle these 3 points make

    I'm getting slightly different results for different combinations I use

    if I take

    CA = a - c = (5 , 13 , -1)
    CB = b - c = (0, 10, -10)

    Taking another combination
    AB = b - a = (-5, -3, -9)
    AC = c - a = (-5, -13, 1)

    Taking the modulus of the cross product and halving yields similar, but different answers. I've check and redone my work

    Why?

    Thanks
    Thomas
     
    Last edited: Jan 24, 2010
  2. jcsd
  3. Jan 24, 2010 #2
    You computed AB wrong, it should be (-5,-3,-9) (added changed 3 to -3 in second coordinate). Then you get the same area.
     
  4. Jan 24, 2010 #3
    sorry that was a typo, not actually what I did :blushing: (i've edited it now)

    When I did it the first combination I get 5 sqrt(246)
    But when I did the second combination I got 5sqrt(233)

    Do you concur?

    Thanks
    Thomas
     
  5. Jan 24, 2010 #4
    No.

    I get the cross product as: (-120,50,50) with both your combinations which gives us an area of 5sqrt(194)
     
  6. Jan 24, 2010 #5
    The correct answer is [-120, 50, 50] for your cross products.

    You are screwing up when you are doing your calculations (I'm assuming you are using a CAS to input the numbers and not doing it all on paper which is why you missed it.)

    If you type in for CB = <0, +10, +10> and do the cross product of CA x CB you will end up with [140, -50, 50] which results in your 5[tex]\sqrt{246}[/tex].

    However if you type in CB correctly = <0, +10, -10> you'll come up with the correct answer.

    Do it all on paper, ignore the calculator or CAS and you'll see where you went wrong. Once done properly you'll see that AB x AC = CA x CB
     
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