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Area of a Triable with three vertices

  1. Apr 27, 2007 #1
    Determine the area of a triangle with vertices A(1,0,-5), B(-2,2,3) and C(-3,1,-1).

    1/2|a x b|

    |a x b| = (1, 0, -5) x (-2, 2, 3)
    = ((0)(3) - (2)(-5), (-5)(-2) - (3)(1), (1)(2) - (-2)(0))
    = (10, 7, 4)

    |(10, 7, 4)|
    = sqr(10^2 + 7^2 + 4^2)
    = sqr(165)

    (1/2)(sqr(165)) = 6.42

    Therefore the area of the triangle is approximately 6.42 square units.
     
    Last edited: Apr 27, 2007
  2. jcsd
  3. Apr 27, 2007 #2
    1. this isn't calculus
    2. you should really explain where you got vectors a and b from.
    3. why are you dotting when you should be crossing?
    4. do you know the definition of area for a triangle with vertices A,B,C?
     
  4. Apr 27, 2007 #3

    HallsofIvy

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    First you have the vector wrong: neither (-3,1,-2) nor (4,-5,2) is a vector between two of the vertices. Second, the area of a triangle where the vectors a, b form two sides is NOT (1/2)|a.b|, it is (1/2)|axb|.
     
  5. Apr 27, 2007 #4
    Sorry, my first post was a complete mess. Please see the revised post above.

    Thanks, Chaotic. I am aware this is not Calculus. However, the geometry posts I made yesterday were moved here so I figured I would save the mods a step...

    I am also aware that triangle is misspelled in the header but I can not correct it at this point.
     
    Last edited: Apr 27, 2007
  6. Apr 27, 2007 #5
    Does this look right? I keep thinking I should be using point C as well.
     
  7. Apr 27, 2007 #6
    yes, you would be right in thinking this.

    look at points A,B,C; maybe try making vectors out of those points? then crossing some of them?
     
  8. Apr 27, 2007 #7
    Ok, how about this:

    |a| = (1, 0, -5) x (-2, 2, 3)
    = ((0)(3) - (2)(-5), (-5)(-2) - (3)(1), (1)(2) - (-2)(0))
    = (10, 7, 4)
    =|(10, 7, 4)|

    |b| = (-2,2,3) x (-3, 1, -1)
    =((2)(-1) - (1)(3), (3)(-3) - (-1)(-2), (-2)(1) - (-3)(2))
    =(-5,-11,4)
    =|(-5, -11, 4)|

    |a x b| = (10, 7, 4) x (-5,-11,4)
    =((7)(4)-(-11)(4), (4)(-5)-(4)(10), (10)(-11)-(-5)(7))
    =(72, -60, -75)
    =(24, -20, -25)
    =|(24, -20, -25)|

    sqr(24^2 + (-20)^2 + (-25)^2) = 40
    (1/2)(40) = 20
     
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