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Find area of parallelogram given vertices

  1. Apr 25, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the are of the parallelogram ABCD where A is (1,2,-3), B is (-1,3,-4) and D is (1,5,-2)


    2. Relevant equations
    Area=[itex]\left|AxB\right|[/itex]
    where A and B are the vectors AD, and AB respectively.


    3. The attempt at a solution
    I have calculated AD to be= (0,-3,-1)
    and AB=(2,-1,1)

    ∴ to find [itex]\left|AxB\right|[/itex] I subbed the two vectors into a 3x3 matrix. Finding the magnitude of th result, I calculated the area to be 2x[itex]\sqrt{11}[/itex]. The Anser is 2x[itex]\sqrt{14}[/itex]. What have I done wrong?
     
    Last edited: Apr 25, 2014
  2. jcsd
  3. Apr 25, 2014 #2

    SammyS

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    You likely made an error in finding the vector product [itex]\vec{A}\times \vec{B}[/itex].

    BTW: Use " \times " for ## \times\ ## .
     
  4. Apr 25, 2014 #3
    Well I will show you how i did it because I cannot find an error:

    i j k
    0-3 1
    2-1 1
    ^that is a matrix

    So i got:
    -2i+2j+6k

    =Sqrt 44
    =2 sqrt 11
     
  5. Apr 25, 2014 #4

    jbunniii

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    Shouldn't the second row of your matrix be ##(0, -3, -1)## instead of ##(0, -3, 1)##?
     
  6. Apr 25, 2014 #5

    Mark44

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    The first vector is actually DA, and the second is BA.
    AD is -(DA) = <0, 3, 1>
    and AB = -(BA) = <-2, 1, -1>.
     
  7. Apr 25, 2014 #6

    SammyS

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    The vectors you have in the OP will work. (They both have the sign error Mark pointed out.) .. but they will give ## 2 \, \sqrt{14} \ ## .
     
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