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Homework Help: Linear Algebra Help, Area of a Parallelogram using vectors

  1. Aug 23, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the area of the parallelogram defined by the vectors

    v = {1 1 3 1}
    w = {-2 -1 2 2}

    2. Relevant equations
    Area = v dot w * sin(theta)
    theta = cos^-1(v dot w / |v|*|w|)

    3. The attempt at a solution
    Solved
    General Solution:

    Area of a parallelogram for non-R^3 vectors = v dot w * sin(cos^-1(v dot w / (|v|*|w|)))
     
    Last edited: Aug 23, 2010
  2. jcsd
  3. Aug 23, 2010 #2
    If v and w are two vectors representing two adjacent sides of the parallelogram, then the area is the magnitude of the cross product of those two vectors.
     
  4. Aug 23, 2010 #3
    You can't perform cross product on vectors outside of R^3
     
  5. Aug 23, 2010 #4
    Whoops! Sorry, I didn't notice they were 4 dimensional vectors.
     
  6. Aug 23, 2010 #5
    I got it, some how. Updated original post.
     
  7. Aug 23, 2010 #6
    What do you mean you got it some how... What did you do? Alegbra mistake?
     
  8. Aug 23, 2010 #7
    I didn't depend on the cross product and started using other equations around the internet.

    I provided them in the original post.
     
  9. Aug 23, 2010 #8
    I don't think that's right. You can let

    [tex] A=\begin{bmatrix} 1 & -2 \\ 1 & -1 \\ 3 & 2 \\ 1 & 2 \end{bmatrix} [/tex]

    and then the volume is:

    [tex] V = \sqrt{det \left(A^TA \right)} [/tex]

    which isn't 5.

    EDIT: Oh, I see you got it with some other formula. Was the answer [itex] \sqrt{131} [/itex] ?
     
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