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Linear Algebra Help, Area and Volume using Vectors

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

    x = {2, 0, 1, 1}
    y = {-1, -1, 1, 1}
    z = {-2, 3, 1, -2}


    2. Relevant equations
    Don't know.


    3. The attempt at a solution
    Don't know where to start.


    The main problem is that most people on the internet only discuss three dimensional volumes, leaving me searching for information regarding this and finding nothing but cross-product and the like. I need to understand the method of finding volume for R^n, not just R^3
     
  2. jcsd
  3. Aug 23, 2010 #2
    Well, if you're interested in the volumes of parallelotopes in dimensions higher than 3, you can look into Gramian matrices and the like.

    However, I don't think you need that. Note that the third element of your four-dimensional vectors is constant, which seems to indicate that maybe the usual three-dimensional techniques will work.
     
  4. Aug 23, 2010 #3
    I tried that just now and it doesn't work. Is there really no way to answer this question?
     
  5. Aug 23, 2010 #4
    What did you try? I got an answer by dotting one vector with the cross product of the other two, ignoring the dimension where all three vectors are constant.

    I mean, if I gave you these two vectors:

    a = {1,1,2}
    b = {3,0,2}

    and said, "give me the area of the parallelogram formed by these two vectors," you would just pretend like the z-dimension didn't exist, because parallelograms are in two dimensions, and everything is happening in a plane here. The same thing is true of your vectors: parallelepipeds are in three dimensions, and everything's happening in three dimensions here.
     
  6. Aug 23, 2010 #5
    I find the answer to be 7, however, Webwork says it's incorrect. I've tried plugging in -7, still nothing.
     
  7. Aug 23, 2010 #6
    Yeah, my visualization was off here. The dimension where they are all the same does matter. Sorry about that.

    Anyway, you can do this, which works for any dimension of vectors:

    Let:

    [tex] A = \begin{bmatrix} a_1 & a_2 & a_3 \end{bmatrix} [/tex]

    Now if the dimension of the vectors equals 3 (ie, A is square), then the absolute value of the determinant of that is the volume.

    If it doesn't, then:

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

    I get [itex] \sqrt{308} [/itex].
     
    Last edited: Aug 24, 2010
  8. Aug 24, 2010 #7
    Thank you very much. That works wonderfully.
     
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