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Linear Algebra, Determine the Volume of the Parallelepiped

  1. Aug 30, 2015 #1
    1. The problem statement, all variables and given/known data
    What is the volume of the parallelepiped with vertices (3, 0, −1), (4, 2, −1), (−1, 1, 0), (3, 1, 5), (0, 3, 0), (4, 3, 5), (−1, 2, 6), and (0, 4, 6)?

    2. Relevant equations
    The volume of a parallelepiped is given by the triple scalar product, (a × b) ⋅ c

    3. The attempt at a solution
    I know how to compute this problem, but I'm not sure how to go from the points given to the correct vectors. Is the best solution to plot the points and then determine the correct vectors for the triple scalar product?

    My other thought is to create all possible vectors and reason through which must be opposing sides.

    Basically I'm asking to see if you guys know any quick tricks for determining the correct vectors.

    Thanks!
    Nick
     
  2. jcsd
  3. Aug 30, 2015 #2

    Geofleur

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    If you just put the points into Wolfram Alpha it will show you a plot of them.
     
  4. Aug 30, 2015 #3
    That's really helpful; thank you. Do you know of a way to label the points in Wolfram?
     
  5. Aug 30, 2015 #4

    Geofleur

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    Not offhand. If you owned Mathematica you could do it, I'm sure. But the plot quality in Alpha isn't bad. You can "eyeball" it.
     
  6. Aug 30, 2015 #5

    Ray Vickson

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    If you have access to Maple, you can use the "textplot3d" command. If your eight points are a1,b1,c1,d1,a2,b2,c2,d2 , where the first four have the four smallest z-values and the last four have the four largest z-values (but with the points otherwise being in the order you listed them), here is what you would get:
    parallelogram_plot.jpg
     
    Last edited: Aug 30, 2015
  7. Aug 30, 2015 #6

    ehild

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    You can choose any point as a corner of the parallelepiped. Then determine all vectors pointing to the vertices from that corners. You need three independent vectors from them, to calculate the volume.
    The volume is equal to the magnitude of the determinant which columns (or rows) consist of the three independent vectors spreading out the parallelepiped.
     
    Last edited: Aug 30, 2015
  8. Aug 30, 2015 #7

    Ray Vickson

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    I was not prepared to believe the result---than any three L.I. vectors emanating from any fixed start -- will do, but having tried it in this example I now see it is true. Is there a simple proof of why that works?
     
  9. Aug 30, 2015 #8

    Geofleur

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    Yes. Take the cross product of two of the vectors. The magnitude of the cross product is the area of the parallelogram spanned by those two vectors. Then dot the cross product into the remaining vector. This projects the remaining vector into the direction perpendicular to the parallelogram. Multiplying the area of the parallelogram by the height of the parallelepiped formed by all three vectors gives the volume. In other words, if A, B, and C are the vectors that make up the edges of the parallelepiped, then the determinant of the matrix you calculated is equal to ## A \cdot (B \times C) ##, and you can understand the result by understanding the geometric interpretations of the dot and cross products.
     
  10. Aug 30, 2015 #9

    ehild

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    The three vectors do no need to be all the edges of the parallelepiped. Let a, b, c the edges, but you choose a, a+b, c for example. Then ##[A\times (A+B)]\cdot C =[A\times A]\cdot C + [A\times B]\cdot C ## as ##A\times A = 0##.
    If you take the determinant of the vectors |a b c| , you can add any column to an other column, the value of the determinant does not change.
     
    Last edited: Aug 30, 2015
  11. Aug 31, 2015 #10

    Ray Vickson

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    Yes, I know all that: ##V = |\bf{(A \times B) \cdot C}|## is the volume, but that was not at all the point of my question. My issue was that in the OP's example there are many different ways of choosing the vectors ##\bf{A, B, C}##. Not even counting permutations, there are more than 12 such triples of linearly-independent vectors, but nevertheless they all have the same value of ##|(\bf{A \times B) \cdot C}|##.
     
  12. Aug 31, 2015 #11

    Geofleur

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    Oh, I see! Well, you learn something new every day...
     
  13. Sep 3, 2015 #12
    Thank you all for your insightful replies and questions.

    Am I interpreting this correctly... If I start with a vertex and create three vectors from the given points, no matter which three vectors, I will get the volume of the parallelepiped? Even if the vectors are, say, the diagonals of the paralellepiped?

    Thanks
     
  14. Sep 3, 2015 #13

    ehild

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    Yes, you can choose any linearly independent three vectors pointing to an other vertex.
     
  15. Sep 4, 2015 #14

    HallsofIvy

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    Geometrically, it is the vectors giving the three edges that intersect at a single vertex. But since, in terms of vectors, there are only three independent vectors in a parallelpiped, you don't need to worry about the "intersection".
     
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