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Volume of parallelpiped

  1. May 29, 2013 #1
    1. The problem statement, all variables and given/known data

    A corner,A , of a parallelepiped ABCDEFGH, has position vector <3,7,4> and the points B,G,D that are neighboring vertices A of have position vectors< 2,9,7>, < 5,10,10> ,< 4,11,9 > , respectively. Find the volume of the parallelepiped in cubic units

    2. Relevant equations

    Triple scalar product in general a cross b dot c

    3. The attempt at a solution
    OK, so I know what to do but I don't know if I did the right things. First I tried to draw it this was kind of hard because I'm not so good at orientating my axes so I can see what I'm looking at.
    I want..I think
    (g-d) cross ( a -b) dot (b-d)

    OK I did

    (g-d) = < 5,10,10> -< 4,11,9 > = < 1 ,-1, 1>
    (a - b) = <3,7,4> -< 2,9,7> = < 1,-2,-3>
    (g-d) cross (a - b) = < 5, 4, -1>
    Then, dot this with vector ( b - d) = < 2,9,7> - <4,11,9> = <-2,-2,-2>
    < 5, 4, -1> dot <-2,-2,-2> = 39 cubic units?


    I feel like if I did this wrong it is because of my drawing maybe. Thanks.
     
  2. jcsd
  3. May 30, 2013 #2

    haruspex

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    That does seem an odd choice. Try thinking in terms of translating all the vectors to make one of the vertices the origin (A, say).
     
  4. May 30, 2013 #3

    HallsofIvy

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    There is a reason why you are given the corner A first, then the other three points!

    The three edges, terminating at A, are the vectors B- A, D- A, and G- A.
     
  5. May 30, 2013 #4
    How do you know this ?

    Since I can't draw if you pause this video about 39 seconds in. If you look at the diagram if you did G-A you would be subtracting across diagonally? I see how you are saying A was given first but how do I know that G-A isn't across the parallelepiped diagonally? thx
     
    Last edited by a moderator: Sep 25, 2014
  6. May 30, 2013 #5
    I guess to rephrase my question is G-A is a vector diagonally across the parallelepiped you can't use this as the length because it will actually be longer then needed?
     
  7. May 30, 2013 #6
    No. G-A is the displacement vector between A and G.

    The key is that you stated that A and G are neighboring vertices.
     
  8. May 30, 2013 #7
    Look, This is my drawing. I know G-A is the displacement vector. I do not know what is meant by neighboring vertices. Look at my drawing and forget about how crappy it is it serves its purpose :)
    I'm given A,B,G,D
    If I want the volume I A-B I need some the length of say A-E or B-F. A to G is a diagonal it is not the same length as A to E which I need. I have the height which is B to D. The diagonal is throwing me off.
     

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  9. May 30, 2013 #8

    haruspex

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    Then that's your problem. It means there is an edge of the object connecting A to each. I.e. AB., AD and AG are edges of the shape.
     
  10. May 30, 2013 #9
    Did you look at my drawing is it correct?
     
  11. May 30, 2013 #10

    SteamKing

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    'Neighboring vertices' means that each vertex which is a neighbor to vertex A has no other vertex between A and it. You know, your neighbor lives next door, not across town.
     
  12. May 30, 2013 #11

    haruspex

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    It came out blank for me.
     
  13. Jun 2, 2013 #12
    OK, So I worked something out I think.
    a = < 3,7,4>
    b = <2,9,7>
    g = <5,10,10>
    d = < 4,11,9>

    So ,
    a' = b-a = <2,9,7>-< 3,7,4> = < -1, 2 ,3>
    b' = g-a = <5,10,10> - < 3,7,4> = < 2,3,6>
    c' = d -a = < 4,11,9> - < 3,7,4> = <1,4,5>

    Then,

    ( b' cross c') dot a' = volume

    ( b' cross c') = <-9,-4,5>
    Now dot with a'
    <-9,-4,5> dot < -1, 2 ,3> = 9 - 8 + 15 = 16

    So volume is 16 cubic units?
    Thanks.
     
  14. Jun 3, 2013 #13

    haruspex

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    Looks right.
     
  15. Jun 3, 2013 #14

    HallsofIvy

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    YOU said that "points B,G,D that are neighboring vertices A". Diagionally opposite points are not "neighboring"..
     
    Last edited by a moderator: Sep 25, 2014
  16. Jun 3, 2013 #15

    epenguin

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    This is a fine site and there are so much riches that if you don't grab something when it appears you may never find it again. I have searched and never found again this one:

    - roughly early last year there was a post from someone saying the volume of a paralellipiped was the key! I am not sure now what it was the key to - whether just determinants or to linear algebra. That this volume was the way to teach determinants - to make them intuitive and easy, which frankly they are not (is it just me?) and could very well do with being made. I don't know whether this was just an once off opinion and maybe the way the poster teaches them, or whether he had some supporting material developed.

    OK, as a journeyman math user I can work out the volume of the ppped with conrners (x1, y1, z1) , 2, 3 without too much difficulty. Given that, I can see how it illustrates the idea of linear dependence. But the formula, if not difficult, is to me anything but self-evident, like most of deterninant theory or linear algebra.

    If that poster is here or anyone remembers the post, and if this idea was ever elaborated it would be useful to at least one person to hear again.
     
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