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Vectors, Volume, and plane equations.

  1. Jun 28, 2013 #1
    1. The problem statement, all variables and given/known data
    Four planes intersect in such a way that there is no common point between all four of them but that each set of three has a single common point. The position vectors of the four common points,A,B,C,D are, <-4,-3,-3> ,<7,5,-2>,<1,-2,5>,<-4,6,-1> respectively.
    (a) Find the volume enclosed by the four planes. (b) Find the cartesian equations of the four planes.


    2. Relevant equations

    When I drew this out I got something that reminded me of a tetrahedron. But it looked like there were two of them. So, in general I for part a i just did V = 1/6[abc] + 1/6[abd]
    I would write it but I don't have my papers with me.

    3. The attempt at a solution

    For the Cartesian equations I don't know what to use. Draw the picture and you will see that there are more then 4 faces so it is confusing me. How do I decifer which ones to use?
    Thanks
     
  2. jcsd
  3. Jun 28, 2013 #2

    verty

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    How did you get that formula for volume?
     
  4. Jun 28, 2013 #3
    When I drew the picture it looked to me like two tetrahedrons. So I just did 1/6[abc]+1/6[abd]. Because the tetrahedron is 1/6th the parrelliped formed when you do a tripple scalar product.
     
  5. Jun 28, 2013 #4

    LCKurtz

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    If you connect the four points all the ways possible you get one tetrahedron with four faces. But you don't use the position vectors to the points in your calculations. You have to use vectors along the sides.
     
  6. Jun 28, 2013 #5
    Ah. So it isn't two then. How did you know that then? I connected them but obviously wrong. That is my problem.
     
  7. Jun 28, 2013 #6

    haruspex

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    If O is the origin, the given vectors are OA, OB, OC, OD. How would you obtain the vector AB from OA and OB?
     
  8. Jun 28, 2013 #7
    OK, I did

    c-a = < 1,-2,5> - < 4,-3,-3> = < -3,1,8>
    b-a = < 7 , 5 , -2 > - < 4,-3,-3> = < 3, 8 ,1>
    d-a = <-4,6,-1> - < 4,-3,-3> = < 8,9,2>


    Then I did the box product and divided by 6.

    < -3,1,8> dot < 8,9,2> cross < 3, 8 ,1> = < -3(9-16) -1(8-6) + 8(64-27)> = 105/2 units^3

    V = [ (c-a)(d-a) (b-a)] = 315/6 = 105/2 units^3


    ?
     
  9. Jun 28, 2013 #8
    Hey this is just a suggestion but check the distance between the points, try to find some type of symmetry which can produce the vol
     
  10. Jun 28, 2013 #9

    haruspex

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    Check that.
     
  11. Jun 28, 2013 #10
    <-8,9,2>

    Thanks. Is the volume correct. The other parts
     
  12. Jun 28, 2013 #11

    LCKurtz

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    Don't you have ##a## wrong in all three?
     
  13. Jun 28, 2013 #12

    haruspex

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    Yes, I didn't check that. I believe there are also errors in the triple product calculation.
     
  14. Jun 29, 2013 #13
    I wrote it wrong origonally it should be a =<4,-3,-3> not a = <-4,-3,-3> Sorry, all the others are correct I double checked though. And I redid the calculation and I got 693/6 units cubed = 231/2 units^3.
     
  15. Jun 29, 2013 #14

    haruspex

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    That's what I got.
     
  16. Jun 29, 2013 #15
    Very cool. I will work on part b now. Thank you for the help haruspex and LCKurtz
     
  17. Jun 29, 2013 #16
    OK, I think I worked out part b
    I will show how I found each of the 4 planes. But in general I need a vector perpendicular to the plane, general points X = < x,y,z> and a point on the plane for each.


    (b-a) = <7,5,-2> - < 4,-3,-3> = <8,8,1>

    (d-a) = <-8,9,2>

    (b-a) cross (d-a) = <7,-14,91>
    A point on the plane is <-4,6,-1> = d

    Then,
    <x,y,z> dot <7,-14,91> = <-4,6,-1> dot <7,-14,91>

    7x -14y+91z = -203
    implies -x +2y -13z = 29

    The next one

    (c-a) = <1,-2,5> - < 4,-3,-3> = <-3,1-8>

    (b-a) = <3,8,1>

    (c-a) cross (b-a) = <-63,27,-27>
    A point on the plane is <1,-2,5> = c

    Then,
    <x,y,z> dot <-63,27,-27> = <1,-2,5> dot <-63,27,-27>
    -63x +27y-27z = -252
    implies 7x -3y+ 3z = 28


    The next one

    (c-a) = <1,-2,5> - < 4,-3,-3> = <-3,1-8>

    (d-a) = <-8,9,2>

    (c-a) cross (d-a) = <-70,-58,-19>
    A point on the plane is <1,-2,5> = d

    Then,
    <x,y,z> dot <-70,-58,-19> = <1,-2,5> dot <-70,-58,-19>

    implies 70x +58y +19z = 49


    The last one

    (c-a) = <1,-2,5> -<-4,6,-1> = <5,-8,6>
    (b-a) = <7,5,-2> - <-4,6,-1> = <11,-1,-1>


    (c-a) cross (b-a) = <14,71,83>

    14x +71y +83z = 287


    Think that is right.
     
    Last edited: Jun 29, 2013
  18. Jun 29, 2013 #17

    haruspex

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    <3,8,1>, which is what you seem to have used later
    Another typo?
    You can check the answers easily enough by seeing whether the given points satisfy the equations for the planes they're in.
     
  19. Jun 29, 2013 #18
    Yeah it was a typo. How do you mean check? Process seem right?
     
  20. Jun 30, 2013 #19

    haruspex

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    Yes, the method's fine.
    If you have an equation for a plane that is supposed to pass through A, B, C, you can substitute the coordinates of A for x, y, z in the equation and see whether the point satisfies the equation. Just a check that you've made no arithmetic errors.
     
  21. Jun 30, 2013 #20
    I found one so far I edited it. I will double check them before again before I have to submit it.
    Thanks
     
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