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Vector geometry - determinant proof

  1. Nov 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Exercise 44 - In the picture attached


    2. Relevant equations

    HINT - expand the expression for n and plug the result into equation (70), then use equation (63)

    n=(C-B)X(B-A)

    n dot (r - A) = 0 (eq. 70)

    A dot (B X C) = det {A B C} (eq. 63)

    [PLAIN]http://postimage.org/image/68xx59akh/ [Broken][/PLAIN]

    [PLAIN]http://postimage.org/image/rzbcga401/ [Broken][/PLAIN]

    [PLAIN]http://postimage.org/image/wtauu6jv9/ [Broken][/PLAIN]


    3. The attempt at a solution

    Like the hint said, I expanded the expression for n , which is (C-B)X(B-A) and I got a giant mess. I have no idea if I did it right or am doing it right or what to do here
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Nov 12, 2012 #2

    tiny-tim

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    hi bossman007! :smile:
    it should come out very simple :confused:

    show us what you did :smile:
     
  4. Nov 12, 2012 #3
    [PLAIN]http://postimage.org/image/j5wfm5rjr/ [Broken][/PLAIN]
     
    Last edited by a moderator: May 6, 2017
  5. Nov 12, 2012 #4

    tiny-tim

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    ohhh! that's why it's a giant mess! :biggrin:

    no, keep it simple

    use the distributive law …

    (C - B) x (B - A) … ? :wink:
     
  6. Nov 12, 2012 #5
    many thanks :D,

    what is the distributive law for 3 vectors?
     
  7. Nov 12, 2012 #6

    tiny-tim

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    P x (Q + R) = (P x Q) + (P x R) :wink:
     
  8. Nov 12, 2012 #7
    thank you :D

    so that means in my case I do B x (C-A) ?
     
  9. Nov 12, 2012 #8
    is this right?
     
  10. Nov 13, 2012 #9

    tiny-tim

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    (just got up :zzz:)
    nooo …

    how did you get that result? :confused:
     
  11. Dec 17, 2012 #10
    What should this be exactly? I'm not sure I understand either
     
  12. Dec 17, 2012 #11

    HallsofIvy

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    ??? There is NO "C- A" in the expression you want to expand!

    [itex](C- B)\times (B- A)= C\times(B- A)- B\times(B- A)[/itex]
    That's one step, now do it again:
    [itex]C\times B- C\times A- B\times B+ B\times A[/itex]
    and, of course, [itex]B\times B= 0[/itex].
     
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