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Dot and cross product?

  1. Aug 11, 2013 #1
    1. The problem statement, all variables and given/known data
    A=(x,3,1) ,B=(x,-x,2)
    Determine the value of x if the vector perpendicular to A and B is given by C=(10,-4,-4)


    2. Relevant equations



    3. The attempt at a solution
    Find A cross B , let A cross B be D . Then D cross C = zero (since they are perpendicular to both A and B) . This sounds logic but there will not be an answer for x. What is wrong?
    Then I tried to use dot product A dot C=0, and B dot C=0. Yet the answer of two x are different.
    Please guide. Thanks
     
  2. jcsd
  3. Aug 11, 2013 #2

    LCKurtz

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    You haven't shown your calculations. But perhaps there is no real value of x that works. Perhaps the problem is mis-printed or copied incorrectly.
     
  4. Aug 11, 2013 #3
    So the both ways to solve the problem are correct?
     
  5. Aug 11, 2013 #4

    LCKurtz

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    I didn't say that. I haven't worked the problem; that's your job. If you want me to comment on what you did you need to show your work.
     
  6. Aug 11, 2013 #5
    The first way
     

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  7. Aug 11, 2013 #6
    Sorry a, my camera not good.
     
  8. Aug 11, 2013 #7
    Please look at the picture, I don't know why I can't upload more picture.
    The second way, A dot C = 0 , x= 1.6
    B dot C = 0 , x= 0.4
    Thanks
     
    Last edited: Aug 11, 2013
  9. Aug 11, 2013 #8

    LCKurtz

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    I won't try to read upside down. But if your arithmetic is correct and you get different values for ##x##, that would tell you there is no solution. Like I said earlier, perhaps the problem is mis-printed or copied incorrectly.
     
  10. Aug 11, 2013 #9
  11. Aug 12, 2013 #10
    I would like to remark that your approach with the cross product is over-complicated. Once you found C = A x B, and assume that C is parallel with D, you can say C = k D, where k is some number. You do not need to cross-multiply C and D, you just need to confirm that C = k D is possible.
     
  12. Aug 12, 2013 #11
    Really easier. But I think the question have some problems. Thank you.
     
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