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Mu and lambda are parameters

  1. Nov 6, 2012 #1
    the lines l2[/SUB and l3
    are given by

    l1 : r = 10i + j + 9k + μ(3i+j+4k)

    l2: x= (y+9)/2= (z-13)/-3

    l3 r= -3i-5j-4k + λ(4i+3j+k)

    where mu and lambda are parameters.

    d) Show that AC=BC

    A(4,-1,1) this was given, I calculated and got B as (5,-1,-2) and C (1,-2,-3)

    This I what I did so far

    2502qvs.png


    I found AC ( -3,-1,-4)

    BC(-4,-1,-1)

    I also found the mods of AC and BC to see if they were equal but, they're not.
     
    Last edited by a moderator: Feb 6, 2013
  2. jcsd
  3. Nov 6, 2012 #2

    SammyS

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    Re: Vectors


    It looks like you haven't given us the whole problem, although we can read between the lines to figure out most of what's missing:
    Two of the lines intersect at point A, which is at (4,-1,1).

    Two of the lines intersect at point B.

    Two of the lines intersect at point C.​
    ...

    Now,

    Calculate vector AC and see if its magnitude matches the magnitude of either of the other two vectors.

    Added in Edit:

    B is at (5,1,-2) not (5,-1,-2)
     
    Last edited: Nov 6, 2012
  4. Nov 6, 2012 #3
    Re: Vectors

    it doesn't Magnitude of AC is root 26

    while magnitude of B is root 30 and C is root 14.

    sorry!!!
    I found BC and the magnitude of it is root 26! Thank you for pointing out the mistake in the coordinates of B!!!
     
  5. Nov 6, 2012 #4

    SammyS

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    Re: Vectors

    Good deal !
     
  6. Nov 6, 2012 #5
    Re: Vectors

    Ummm there's is one more part to the question , It says write down the coordinates of the point D on AB such that CB is perpendicular to AB.

    I'm not sure what to do... do I do something with the dot product?
     
  7. Nov 6, 2012 #6

    SammyS

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    Re: Vectors

    Well, using the dot product will be a good way to check your answer, but think about the overall situation.

    Triangle ABC has
    length AC = length BC .​
    Where should point D be on AB so that ... looks like a typo ????
     
  8. Nov 6, 2012 #7
    Re: Vectors

    write down the coordinates of the point D on AB such that CD is perpendicular to AB.
     
  9. Nov 7, 2012 #8

    SammyS

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    Re: Vectors

    OK! That makes sense.


    Now,

    Triangle ABC has
    length AC = length BC .​
    Where should point D be on AB so that CD is perpendicular to AB.
     
  10. Nov 7, 2012 #9
    Re: Vectors

    In the middle of AB?
     
  11. Nov 7, 2012 #10

    SammyS

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    Re: Vectors

    Yes. D is the midpoint of AB .
     
  12. Nov 7, 2012 #11
    Re: Vectors

    LOL so wait I can find find AD then just get D and that's it?
     
  13. Nov 7, 2012 #12

    SammyS

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    Re: Vectors

    Yes. That's it.

    You can check your result by seeing if the appropriate dot product is zero.
     
  14. Nov 7, 2012 #13
    Re: Vectors

    Thank you again for the help. Homework is now finished!
     
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