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Angles between vectors

  1. Jul 30, 2008 #1
    how is this done? say angles between 2i -3j -k and 4i+2k-3j got an exam coming up soon and cant find the formula for it anywhere help please?
     
  2. jcsd
  3. Jul 30, 2008 #2
    and also is the cross product just the determinant?
     
  4. Jul 30, 2008 #3

    Dick

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    A.B=|A||B|cos(t) (dot product). And you can USE a determinant to find the cross product. I wouldn't think of it as being primarily a determinant, it's a vector.
     
  5. Jul 30, 2008 #4
    so angles between is just |A||B|cos(t) and cross product is just using the det so for A=2i-3j-k and B=4i+2k-3j i would have angle between of root 14 root 29 cos(T) would that just be the answer? still confused on that one and for the cross product would the answer be 12i-2j+16k?
     
  6. Jul 30, 2008 #5

    Dick

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    Yes. A.B=sqrt(14)*sqrt(29)*cos(T). So now if you can figure out A.B then you can find cos(T) and then T. I don't get what you get for the cross product. For example for the i component I figure (-3)(2)-(-3)(-1)=(-9).
     
  7. Jul 30, 2008 #6
    not a clue where that determinants coming from, how do you do the determinant of a 3 by 2 matrix
     
  8. Jul 30, 2008 #7

    Dick

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    There is no such thing as the determinant of a 3x2 matrix. Check out http://en.wikipedia.org/wiki/Cross_product
     
  9. Jul 30, 2008 #8
    ahah got ya! right now just give me 5 minutes to get stuck on finding A.B and i'l be right back!
     
  10. Jul 30, 2008 #9
    sorry i still have no clue where your plucking them numbers from for the determinant isnt the determinant AD-BC but on your answer you seen to have done AB-CD
     
  11. Jul 30, 2008 #10

    Dick

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    What are the second and third rows of your determinant? I can guess what you are doing wrong. Notice the components in vector B are written in the order ikj. Not ijk. I was wondering if it's a typo, but you repeated it twice, so I'm guessing that's how it's written in the problem.
     
  12. Jul 30, 2008 #11
    ye just me being dum thats a typo ment to be ijk so it is AD-BC ye?
     
  13. Jul 30, 2008 #12

    Dick

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    I get [2, - 3, - 1] x [4, 2, - 3]=[11, 2, 16], if that's the problem you are trying to solve...
     
  14. Aug 2, 2008 #13
    Well, if the two vectors u and v that are nonzero (in [tex]R^{2}[/tex] or [tex]R^{3}[/tex]) the angle between them is given by the formula;

    [tex]\theta = cos^{-1}\left(\frac{u.v}{\left\|u\right\| \left\|v\right\|}\right)[/tex]

    In your question you have u = [tex]\left(2, - 3, - 1\right)[/tex] & v = [tex]\left(4, 2, -3)[/tex]

    But you have to put your equation system in the form of an argumented matrix!
     
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