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Homework Help: Vector angle

  1. Jan 26, 2010 #1
    1. Three points P, Q and R have position vectors p, q and r respectively, where:
    p=8i+11j, q=7i-5j and r=2i+4j.
    Write down the vectors QP and QR and show that they are not perpendicular. Hence determine the angle PQR.




    2. |QP|.|QR|cos(theta)




    3. QP= QO+OP=i+16j
    QR=QO+OR=-5i+9j
    |QP|=root257
    |QR|=root106

    If the vectors are not perpendicular then a.b=0
    a.b= QP=i+16j x QR=-5i+9j = (1)(-5)+(16)(9)=-5+144=139 - not perpendicular

    a.b=|QP|.|QR|cos(theta)
    cos(theta)=a.b/|QP|.|QR|=139/root257 x root106 = 32.6


    Could anybody check to see how pathetic this attempt is, please?
     
  2. jcsd
  3. Jan 27, 2010 #2

    tiny-tim

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    Hi lemon! :smile:

    Yes, that looks fine, except for the very last bit (the 32.6). :wink:
     
  4. Jan 27, 2010 #3
    errr. cant see it
    thats the same figure i keep getting out.

    ?
     
  5. Jan 27, 2010 #4

    tiny-tim

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    i don't know how you get that :redface:

    139/√257√106 is approximately 139/16*10 < 1 :confused:
     
  6. Jan 27, 2010 #5
    yeah but that is cos(theta), right?
    so to find theta i need to inverse cos - 32.6
     
  7. Jan 27, 2010 #6

    tiny-tim

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    But it isn't 32.6! :cry:
     
  8. Jan 27, 2010 #7
    it is on my calculator. I get 0.8421613497
    inverse cos = 32.63093847
     
  9. Jan 27, 2010 #8
    errr. cant see it
    thats the same figure i keep getting out.

    ?
     
  10. Jan 27, 2010 #9

    tiny-tim

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    ohhh! you wrote cos(theta) = 32.6 …
    yes, theta = 32.6º is fine. :smile:
     
  11. Jan 27, 2010 #10
    thanks tT
     
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