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Calc Vector (dot product)

  1. May 19, 2009 #1
    1. The problem statement, all variables and given/known data
    The cosine of the angle between vector a and vector b is 4/21. Find p.

    vector a = 6i + 3j - 2k
    vector b = -2i + pj - 4k

    2. Relevant equations

    (Vector a) . (Vector b) = abcos(theta)
    (Vector a) . (Vector b) = axbx + ayby + azbz
    mag(a) = root(ax^2 + ay^2 + az^2)

    3. The attempt at a solution

    mag(a) = root(36 + 9 + 4) = 7
    mag(b) = root(4 + p^2 + 16) = root(p^2 + 20)

    (Vector a) . (Vector b) = -12 + 3p + 8 = 3p - 4

    (Vector a) . (Vector b) = abcos(theta)
    3p - 4 = 7[root(p^2 + 20)](4/21)
    3p - 4 = [4/3][root(p^2 + 20)]

    Square both sides

    9p^2 - 24p + 16 = 16/9p^2 + 320/p

    Collect like terms, and set equation to 0

    (65/9)p^2 - 24p - 176/9

    Sub into quadratic equation

    p = 4
    p = -44/65

    [CORRECT ANSWER: P = 4]

    Thanks for the help ;)
     
    Last edited: May 19, 2009
  2. jcsd
  3. May 19, 2009 #2

    jbunniii

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    What you have done so far looks OK. Why are you surprised to obtain a quadratic? There are two angles such that cos(theta) = 4/17, so you should expect that there will be two valid solutions.
     
  4. May 19, 2009 #3
    Well its not so much the quadratic that troubles me, as the solution. When i solve it i get a number with decimals, and it is not near the final answer which the book gives. (4)

    If someone could finish off my solution that would be great, as when i did it, it was incorrect.
     
  5. May 19, 2009 #4

    Cyosis

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    Show your work so we can see where you went wrong.
     
  6. May 19, 2009 #5
    I just tried it again, edited the first post, no quadratic but i could have done something wrong.
     
  7. May 19, 2009 #6

    Cyosis

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    There will be a quadratic. You have a p on one side and a sqrt(p) on the other side. To solve it you raise both sides to the power two. As a result you will have a p^2 on one side and a p on the other. If you don't show us your work we can't tell you where you go wrong.
     
  8. May 19, 2009 #7
    I showed my work in the first post
     
  9. May 19, 2009 #8

    Cyosis

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    Seems I am going blind, sorry. I'll check it out.

    Edit: You edited didn't you!
     
  10. May 19, 2009 #9

    Cyosis

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    You're squaring incorrectly, [itex](3p - 4)^2=9p^2-24p+16[/itex].
     
  11. May 19, 2009 #10
    heh, yeah i did square it wrong that time... but on my paper i did it jsut like that, but after i use the quadratic formula on the final outcome, its still wrong... maybe you have better luck?
     
  12. May 19, 2009 #11

    Cyosis

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    Well I have calculated it and i get p=-44/65 and p=4. It is not me who needs to practice this though so regardless whether I do it correctly or not we still need to see where you went wrong! It is a bit of a messy calculation but still.
     
  13. May 19, 2009 #12
    EDIT: nvm i got it, dumb mistake.
     
    Last edited: May 19, 2009
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