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Homework Help: Parallel vectors and scalar product rule

  1. Sep 11, 2006 #1
    hey,
    ive been given a problem where vector a = 2i + 3j and vector b = [tex]\lambda[/tex]i + 12j and also told that these vectors are parallel of each other. i understand since the vectors are parallel of each other, the angle between them would be equal to zero, thus i could apply the scalar product rule to help solve this problem, which in this case i did.

    My working:

    [tex]
    \begin{array}{c}
    {\bf{a}} = 2{\bf{i}} + 3{\bf{j}} \\
    {\bf{b}} = \lambda {\bf{i}} + 12{\bf{j}} \\
    {\bf{a}}\parallel {\bf{b}} \\
    \theta = 0 \\
    \cos \theta = 1 \\
    {\bf{a}} \cdot {\bf{b}} = \left| {\bf{a}} \right|\left| {\bf{b}} \right|\cos \theta \\
    \left( {2{\bf{i}} + 3{\bf{j}}} \right) \cdot \left( {\lambda {\bf{i}} + 12{\bf{j}}} \right) = \left( {\sqrt {13} } \right)\left( {\sqrt {144 + \lambda ^2 } } \right)\left( 1 \right) \\
    2\lambda + 36 = \left( {\sqrt {13} } \right)\left( {\sqrt {144 + \lambda ^2 } } \right) \\
    \left( {2\lambda + 36} \right)^2 = 13\left( {144 + \lambda ^2 } \right) \\
    4\lambda ^2 + 144\lambda + 1296 = 1872 + 13\lambda ^2 \\
    - 9\lambda ^2 + 144\lambda - 576 = 0 \\
    x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} \\
    \lambda = \frac{{ - 144 \pm \sqrt {144^2 - 4\left( { - 9} \right)\left( { - 576} \right)} }}{{2\left( { - 576} \right)}} \\
    \lambda = \frac{1}{8} \\
    \end{array}
    [/tex]

    this is the wrong answer i know, as the correct answer would result in [tex]\lambda[/tex] being equal to eight, however i do not know how to obtain this answer, any help is greatly appriciated, thanks. sorry for the many post lately but i have a test comeing up which i realy wish to do well in to prove the teacher that i can do her subject. thanks once again,
    Pavadrin
     
  2. jcsd
  3. Sep 11, 2006 #2
    Check your working, it should be OK. The second last equation you quoted is right, but you didn't put the numbers in correctly.

    Alternately, you can divide your quadratic by -9, then factorise it.
     
  4. Sep 11, 2006 #3
    okay thanks for your help ill try that right now
     
  5. Sep 11, 2006 #4
    hey that worked thanks~! i got the answer of 8 and -8. can i ask you why it didn't work when i used the quadratic formula? thanks
     
  6. Sep 11, 2006 #5

    Astronuc

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    Staff Emeritus
    Science Advisor

    [tex]\lambda = \frac{{ - 144 \pm \sqrt {144^2 - 4\left( { - 9} \right)\left( { - 576} \right)} }}{{2\left( { - 9} \right)}}[/tex]
    One divided by 2c instead of 2a.

    As Tomsk suggested, dividing through by -9 would have simplied the solution.

    [tex]\lambda ^2 - 16\lambda + 64 = 0[/tex]

    Two vectors are parallel if one is a multiple of the other, i.e.

    a i + b j = [itex]\lambda[/itex]a i + [itex]\lambda[/itex]b j
     
    Last edited: Sep 11, 2006
  7. Sep 11, 2006 #6
    oh i see thanks, i feel so stupid now :blushing:
    thanks once again,
    Pavadrin
     
  8. Sep 11, 2006 #7

    0rthodontist

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    Science Advisor

    It would have been a lot simpler to use the fact that parallel vectors are multiples of each other from the start. You have:
    2xi + 3xj = yi + 12j
    3x = 12
    x = 4
    2x = y
    y = 8
     
  9. Sep 12, 2006 #8
    okay thanks, i was thinking that perhaps there was an easier way of solving the question but it never quite clicked, thanks
     
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