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Are these vectors orthogonal?

  1. Feb 2, 2008 #1

    tony873004

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    1. The problem statement, all variables and given/known data
    Determine whether the given vectors are orthogonal, parallel, or neither.

    2. Relevant equations
    [tex]
    \cos \theta = \frac{{\overrightarrow {\rm{a}} \cdot \overrightarrow {\rm{b}} }}{{\left| {\overrightarrow {\rm{a}} } \right|\left| {\overrightarrow {\rm{b}} } \right|}}\,\, \Rightarrow \,\,\theta = \cos ^{ - 1} \left( {\frac{{\overrightarrow {\rm{a}} \cdot \overrightarrow {\rm{b}} }}{{\left| {\overrightarrow {\rm{a}} } \right|\left| {\overrightarrow {\rm{b}} } \right|}}} \right)
    [/tex]

    3. The attempt at a solution
    [tex]
    \begin{array}{l}
    \overrightarrow {\rm{a}} = 2i + 6j - 4k,\,\,\,\,\overrightarrow {\rm{b}} = - 3{\rm{\hat i}} - {\rm{9\hat j}}\,{\rm{ + }}\,{\rm{6\hat k}} \\
    \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overrightarrow {\rm{a}} \cdot \overrightarrow {\rm{b}} = \left( {2 \cdot - 3} \right) + \left( {6 \cdot - 9} \right) + \left( { - 4 \cdot 6} \right) = - 6 + \left( { - 54} \right) + \left( { - 10} \right) = - 58 \ne 0{\rm{__not_ orthogonal}} \\
    \\
    \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos ^{ - 1} \left( {\frac{{\overrightarrow {\rm{a}} \cdot \overrightarrow {\rm{b}} }}{{\left| {\overrightarrow {\rm{a}} } \right|\left| {\overrightarrow {\rm{b}} } \right|}}} \right) = \cos ^{ - 1} \left( {\frac{{ - 58}}{{\sqrt {2^2 + 6^2 + \left( { - 4} \right)^2 } \sqrt {\left( { - 3} \right)^2 + \left( { - 9} \right)^2 + 6^2 } }}} \right) = \\
    \\
    \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos ^{ - 1} \left( {\frac{{ - 58}}{{\sqrt {4 + 36 + 16} \sqrt {9 + 81 + 36} }}} \right) = \\
    \\
    \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos ^{ - 1} \left( {\frac{{ - 58}}{{\sqrt {56} \sqrt {126} }}} \right) \approx 133.67^\circ \ne 0^\circ \,{\rm{not_ parallel,}}\,\, \\
    \end{array}
    [/tex]

    But the back of the book says parallel.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 2, 2008 #2

    Dick

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    -4*6=(-24). -6-54-24=-84. That's the dot product of a and b.
     
  4. Feb 2, 2008 #3

    tony873004

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    :redface:oops. back to 2nd grade for me:rofl:. Thanks, Dick.
     
  5. Feb 2, 2008 #4

    Dick

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    BTW you can also see that they are parallel by inspection. (-3/2)*a=b.
     
  6. Feb 2, 2008 #5
    [tex]
    \begin{align*}
    \textbf{a}\cdot\textbf{b} = 0 & &\Rightarrow & &\textbf{a} \perp \textbf{b}\\
    \textbf{a}\times\textbf{b} = \textbf{0} & &\Rightarrow & &\textbf{a} \parallel \textbf{b}
    \end{align*}
    [/tex]
     
  7. Feb 2, 2008 #6

    tony873004

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    Another student did it this way, but I don't see it. Where does -3 and 2 come from?

    Cross product is next chapter, but thanks, that gives me a good preview of what's to come!
     
  8. Feb 2, 2008 #7
    Solve each of these for x:

    [tex]
    \begin{align*}
    2x &= -3\\
    6x &= -9\\
    -4x &= 6
    \end{align*}
    [/tex]
     
  9. Feb 2, 2008 #8

    tony873004

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    I still don't get it. How does solving them demonstrate that a and b are parallel? Sorry, but the book does not explain this method.
     
  10. Feb 2, 2008 #9

    tony873004

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    now I get it. The -3 is factored out of b, and the 2 is factored out of a, The remaining parts of a and b are equal, therefore parallel.
     
  11. Feb 2, 2008 #10
    Exactly. In order for two vectors to be parallel, one must be a constant multiple (in this case [tex]-\frac{3}{2}[/tex]) of the other.
     
  12. Feb 2, 2008 #11

    tony873004

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    Now I'm glad I made my original dumb mistake of 6*4=10, or I wouldn't have posted and learned this easier method.
     
  13. Feb 3, 2008 #12
    They say the only way to learn is to make mistakes. ;)
     
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