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Math Help - Verctors in R3: cross product

  1. #1
    Junior Member Freaky-Person's Avatar
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    Verctors in R3: cross product

    I have this problem:

    d,e,f are vectors in R3

    (d\times f)\bullet e

    I don't need to hear the solution, I just need some pointers about where to start. I mean, I did the cross multiplication and now I have letters everywhere.

    Is there a cross product rule I'm missing? or a dot product rule?
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  2. #2
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    Quote Originally Posted by Freaky-Person View Post
    I have this problem:
    d,e,f are vectors in R3
    (d\times f)\bullet e
    I don't need to hear the solution, I just need some pointers about where to start. I mean, I did the cross multiplication and now I have letters everywhere.
    Is there a cross product rule I'm missing? or a dot product rule?
    First you find g=d\times f.
    Then you find g\cdot e.
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  3. #3
    Junior Member Freaky-Person's Avatar
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    using this, should I still have a couple of lines full of letters? Cause that's what I'm getting to >.>
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  4. #4
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    Quote Originally Posted by Freaky-Person View Post
    using this, should I still have a couple of lines full of letters? Cause that's what I'm getting to >.>
    \left( {d_2 f_3  - d_3 f_2 } \right)e_1  - \left( {d_1 f_3  - d_3 f_1 } \right)e_2  + \left( {d_1 f_2  - d_2 f_1 } \right)e_3
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  5. #5
    Junior Member Freaky-Person's Avatar
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    ooooh i,j,k disappear! And oh! Wow! I totally get it now! ha...

    Thank you!
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  6. #6
    Junior Member Freaky-Person's Avatar
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    Quote Originally Posted by Plato View Post
    \left( {d_2 f_3  - d_3 f_2 } \right)e_1  - \left( {d_1 f_3  - d_3 f_1 } \right)e_2  + \left( {d_1 f_2  - d_2 f_1 } \right)e_3
    waaaait why is it \left( {d_2 f_3  - d_3 f_2 } \right)e_1  - \left( {d_1 f_3  - d_3 f_1 } \right)e_2? That minus, is it a typo?
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  7. #7
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    Quote Originally Posted by Freaky-Person View Post
    waaaait why is it \left( {d_2 f_3  - d_3 f_2 } \right)e_1  - \left( {d_1 f_3  - d_3 f_1 } \right)e_2? That minus, is it a typo?
    NO!
    \left| {\begin{array}{ccc}<br />
   i & j & k  \\<br />
   {d_1 } & {d_2 } & {d_3 }  \\<br />
   {f_1 } & {f_2 } & {f_3 }  \\<br /> <br />
 \end{array} } \right| = \left( {d_2 f_3  - d_3 f_2 } \right)i - \left( {d_1 f_3  - d_3 f_1 } \right)j + \left( {d_1 f_2  - d_2 f_1 } \right)k
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  8. #8
    Junior Member Freaky-Person's Avatar
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    my book and wikipedia say otherwise.

    Wikipedia:
    This equation is the sum of nine simple cross products. After all the multiplication is carried out using the basic cross product relationships between i, j, and k defined above,

    a b = a1b1(0) + a1b2(k) + a1b3(−j) + a2b1(−k) + a2b2(0) + a2b3(i) + a3b1(j) + a3b2(−i) + a3b3(0).

    This equation can be factored to form

    a b = (a2b3 − a3b2) i + (a3b1 − a1b3) j + (a1b2 − a2b1) k = (a2b3 − a3b2, a3b1 − a1b3, a1b2 − a2b1).
    Oh wait, I see it. I go: 23, 31, 12. You go: 23, 13, 12.
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  9. #9
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    Your book is not wrong. I am not wrong.
    You are wrong for not reading carefully.
     - \left( {d_1 f_3  - d_3 f_1 } \right) = \left( {d_3 f_1  - d_1 f_3 } \right)
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