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Thread: Vectors

  1. #1
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    Post Vectors

    justift whether ( u+v)dot(u cross v)=0 for all non-zero vectors u and v.
    need help with this question, managed to finish rest in problem set. thanks for the help
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  2. #2
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    let u = ai + bj + ck

    v = di + ej + fk

    u + v = (a+d)i + (b+e)j + (c+f)k

    u X v = (bf - ce)i - (af - cd)j + (ae - bd)k

    (u + v) dot (u X v) =

    (a + d)(bf - ce) - (b + e)(af - cd) + (c + f)(ae - bd)

    (abf + bdf - ace - cde) - (abf + aef - bcd - cde) + (ace + aef - bcd - bdf)

    abf + bdf - ace - cde - abf - aef + bcd + cde + ace + aef - bcd - bdf

    believe all these cancel out to 0.
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  3. #3
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    $\displaystyle \left( {A + B} \right) \cdot C = A \cdot C + B \cdot C$

    $\displaystyle A \cdot \left( {A \times B} \right) = 0\,\& \,B \cdot \left( {A \times B} \right)= 0$

    Applying these two will answer your question.
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  4. #4
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    It also helps to know that $\displaystyle <x_1,y_1,z_1>\cdot(<x_2,y_2,z_2>\times<x_3,y_3,z_3 >)$, the "triple product", can be written in the determinant form:
    $\displaystyle \left|\begin{array}{ccc}x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3\end{array}\right|$

    In the case of either $\displaystyle u\cdot(u\times v)$ or $\displaystyle v\cdot(u\times v)$, two of the rows of the determinant are the same and so the determinant is 0.
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