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Math Help - Vector Dot Problem

  1. #1
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    Tehran
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    Vector Dot Problem

    -----
    hi .
    please help to solve : Vector Dot Problem

    if we know that \vec{A}.\vec{B_1} = \vec{A}.\vec{B_2} can we infer from that : \vec{B_1} = \vec{B_2} ?

    thanks in advance ...
    Last edited by CaptainBlack; December 22nd 2010 at 12:00 AM. Reason: remove routine invocation of a diety
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  2. #2
    MHF Contributor Unknown008's Avatar
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    No, I don't think so...

    \left(\begin{array}{c} 1 \\ 2 \\ 1 \end{array} \right). \left(\begin{array}{c} 2 \\ 1 \\ 2 \end{array} \right) =  \left(\begin{array}{c} 1 \\ 2 \\ 1 \end{array} \right) . \left(\begin{array}{c} 4 \\ 0 \\ 2 \end{array} \right)

    But

    \left(\begin{array}{c} 2 \\ 1 \\ 2 \end{array} \right) \neq \left(\begin{array}{c} 4 \\ 0 \\ 2 \end{array} \right)

    EDIT: I'm not sure of that... yet.

    EDIT2: Confirmed!
    Last edited by Unknown008; October 18th 2010 at 11:31 AM.
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  3. #3
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    No, assuming \vec{A} \neq \vec{0}.

    Let \vec{A} = <a_1,a_2>, \vec{B_1} = <u_1,u_2>, and \vec{B_2} = <v_1,v_2>. Then, \vec{A} \cdot \vec{B_1} = a_1u_1 + a_2u_2 and \vec{A} \cdot \vec{B_2} = a_1v_1 + a_2v_2.

    If \vec{A} \cdot \vec{B_1} = \vec{A} \cdot \vec{B_2}, then a_1u_1 + a_2u_2 = a_1v_1 + a_2v_2. The preceding equation cannot be solved; therefore, the answer is no.

    I found a disproof on Wikipedia: Dot product - Wikipedia, the free encyclopedia
    Last edited by NOX Andrew; October 18th 2010 at 11:23 AM. Reason: Flawed reasoning
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Hm... maybe I'm wrong... or is my example an exception?
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  5. #5
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    No, you are correct. My reasoning was flawed. Here is a disproof on Wikipedia: Dot product - Wikipedia, the free encyclopedia
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