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Math Help - vector problem - exam revision

  1. #1
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    vector problem - exam revision

    I have 3 vectors a b and c such that a+b+c=0. This implies they are all the same length as they finish where they start. The question then asks me to prove that axb = bxc = cxa. This seems obvious as if the values are the same then the cross products will be too. However i need to prove this and am not sure how to. Any help would be appreciated.
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  2. #2
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    Quote Originally Posted by schteve View Post
    I have 3 vectors a b and c such that a+b+c=0. This implies they are all the same length as they finish where they start. <<<<<<<< are you sure?
    The question then asks me to prove that axb = bxc = cxa. This seems obvious as if the values are the same then the cross products will be too. However i need to prove this and am not sure how to. Any help would be appreciated.
    If the vector sum equals zero you have a closed curve.

    Your 3 vectors form a triangle which determines allways a plane. The cross-product of such 2 vectors is the normal vector of the plane.
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  3. #3
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    Hello, schteve!

    I have 3 vectors \vec a, \vec b, \vec c such that: . a+b+c\:=\:0

    This implies they are all the same length . . . . not true
    and they finish where they start.

    The question then asks me to prove that: . \vec a\times\vec b \:=\: \vec b\times\vec c \:=\: \vec c\times\vec a

    Earboth showed you why the three cross products have the same direction.
    Now we must prove they have the same magnitude (length).


    The three vectors form a triangle.
    Code:
                C
                *
               *  *
          |b| *     *  |a|
             *        *
            *           *
           *              *
        A *  *  *  *  *  *  * B
                  |c|
    Law of Sines: . \frac{|a|}{\sin A} \:=\:\frac{|b|}{\sin B} \quad\Rightarrow\quad|a|\sin B \:=\:|b|\sin A


    Multiply both sides by |c|\!:\;\;|a||c|\sin B \:=\:|b||c|\sin A .[1]



    Recall that the magnitude of a cross product is the area
    . . of the parallelogram determined by the two vectors.
    Code:
              * - - - - - - *
             /:::::::::::::/
          v /:::::::::::::/
           /:θ:::::::::::/
          * - - - - - - *
                 u
    That is: . \text{Area} \:=\:|\vec u \times \vec v| \:=\:|u||v|\sin\theta


    Hence: . \begin{array}{ccc}|\vec a \times \vec c| &=& |a||c|\sin B \\ \\[-4mm] |\vec b \times \vec c| &=& |b||c|sin A \end{array}


    According to [1], these two magnitudes are equal: . |\vec a \times \vec c| \:=\:|\vec b \times \vec c|


    In a similar fashion, we can prove that all three magnitudes are equal.


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  4. #4
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    Thank you to both of the above respondents - it seems obvious now and i feel better for now knowing the answer but also slightly more stupid at the same time because i didn't get it straight away.
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