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Math Help - Alternative solution for problem on Vectors

  1. #1
    Junior Member
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    Alternative solution for problem on Vectors

    Hello everyone,

    Could anyone please tell me how to solve the following problem without knowledge of dot product or cross product?

    Thank you very much.

    ---

    1. If  |\vec x|= 11, |\vec y| = 23, and  |\vec x - \vec y | = 30 , find  |\vec x + \vec y|.

    ---

    My current solution:

     30 = \sqrt{11^2 + 23^2 - 2(11)(23)\cos \theta}  \Rightarrow \cos \theta = -0.49407

    Since  |\vec x + \vec y| = \sqrt{x^2 + y^2 + 2xy\cos \theta} ,  |\vec x + \vec y| = 20.

    As seen, my current solution involves the fact that  \vec x \cdot \vec y = |\vec x||\vec y|\cos \theta .
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, scherz0!

    Could anyone please tell me how to solve the following
    problem without knowledge of dot product or cross product?

    1. If |\vec x|= 11,\;|\vec y| = 23, and |\vec x - \vec y | = 30 , find |\vec x + \vec y|.

    It can be solved with just the Law of Cosines.


    Code:
        D * * * * * * * * * * C
           *  *              *
            *     *   30      *
          23 *        *        *
              *           *     *
               * 119.6       *  *
              A * * * * * * * * * * B
                        11

    I used this variation to find angle A.

    . . \cos A \:=\:\frac{11^2 + 23^2 - 30^2}{2(11)(23)} \:=\:-0.494071146 \quad\Rightarrow\quad A \:\approx\:119.6^o

    Then in parallelogram ABCD,\;\angle B \,=\,60.4^o


    Code:
        D * * * * * * * * * * C
           *              *  *
            *           *     *
             *        *        * 23
              *     *           *
               *  *        60.4 *
              A * * * * * * * * * * B
                         11

    In \Delta ABC\!:\;\;AC^2 \;=\;11^2 + 24^2 - 2(11)(23)\cos60.4^o \;=\;400.0654155

    Therefore: . |\vec x + \vec y| \;=\;AC \;\approx\;20

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