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Math Help - Vector Problem

  1. #1
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    Vector Problem

    So I have this small problem. Two vectors A and B added together give the vector S. Show that S (which I'm assuming is the scalar of S) is equal to (A^2 + B^2 + 2AB Cos (theta) ) / 2 remembering that S . S = S^2 and S= A+B. I'm really not sure how to make a start on it. I'm thinking that it has something to do with the rule a.b= [a][b]cos theta, but I can't seem to make the first leap. Any suggestions
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  2. #2
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    Hi,

    I'm not sure the question quite makes sense - vectors can't usually be "squared" - you can multiply them with the dot or cross product, but the notation in your question suggests otherwise.

    Does the answer refer to the magnitude of the two vectors?
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  3. #3
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    I think your right about the squaring of the vector, not being possible. However I don't theink the squraed parts are refering to the vectors. They weren't in bold, which I believe is the usual convention for vectors. So I'm assuming that the are the values or the scalar. ie Ai + Aj or possible the scalar product. Sorry I'm really at a loss about this stuff.
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  4. #4
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    It follows at once from this.
    S = A + B \Rightarrow \left( {S \cdot S} \right) = \left( {A + B} \right) \cdot \left( {A + B} \right) = A \cdot A + 2A \cdot B + B \cdot B
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  5. #5
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    So the complete derivation would be something like this:

    S.S= (A+B).(A+B)
    => A.A + B.B + 2(A.B)

    => A.A = [A][A] cos theta = [A]^2 because cos 0 = 1
    => B.B = [B][B] cos theta = [B]^2 as above
    => 2(A.B) = 2[A][B] cos theta

    putting all the parts together

    A^2 + B^2 + 2AB cos theta

    then taking the square root bcause S.S = S^2

    giving S = (A^2 + B^2 + 2AB cos theta)^1/2

    Is that how it goes? Thanks for the help
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