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Math Help - Vectors

  1. #1
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    Vectors

    Hi, does anyone know if there is/what the general formula for the addition of vectors is?

    In particular, I'm trying to work out the problem: Two vectors and have precisely equal magnitudes. In order for the magnitude of + to be larger than the magnitude of - by the factor n, what must be the angle between them?

    A friend offered the formula: sum^2 = a^2 + a^2 + a^2cos(t)
    diff^2 = a^2 + a^2 - a^2cos(t)

    However, I've never seen these formulas before and I don't understand the premise behind them so cannot determine if they are correct...

    Can someone please help explain these formulas? Or offer another method of solving this problem?

    Thank you!
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  2. #2
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    Quote Originally Posted by strgrl View Post
    Two vectors and have precisely equal magnitudes. In order for the magnitude of + to be larger than the magnitude of - by the factor n, what must be the angle between them?
    I find this question a bit vague.
    We assume that from the wording that n >1.
    In which case in order for \left\| {A + B} \right\| to be greater than \left\| {A - B} \right\| the angle between A~\&~B must be acute.
    (The length of A~\&~B is irrelevant to the problem.)

    Look at what follows.
    \left\| {A + B} \right\|^2  = \left\| A \right\|^2  + 2A \cdot B + \left\| B \right\|^2
    \left\| {A - B} \right\|^2  = \left\| A \right\|^2  - 2A \cdot B + \left\| B \right\|^2
    So in order for \left\| {A + B} \right\| > \left\| {A - B} \right\| it must be the case that A \cdot B > 0.
    That happens only if the angle between them is acute.
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  3. #3
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    [size=3]Hello, strgrl!

    Two vectors \vec A \text{ and }\vec B have equal magnitudes.
    In order for the magnitude of \vec A + \vec B to be larger than the magnitude of \vec A - \vec B
    by the factor n, what must be the angle between them?


    A friend offered the formula: . \begin{array}{ccc}\text{sum}^2 &=& a^2 + a^2 + a^2\cos\theta \\\text{diff}^2 &=& a^2 + a^2 - a^2\cos\theta \end{array}

    However, I've never seen these formulas before
    and I don't understand the premise behind them
    so cannot determine if they are correct...

    Can someone please help explain these formulas?
    Or offer another method of solving this problem?
    I believe your friend is using the Law of Cosines ... but incorrectly.


    The sum of the vectors, \vec A + \vec B, is the diagonal of their parallelogram.

    Code:
                .  .  .  .  .  .  *
               .              *  *
              .    A+B    *     *
             .        *        * B
            .     *           *
           .  *            θ *
          *  *  *  *  *  *  *
                   A
    Law of Cosines: . |\vec A + \vec B|^2 \;=\;|\vec A|^2 + |\vec B|^2 - 2\!\cdot\!|\vec A|\!\cdot\!|\vec B|\cos\theta



    The difference of the vectors, \vec A - \vec B, is the other diagonal.
    Code:
                *  .  .  .  .  .  .
               *  *              .
              *     *  A-B      .
          -B *        *        .
            *           *     .
           * 180-θ       *  .
          *  *  *  *  *  *  *
                   A
    Law of Cosines: . |A + (\text{-}B)|^2 \;=\;|A|^2 + |\text{-}B|^2 - 2\!\cdot\!|A|\!\cdot\!|\text{-}B|\cos(180^o -\theta)

    . . which simplifies to: . |A - B|^2\;=\;|A|^2 + |B|^2 + 2\!\cdot\!|A|\!\cdot\!|B|\cos\theta


    Your turn . . .

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  4. #4
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    Quote Originally Posted by Soroban View Post

    The sum of the vectors, \vec A + \vec B, is the diagonal of their parallelogram.
    Code:
                .  .  .  .  .  .  *
               .              *  *
              .    A+B    *     *
             .        *        * B
            .     *           *
           .  *            θ *
          *  *  *  *  *  *  *
                   A
    Law of Cosines: . |\vec A + \vec B|^2 \;=\;|\vec A|^2 + |\vec B|^2 - 2\!\cdot\!|\vec A|\!\cdot\!|\vec B|\cos\theta
    If you look carefully at your own diagram, the diagonal you have described is B-A and not A+B
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  5. #5
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    Quote Originally Posted by Plato View Post
    If you look carefully at your own diagram, the diagonal you have described is B-A and not A+B
    Without arrow heads I suppose there is a certain ambiguity. However, I think all the posts in this thread collectively give enough help for the OP to work out the answer to the question.
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