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Math Help - vectors!!

  1. #1
    Newbie sir nerdalot's Avatar
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    vectors!!

    hello!!

    i'm having a little trouble converting between vector forms!!
    so how would you convert the vector V = (3,40 degrees) into cartesian form
    and convert v = 2i + 5j into polar form...

    thanks!!
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  2. #2
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    Hello, sir nerdalot!

    How would you convert the vector v \:= \:(3,\,40^o) into cartesian form
    and convert v \:= \:2i + 5j into polar form?

    If you sketch this diagram, the pieces should come together.
    Code:
            |
            |     *
            |    /|
            | r / |
            |  /  |y
            | /   |
            |/θ   |
        - - * - - + - -
            |  x
    We can see that: . \begin{array}{c}x \: =\:  r\cos\theta \\ y \: =\: r\sin\theta\end{array}

    . . . and that: . \begin{array}{ccc}r & = & \sqrt{x^2+y^2} \\ \theta & = & \tan^{-1}\!\left(\frac{y}{x}\right)\end{array}


    Then use those conversion formulas . . .

    \begin{Bmatrix}x & = & 3\cos40^o & \approx & 2.3 \\<br />
y & = & 3\sin40^o & \approx & 1.9\end{Bmatrix}\qquad (3,\,40^o)\;\Rightarrow\;2.3i + 1.9j

    \begin{Bmatrix}r & = & \sqrt{2^2+5^2} & = & \sqrt{29} \\<br />
\theta & = & \tan^{-1}\!\left(\frac{5}{2}\right) & \approx & 68.2^o\end{Bmatrix}\qquad2i + 5j\;\Rightarrow\;\left(\sqrt{29},\,68.2^o\right)<br />

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  3. #3
    Newbie sir nerdalot's Avatar
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    thanks for that....

    but how would i find the angle between these two vectors???
    i thought about trying to use the cosine/sine rule but found that there wasn't enough information to do this.

    am i going about this the right way???
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by sir nerdalot View Post
    thanks for that....

    but how would i find the angle between these two vectors???
    i thought about trying to use the cosine/sine rule but found that there wasn't enough information to do this.

    am i going about this the right way???
    use the dot product formula

    Recall: If \bold {a,b} are two vectors, then \boxed { \bold {a} \cdot \bold {b} = |\bold {a}||\bold {b}|\cos \theta } where \theta is the angle between them.

    you have enough information to solve that formula for \theta

    do you remember how to take the dot product?
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  5. #5
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    Hello, sir nerdalot!

    I thought about trying to use the cosine/sine rule . . . . Good thinking!
    but found that there wasn't enough information to do this.
    . .
    Well, there is . . . but it takes some Olympic-level gymnastics.

    Given two vectors \vec{u} and \vec{v}, let \theta be the angle between them.
    Code:
                      Q
                      *
                    *  \
               v  *     \
                *        \ w
              *           \
            * θ            \
          * * * * * * * * * *
         O          u         P
    We have: . \vec{u} \,= \,\overrightarrow{OP},\;\;\vec{v} \,=\, \overrightarrow{OQ},\;\;\theta \,= \,\angle POQ

    . . If \vec{w} \,= \,\overrightarrow{PQ}, then: . \vec{w} \:= \:\vec{u} - \vec{v}


    Using the Law of Cosines, we have: . |w|^2 \;=\;|u|^2 + |v|^2 - 2|u||v|\cos\theta .[1]


    A small detour:

    . . |\vec{w}|^2 \;=\;|\vec{u} - \vec{v}|^2 \;=\;(\vec{u}-\vec{v})\cdot(\vec{u}-\vec{v}) \;=\;\vec{u}\cdot\vec{u} - \vec{u}\cdot\vec{v} - \vec{v}\cdot\vec{u} + \vec{v}\cdot\vec{v}

    . . |w|^2\;=\;|\vec{u}|^2 - 2(\vec{u}\cdot\vec{v}) + |\vec{v}|^2 .[2]


    Substitute [2] into [1]: . |\vec{u}|^2 - 2(\vec{u}\cdot\vec{v}) + |\vec{v}|^2       \;=\;|\vec{u}|^2 + |\vec{v}|^2 - 2|\vec{u}||\vec{v}|\cos\theta

    This simplifies to: . 2|\vec{u}||\vec{v}|\cos\theta \:=\:2(\vec{u}\cdot\vec{v})\quad\Rightarrow\quad\b  oxed{\cos\theta \;=\;\frac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|  }} . . . . There!

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  6. #6
    Newbie sir nerdalot's Avatar
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    coolies and so i substitute the two sides into u and v

    thanks all!!!
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