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Thread: 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 $\displaystyle v \:= \:(3,\,40^o)$ into cartesian form
    and convert $\displaystyle v \:= \:2i + 5j$ into polar form?

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

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


    Then use those conversion formulas . . .

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

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

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

    you have enough information to solve that formula for $\displaystyle \theta$

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

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


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


    A small detour:

    . . $\displaystyle |\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} $

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


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

    This simplifies to: .$\displaystyle 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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