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Math Help - Vectors

  1. #1
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    Vectors

    Let p=(2,k) and q=(3,5), find k such that the angle between p and q is π/4.

    The question is under the vector section of the course elementary linear algebra. It's the first math course I've taken at university level. According to the key, the answer is 1/2,
    but I see no way of getting there. Any help is appreciated.

    Last edited by Cloudhorn; January 30th 2013 at 08:21 AM.
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  2. #2
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    Re: Vectors

    If the angle between the two vectors is \pi/4, then the dot product of the two vectors is |p||q|\sin(\pi/4). Do you know how to calculate the dot product of vectors (2,k) and (3,5)?
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  3. #3
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    Re: Vectors

    Hello, Cloudhorn!

    \text{Let } \vec{p}=\langle 2,k\rangle\,\text{ and }\,\vec{q} = \langle3,5\rangle
    \text{Find }k\text{ such that the angle between }\vec{p}\text{ and }\vec{q}\text{ is }\tfrac{\pi}{4}
    You are expected to know this formula.
    The angle \theta between vectors \vec{u} and \vec{v} is given by: . \cos\theta \;=\;\frac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|  }

    In our problem: . \frac{\vec{p}\cdot\vec{q}}{|\vec{p}||\vec{q}|} \;=\;\cos\tfrac{\pi}{4}

    We have: . \begin{Bmatrix}\vec{p}\cdot\vec{q} &=&\langle2,k\rangle\cdot\langle3,5\rangle &=& 6 + 5k \\ |\vec{p}| &=& \sqrt{2^2+k^2 &=& \sqrt{k^2+4} \\ |\vec{q}| &=& \sqrt{3^2+5^2} &=& \sqrt{34} \end{Bmatrix}

    Substitute: . \dfrac{5k+6}{\sqrt{k^2+4}\sqrt{34}} \:=\:\dfrac{1}{\sqrt{2}} \quad\Rightarrow\quad \sqrt{2}(5k+6) \;=\;\sqrt{34}\sqrt{k^2+4}

    Square both sides: . 2(5k+6)^2 \;=\;34(k^2+4) \quad\Rightarrow\quad (5k+6)^2 \;=\;17(k^2+4)

    . . . . . 25k^2 + 60k + 36 \:=\:17k^2 + 68 \quad\Rightarrow\quad 8k^2 + 60k - 32 \:=\:0

    . . . . . 2k^2 + 15k - 8 \:=\:0 \quad\Rightarrow\quad (2k-1)(k+8) \:=\:0

    We have: . \begin{Bmatrix}2k-1 \:=\:0 & \Rightarrow & k \:=\:\frac{1}{2} \\ k+8 \:=\:0 & \Rightarrow & k \:=\:\text{-}8 \end{Bmatrix}

    If k = \text{-}8, we find that \theta \,=\,\tfrac{3\pi}{4}

    Therefore: . k \,=\,\tfrac{1}{2}
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    Re: Vectors

    Quote Originally Posted by Cloudhorn View Post
    Let p=(2,k) and q=(3,5), find k such that the angle between p and q is π/4.
    The question is under the vector section of the course elementary linear algebra. It's the first math course I've taken at university level. According to the key, the answer is 1/2, but I see no way of getting there. Any help is appreciated.
    The angle between to vectors is \cos \left( {\frac{\pi }{4}} \right) = \frac{{p \cdot q}}{{\left\| p \right\|\left\| q \right\|}}.
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  5. #5
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    Re: Vectors

    This is perfectly clear to me now, thank you all for your answers.
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