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Thread: easy vector problem

  1. #1
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    easy vector problem

    Hi! Given $\displaystyle \vec a=<2x,1,1-x>$, $\displaystyle \vec b=<-1,3,0>$ and $\displaystyle \vec c=<5,-1,8>$ as three vectors, find x so that $\displaystyle \vec a$ forms equal angles with the vectors $\displaystyle \vec b$ and $\displaystyle \vec c$.

    I tried this $\displaystyle a \cdot b = a \cdot c$ ($\displaystyle \cdot$ is a dot product). Then I isolated $\displaystyle cos\alpha$ and $\displaystyle cos\beta$ and equated them. Then I solved by x. Is this correct?
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Um... not quite.

    The general formula:

    $\displaystyle \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos\theta$

    Here, the angles are equal, so we get:

    $\displaystyle \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos\theta$

    $\displaystyle \vec{a} \cdot \vec{c} = |\vec{a}||\vec{c}| \cos\theta$

    Equate cos theta;

    $\displaystyle \dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \dfrac{\vec{a} \cdot \vec{c}}{|\vec{a}||\vec{c}|} $
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  3. #3
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    Quote Originally Posted by Unknown008 View Post
    Um... not quite.

    The general formula:

    $\displaystyle \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos\theta$

    Here, the angles are equal, so we get:

    $\displaystyle \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos\theta$

    $\displaystyle \vec{a} \cdot \vec{c} = |\vec{a}||\vec{c}| \cos\theta$

    Equate cos theta;

    $\displaystyle \dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \dfrac{\vec{a} \cdot \vec{c}}{|\vec{a}||\vec{c}|} $
    That's what I was talking about. What wasn't right then?
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  4. #4
    MHF Contributor Unknown008's Avatar
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    From what I understood, you didn't divide by the product of the magnitudes of the vectors.
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  5. #5
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    I did. I wasn't clear, my fault. Thank you very much.
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  6. #6
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    Just solve $\displaystyle 3\vec{a}\cdot\vec{b}=\vec{a}\cdot\vec{c} $ for $\displaystyle x.$
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  7. #7
    MHF Contributor Unknown008's Avatar
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    Quote Originally Posted by Plato View Post
    Just solve $\displaystyle 3\vec{a}\cdot\vec{b}=\vec{a}\cdot\vec{c} $ for $\displaystyle x.$
    Could you explain why you used 3a?

    I admit that solving from the formula isn't that easy with the quadratic equations
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  8. #8
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    Quote Originally Posted by Unknown008 View Post
    Could you explain why you used 3a?
    $\displaystyle \left\| {\overrightarrow b } \right\| = \sqrt {10} \;\& \,\left\| {\overrightarrow c } \right\| = 3\sqrt {10} $
    The 3 is left after division.
    $\displaystyle \dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \dfrac{\vec{a} \cdot \vec{c}}{|\vec{a}||\vec{c}|}$
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  9. #9
    MHF Contributor Unknown008's Avatar
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    Oh gosh, right, I don't know how I missed that part >.< Thanks for that
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