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Thread: Need help with vectors question

  1. #1
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    Need help with vectors question

    Sorry, pretty much in a rush to finish this soon as I have been stuck at understanding this vectors chapter for a really long time. I hope someone will be able to help with it...I've tried but I really cannot figure it out alone.

    The line L has cartesian equations $\displaystyle x - 2 = \frac{y - 5}{2} = -z$. Find the perpendicular distance from the origin to L.

    Find also the acute angle between L and the z-axis.

    I don't know what concepts I am missing out on but I am stuck.

    Help will be much appreciated. I'm trying to learn and not just get answers. Rushing on revision as I don't have much time...planning to finish my math revision in 2 months.
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  2. #2
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    Workings:





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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Puzzled View Post
    Sorry, pretty much in a rush to finish this soon as I have been stuck at understanding this vectors chapter for a really long time. I hope someone will be able to help with it...I've tried but I really cannot figure it out alone.

    The line L has cartesian equations $\displaystyle x - 2 = \frac{y - 5}{2} = -z$. Find the perpendicular distance from the origin to L.

    Find also the acute angle between L and the z-axis.

    I don't know what concepts I am missing out on but I am stuck.

    Help will be much appreciated. I'm trying to learn and not just get answers. Rushing on revision as I don't have much time...planning to finish my math revision in 2 months.

    The parametric form of the line is:

    $\displaystyle
    L(\lambda ) = \left[ {\begin{array}{*{20}c}
    {\lambda - 2} \\
    {2\lambda + 5} \\
    { - \lambda } \\
    \end{array}} \right] = \left[ {\begin{array}{*{20}c}
    { - 2} \\
    5 \\
    0 \\
    \end{array}} \right] + \left[ {\begin{array}{*{20}c}
    \lambda \\
    {2\lambda } \\
    { - \lambda } \\
    \end{array}} \right]
    $

    RonL
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  4. #4
    MHF Contributor

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    The classic formula for distance of a point to a line is:
    $\displaystyle \begin{array}{l}
    l(t) = Q + tD\,\& \,P \notin l(t) \\
    d\left( {l,P} \right) = \frac{{\left\| {\overrightarrow {QP} \times D} \right\|}}{{\left\| D \right\|}} \\
    \end{array}$.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by Puzzled View Post
    Sorry, pretty much in a rush to finish this soon as I have been stuck at understanding this vectors chapter for a really long time. I hope someone will be able to help with it...I've tried but I really cannot figure it out alone.

    The line L has cartesian equations $\displaystyle x - 2 = \frac{y - 5}{2} = -z$. Find the perpendicular distance from the origin to L.

    Find also the acute angle between L and the z-axis.

    I don't know what concepts I am missing out on but I am stuck.

    Help will be much appreciated. I'm trying to learn and not just get answers. Rushing on revision as I don't have much time...planning to finish my math revision in 2 months.
    The perpendicular distance from the origin to the line can be found in lots of
    ways. I will resolve $\displaystyle x=[-2,5,0]^t$ into a component parrallel to the line and
    subtract this from $\displaystyle x$ to find the component normal $\displaystyle n$ to the line and the answer
    will be the norm of this component.

    $\displaystyle
    \begin{array}{l}
    l = \left[ {\begin{array}{*{20}c}
    1 \\
    2 \\
    { - 1} \\
    \end{array}} \right]\\ \\
    n = x - (x.l/\left\| l \right\|)l \\
    \end{array}
    $

    RonL
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