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Math Help - 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 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 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:

    <br />
L(\lambda ) = \left[ {\begin{array}{*{20}c}<br />
   {\lambda  - 2}  \\<br />
   {2\lambda  + 5}  \\<br />
   { - \lambda }  \\<br />
\end{array}} \right] = \left[ {\begin{array}{*{20}c}<br />
   { - 2}  \\<br />
   5  \\<br />
   0  \\<br />
\end{array}} \right] + \left[ {\begin{array}{*{20}c}<br />
   \lambda   \\<br />
   {2\lambda }  \\<br />
   { - \lambda }  \\<br />
\end{array}} \right]<br />

    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:
    \begin{array}{l}<br />
 l(t) = Q + tD\,\& \,P \notin l(t) \\ <br />
 d\left( {l,P} \right) = \frac{{\left\| {\overrightarrow {QP}  \times D} \right\|}}{{\left\| D \right\|}} \\ <br />
 \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 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 x=[-2,5,0]^t into a component parrallel to the line and
    subtract this from x to find the component normal n to the line and the answer
    will be the norm of this component.

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

    RonL
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