# Thread:

1. ## 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.

2. Workings:

3. Originally Posted by Puzzled
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

4. The classic formula for distance of a point to a line is:
$\begin{array}{l}<br /> l(t) = Q + tD\,\& \,P \notin l(t) \\ <br /> d\left( {l,P} \right) = \frac{{\left\| {\overrightarrow {QP} \times D} \right\|}}{{\left\| D \right\|}} \\ <br /> \end{array}$.

5. Originally Posted by Puzzled
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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