Need help with vectors question

**Puzzled** · March 21st 2008, 01:24 AM

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.

**Puzzled** · March 21st 2008, 05:37 AM

Workings:

**CaptainBlack** · March 21st 2008, 07:27 AM

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:

$ 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

**Plato** · March 21st 2008, 07:43 AM

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

**CaptainBlack** · March 21st 2008, 07:45 AM

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.

$ \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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