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$\mbox{where } \vec{u_{0}} \mbox{ is a unit vector}$

$g: \vec{x} = \vec{p} + \mu \vec{u_{0}} $

$(1):\mbox{    }cos{\alpha} = \frac{\displaystyle|\vec{f}-\vec{p}|}{\displaystyle|\vec{r}-\vec{p}|}\mbox{    }\to\mbox{    } |\vec{f}-\vec{p}| =\ cos{\alpha}|\vec{r}-\vec{p}| $

$(2):\mbox{    }\cos{\alpha} = \frac{\displaystyle|\vec{a}\cdot\vec{b}|}{\displaystyle|\vec{a}|\cdot|\vec{b}|}=\vec{a_{0}}\cdot\vec{b_{0}}$

$(1),\;(2):\mbox{    } |\vec{f}-\vec{p}| = \frac{\displaystyle(\vec{f}-\vec{p})}{\displaystyle|\vec{a}-\vec{b}|}\cdot\vec{u_{0}}\cdot|\vec{r}-\vec{p}| = |(\vec{f}-\vec{p})\cdot\vec{u_{0}}|$

Phytagoras' thorem gives:

$d=\sqrt{\displaystyle(\vec{r}-\vec{p})^{2}-(\displaystyle(\vec{r}-\vec{p})\cdot\vec{u_{0}})^{2}}$

...Back to Vertex Finding

-- PeterZumbruch - 23 Feb 2004
Topic attachments
I Attachment Action Size Date Who Comment
distance.drawdraw distance.draw manage 7.5 K 2004-02-23 - 17:08 PeterZumbruch TWikiDraw draw file
distance.gifgif distance.gif manage 2.5 K 2004-02-23 - 17:09 PeterZumbruch TWikiDraw GIF file
Topic revision: r6 - 2005-06-29, PeterZumbruch
 
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