%DRAWING{"distance"}%

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

%$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