Mathematics-Online course: Preparatory Course Mathematics-Linear Algebra and Geometry-Lines and Planes-Distance Point-Line

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	Mathematics-Online course: Preparatory Course Mathematics - Linear Algebra and Geometry - Lines and Planes
	Distance Point-Line

The projection

of a point

on a line through

with direction $\vec{u}$ satisfies

$\displaystyle \overrightarrow{PX} = t\vec{u},\quad t = \frac{(\vec{q}-\vec{p})\cdot\vec{u}}{\vert\vec{u}\vert^2} \,.$

From this we obtain the distance of the point

from the line by

$\displaystyle d = \frac{\left\vert(\vec{q}-\vec{p})\times\vec{u}\right\vert} {\vert\vec{u}\vert}\, .$

$\includegraphics[width=12cm]{a_abstand_punkt_gerade_bild}$

(temporary unavailable)

automatically generated 1/9/2017