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Mathematics-Online course: Prepcourse Mathematics - Linear Algebra and Geometry - Lines and Planes

Distance Point-Line

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The projection $ X$ of a point $ Q$ on a line through $ P$ 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 $ Q$ 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}
The projection of point $ Q=(3,3,3)$ onto the line

$\displaystyle g: \left(\begin{array}{c}2\\ 1\\ 3\end{array}\right)+ t\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right) $

is the point with position vector
$\displaystyle \vec{x} = \vec{p}+\frac{(\vec{q}-\vec{p})\cdot\vec{u}}{\vert\vec{u}\vert^2}\,\vec{u}$ $\displaystyle =$ $\displaystyle \left(\begin{array}{r}2\\ 1\\ 3\end{array}\right) + \frac{ \left(... ...rray}\right)}{1^2+1^2+1^2} \, \left(\begin{array}{r}1\\ 1\\ 1\end{array}\right)$  
  $\displaystyle =$ $\displaystyle \left(\begin{array}{r}2\\ 1\\ 3\end{array}\right) +\frac{1\cdot 1... ... 1\\ 1\end{array}\right) = \left(\begin{array}{r}3\\ 2\\ 4\end{array}\right)\,.$  

The disctance is

$\displaystyle d= \frac{\left\vert\left(\begin{array}{r}1\\ 2\\ 0\end{array}\ri... ...right\vert}{\sqrt{3}} = \sqrt{\frac{(2+0)^2+(0-1)^2+(1-2)^2}{3}}=\sqrt{2}\,, $

in accordance with

$\displaystyle \left\vert\overrightarrow{XQ}\right\vert = \sqrt{(3-3)^2+(3-2)^2+(3-4)^2}=\sqrt{2}\,. $

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  automatically generated 9/18/2007