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

Momentum Form

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The points $ X$ on a line through $ P$ in direction $ \vec{u}$ can be described by

$\displaystyle \overrightarrow{PX}\times \vec{u} = \vec{0} \,. $

\includegraphics[width=12cm]{a_momentenform_bild}

Correspondingly, we have

$\displaystyle \vec{x}\times\vec{u} = \vec{c},\quad \vec{c} = \vec{p}\times\vec{u}\, . $

The momentum form of a line can be used to verify whether the point $ X$ lies on a given line. For the depicted figure we obtain for the point $ X_1$
$\displaystyle \overrightarrow{PX_1} \times \vec{u}$ $\displaystyle =$ $\displaystyle \left( \left(\begin{array}{r}5\\ 1\\ 0\end{array}\right)- \left(\... ...{array}\right) \right)\times \left(\begin{array}{r}2\\ -1\\ 0\end{array}\right)$  
  $\displaystyle =$ $\displaystyle \left(\begin{array}{r}4\\ -2\\ 0\end{array}\right) \times \left(\begin{array}{r}2\\ -1\\ 0\end{array}\right)$  
  $\displaystyle =$ $\displaystyle \left(\begin{array}{c}0\\ 0\\ 4\cdot (-1)-2\cdot(-2) \end{array}\right)=\vec{0}\,.$  

Whereas we have for the point $ X_2$
$\displaystyle \overrightarrow{PX_2} \times \vec{u}$ $\displaystyle =$ $\displaystyle \left(\begin{array}{r}2.5\\ -2\\ 0\end{array}\right) \times \left(\begin{array}{r}2\\ -1\\ 0\end{array}\right)$  
  $\displaystyle =$ $\displaystyle \left(\begin{array}{c}0\\ 0\\ 2.5\cdot (-1)-2\cdot(-2) \end{array}\right)= \left(\begin{array}{c}0\\ 0\\ 1.5 \end{array}\right) \neq\vec{0}\,.$  

\includegraphics[width=12.4cm]{b_momentenform_bild}
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  automatically generated 9/18/2007