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

Parametrical Representation of Planes

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The points in a plane through $ P$ spanned by two nonparallel directions $ \vec{u}$ und $ \vec{v}$ satisfy

$\displaystyle \overrightarrow{PX} = s\vec{u}+t\vec{v},\quad s,t\in\mathbb{R}\, . $

\includegraphics[width=10cm]{param_darst}

Correspondingly, we have

$\displaystyle x_i = p_i + su_i+tv_i,\quad i=1,2,3\, . $

for the coordinates of the position vector $ \vec{x}$ .
Given a plane through point $ P=(1, 2, 3)$ spanned by the vectors $ \vec{u} = \begin{pmatrix}2\\ 0\\ 0\\ \end{pmatrix}$ , $ \vec{v} = \begin{pmatrix}1\\ 1\\ 1\\ \end{pmatrix}$ .

Then $ X = (1,1,2)$ lies in this plane, because we have

$\displaystyle \overrightarrow{PX} = \begin{pmatrix}0\\ -1 \\ -1\\ \end{pmatrix} = \frac{1}{2} \vec{u} + (-1) \vec{v} $

On the other hand, $ X = (0,0,0)$ does not lie in this plane, because we have $ \overrightarrow{PX} = \begin{pmatrix}-1\\ -2 \\ -3\\ \end{pmatrix}$ and the system of equations

$\displaystyle s \begin{pmatrix}2\\ 0 \\ 0\\ \end{pmatrix} + t \begin{pmatrix}1\\ 1 \\ 1\\ \end{pmatrix} = \begin{pmatrix}-1\\ -2 \\ -3\\ \end{pmatrix} $

has no solution as we can see from the second and the third line.

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  automatically generated 9/18/2007