Mo logo [home] [lexicon] [problems] [tests] [courses] [auxiliaries] [notes] [staff] german flag
Mathematics-Online course: Prepcourse Mathematics - Linear Algebra and Geometry - Lines and Planes

Intersection of Planes

[previous page] [next page] [table of contents][page overview]
The smaller one, $ \varphi\in[0,\pi/2]$ , of the two angles between two planes with normal vectors $ \vec{n}_i$ is uniquely determined by

$\displaystyle \cos(\varphi) = \frac{\vert\vec{n}_1\cdot \vec{n}_2\vert} {\vert\vec{n}_1\vert\vert \vec{n}_2\vert} $

and

$\displaystyle \vec{n}_1\times\vec{n}_2 $

is the direction of the line of intersection.

\includegraphics[clip,width=.6\linewidth]{schnitt}
Given two planes with the respective normal vectors

$\displaystyle \vec{n}_1= \begin{pmatrix}1\\ -1\\ 0\\ \end{pmatrix}, \quad \vec{n}_2= \begin{pmatrix}0\\ -1\\ 1\\ \end{pmatrix}$

through the respective points $ P_1=( 1,2 ,0)$ , $ P_2=(0,1,3)$ . We want to compute the angle $ \varphi$ between these planes and the line of intersection.


We have

$\displaystyle \cos(\varphi) = \frac{\vert\vec{n}_1\cdot\vec{n}_2\vert} {\vert\vec{n}_1\vert \vert\vec{n}_2\vert} = \frac{1}{2} $

and therefore

$\displaystyle \varphi= 60^\circ \, .$

The computation

$\displaystyle \vec{r} = \vec{n}_1 \times \vec{n}_2 = \begin{pmatrix}-1\\ -1\\ -1\\ \end{pmatrix}$

yields the direction of the line of intersection.

To find a point of intersection we need

$\displaystyle d_1 = \vec{p}_1 \cdot \vec{n}_1 = -1, \quad d_2 = \vec{p}_2 \cdot \vec{n}_2 = 2$

and we determine a point $ S$ satisfying the equations

$\displaystyle \vec{s} \cdot \vec{n}_1 = d_1, \quad \vec{s} \cdot \vec{n}_2 = d_2\,.$

We find, for example, $ S=(2,3 ,5)$ and so we obtain the line of intersection

$\displaystyle g:\vec{x} = \vec{s} + t \vec{r} = \begin{pmatrix}2\\ 3\\ 5\\ \en... ...atrix} + t \begin{pmatrix}1\\ 1\\ 1\\ \end{pmatrix} \quad t \in \mathbb{R}\, .$

[previous page] [next page] [table of contents][page overview]
  automatically generated 9/18/2007