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SerloDie freie Lernplattform

Berechne den Schnittwinkel zwischen den beiden Ebenen.

  1. E1:  (2−13)∘[x→−(111)]\displaystyle {\mathrm E}_1:\;\begin{pmatrix}2\\-1\\3\end{pmatrix}\circ\left[\overrightarrow{\mathrm x}-\begin{pmatrix}1\\1\\1\end{pmatrix}\right]==0\displaystyle 0
    E2:  (12−1)⋅[x→−(−21−2)]\displaystyle {\mathrm E}_2:\;\begin{pmatrix}1\\2\\-1\end{pmatrix}\cdot\left[\overrightarrow{\mathrm x}-\begin{pmatrix}-2\\1\\-2\end{pmatrix}\right]==0\displaystyle 0
  2. E1:  x→\displaystyle {\mathrm E}_1:\;\overrightarrow{\mathrm x}==(40−3)+r⋅(0−10)+s⋅(−203)\displaystyle \begin{pmatrix}4\\0\\-3\end{pmatrix}+\mathrm r\cdot\begin{pmatrix}0\\-1\\0\end{pmatrix}+\mathrm s\cdot\begin{pmatrix}-2\\0\\3\end{pmatrix}
    E2:  x→\displaystyle {\mathrm E}_2:\;\overrightarrow{\mathrm x}==(−230)+r⋅(00−1)+s⋅(2−13)\displaystyle \begin{pmatrix}-2\\3\\0\end{pmatrix}+\mathrm r\cdot\begin{pmatrix}0\\0\\-1\end{pmatrix}+\mathrm s\cdot\begin{pmatrix}2\\-1\\3\end{pmatrix}
  3. E1:  (−3−21)∘x→−2\displaystyle {\mathrm E}_1:\;\begin{pmatrix}-3\\-2\\1\end{pmatrix}\circ\overrightarrow{\mathrm x}-2==0\displaystyle 0
    E2:  (−3−2−1)∘x→+1\displaystyle {\mathrm E}_2:\;\begin{pmatrix}-3\\-2\\-1\end{pmatrix}\circ\overrightarrow{\mathrm x}+1==0\displaystyle 0
  4. E1:  x→\displaystyle {\mathrm E}_1:\;\overrightarrow{\mathrm x}==(1−21)+r⋅(2−11)+s⋅(1−21)\displaystyle \begin{pmatrix}1\\-2\\1\end{pmatrix}+\mathrm r\cdot\begin{pmatrix}2\\-1\\1\end{pmatrix}+\mathrm s\cdot\begin{pmatrix}1\\-2\\1\end{pmatrix}
    E2:  x→\displaystyle {\mathrm E}_2:\;\overrightarrow{\mathrm x}==(21−3)+r⋅(10−2)+s⋅(−110)\displaystyle \begin{pmatrix}2\\1\\-3\end{pmatrix}+\mathrm r\cdot\begin{pmatrix}1\\0\\-2\end{pmatrix}+\mathrm s\cdot\begin{pmatrix}-1\\1\\0\end{pmatrix}
  5. E1:  5⋅x1+2⋅x2+3⋅x3−30\displaystyle {\mathrm E}_1:\;5\cdot{\mathrm x}_1+2\cdot{\mathrm x}_2+3\cdot{\mathrm x}_3-30==0\displaystyle 0
    E2:  10⋅x1+7⋅x2−12⋅x3−45\displaystyle {\mathrm E}_2:\;10\cdot{\mathrm x}_1+7\cdot{\mathrm x}_2-12\cdot{\mathrm x}_3-45==0\displaystyle 0
  6. E1:  (10−1)∘x→−2\displaystyle {\mathrm E}_1:\;\begin{pmatrix}1\\0\\-1\end{pmatrix}\circ\overrightarrow{\mathrm x}-2==0\displaystyle 0
    E2:  −x1+x2−x3−1\displaystyle {\mathrm E}_2:\;-{\mathrm x}_1+{\mathrm x}_2-{\mathrm x}_3-1==0\displaystyle 0
  7. E1:  x→\displaystyle {\mathrm E}_1:\;\overrightarrow{\mathrm x}==(111)+r⋅(0−11)+s⋅(2−2−2)\displaystyle \begin{pmatrix}1\\1\\1\end{pmatrix}+\mathrm r\cdot\begin{pmatrix}0\\-1\\1\end{pmatrix}+\mathrm s\cdot\begin{pmatrix}2\\-2\\-2\end{pmatrix}
    E2:  x→\displaystyle {\mathrm E}_2:\;\overrightarrow{\mathrm x}==(011)+r⋅(2−1−3)+s⋅(2−2−1)\displaystyle \begin{pmatrix}0\\1\\1\end{pmatrix}+\mathrm r\cdot\begin{pmatrix}2\\-1\\-3\end{pmatrix}+\mathrm s\cdot\begin{pmatrix}2\\-2\\-1\end{pmatrix}