Vektorrechnung

Bestimmung eines Normalvektors ${\vec {n}}$, der senkrecht zu den Spannvektoren ${\vec {u}}$ und ${\vec {v}}$ ist

Es gelten die folgenden Bedingungen:

${\vec {u}}\cdot {\vec {n}}=0$

1) $u_{1}\cdot n_{1}+u_{2}\cdot n_{2}+u_{3}\cdot n_{3}=0$

${\vec {v}}\cdot {\vec {n}}=0$

2) $v_{1}\cdot n_{1}+v_{2}\cdot n_{2}+v_{3}\cdot n_{3}=0$

Fall 1: $n_{1}=K$

aus 1) folgt:

3) $n_{2}=-{\frac {u_{1}\cdot K+u_{3}\cdot n_{3}}{u_{2}}}$

$n_{2}$ in 2):

$v_{1}\cdot K-v_{2}\cdot {\frac {u_{1}\cdot K+u_{3}\cdot n_{3}}{u_{2}}}+v_{3}\cdot n_{3}=0$

daraus folgt:

$K\cdot \left(v_{1}-v_{2}\cdot {\frac {u_{1}}{u_{2}}}\right)+n_{3}\cdot \left(v_{3}-v_{2}\cdot {\frac {u_{3}}{u_{2}}}\right)=0$

$n_{3}=-K\cdot {\frac {v_{1}-v_{2}\cdot {\frac {u_{1}}{u_{2}}}}{v_{3}-v_{2}\cdot {\frac {u_{3}}{u_{2}}}}}=K\cdot {\frac {u_{1}\cdot v_{2}-u_{2}\cdot v_{1}}{u_{2}\cdot v_{3}-u_{3}\cdot v_{2}}}$

$n_{3}$ in 3):

$n_{2}=-{\frac {u_{1}}{u_{2}}}\cdot K-{\frac {u_{3}}{u_{2}}}\cdot K\cdot {\frac {u_{1}\cdot v_{2}-u_{2}\cdot v_{1}}{u_{2}\cdot v_{3}-u_{3}\cdot v_{2}}}=K\cdot {\frac {-{\frac {u_{1}}{u_{2}}}\cdot \left(u_{2}\cdot v_{3}-u_{3}\cdot v_{2}\right)-{\frac {u_{3}}{u_{2}}}\cdot \left(u_{1}\cdot v_{2}-u_{2}\cdot v_{1}\right)}{u_{2}\cdot v_{3}-u_{3}\cdot v_{2}}}=K\cdot {\frac {-u_{1}\cdot v_{3}+{\frac {u_{1}}{u_{2}}}\cdot u_{3}\cdot v_{2}-{\frac {u_{3}}{u_{2}}}\cdot u_{1}\cdot v_{2}+u_{3}\cdot v_{1}}{u_{2}\cdot v_{3}-u_{3}\cdot v_{2}}}=K\cdot {\frac {u_{3}\cdot v_{1}-u_{1}\cdot v_{3}}{u_{2}\cdot v_{3}-u_{3}\cdot v_{2}}}$

Zusammengefasst:

$n_{1}=K$

$n_{2}=K\cdot {\frac {u_{3}\cdot v_{1}-u_{1}\cdot v_{3}}{u_{2}\cdot v_{3}-u_{3}\cdot v_{2}}}$

$n_{3}=K\cdot {\frac {u_{1}\cdot v_{2}-u_{2}\cdot v_{1}}{u_{2}\cdot v_{3}-u_{3}\cdot v_{2}}}$

Fall 2: $n_{2}=K$

aus 1) folgt:

3) $n_{1}=-{\frac {u_{2}\cdot K+u_{3}\cdot n_{3}}{u_{1}}}$

$n_{1}$ in 2):

$-v_{1}\cdot {\frac {u_{2}\cdot K+u_{3}\cdot n_{3}}{u_{1}}}+v_{2}\cdot K+v_{3}\cdot n_{3}=0$

daraus folgt:

$K\cdot \left(v_{2}-v_{1}\cdot {\frac {u_{2}}{u_{1}}}\right)+n_{3}\cdot \left(v_{3}-v_{1}\cdot {\frac {u_{3}}{u_{1}}}\right)=0$

$n_{3}=-K\cdot {\frac {v_{2}-v_{1}\cdot {\frac {u_{2}}{u_{1}}}}{v_{3}-v_{1}\cdot {\frac {u_{3}}{u_{1}}}}}=K\cdot {\frac {u_{2}\cdot v_{1}-u_{1}\cdot v_{2}}{u_{1}\cdot v_{3}-u_{3}\cdot v_{1}}}$

$n_{3}$ in 3):

$n_{1}=-{\frac {u_{2}}{u_{1}}}\cdot K-{\frac {u_{3}}{u_{1}}}\cdot K\cdot {\frac {u_{2}\cdot v_{1}-u_{1}\cdot v_{2}}{u_{1}\cdot v_{3}-u_{3}\cdot v_{1}}}=K\cdot {\frac {-{\frac {u_{2}}{u_{1}}}\cdot \left(u_{1}\cdot v_{3}-u_{3}\cdot v_{1}\right)-{\frac {u_{3}}{u_{1}}}\cdot \left(u_{2}\cdot v_{1}-u_{1}\cdot v_{2}\right)}{u_{1}\cdot v_{3}-u_{3}\cdot v_{1}}}=K\cdot {\frac {-u_{2}\cdot v_{3}+{\frac {u_{2}}{u_{1}}}\cdot u_{3}\cdot v_{1}-{\frac {u_{3}}{u_{1}}}\cdot u_{2}\cdot v_{1}+u_{3}\cdot v_{2}}{u_{1}\cdot v_{3}-u_{3}\cdot v_{1}}}=K\cdot {\frac {u_{3}\cdot v_{2}-u_{2}\cdot v_{3}}{u_{1}\cdot v_{3}-u_{3}\cdot v_{1}}}$

Zusammengefasst:

$n_{1}=K\cdot {\frac {u_{3}\cdot v_{2}-u_{2}\cdot v_{3}}{u_{1}\cdot v_{3}-u_{3}\cdot v_{1}}}$

$n_{2}=K$

$n_{3}=K\cdot {\frac {u_{2}\cdot v_{1}-u_{1}\cdot v_{2}}{u_{1}\cdot v_{3}-u_{3}\cdot v_{1}}}$

Fall 3: $n_{3}=K$

aus 1) folgt:

3) $n_{1}=-{\frac {u_{2}\cdot n_{2}+u_{3}\cdot K}{u_{1}}}$

$n_{1}$ in 2):

$-v_{1}\cdot {\frac {u_{2}\cdot n_{2}+u_{3}\cdot K}{u_{1}}}+v_{2}\cdot n_{2}+v_{3}\cdot K=0$

daraus folgt:

$K\cdot \left(v_{3}-v_{1}\cdot {\frac {u_{3}}{u_{1}}}\right)+n_{2}\cdot \left(v_{2}-v_{1}\cdot {\frac {u_{2}}{u_{1}}}\right)=0$

$n_{2}=-K\cdot {\frac {v_{3}-v_{1}\cdot {\frac {u_{3}}{u_{1}}}}{v_{2}-v_{1}\cdot {\frac {u_{2}}{u_{1}}}}}=K\cdot {\frac {u_{3}\cdot v_{1}-u_{1}\cdot v_{3}}{u_{1}\cdot v_{2}-u_{2}\cdot v_{1}}}$

$n_{2}$ in 3):

$n_{1}=-{\frac {u_{2}}{u_{1}}}\cdot K\cdot {\frac {u_{3}\cdot v_{1}-u_{1}\cdot v_{3}}{u_{1}\cdot v_{2}-u_{2}\cdot v_{1}}}-{\frac {u_{3}}{u_{1}}}\cdot K=K\cdot {\frac {-{\frac {u_{2}}{u_{1}}}\cdot \left(u_{3}\cdot v_{1}-u_{1}\cdot v_{3}\right)-{\frac {u_{3}}{u_{1}}}\cdot \left(u_{1}\cdot v_{2}-u_{2}\cdot v_{1}\right)}{u_{1}\cdot v_{2}-u_{2}\cdot v_{1}}}=K\cdot {\frac {-{\frac {u_{2}}{u_{1}}}\cdot u_{3}\cdot v_{1}+u_{2}\cdot v_{3}-u_{3}\cdot v_{2}+{\frac {u_{3}}{u_{1}}}\cdot u_{2}\cdot v_{1}}{u_{1}\cdot v_{2}-u_{2}\cdot v_{1}}}=K\cdot {\frac {u_{2}\cdot v_{3}-u_{3}\cdot v_{2}}{u_{1}\cdot v_{2}-u_{2}\cdot v_{1}}}$

Zusammengefasst:

$n_{1}=K\cdot {\frac {u_{2}\cdot v_{3}-u_{3}\cdot v_{2}}{u_{1}\cdot v_{2}-u_{2}\cdot v_{1}}}$

$n_{2}=K\cdot {\frac {u_{3}\cdot v_{1}-u_{1}\cdot v_{3}}{u_{1}\cdot v_{2}-u_{2}\cdot v_{1}}}$

$n_{3}=K$

Fall 4: $|{\vec {n}}|=K$

3) $|{\vec {n}}|^{2}=K^{2}=n_{1}^{2}+n_{2}^{2}+n_{3}^{2}$