Als Probe berechnen wir
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{\displaystyle {}{\begin{aligned}{\frac {3}{2}}x_{1}^{2}+2x_{2}^{2}+2x_{1}x_{2}-2x_{2}x_{3}&={\frac {3}{2}}{\left({\frac {2}{\sqrt {6}}}y_{1}+{\frac {2}{\sqrt {14}}}y_{2}-{\frac {1}{4{\sqrt {21}}}}y_{3}\right)}^{2}+2{\left(-{\frac {1}{\sqrt {6}}}y_{1}+{\frac {3}{\sqrt {14}}}y_{2}+{\frac {1}{2{\sqrt {21}}}}y_{3}\right)}^{2}+2{\left({\frac {2}{\sqrt {6}}}y_{1}+{\frac {2}{\sqrt {14}}}y_{2}-{\frac {1}{4{\sqrt {21}}}}y_{3}\right)}{\left(-{\frac {1}{\sqrt {6}}}y_{1}+{\frac {3}{\sqrt {14}}}y_{2}+{\frac {1}{2{\sqrt {21}}}}y_{3}\right)}-2{\left(-{\frac {1}{\sqrt {6}}}y_{1}+{\frac {3}{\sqrt {14}}}y_{2}+{\frac {1}{2{\sqrt {21}}}}y_{3}\right)}{\left({\frac {1}{\sqrt {6}}}y_{1}-{\frac {1}{\sqrt {14}}}y_{2}+{\frac {1}{\sqrt {21}}}y_{3}\right)}\\\,\end{aligned}}}
euklidischer Raum
ist insbesondere ein Hilbert-Raum. Diese beiden Begriffe werden vor allem für unendlich dimensionale Vektorräume eingesetzt.
Fakt
lässt sich der Abstand
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{\displaystyle {}{\overrightarrow {AB}}}
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{\displaystyle {}{\begin{aligned}d_{2}=\left\langle {\begin{pmatrix}a_{1}\\a_{2}\end{pmatrix}},{\frac {1}{c}}{\begin{pmatrix}b_{2}-a_{2}\\-b_{1}+a_{1}\end{pmatrix}}\right\rangle &={\frac {1}{c}}{\left(a_{1}b_{2}-a_{1}a_{2}-a_{2}b_{1}+a_{1}a_{2}\right)}\\&={\frac {1}{c}}{\left(a_{1}b_{2}-a_{2}b_{1}\right)},\,\end{aligned}}}
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{\displaystyle {}{\begin{aligned}d_{1}=\left\langle {\frac {1}{a+b+c}}{\begin{pmatrix}aa_{1}+bb_{1}+cc_{1}\\aa_{2}+bb_{2}+cc_{2}\end{pmatrix}},{\frac {1}{c}}{\begin{pmatrix}b_{2}-a_{2}\\-b_{1}+a_{1}\end{pmatrix}}\right\rangle &={\frac {1}{a+b+c}}{\frac {1}{c}}{\left(aa_{1}b_{2}-ba_{2}b_{1}+cc_{1}b_{2}-cc_{1}a_{2}-aa_{2}b_{1}+ba_{1}b_{2}-cb_{1}c_{2}+ca_{1}c_{2}\right)}\\&={\frac {1}{a+b+c}}{\frac {1}{c}}{\left(a(a_{1}b_{2}-a_{2}b_{1})+b(a_{1}b_{2}-a_{2}b_{1})+c(c_{1}b_{2}-b_{1}c_{2})+c(-c_{1}a_{2}+a_{1}c_{2})\right)}\\&={\frac {1}{a+b+c}}{\frac {1}{c}}{\left((a+b+c)(a_{1}b_{2}-a_{2}b_{1})+c(-a_{1}b_{2}+a_{2}b_{1}+c_{1}b_{2}-b_{1}c_{2}-c_{1}a_{2}+a_{1}c_{2})\right)}\\&={\frac {1}{c}}{\left(a_{1}b_{2}-a_{2}b_{1}\right)}+{\frac {1}{a+b+c}}{\left(\left\langle {\begin{pmatrix}a_{1}\\a_{2}\end{pmatrix}},{\begin{pmatrix}-b_{2}\\b_{1}\end{pmatrix}}\right\rangle +\left\langle {\begin{pmatrix}b_{1}\\b_{2}\end{pmatrix}},{\begin{pmatrix}-c_{2}\\c_{1}\end{pmatrix}}\right\rangle +\left\langle {\begin{pmatrix}c_{1}\\c_{2}\end{pmatrix}},{\begin{pmatrix}-a_{2}\\a_{1}\end{pmatrix}}\right\rangle \right)}\\&=d_{2}+{\frac {1}{a+b+c}}{\left(\left\langle {\begin{pmatrix}a_{1}\\a_{2}\end{pmatrix}},{\begin{pmatrix}-b_{2}\\b_{1}\end{pmatrix}}\right\rangle +\left\langle {\begin{pmatrix}b_{1}\\b_{2}\end{pmatrix}},{\begin{pmatrix}-c_{2}\\c_{1}\end{pmatrix}}\right\rangle +\left\langle {\begin{pmatrix}c_{1}\\c_{2}\end{pmatrix}},{\begin{pmatrix}-a_{2}\\a_{1}\end{pmatrix}}\right\rangle \right)}.\,\end{aligned}}}
Der rechte Summand ist
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{\displaystyle {}d}
Man kann auch zur Berechnung des Schnittpunkts die obige Bedingung mit
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{\displaystyle s(ab(b_{1}-a_{1})+ac(c_{1}-a_{1}))-t(ab(a_{1}-b_{1})+bc(c_{1}-b_{1}))=abc(b_{1}-a_{1})}
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{\displaystyle s(ab(b_{2}-a_{2})+ac(c_{2}-a_{2}))-t(ab(a_{2}-b_{2})+bc(c_{2}-b_{2}))=abc(b_{2}-a_{2})}
schreiben.
Mit Cramer:
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{\displaystyle {}{\begin{aligned}s&={\frac {[[:Vorlage:Op:det]]}{[[:Vorlage:Op:det]]}}\\&={\frac {bc[[:Vorlage:Op:det]]}{[[:Vorlage:Op:det]]}}\\&=bc{\frac {(b_{1}-a_{1})(a(a_{2}-b_{2})+c(c_{2}-b_{2}))-(b_{2}-a_{2})(a(a_{1}-b_{1})+c(c_{1}-b_{1}))}{(b(b_{1}-a_{1})+c(c_{1}-a_{1}))(a(a_{2}-b_{2})+c(c_{2}-b_{2}))-(a(a_{1}-b_{1})+c(c_{1}-b_{1}))(b(b_{2}-a_{2})+c(c_{2}-a_{2}))}}\\&=bc{\frac {a(b_{1}-a_{1})(a_{2}-b_{2})+c(b_{1}-a_{1})(c_{2}-b_{2})-a(b_{2}-a_{2})(a_{1}-b_{1})-c(b_{2}-a_{2})(c_{1}-b_{1})}{ab(b_{1}-a_{1})(a_{2}-b_{2})+bc(b_{1}-a_{1})(c_{2}-b_{2})+ac(c_{1}-a_{1})(a_{2}-b_{2})+c^{2}(c_{1}-a_{1})(c_{2}-b_{2})-ab(a_{1}-b_{1})(b_{2}-a_{2})-ac(a_{1}-b_{1})(c_{2}-a_{2})+bc(c_{1}-b_{1})(b_{2}-a_{2})-ac(c_{1}-b_{1})(c_{2}-a_{2})}}\\&=bc{\frac {(b_{1}-a_{1})(c_{2}-b_{2})-(b_{2}-a_{2})(c_{1}-b_{1})}{b(b_{1}-a_{1})(c_{2}-b_{2})+a(c_{1}-a_{1})(a_{2}-b_{2})+c(c_{1}-a_{1})(c_{2}-b_{2})-a(a_{1}-b_{1})(c_{2}-a_{2})+b(c_{1}-b_{1})(b_{2}-a_{2})-a(c_{1}-b_{1})(c_{2}-a_{2})}}\\&=bc{\frac {(b_{1}-a_{1})(c_{2}-b_{2})-(b_{2}-a_{2})(c_{1}-b_{1})}{(a+b+c)((b_{1}-a_{1})(c_{2}-b_{2})-(b_{2}-a_{2})(c_{1}-b_{1}))}}\\&={\frac {bc}{a+b+c}}.\end{aligned}}}
![{\displaystyle {}{\begin{aligned}s&={\frac {[[:Vorlage:Op:det]]}{[[:Vorlage:Op:det]]}}\\&={\frac {bc[[:Vorlage:Op:det]]}{[[:Vorlage:Op:det]]}}\\&=bc{\frac {(b_{1}-a_{1})(a(a_{2}-b_{2})+c(c_{2}-b_{2}))-(b_{2}-a_{2})(a(a_{1}-b_{1})+c(c_{1}-b_{1}))}{(b(b_{1}-a_{1})+c(c_{1}-a_{1}))(a(a_{2}-b_{2})+c(c_{2}-b_{2}))-(a(a_{1}-b_{1})+c(c_{1}-b_{1}))(b(b_{2}-a_{2})+c(c_{2}-a_{2}))}}\\&=bc{\frac {a(b_{1}-a_{1})(a_{2}-b_{2})+c(b_{1}-a_{1})(c_{2}-b_{2})-a(b_{2}-a_{2})(a_{1}-b_{1})-c(b_{2}-a_{2})(c_{1}-b_{1})}{ab(b_{1}-a_{1})(a_{2}-b_{2})+bc(b_{1}-a_{1})(c_{2}-b_{2})+ac(c_{1}-a_{1})(a_{2}-b_{2})+c^{2}(c_{1}-a_{1})(c_{2}-b_{2})-ab(a_{1}-b_{1})(b_{2}-a_{2})-ac(a_{1}-b_{1})(c_{2}-a_{2})+bc(c_{1}-b_{1})(b_{2}-a_{2})-ac(c_{1}-b_{1})(c_{2}-a_{2})}}\\&=bc{\frac {(b_{1}-a_{1})(c_{2}-b_{2})-(b_{2}-a_{2})(c_{1}-b_{1})}{b(b_{1}-a_{1})(c_{2}-b_{2})+a(c_{1}-a_{1})(a_{2}-b_{2})+c(c_{1}-a_{1})(c_{2}-b_{2})-a(a_{1}-b_{1})(c_{2}-a_{2})+b(c_{1}-b_{1})(b_{2}-a_{2})-a(c_{1}-b_{1})(c_{2}-a_{2})}}\\&=bc{\frac {(b_{1}-a_{1})(c_{2}-b_{2})-(b_{2}-a_{2})(c_{1}-b_{1})}{(a+b+c)((b_{1}-a_{1})(c_{2}-b_{2})-(b_{2}-a_{2})(c_{1}-b_{1}))}}\\&={\frac {bc}{a+b+c}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/64fa5881ed8c015928ff5f6a98696155d69ce100)