• Pythagoras und Skalarprodukt
  • anonym
  • 30.06.2020
  • Mathematik
  • 11
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anonym

Zueinander orthogonale Vektoren

γ=\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \gamma =
ab\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \vec{a}\perp \vec{b}

Es gilt:

Es gilt:

a2+b2=\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} a^{2}+b^{2} =
a2+b2=ab2\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \mid\vec{a}\mid ^{2}+\mid\vec{b}\mid ^{2}=\mid\vec{a}-\vec{b}\mid ^{2}

Daher gilt auch:

Daher soll gelten:

a=\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \mid\vec{a}\mid =
=>a2=\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} => \mid\vec{a}\mid ^{2} =
=>b2=\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} => \mid\vec{b}\mid ^{2} =
=>ab2=(a1b1)2+\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} => \mid\vec{a}-\vec{b}\mid ^{2} = \left ( {a_{1}}-{b_{1}} \right )^{2}+
=>ab2=a122a1b1+b12+\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} => \mid\vec{a}-\vec{b}\mid ^{2} = {a_{1}}^{2}-2{a_{1}}{b_{1}}+{b_{1}}^{2} +

Letztendlich gilt:

ab2=a12+a22+b12+b222(a1b1+a2b2)\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \mid\vec{a}-\vec{b}\mid ^{2} = {a_{1}}^{2}+{a_{2}}^{2}+{b_{1}}^{2}+{b_{2}}^{2} -2({a_{1}}{b_{1}}+{a_{2}}{b_{2}})

Ergänze (siehe oben):

a2+b2=\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \mid\vec{a}\mid ^{2} +\mid\vec{b}\mid ^{2}=
1
Wann ist die Gleichung


erfüllt?
a2+b2=ab2\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \mid\vec{a}\mid ^{2}+\mid\vec{b}\mid ^{2}=\mid\vec{a}-\vec{b}\mid ^{2}
Skalarprodukt

Die Vektoren                      sind genau dann                                  zueinander, wenn gilt:

                                                                   

                                           wird als Skalarprodukt der Vektoren                    bezeichnet.

a  und  b\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \vec{a} \; und \; \vec{b}
ab=a1b1+a2b2=\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \vec{a}\cdot\vec{b} = {a_{1}}{b_{1}}+{a_{2}}{b_{2}} =
ab=a1b1+a2b2\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \vec{a}\cdot\vec{b} = {a_{1}}{b_{1}}+{a_{2}}{b_{2}}
a  und  b\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \vec{a} \;und \;\vec{b}