$E:\left(\begin{array}{c}6\\ -1\\ -4\end{array}\right)\circ [\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)-\left(\begin{array}{c}1\\ 0\\ 3\end{array}\right)]=0E:\backslash begin\{pmatrix\}6\backslash \backslash -1\backslash \backslash -4\backslash end\{pmatrix\}\backslash circ\backslash left[\backslash begin\{pmatrix\}x\_1\backslash \backslash x\_2\backslash \backslash \{x\}\_3\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}1\backslash \backslash 0\backslash \backslash 3\backslash end\{pmatrix\}\backslash right]=0$

Multipliziere die Klammern mit Hilfe des  Distributivgesetzes aus und berechne das Skalarprodukt.

$E:6x_1-6-x_2-4x_3+12=0E:\backslash ;6\{x\}\_1-6-\{x\}\_2-4\{x\}\_3+12=0$

Fasse zusammen.

$E:6x_1-x_2-4x_3+6=0E:6\{x\}\_1-\{x\}\_2-4\{x\}\_3+6=0$

$E:\left(\begin{array}{c}4\\ -7\\ 2\end{array}\right)\circ [\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)-\left(\begin{array}{c}0\\ 1\\ 2\end{array}\right)]=0E:\backslash begin\{pmatrix\}4\backslash \backslash -7\backslash \backslash 2\backslash end\{pmatrix\}\backslash circ\backslash left[\backslash begin\{pmatrix\}\{x\}\_1\backslash \backslash \{x\}\_2\backslash \backslash \{x\}\_3\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}0\backslash \backslash 1\backslash \backslash 2\backslash end\{pmatrix\}\backslash right]=0$

Berechne zunächst das Skalarprodukt :

$E:4\cdot x_1+(-7)\cdot (x_2-1)+2\cdot (x_3-2)=0E:\; 4\backslash cdot\{x\}\_1+\backslash left(-7\backslash right)\backslash cdot\backslash left(\{x\}\_2-1\backslash right)+2\backslash cdot\backslash left(\{x\}\_3-2\backslash right)=0$

Multipliziere nun die Klammern mit Hilfe des  Distributivgesetzes aus:

$E:4x_1-7x_2+7+2x_3-4=0E:\backslash ;4\{x\}\_1-7\{x\}\_2+7+2\{x\}\_3-4=0$

Fasse zusammen:

$E:4x_1-7x_2+2x_3+3=0E:\; 4\{x\}\_1-7\{x\}\_2+2\{x\}\_3+3=0$

$E:\left(\begin{array}{c}-6\\ -3\\ 2\end{array}\right)\circ \left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)=0E:\backslash ;\backslash begin\{pmatrix\}-6\backslash \backslash -3\backslash \backslash 2\backslash end\{pmatrix\}\backslash circ\backslash begin\{pmatrix\}\{x\}\_1\backslash \backslash \{x\}\_2\backslash \backslash \{x\}\_3\backslash end\{pmatrix\}=0$

Der zweite Faktor enthält keine Differenz. Berechne daher nur das Skalarprodukt :

$E:-6x_1-3x_2+2x_3=0\mathrm{\mid }\cdot (-1)E:-6\{x\}\_1-3\{x\}\_2+2\{x\}\_3=0\backslash ;\; |\; \backslash cdot\; (-1)$

$E:\phantom{-}6x_1+3x_2-2x_3=0E:\backslash phantom\{-\}6\{x\}\_1+3\{x\}\_2-2\{x\}\_3=0$

$E:\left(\begin{array}{c}-4\\ -2\\ 1\end{array}\right)\circ [\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)-\left(\begin{array}{c}1\\ 1\\ -1\end{array}\right)]=0E:\; \backslash begin\{pmatrix\}-4\backslash \backslash -2\backslash \backslash 1\backslash end\{pmatrix\}\backslash circ\backslash left[\backslash begin\{pmatrix\}\{x\}\_1\backslash \backslash \{x\}\_2\backslash \backslash \{x\}\_3\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}1\backslash \backslash 1\backslash \backslash -1\backslash end\{pmatrix\}\backslash right]=0$

Berechne das Skalarprodukt :

$E:-4\cdot (x_1-1)+(-2)\cdot (x_2-1)+1\cdot (x_3+1)=0E:\; -4\backslash cdot\backslash left(\{x\}\_1-1\backslash right)+\backslash left(-2\backslash right)\backslash cdot\backslash left(\{x\}\_2-1\backslash right)+1\backslash cdot\backslash left(\{x\}\_3+1\backslash right)=0$

Multipliziere die Klammern mit Hilfe des  Distributivgesetzes aus:

$E:-4x_1+4-2x_2+2+x_3+1=0E:\; -4\{x\}\_1+4-2\{x\}\_2+2+\{x\}\_3+1=0$

Fasse zusammen:

$E:-4x_1-2x_2+x_3+7=0E:\; -4\{x\}\_1-2\{x\}\_2+\{x\}\_3+7=0$

$E:\left(\begin{array}{c}1\\ 0\\ -1\end{array}\right)\circ [\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)-\left(\begin{array}{c}3\\ 1\\ 2\end{array}\right)]=0E:\; \backslash begin\{pmatrix\}1\backslash \backslash 0\backslash \backslash -1\backslash end\{pmatrix\}\backslash circ\backslash left[\backslash begin\{pmatrix\}\{x\}\_1\backslash \backslash \{x\}\_2\backslash \backslash \{x\}\_3\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}3\backslash \backslash 1\backslash \backslash 2\backslash end\{pmatrix\}\backslash right]=0$

Berechne das Skalarprodukt :

$E:1\cdot (x_1-3)+(-1)\cdot (x_3-2)=0E:\; 1\backslash cdot\backslash left(\{x\}\_1-3\backslash right)+\backslash left(-1\backslash right)\backslash cdot\backslash left(\{x\}\_3-2\backslash right)=0$

Multipliziere die Klammern mit Hilfe des  Distributivgesetzes aus:

$E:x_1-3-x_3+2=0E:\; x\_1-3-\{x\}\_3+2=0$

Fasse zusammen:

$E:x_1-x_3-1=0E:\; \{x\}\_1-\{x\}\_3-1=0$

$E:\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)\circ [\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)-\left(\begin{array}{c}2\\ 2\\ 2\end{array}\right)]=0E:\backslash begin\{pmatrix\}0\backslash \backslash 1\backslash \backslash 0\backslash end\{pmatrix\}\backslash circ\backslash left[\backslash begin\{pmatrix\}\{x\}\_1\backslash \backslash \{x\}\_2\backslash \backslash \{x\}\_3\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}2\backslash \backslash 2\backslash \backslash 2\backslash end\{pmatrix\}\backslash right]=0$

Berechne das Skalarprodukt :

$E:1\cdot (x_2-2)=0E:\; 1\backslash cdot\backslash left(\{x\}\_2-2\backslash right)=0$

$E:x_2-2=0E:\; \{x\}\_2-2=0$

(Hinweis: Die Ebene ist parallel zur  $x_1-x_3-\text{Ebene}\{x\}\_1-\{x\}\_3-\backslash text\{Ebene\}$ )

$E:\left(\begin{array}{c}-500\\ 550\\ 100\end{array}\right)\circ [\left(\begin{array}{c}{\mathrm{x}}_1\\ x_2\\ x_3\end{array}\right)-\left(\begin{array}{c}40\\ 80\\ 0\end{array}\right)]=0E:\; \backslash begin\{pmatrix\}-500\backslash \backslash 550\backslash \backslash 100\backslash end\{pmatrix\}\backslash circ\backslash left[\backslash begin\{pmatrix\}\{\backslash mathrm\; x\}\_1\backslash \backslash \{x\}\_2\backslash \backslash \{x\}\_3\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}40\backslash \backslash 80\backslash \backslash 0\backslash end\{pmatrix\}\backslash right]=0$

Berechne das Skalarprodukt .

$E:-500\cdot (x_1-40)+550\cdot (x_2-80)+100\cdot x_3=0E:\; -500\backslash cdot\backslash left(\{x\}\_1-40\backslash right)+550\backslash cdot\backslash left(\{\; x\}\_2-80\backslash right)+100\backslash cdot\{x\}\_3=0$

Multipliziere die Klammern mit Hilfe des  Distributivgesetzes aus.

$E:-500x_1+20000+550x_2-44000+100x_3=0E:\; -500\{x\}\_1+20000+550\; \{x\}\_2-44000+100\; \{x\}\_3=0$

Fasse zusammen.

$E:-500x_1+550x_2+100x_3-24000=0E:\; -500\{x\}\_1+550\{x\}\_2+100\{x\}\_3-24000=0$

Um die Gleichung noch zu vereinfachen, kann man sie auf beide Seiten durch  $5050$  dividieren.

$E:-10x_1+11x_2+2x_3-480=0E:\; -10\{x\}\_1+11\{x\}\_2+2\{x\}\_3-480=0$

Alternative Lösung: 

Meistens ist es einfacher, zuerst den Normalenvektor der Ebene zu vereinfachen und dann erst das Skalarprodukt zu berechnen.

Klammere  $5050$  im Normalenvektor aus, teile durch $5050$ und berechne dann das Skalarprodukt:

$E:\left(\begin{array}{c}-10\\ 11\\ 22\end{array}\right)\circ [\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)-\left(\begin{array}{c}40\\ 80\\ 0\end{array}\right)]=0E:\; \backslash begin\{pmatrix\}-10\backslash \backslash 11\backslash \backslash 22\backslash end\{pmatrix\}\backslash circ\backslash left[\backslash begin\{pmatrix\}\{x\}\_1\backslash \backslash \{x\}\_2\backslash \backslash \{x\}\_3\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}40\backslash \backslash 80\backslash \backslash 0\backslash end\{pmatrix\}\backslash right]=0$

Multipliziere die Klammern mit Hilfe des  Distributivgesetzes aus:

$E:-10\cdot (x_1-40)+11\cdot (x_2-80)+2\cdot x_3=0_{}^{}E:-10\backslash cdot\backslash left(x\_1-40\backslash right)+11\backslash cdot\backslash left(x\_2-80\backslash right)+2\backslash cdot\; x\_3=0\_\{\; \}^\{\; \}$

Multipliziere die Klammern mit Hilfe des  Distributivgesetzes aus:

$E:-10x_1+400+11x_2-880+2x_3=0E:\; -10\{x\}\_1+400+11\{x\}\_2-880+2\{x\}\_3=0$

Fasse zusammen.

$E:-10x_1+11x_2+2x_3-480=0E:\; -10\{x\}\_1+11\{x\}\_2+2\{x\}\_3-480=0$