Die Kantenlänge des Würfels soll 12 betragen

Berechne den Betrag des Vektors $\overrightarrow{AC}\backslash overrightarrow\{AC\}$.

$\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=\left(\begin{array}{c}-3\\ -6\\ 9\end{array}\right)-\left(\begin{array}{c}1\\ 2\\ 1\end{array}\right)=\left(\begin{array}{c}-4\\ -8\\ 8\end{array}\right)\backslash overrightarrow\{AC\}=\backslash overrightarrow\{OC\}-\backslash overrightarrow\{OA\}=\backslash begin\{pmatrix\}-3\backslash \backslash -6\backslash \backslash 9\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}1\backslash \backslash 2\backslash \backslash 1\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}-4\backslash \backslash -8\backslash \backslash 8\backslash end\{pmatrix\}$

$\mid \overrightarrow{AC}\mid =\sqrt{(-4{)}^2+(-8{)}^2+8^2}=\sqrt{16+64+64}=\sqrt{144}=12\backslash left|\backslash overrightarrow\{AC\}\backslash right|=\backslash sqrt\{(-4)^2+(-8)^2+8^2\}=\backslash sqrt\{16+64+64\}=\backslash sqrt\{144\}=12$

Die Kantenlänge des Würfels beträgt $12\text{LE}12\backslash ;\backslash text\{LE\}$.

Die Koordinaten eines der beiden Eckpunkte des Oktaeders, die nicht in H liegen.

Die Spitze $SS$ der oberen Pyramide liegt senkrecht über dem Mittelpunkt von $\overline{AC}\backslash overline\{AC\}$.

$\overrightarrow{O{M}_{AC}}={\displaystyle \frac{1}{2}}(\overrightarrow{OA}+\overrightarrow{OC})={\displaystyle \frac{1}{2}}\left(\begin{array}{c}1-3\\ 2-6\\ 1+9\end{array}\right)={\displaystyle \frac{1}{2}}\left(\begin{array}{c}-2\\ -4\\ 10\end{array}\right)=\left(\begin{array}{c}-1\\ -2\\ 5\end{array}\right)\backslash overrightarrow\{OM\_\{AC\}\}=\backslash dfrac\{1\}\{2\}\backslash left(\backslash overrightarrow\{OA\}+\backslash overrightarrow\{OC\}\backslash right)=\backslash dfrac\{1\}\{2\}\backslash begin\{pmatrix\}1-3\backslash \backslash 2-6\backslash \backslash 1+9\backslash end\{pmatrix\}=\backslash dfrac\{1\}\{2\}\backslash begin\{pmatrix\}-2\backslash \backslash -4\backslash \backslash 10\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}-1\backslash \backslash -2\backslash \backslash 5\backslash end\{pmatrix\}$

Weiterhin kann der Normalenvektor der Ebene $HH$ abgelesen werden:

${\vec{n}}_H=\left(\begin{array}{c}2\\ 1\\ 2\end{array}\right)\backslash vec\; n\_H=\backslash begin\{pmatrix\}2\backslash \backslash 1\backslash \backslash 2\backslash end\{pmatrix\}$

Sein Betrag ist $\mid {\vec{n}}_H\mid =\sqrt{2^2+1^2+2^2}=\sqrt{4+1+4}=\sqrt{9}=3\backslash left|\backslash vec\; n\_H\backslash right|=\backslash sqrt\{2^2+1^2+2^2\}=\backslash sqrt\{4+1+4\}=\backslash sqrt\{9\}=3$.

Da die Kantenlänge des Würfels $1212$ ist, ist die Höhe einer Pyramide gleich $66$.

Da der Betrag des Normalenvektors $33$ beträgt, muss er zweimal zum Vektor $\overrightarrow{O{M}_{AC}}\backslash overrightarrow\{OM\_\{AC\}\}$ addiert werden, um zur Pyramidenspitze $SS$ zu gelangen ($h=6=2\cdot \mid {\vec{n}}_H\mid h=6=2\backslash cdot\; \backslash left|\backslash vec\; n\_H\backslash right|$).

Dann folgt für den Vektor $\overrightarrow{OS}\backslash overrightarrow\{OS\}$:

$\overrightarrow{OS}=\overrightarrow{O{M}_{AC}}+2\cdot {\vec{n}}_H=\left(\begin{array}{c}-1\\ -2\\ 5\end{array}\right)+2\cdot \left(\begin{array}{c}2\\ 1\\ 2\end{array}\right)=\left(\begin{array}{c}-1+4\\ -2+2\\ 5+4\end{array}\right)=\left(\begin{array}{c}3\\ 0\\ 9\end{array}\right)\backslash overrightarrow\{OS\}=\backslash overrightarrow\{OM\_\{AC\}\}+2\backslash cdot\; \backslash vec\; n\_H=\backslash begin\{pmatrix\}-1\backslash \backslash -2\backslash \backslash 5\backslash end\{pmatrix\}+2\backslash cdot\backslash begin\{pmatrix\}2\backslash \backslash 1\backslash \backslash 2\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}-1+4\backslash \backslash -2+2\backslash \backslash 5+4\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}3\backslash \backslash 0\backslash \backslash 9\backslash end\{pmatrix\}$

Der eine Eckpunkt $SS$ des Oktaeders hat die Koordinaten $S(3\mathrm{\mid }0\mathrm{\mid }9)S(3|0|9)$.

Ergänzung (nicht in der Aufgabenstellung gefordert):

Der andere Eckpunkt $S^{\mathrm{\prime }}S\text{'}$, der nicht in $HH$ liegt, ist dann:

$\overrightarrow{O{S}^{\mathrm{\prime }}}=\overrightarrow{O{M}_{AC}}-2\cdot {\vec{n}}_H=\left(\begin{array}{c}-1\\ -2\\ 5\end{array}\right)-2\cdot \left(\begin{array}{c}2\\ 1\\ 2\end{array}\right)=\left(\begin{array}{c}-1-4\\ -2-2\\ 5-4\end{array}\right)=\left(\begin{array}{c}-5\\ -4\\ 1\end{array}\right)\backslash overrightarrow\{OS\text{'}\}=\backslash overrightarrow\{OM\_\{AC\}\}-2\backslash cdot\; \backslash vec\; n\_H=\backslash begin\{pmatrix\}-1\backslash \backslash -2\backslash \backslash 5\backslash end\{pmatrix\}-2\backslash cdot\backslash begin\{pmatrix\}2\backslash \backslash 1\backslash \backslash 2\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}-1-4\backslash \backslash -2-2\backslash \backslash 5-4\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}-5\backslash \backslash -4\backslash \backslash 1\backslash end\{pmatrix\}$

Der andere Eckpunkt des Oktaeders $S^{\mathrm{\prime }}S\text{'}$ hat die Koordinaten $S^{\mathrm{\prime }}(-5\mathrm{\mid }-4\mathrm{\mid }1)S\text{'}(-5|-4|1)$.