Die beiden sich außen berührenden Kugeln können auf zwei verschiedene Arten angeordnet sein.

Fall 1: Die größere Kugel liegt rechts von der kleineren Kugel.

Fall 2: Die größere Kugel liegt links von der kleineren Kugel.

Lösung für den Fall 1

Berechnung von $\overrightarrow{B{M}_2}\backslash overrightarrow\{BM\_2\}$ für Kugel $K_{2}K\_2$

Aus dem gegebenen Richtungsvektor ${\vec{v}}_{M_1{M}_2}=\left(\begin{array}{c}-1\\ 2\\ -2\end{array}\right)\backslash vec\; v\_\{M\_1M\_2\}=\backslash begin\{pmatrix\}-1\backslash \backslash 2\backslash \backslash -2\backslash end\{pmatrix\}$ erstellst du einen Einheitsvektor ${\vec{n}}_0={\displaystyle \frac{{\vec{v}}_{M_1{M}_2}}{\mathrm{\mid }{\vec{v}}_{M_1{M}_2}\mathrm{\mid }}}\backslash vec\; n\_0=\backslash dfrac\{\backslash vec\; v\_\{M\_1M\_2\}\}\{|\backslash vec\; v\_\{M\_1M\_2\}|\}$. Dazu muss der Betrag von ${\vec{v}}_{M_1{M}_2}\backslash vec\; v\_\{M\_1M\_2\}$ berechnet werden:

$\mathrm{\mid }{\vec{v}}_{M_1{M}_2}\mathrm{\mid }=\sqrt{(-1{)}^2+2^2+(-2{)}^2}=\sqrt{9}=3\Rightarrow {\vec{n}}_0={\displaystyle \frac{1}{3}}\cdot \left(\begin{array}{c}-1\\ 2\\ -2\end{array}\right)|\backslash vec\; v\_\{M\_1M\_2\}|=\backslash sqrt\{(-1)^2+2^2+(-2)^2\}=\backslash sqrt\{9\}=3\backslash ;\backslash Rightarrow\backslash ;\backslash vec\; n\_0=\backslash dfrac\{1\}\{3\}\backslash cdot\; \backslash begin\{pmatrix\}-1\backslash \backslash 2\backslash \backslash -2\backslash end\{pmatrix\}$

Für den Vektor $\overrightarrow{B{M}_2}\backslash overrightarrow\{BM\_2\}$ gilt mit dem berechneten Einheitsvektor ${\vec{n}}_0\backslash vec\; n\_0$ und $r_2=4r\_2=4$:

$\overrightarrow{B{M}_2}=r_2\cdot {\vec{n}}_0=4\cdot {\displaystyle \frac{1}{3}}\cdot \left(\begin{array}{c}-1\\ 2\\ -2\end{array}\right)={\displaystyle \frac{4}{3}}\cdot \left(\begin{array}{c}-1\\ 2\\ -2\end{array}\right)\backslash overrightarrow\{BM\_2\}=r\_2\; \backslash cdot\; \backslash vec\; n\_0=4\backslash cdot\backslash dfrac\{1\}\{3\}\backslash cdot\; \backslash begin\{pmatrix\}-1\backslash \backslash 2\backslash \backslash -2\backslash end\{pmatrix\}=\backslash dfrac\{4\}\{3\}\backslash cdot\; \backslash begin\{pmatrix\}-1\backslash \backslash 2\backslash \backslash -2\backslash end\{pmatrix\}$

Berechnung von $\overrightarrow{O{M}_2}\backslash overrightarrow\{OM\_2\}$

$\overrightarrow{O{M}_2}=\overrightarrow{OB}+\overrightarrow{B{M}_2}\backslash overrightarrow\{OM\_2\}=\backslash overrightarrow\{OB\}+\backslash overrightarrow\{BM\_2\}$

$\overrightarrow{O{M}_2}=\left(\begin{array}{c}1\\ -1\\ 1\end{array}\right)+{\displaystyle \frac{4}{3}}\cdot \left(\begin{array}{c}-1\\ 2\\ -2\end{array}\right)=\left(\begin{array}{c}1-{\displaystyle \frac{4}{3}}\\ -1+{\displaystyle \frac{8}{3}}\\ 1-{\displaystyle \frac{8}{3}}\end{array}\right)=\left(\begin{array}{c}-{\displaystyle \frac{1}{3}}\\ {\displaystyle \frac{5}{3}}\\ -{\displaystyle \frac{5}{3}}\end{array}\right)\backslash overrightarrow\{OM\_2\}=\backslash begin\{pmatrix\}1\backslash \backslash -1\backslash \backslash 1\backslash end\{pmatrix\}+\backslash dfrac\{4\}\{3\}\backslash cdot\; \backslash begin\{pmatrix\}-1\backslash \backslash 2\backslash \backslash -2\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}1-\backslash dfrac\{4\}\{3\}\backslash \backslash [2ex]-1+\backslash dfrac\{8\}\{3\}\backslash \backslash [2ex]1-\backslash dfrac\{8\}\{3\}\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}-\backslash dfrac\{1\}\{3\}\backslash \backslash [2ex]\backslash dfrac\{5\}\{3\}\backslash \backslash [2ex]-\backslash dfrac\{5\}\{3\}\backslash end\{pmatrix\}$

Der Mittelpunkt der Kugel $K_{2}K\_2$ hat die Koordinaten:

$M_2(-{\displaystyle \frac{1}{3}}\mid {\displaystyle \frac{5}{3}}\mid -{\displaystyle \frac{5}{3}})M\_2\backslash left(-\backslash dfrac\{1\}\{3\}\backslash Big\backslash vert\backslash dfrac\{5\}\{3\}\backslash Big\backslash vert-\backslash dfrac\{5\}\{3\}\backslash right)$.

Berechnung von $\overrightarrow{B{M}_1}\backslash overrightarrow\{BM\_1\}$ für Kugel $K_{1}K\_1$

Für den Vektor $\overrightarrow{B{M}_1}\backslash overrightarrow\{BM\_1\}$ gilt mit dem oben berechneten Einheitsvektor $(-{\vec{n}}_0)(-\backslash vec\; n\_0)$ und $r_1=2r\_1=2$:

$\overrightarrow{B{M}_1}=r_1\cdot (-{\vec{n}}_0)=2\cdot {\displaystyle \frac{1}{3}}\cdot \left(\begin{array}{c}1\\ -2\\ 2\end{array}\right)={\displaystyle \frac{2}{3}}\cdot \left(\begin{array}{c}1\\ -2\\ 2\end{array}\right)\backslash overrightarrow\{BM\_1\}=r\_1\; \backslash cdot\; (-\backslash vec\; n\_0)=2\backslash cdot\backslash dfrac\{1\}\{3\}\backslash cdot\; \backslash begin\{pmatrix\}1\backslash \backslash -2\backslash \backslash 2\backslash end\{pmatrix\}=\backslash dfrac\{2\}\{3\}\backslash cdot\; \backslash begin\{pmatrix\}1\backslash \backslash -2\backslash \backslash 2\backslash end\{pmatrix\}$

Berechnung von $\overrightarrow{O{M}_1}\backslash overrightarrow\{OM\_1\}$

$\overrightarrow{O{M}_1}=\overrightarrow{OB}+\overrightarrow{B{M}_1}\backslash overrightarrow\{OM\_1\}=\backslash overrightarrow\{OB\}+\backslash overrightarrow\{BM\_1\}$

$\overrightarrow{O{M}_1}=\left(\begin{array}{c}1\\ -1\\ 1\end{array}\right)+{\displaystyle \frac{2}{3}}\cdot \left(\begin{array}{c}1\\ -2\\ 2\end{array}\right)=\left(\begin{array}{c}1+{\displaystyle \frac{2}{3}}\\ -1-{\displaystyle \frac{4}{3}}\\ 1+{\displaystyle \frac{4}{3}}\end{array}\right)=\left(\begin{array}{c}{\displaystyle \frac{5}{3}}\\ -{\displaystyle \frac{7}{3}}\\ {\displaystyle \frac{7}{3}}\end{array}\right)\backslash overrightarrow\{OM\_1\}=\backslash begin\{pmatrix\}1\backslash \backslash -1\backslash \backslash 1\backslash end\{pmatrix\}+\backslash dfrac\{2\}\{3\}\backslash cdot\; \backslash begin\{pmatrix\}1\backslash \backslash -2\backslash \backslash 2\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}1+\backslash dfrac\{2\}\{3\}\backslash \backslash [2ex]-1-\backslash dfrac\{4\}\{3\}\backslash \backslash [2ex]1+\backslash dfrac\{4\}\{3\}\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}\backslash dfrac\{5\}\{3\}\backslash \backslash [2ex]-\backslash dfrac\{7\}\{3\}\backslash \backslash [2ex]\backslash dfrac\{7\}\{3\}\backslash end\{pmatrix\}$

Der Mittelpunkt der Kugel $K_{1}K\_1$ hat die Koordinaten:

$M_1({\displaystyle \frac{5}{3}}\mid -{\displaystyle \frac{7}{3}}\mid {\displaystyle \frac{7}{3}})M\_1\backslash left(\backslash dfrac\{5\}\{3\}\backslash Big\backslash vert-\backslash dfrac\{7\}\{3\}\backslash Big\backslash vert\backslash dfrac\{7\}\{3\}\backslash right)$.

Antwort: Damit können die beiden Kugelgleichungen angegeben werden:

$K_1:{(\vec{x}-\left(\begin{array}{c}{\displaystyle \frac{5}{3}}\\ -{\displaystyle \frac{7}{3}}\\ {\displaystyle \frac{7}{3}}\end{array}\right))}^2=2^2=4K\_1:\backslash ;\; \backslash left(\backslash vec\; x-\backslash begin\{pmatrix\}\backslash dfrac\{5\}\{3\}\backslash \backslash [2ex]-\backslash dfrac\{7\}\{3\}\backslash \backslash [2ex]\backslash dfrac\{7\}\{3\}\backslash end\{pmatrix\}\backslash right)^2=2^2=4$ und

$K_2:{(\vec{x}-\left(\begin{array}{c}-{\displaystyle \frac{1}{3}}\\ {\displaystyle \frac{5}{3}}\\ -{\displaystyle \frac{5}{3}}\end{array}\right))}^2=4^2=16K\_2:\backslash ;\; \backslash left(\backslash vec\; x-\backslash begin\{pmatrix\}-\backslash dfrac\{1\}\{3\}\backslash \backslash [2ex]\backslash dfrac\{5\}\{3\}\backslash \backslash [2ex]-\backslash dfrac\{5\}\{3\}\backslash end\{pmatrix\}\backslash right)^2=4^2=16$

Lösung für den Fall 2

Berechnung von $\overrightarrow{B{M}_2^{\mathrm{\prime }}}\backslash overrightarrow\{BM\_2\text{'}\}$ für Kugel $K_{2}K\_2$

Für den Vektor $\overrightarrow{B{M}_2^{\mathrm{\prime }}}\backslash overrightarrow\{BM\_2\text{'}\}$ gilt mit dem oben berechneten Einheitsvektor $(-{\vec{n}}_0)(-\backslash vec\; n\_0)$ und $r_2=4r\_2=4$:

$\overrightarrow{B{M}_2^{\mathrm{\prime }}}=r_2\cdot (-{\vec{n}}_0)=4\cdot {\displaystyle \frac{1}{3}}\cdot \left(\begin{array}{c}1\\ -2\\ 2\end{array}\right)={\displaystyle \frac{4}{3}}\cdot \left(\begin{array}{c}1\\ -2\\ 2\end{array}\right)\backslash overrightarrow\{BM\_2\text{'}\}=r\_2\; \backslash cdot(-\backslash vec\; n\_0)=4\backslash cdot\backslash dfrac\{1\}\{3\}\backslash cdot\; \backslash begin\{pmatrix\}1\backslash \backslash -2\backslash \backslash 2\backslash end\{pmatrix\}=\backslash dfrac\{4\}\{3\}\backslash cdot\; \backslash begin\{pmatrix\}1\backslash \backslash -2\backslash \backslash 2\backslash end\{pmatrix\}$

Berechnung von $\overrightarrow{O{M}_2^{\mathrm{\prime }}}\backslash overrightarrow\{OM\_2\text{'}\}$

$\overrightarrow{O{M}_2^{\mathrm{\prime }}}=\overrightarrow{OB}+\overrightarrow{B{M}_2^{\mathrm{\prime }}}\backslash overrightarrow\{OM\_2\text{'}\}=\backslash overrightarrow\{OB\}+\backslash overrightarrow\{BM\_2\text{'}\}$

$\overrightarrow{O{M}_2^{\mathrm{\prime }}}=\left(\begin{array}{c}1\\ -1\\ 1\end{array}\right)+{\displaystyle \frac{4}{3}}\cdot \left(\begin{array}{c}1\\ -2\\ 2\end{array}\right)=\left(\begin{array}{c}1+{\displaystyle \frac{4}{3}}\\ -1-{\displaystyle \frac{8}{3}}\\ 1+{\displaystyle \frac{8}{3}}\end{array}\right)=\left(\begin{array}{c}{\displaystyle \frac{7}{3}}\\ -{\displaystyle \frac{11}{3}}\\ {\displaystyle \frac{11}{3}}\end{array}\right)\backslash overrightarrow\{OM\_2\text{'}\}=\backslash begin\{pmatrix\}1\backslash \backslash -1\backslash \backslash 1\backslash end\{pmatrix\}+\backslash dfrac\{4\}\{3\}\backslash cdot\; \backslash begin\{pmatrix\}1\backslash \backslash -2\backslash \backslash 2\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}1+\backslash dfrac\{4\}\{3\}\backslash \backslash [2ex]-1-\backslash dfrac\{8\}\{3\}\backslash \backslash [2ex]1+\backslash dfrac\{8\}\{3\}\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}\backslash dfrac\{7\}\{3\}\backslash \backslash [2ex]-\backslash dfrac\{11\}\{3\}\backslash \backslash [2ex]\backslash dfrac\{11\}\{3\}\backslash end\{pmatrix\}$

Der Mittelpunkt der Kugel $K_{2}K\_2$ hat die Koordinaten:

$M_2^{\mathrm{\prime }}({\displaystyle \frac{7}{3}}\mid -{\displaystyle \frac{11}{3}}\mid {\displaystyle \frac{11}{3}})M\_2\text{'}\backslash left(\backslash dfrac\{7\}\{3\}\backslash Big\backslash vert-\backslash dfrac\{11\}\{3\}\backslash Big\backslash vert\backslash dfrac\{11\}\{3\}\backslash right)$.

Berechnung von $\overrightarrow{B{M}_1^{\mathrm{\prime }}}\backslash overrightarrow\{BM\_1\text{'}\}$ für Kugel $K_{1}K\_1$

Für den Vektor $\overrightarrow{B{M}_1^{\mathrm{\prime }}}\backslash overrightarrow\{BM\_1\text{'}\}$ gilt mit dem oben berechneten Einheitsvektor ${\vec{n}}_0\backslash vec\; n\_0$ und $r_1=2r\_1=2$:

$\overrightarrow{B{M}_1^{\mathrm{\prime }}}=r_1\cdot {\vec{n}}_0=2\cdot {\displaystyle \frac{1}{3}}\cdot \left(\begin{array}{c}-1\\ 2\\ -2\end{array}\right)={\displaystyle \frac{2}{3}}\cdot \left(\begin{array}{c}-1\\ 2\\ -2\end{array}\right)\backslash overrightarrow\{BM\_1\text{'}\}=r\_1\; \backslash cdot\; \backslash vec\; n\_0=2\backslash cdot\backslash dfrac\{1\}\{3\}\backslash cdot\; \backslash begin\{pmatrix\}-1\backslash \backslash 2\backslash \backslash -2\backslash end\{pmatrix\}=\backslash dfrac\{2\}\{3\}\backslash cdot\; \backslash begin\{pmatrix\}-1\backslash \backslash 2\backslash \backslash -2\backslash end\{pmatrix\}$

Berechnung von $\overrightarrow{O{M}_1^{\mathrm{\prime }}}\backslash overrightarrow\{OM\_1\text{'}\}$

$\overrightarrow{O{M}_1^{\mathrm{\prime }}}=\overrightarrow{OB}+\overrightarrow{B{M}_1^{\mathrm{\prime }}}\backslash overrightarrow\{OM\_1\text{'}\}=\backslash overrightarrow\{OB\}+\backslash overrightarrow\{BM\_1\text{'}\}$

$\overrightarrow{O{M}_1}=\left(\begin{array}{c}1\\ -1\\ 1\end{array}\right)+{\displaystyle \frac{2}{3}}\cdot \left(\begin{array}{c}-1\\ 2\\ -2\end{array}\right)=\left(\begin{array}{c}1-{\displaystyle \frac{2}{3}}\\ -1+{\displaystyle \frac{4}{3}}\\ 1-{\displaystyle \frac{4}{3}}\end{array}\right)=\left(\begin{array}{c}{\displaystyle \frac{1}{3}}\\ {\displaystyle \frac{1}{3}}\\ -{\displaystyle \frac{1}{3}}\end{array}\right)\backslash overrightarrow\{OM\_1\}=\backslash begin\{pmatrix\}1\backslash \backslash -1\backslash \backslash 1\backslash end\{pmatrix\}+\backslash dfrac\{2\}\{3\}\backslash cdot\; \backslash begin\{pmatrix\}-1\backslash \backslash 2\backslash \backslash -2\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}1-\backslash dfrac\{2\}\{3\}\backslash \backslash [2ex]-1+\backslash dfrac\{4\}\{3\}\backslash \backslash [2ex]1-\backslash dfrac\{4\}\{3\}\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}\backslash dfrac\{1\}\{3\}\backslash \backslash [2ex]\backslash dfrac\{1\}\{3\}\backslash \backslash [2ex]-\backslash dfrac\{1\}\{3\}\backslash end\{pmatrix\}$

Der Mittelpunkt der Kugel $K_{1}K\_1$ hat die Koordinaten:

$M_1({\displaystyle \frac{1}{3}}\mid {\displaystyle \frac{1}{3}}\mid -{\displaystyle \frac{1}{3}})M\_1\backslash left(\backslash dfrac\{1\}\{3\}\backslash Big\backslash vert\backslash dfrac\{1\}\{3\}\backslash Big\backslash vert-\backslash dfrac\{1\}\{3\}\backslash right)$.

Antwort: Damit können die beiden Kugelgleichungen angegeben werden

$K_1^{\mathrm{\prime }}:{(\vec{x}-\left(\begin{array}{c}{\displaystyle \frac{1}{3}}\\ {\displaystyle \frac{1}{3}}\\ -{\displaystyle \frac{1}{3}}\end{array}\right))}^2=2^2=4K\_1\text{'}:\backslash ;\; \backslash left(\backslash vec\; x-\backslash begin\{pmatrix\}\backslash dfrac\{1\}\{3\}\backslash \backslash [2ex]\backslash dfrac\{1\}\{3\}\backslash \backslash [2ex]-\backslash dfrac\{1\}\{3\}\backslash end\{pmatrix\}\backslash right)^2=2^2=4$ und

$K_2^{\mathrm{\prime }}:{(\vec{x}-\left(\begin{array}{c}{\displaystyle \frac{7}{3}}\\ -{\displaystyle \frac{11}{3}}\\ {\displaystyle \frac{11}{3}}\end{array}\right))}^2=4^2=16K\_2\text{'}:\backslash ;\; \backslash left(\backslash vec\; x-\backslash begin\{pmatrix\}\backslash dfrac\{7\}\{3\}\backslash \backslash [2ex]-\backslash dfrac\{11\}\{3\}\backslash \backslash [2ex]\backslash dfrac\{11\}\{3\}\backslash end\{pmatrix\}\backslash right)^2=4^2=16$