Der Mittelpunkt $M(m_1\mathrm{\mid }{m}_2\mathrm{\mid }{m}_3)M(m\_1|m\_2|m\_3)$ liegt auf der Geraden $gg$. Setze ein.

$\left(\begin{array}{c}{m}_1\\ m_2\\ m_3\end{array}\right)=\left(\begin{array}{c}2\\ 6\\ 1\end{array}\right)+t\cdot \left(\begin{array}{c}0\\ -4\\ 0\end{array}\right)=\left(\begin{array}{c}2\\ 6-4t\\ 1\end{array}\right)\backslash begin\{pmatrix\}m\_1\backslash \backslash m\_2\backslash \backslash m\_3\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}2\backslash \backslash 6\backslash \backslash 1\backslash end\{pmatrix\}+t\backslash cdot\backslash begin\{pmatrix\}0\; \backslash \backslash -4\; \backslash \backslash \; 0\; \backslash end\{pmatrix\}=\backslash begin\{pmatrix\}2\backslash \backslash \backslash textcolor\{ff6600\}\{6-4t\}\backslash \backslash 1\backslash end\{pmatrix\}$

Erstelle von der Ebene $E:x_2=6E:\backslash ;\; x\_2=6$ die Hessesche Normalenform:

Allgemein gilt: $E_{HNF}:{\displaystyle \frac{a\cdot x_1+b\cdot x_2+c\cdot x_3-d}{\sqrt{a^2+b^2+c^2}}}=0E\_\{HNF\}:\; \backslash ;\backslash dfrac\{a\backslash cdot\; x\_1+b\backslash cdot\; x\_2+c\backslash cdot\; x\_3-d\}\{\backslash sqrt\{a^2+b^2+c^2\}\}=0$

$E_{HNF}:{\displaystyle \frac{0\cdot x_1+1\cdot x_2+0\cdot x_3-6}{\sqrt{0^2+1^2+0^2}}}=x_2-6=0E\_\{HNF\}:\; \backslash ;\backslash dfrac\{0\backslash cdot\; x\_1+1\backslash cdot\; x\_2+0\backslash cdot\; x\_3-6\}\{\backslash sqrt\{0^2+1^2+0^2\}\}=\backslash textcolor\{ff6600\}\{x\_2\}-6=0$

Der Abstand des Punktes $MM$ von der Ebene $EE$ soll gleich dem Kugelradius $r=4\backslash textcolor\{006400\}\{r=4\}$ sein.

$d(M,E)=\mid 6-4t-6\mid =\mid -4t\mid =4d(M,E)=\backslash left|\backslash textcolor\{ff6600\}\{6-4t\}-6\backslash right|=\backslash left|-4t\backslash right|=\backslash textcolor\{006400\}\{4\}$

Löse den Betrag auf:

Fall -

$4t=4\Rightarrow t_1=14t=4\; \backslash Rightarrow\; \backslash ;t\_1=1$

Fall +

$-4t=4\Rightarrow t_2=-1-4t=4\; \backslash Rightarrow\; \backslash ;t\_2=-1$

Du hast zwei Lösungen für den Parameter $tt$ erhalten. Demzufolge gibt es auch zwei Kugelmittelpunkte.

Setze $t_1=1t\_1=1$ in die Geradengleichung ein:

${\vec{X}}_{M_1}=\left(\begin{array}{c}2\\ 6\\ 1\end{array}\right)+1\cdot \left(\begin{array}{c}0\\ -4\\ 0\end{array}\right)=\left(\begin{array}{c}2\\ 2\\ 1\end{array}\right)\Rightarrow M_1(2\mathrm{\mid }2\mathrm{\mid }1)\backslash vec\; X\_\{M\_1\}=\backslash begin\{pmatrix\}2\backslash \backslash 6\backslash \backslash 1\backslash end\{pmatrix\}+1\backslash cdot\backslash begin\{pmatrix\}0\; \backslash \backslash -4\; \backslash \backslash \; 0\; \backslash end\{pmatrix\}=\backslash begin\{pmatrix\}2\backslash \backslash 2\backslash \backslash 1\backslash end\{pmatrix\}\backslash ;\backslash Rightarrow\backslash :M\_1(2|2|1)$

Setze $t_2=-1t\_2=-1$ in die Geradengleichung ein:

${\vec{X}}_{M_2}=\left(\begin{array}{c}2\\ 6\\ 1\end{array}\right)+(-1)\cdot \left(\begin{array}{c}0\\ -4\\ 0\end{array}\right)=\left(\begin{array}{c}2\\ 10\\ 1\end{array}\right)\Rightarrow M_2(2\mathrm{\mid }10\mathrm{\mid }1)\backslash vec\; X\_\{M\_2\}=\backslash begin\{pmatrix\}2\backslash \backslash 6\backslash \backslash 1\backslash end\{pmatrix\}+(-1)\backslash cdot\backslash begin\{pmatrix\}0\; \backslash \backslash -4\; \backslash \backslash \; 0\; \backslash end\{pmatrix\}=\backslash begin\{pmatrix\}2\backslash \backslash 10\backslash \backslash 1\backslash end\{pmatrix\}\backslash ;\backslash Rightarrow\backslash :M\_2(2|10|1)$

Antwort: Die beiden Kugeln mit den Mittelpunkten $M_1(2\mathrm{\mid }2\mathrm{\mid }1)M\_1(2|2|1)$ und $M_2(2\mathrm{\mid }10\mathrm{\mid }1)M\_2(2|10|1)$ und dem Radius $r=4r=4$ liegen auf der Geraden $gg$ und die Ebene ist eine Tangentialebene.