Lies aus der gegebenen Kugelgleichung die Koordinaten des Mittelpunktes $MM$ und den Radius $rr$ ab:

$M(2\mathrm{\mid }3\mathrm{\mid }1)M(2|3|1)$ und $r=\sqrt{16}=4r=\backslash sqrt\{16\}=4$.

Berechne den Vektor $\overrightarrow{PM}=\overrightarrow{OM}-\overrightarrow{OP}=\left(\begin{array}{c}2\\ 3\\ 1\end{array}\right)-\left(\begin{array}{c}2\\ 5\\ -3\end{array}\right)=\left(\begin{array}{c}0\\ -2\\ 4\end{array}\right)\backslash overrightarrow\{PM\}=\backslash overrightarrow\{OM\}-\backslash overrightarrow\{OP\}=\backslash begin\{pmatrix\}2\backslash \backslash 3\backslash \backslash 1\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}2\backslash \backslash 5\backslash \backslash -3\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}0\backslash \backslash -2\backslash \backslash 4\backslash end\{pmatrix\}$.

Dann folgt für die gesuchte Bedingung: $\mathrm{\mid }\overrightarrow{PM}\circ {\vec{n}}_0\mathrm{\mid }=r\Rightarrow \mid \left(\begin{array}{c}0\\ -2\\ 4\end{array}\right)\circ {\vec{n}}_0\mid =4|\backslash overrightarrow\{PM\}\backslash circ\; \backslash vec\{n\}\_0|=r\backslash ;\backslash ;\backslash Rightarrow\backslash ;\backslash ;\backslash left|\backslash begin\{pmatrix\}0\backslash \backslash -2\backslash \backslash 4\backslash end\{pmatrix\}\backslash circ\; \backslash vec\{n\}\_0\backslash right|=4$

Antwort: Erfüllt der Normaleneinheitsvektor ${\vec{n}}_0\backslash vec\{n\}\_0$ der Ebene $EE$ die folgende Bedingung $\mid \left(\begin{array}{c}0\\ -2\\ 4\end{array}\right)\circ {\vec{n}}_0\mid =4\backslash left|\backslash begin\{pmatrix\}0\backslash \backslash -2\backslash \backslash 4\backslash end\{pmatrix\}\backslash circ\; \backslash vec\{n\}\_0\backslash right|=4$, dann ist die Ebene $EE$ eine Tangentialebene an die Kugel $KK$ und enthält den Punkt $PP$.