Im rechtwinkligen Dreieck $MPBMPB$ gilt der Kathetensatz: $b^2=q\cdot cb^2=q\backslash cdot\; c$

$b=rb=r$, $q=\overline{MF}q=\backslash overline\{MF\}$ und $c=\overline{MP}c=\; \backslash overline\{MP\}$

$\Rightarrow r^2=\overline{MF}\cdot \overline{MP}\Rightarrow \overline{MP}={\displaystyle \frac{r^2}{\overline{MF}}}\backslash Rightarrow\backslash ;\backslash ;\; r^2=\; \backslash overline\{MF\}\; \backslash cdot\; \backslash overline\{MP\}\backslash ;\backslash ;\backslash Rightarrow\backslash ;\backslash ;\; \backslash overline\{MP\}=\backslash dfrac\{r^2\}\{\backslash overline\{MF\}\}$

Erstelle die Hessesche Normalenform der Ebene $EE$ und berechne den Abstand des Punktes $MM$ von $EE$.

$E_{HNF}:{\displaystyle \frac{-6x_1-x_3-11}{\sqrt{(-6{)}^2+0^2+(-1{)}^2}}}=0\Rightarrow {\displaystyle \frac{-6x_1-x_3-11}{\sqrt{37}}}=0E\_\{HNF\}:\backslash ;\backslash dfrac\{-6x\_1-x\_3-11\}\{\backslash sqrt\{(-6)^2+0^2+(-1)^2\}\}=0\backslash Rightarrow\backslash ;\backslash ;\backslash dfrac\{-6x\_1-x\_3-11\}\{\backslash sqrt\{37\}\}=0$

Abstand des Punktes $MM$ von $EE$:

$\overline{MF}=d(M,E)=\mid {\displaystyle \frac{-6\cdot (-4)-1\cdot 5-11}{\sqrt{37}}}\mid =\mid {\displaystyle \frac{8}{\sqrt{37}}}\mid ={\displaystyle \frac{8}{\sqrt{37}}}\backslash overline\{MF\}=d(M,E)=\backslash left|\backslash dfrac\{-6\backslash cdot\; (-4)-1\backslash cdot\; 5-11\}\{\backslash sqrt\{37\}\}\backslash right|=\backslash left|\backslash dfrac\{8\}\{\backslash sqrt\{37\}\}\backslash right|=\backslash dfrac\{8\}\{\backslash sqrt\{37\}\}$

Der Abstand des Punktes $MM$ von der Ebene $EE$ beträgt $\overline{MF}={\displaystyle \frac{8}{\sqrt{37}}}\backslash overline\{MF\}=\backslash dfrac\{8\}\{\backslash sqrt\{37\}\}$ .

Berechne nun $\overline{MP}={\displaystyle \frac{r^2}{\overline{MF}}}\backslash overline\{MP\}=\backslash dfrac\{r^2\}\{\backslash overline\{MF\}\}$, dabei ist $r=4r=4$ und $\overline{MF}={\displaystyle \frac{8}{\sqrt{37}}}\backslash overline\{MF\}=\backslash dfrac\{8\}\{\backslash sqrt\{37\}\}$

$\Rightarrow \overline{MP}={\displaystyle \frac{r^2}{\overline{MF}}}={\displaystyle \frac{16}{{\displaystyle \frac{8}{\sqrt{37}}}}}=2\cdot \sqrt{37}\backslash Rightarrow\backslash ;\backslash ;\; \backslash overline\{MP\}=\backslash dfrac\{r^2\}\{\backslash overline\{MF\}\}=\backslash dfrac\{16\}\{\backslash dfrac\{8\}\{\backslash sqrt\{37\}\}\}=2\backslash cdot\; \backslash sqrt\{37\}$

Berechne den Vektor $\overrightarrow{MP}\backslash overrightarrow\{MP\}$, indem du seine Länge $2\cdot \sqrt{37}2\backslash cdot\; \backslash sqrt\{37\}$ mit dem Einheitsvektor ${\vec{n}}_0\backslash vec\; n\_0$ multiplizierst.

Berechne nun den Vektor $\overrightarrow{MP}\backslash overrightarrow\{MP\}$ mit $\overline{MP}=2\cdot \sqrt{37}\backslash overline\{MP\}=2\backslash cdot\; \backslash sqrt\{37\}$ und ${\vec{n}}_0={\displaystyle \frac{1}{\sqrt{37}}}\cdot \left(\begin{array}{c}6\\ 0\\ 1\end{array}\right)\backslash vec\; n\_0=\backslash dfrac\{1\}\{\backslash sqrt\{37\}\}\backslash cdot\; \{\backslash begin\{pmatrix\}6\backslash \backslash 0\backslash \backslash 1\backslash end\{pmatrix\}\}$

$\overrightarrow{MP}=\overline{MP}\cdot {\vec{n}}_0=2\cdot \sqrt{37}\cdot {\displaystyle \frac{1}{\sqrt{37}}}\cdot \left(\begin{array}{c}6\\ 0\\ 1\end{array}\right)=\left(\begin{array}{c}12\\ 0\\ 2\end{array}\right)\backslash overrightarrow\{MP\}=\backslash overline\{MP\}\backslash cdot\; \backslash vec\; n\_0=2\backslash cdot\; \backslash sqrt\{37\}\; \backslash cdot\; \backslash dfrac\{1\}\{\backslash sqrt\{37\}\}\backslash cdot\; \{\backslash begin\{pmatrix\}6\backslash \backslash 0\backslash \backslash 1\backslash end\{pmatrix\}\}=\backslash begin\{pmatrix\}12\backslash \backslash 0\backslash \backslash 2\backslash end\{pmatrix\}$

Der gesuchte Punkt $PP$ kann nun berechnet werden:

$\overrightarrow{OP}=\overrightarrow{OM}+\overrightarrow{MP}=\left(\begin{array}{c}-4\\ 1\\ 5\end{array}\right)+\left(\begin{array}{c}12\\ 0\\ 2\end{array}\right)=\left(\begin{array}{c}8\\ 1\\ 7\end{array}\right)\backslash overrightarrow\{OP\}=\backslash overrightarrow\{OM\}+\backslash overrightarrow\{MP\}=\; \backslash begin\{pmatrix\}-4\backslash \backslash 1\backslash \backslash 5\backslash end\{pmatrix\}+\backslash begin\{pmatrix\}12\backslash \backslash 0\backslash \backslash 2\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}8\backslash \backslash 1\backslash \backslash 7\backslash end\{pmatrix\}$

Antwort: Der Punkt $PP$ hat die Koordinaten $P(8\mathrm{\mid }1\mathrm{\mid }7)P(8|1|7)$.