Schnittwinkel in der analytischen Geometrie

Schnittwinkel zwischen zwei Geraden

Seien $,\backslash overset\{\backslash boldsymbol\backslash rightarrow\}\{\backslash mathbf\; u\}\backslash boldsymbol,\backslash overset\{\backslash boldsymbol\backslash rightarrow\}\{\backslash mathbf\; v\}$ die Richtungsvektoren der Geraden.

Dann lässt sich der Schnittwinkel $\alpha \backslash alpha$ so berechnen:

$\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\alpha }=\frac{\mid \mathit{\circ }\mid }{\mid \mid \mathit{\cdot }\mid \mid }\{\backslash mathbf\{cos\}\backslash ,\backslash mathbf\backslash alpha\backslash boldsymbol=\backslash frac\{\backslash left|\backslash overset\{\backslash boldsymbol\backslash rightarrow\}\{\backslash mathbf\; u\}\backslash boldsymbol\backslash circ\backslash overset\{\backslash boldsymbol\backslash rightarrow\}\{\backslash mathbf\; v\}\backslash right|\}\{\backslash left|\backslash overset\{\backslash boldsymbol\backslash rightarrow\}\{\backslash mathbf\; u\}\backslash right|\backslash boldsymbol\backslash cdot\backslash left|\backslash overset\{\backslash boldsymbol\backslash rightarrow\}\{\backslash mathbf\; v\}\backslash right|\}\}$

Beispiel

Gegeben sind zwei sich schneidende Geraden g und h:

$g:\vec{X}=\left(\begin{array}{c}2\\ 0\\ 5\end{array}\right)+r\cdot \left(\begin{array}{c}1\\ 3\\ 5\end{array}\right)g:\; \backslash vec\; X=\backslash begin\{pmatrix\}2\backslash \backslash 0\backslash \backslash 5\backslash end\{pmatrix\}+r\backslash cdot\backslash begin\{pmatrix\}1\; \backslash \backslash 3\; \backslash \backslash \; 5\backslash end\{pmatrix\}$und $h:\vec{X}=\left(\begin{array}{c}2\\ 0\\ 5\end{array}\right)+s\cdot \left(\begin{array}{c}2\\ -3\\ 4\end{array}\right)h:\; \backslash vec\; X=\backslash begin\{pmatrix\}2\backslash \backslash 0\backslash \backslash 5\backslash end\{pmatrix\}+s\backslash cdot\backslash begin\{pmatrix\}2\; \backslash \backslash -3\; \backslash \backslash \; 4\backslash end\{pmatrix\}$:

Berechne den Schnittwinkel $\alpha \backslash alpha$.

Für die Winkelberechnung der beiden Geraden benötigst du ihre Richtungsvektoren und deren Beträge.

$g:\vec{u}=\left(\begin{array}{c}1\\ 3\\ 5\end{array}\right)g:\backslash ;\backslash vec\; u=\backslash begin\{pmatrix\}1\; \backslash \backslash 3\; \backslash \backslash \; 5\backslash end\{pmatrix\}$, $\mathrm{\mid }\vec{u}\mathrm{\mid }=\sqrt{1^2+3^2+5^2}=\sqrt{35}|\backslash vec\{u\}|=\backslash sqrt\{1^2+3^2+5^2\}=\backslash sqrt\{35\}$

$h:\vec{v}=\left(\begin{array}{c}2\\ -3\\ 4\end{array}\right)h:\backslash ;\backslash vec\; v=\backslash begin\{pmatrix\}2\; \backslash \backslash -3\; \backslash \backslash \; 4\backslash end\{pmatrix\}$, $\mathrm{\mid }\vec{v}\mathrm{\mid }=\sqrt{2^2+(-3{)}^2+4^2}=\sqrt{29}|\backslash vec\{v\}|=\backslash sqrt\{2^2+(-3)^2+4^2\}=\backslash sqrt\{29\}$

Du hast die Gleichung $\cos \alpha =\mathrm{0,4080}\backslash cos\backslash ;\backslash alpha=0\{,\}4080$ erhalten. Durch Anwendung der Umkehrfunktion des Kosinus kannst du den Winkel $\alpha \backslash alpha$ berechnen. Benutze auf dem Taschenrechner die Funktion ${\cos }^{-1}(x)\backslash cos^\{-1\}(x)$.

$\alpha =\arccos (\mathrm{0,4080})\approx {\mathrm{65,92}}^{\circ }\backslash alpha=\backslash arccos(0\{,\}4080)\backslash approx\; 65\{,\}92^\backslash circ$

Antwort: Der Schnittwinkel zwischen den beiden Geraden beträgt rund ${\mathrm{65,9}}^{\circ }65\{,\}9^\backslash circ$.

Schnittwinkel zwischen Ebene und Gerade

Sei $\overrightarrow{n}\backslash overrightarrow\; n$ der Normalenvektor der Ebene und $\overrightarrow{u}\backslash overrightarrow\; u$ der Richtungsvektor der Gerade.

Dann kann der Schnittwinkel $\alpha \backslash alpha$ so berechnet werden:

$\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\alpha }=\frac{\mid \mathit{\circ }\mid }{\mid \mid \mathit{\cdot }\mid \mid }\{\backslash mathbf\{sin\}\backslash ,\backslash mathbf\backslash alpha\backslash boldsymbol=\backslash frac\{\backslash left|\backslash overset\{\backslash boldsymbol\backslash rightarrow\}\{\backslash mathbf\; n\}\backslash boldsymbol\backslash circ\backslash overset\{\backslash boldsymbol\backslash rightarrow\}\{\backslash mathbf\; u\}\backslash right|\}\{\backslash left|\backslash overset\{\backslash boldsymbol\backslash rightarrow\}\{\backslash mathbf\; n\}\backslash right|\backslash boldsymbol\backslash cdot\backslash left|\backslash overset\{\backslash boldsymbol\backslash rightarrow\}\{\backslash mathbf\; u\}\backslash right|\}\}$

Eine weitere Möglichkeit ist, $\mathbf{c}\mathbf{o}\mathbf{s}({\mathbf{90}}^{\circ }\mathit{-}\mathbf{\alpha })=\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\alpha }=\frac{\mid \mathit{\circ }\mid }{\mid \mid \mathit{\cdot }\mid \mid }\backslash mathbf\{cos\}\backslash boldsymbol(\backslash mathbf\{90^\backslash circ\}\backslash boldsymbol-\backslash mathbf\backslash alpha\backslash boldsymbol)\backslash boldsymbol=\backslash mathbf\{sin\}\backslash mathbf\backslash alpha\backslash boldsymbol=\backslash frac\{\backslash left|\backslash overset\{\backslash boldsymbol\backslash rightarrow\}\{\backslash mathbf\; n\}\backslash boldsymbol\backslash circ\backslash overset\{\backslash boldsymbol\backslash rightarrow\}\{\backslash mathbf\; u\}\backslash right|\}\{\backslash left|\backslash overset\{\backslash boldsymbol\backslash rightarrow\}\{\backslash mathbf\; n\}\backslash right|\backslash boldsymbol\backslash cdot\backslash left|\backslash overset\{\backslash boldsymbol\backslash rightarrow\}\{\backslash mathbf\; u\}\backslash right|\}$ auszurechnen.

Beispiel

Gegeben sind eine Geraden g und eine Ebene $EE$:

$g:\vec{X}=\left(\begin{array}{c}2\\ 0\\ 5\end{array}\right)+s\cdot \left(\begin{array}{c}2\\ -3\\ 4\end{array}\right)g:\; \backslash vec\; X=\backslash begin\{pmatrix\}2\backslash \backslash 0\backslash \backslash 5\backslash end\{pmatrix\}+s\backslash cdot\backslash begin\{pmatrix\}2\; \backslash \backslash -3\; \backslash \backslash \; 4\backslash end\{pmatrix\}$ und $E:2x_1-x_2+3x_3=4E:\; \backslash ;2x\_1-x\_2+3x\_3=4$

Berechne den Schnittwinkel $\alpha \backslash alpha$.

Für die Winkelberechnung zwischen Gerade $gg$ und Ebene $EE$ benötigst du von der Geraden den Richtungsvektor und dessen Betrag und von der Ebene den Normalenvektor $\vec{n}\backslash vec\; n$ und dessen Betrag.

$g:\vec{v}=\left(\begin{array}{c}2\\ -3\\ 4\end{array}\right)g:\backslash ;\backslash vec\; v=\backslash begin\{pmatrix\}2\; \backslash \backslash -3\; \backslash \backslash \; 4\backslash end\{pmatrix\}$, $\mathrm{\mid }\vec{v}\mathrm{\mid }=\sqrt{2^2+(-3{)}^2+4^2}=\sqrt{29}|\backslash vec\{v\}|=\backslash sqrt\{2^2+(-3)^2+4^2\}=\backslash sqrt\{29\}$

$E:\vec{n}=\left(\begin{array}{c}2\\ -1\\ 3\end{array}\right)E:\backslash ;\backslash vec\; n=\backslash begin\{pmatrix\}2\backslash \backslash -1\backslash \backslash 3\backslash end\{pmatrix\}$, $\mathrm{\mid }\vec{n}\mathrm{\mid }=\sqrt{2^2+(-1{)}^2+3^2}=\sqrt{4+1+9}=\sqrt{14}|\backslash vec\; n|=\backslash sqrt\{2^2+(-1)^2+3^2\}\; =\backslash sqrt\{4+1+9\}=\; \backslash sqrt\{14\}$

Setze in die oben genannte Formel ein:

Du hast die Gleichung $\sin \alpha =\mathrm{0,9430}\backslash sin\backslash ;\backslash alpha=0\{,\}9430$ erhalten. Durch Anwendung der Umkehrfunktion des Sinus kannst du den Winkel $\alpha \backslash alpha$ berechnen. Benutze auf dem Taschenrechner die Funktion ${\sin }^{-1}(x)\backslash sin^\{-1\}(x)$.

$\alpha =\arcsin \left(\mathrm{0,9430}\right)\approx {\mathrm{70,56}}^{\circ }\backslash alpha=\backslash arcsin\backslash left(0\{,\}9430\backslash right)\backslash approx\; 70\{,\}56^\backslash circ$

Antwort: Der Schnittwinkel $\alpha \backslash alpha$ zwischen der Geraden und der Ebene beträgt rund ${\mathrm{70,6}}^{\circ }70\{,\}6^\backslash circ$.

Schnittwinkel zwischen zwei Ebenen

Seien  $\overrightarrow{n},\overrightarrow{m}\backslash overrightarrow\; n,\backslash ;\backslash overrightarrow\; m$ die  Normalenvektoren der Ebenen.

Dann lässt sich der Schnittwinkel $\alpha \backslash alpha$ so berechnen:

$\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\alpha }=\frac{\mid \mathit{\circ }\mid }{\mid \mid \mathit{\cdot }\mid \mid }\{\backslash mathbf\{cos\}\backslash ;\backslash mathbf\backslash alpha\backslash ;\backslash boldsymbol=\backslash ;\backslash frac\{\backslash left|\backslash overset\{\backslash boldsymbol\backslash rightarrow\}\{\backslash mathbf\; n\}\backslash ;\backslash boldsymbol\backslash circ\backslash ;\backslash overset\{\backslash boldsymbol\backslash rightarrow\}\{\backslash mathbf\; m\}\backslash right|\}\{\backslash left|\backslash overset\{\backslash boldsymbol\backslash rightarrow\}\{\backslash mathbf\; n\}\backslash right|\backslash boldsymbol\backslash cdot\backslash left|\backslash overset\{\backslash boldsymbol\backslash rightarrow\}\{\backslash mathbf\; m\}\backslash right|\}\}$

Beispiel

Gegeben sind zwei sich schneidende Ebenen $E:2x_1-x_2+3x_3=4E:\; \backslash ;2x\_1-x\_2+3x\_3=4$ und $F:x_1-x_2-x_3=2F:\; \backslash ;x\_1-x\_2-x\_3=2$. Berechne den Schnittwinkel $\alpha \backslash alpha$.

Für die Winkelberechnung zwischen zwei Ebenen benötigst du von den Ebenen deren Normalenvektoren und deren Beträge.

Lies die Normalenvektoren aus den Koordinatengleichungen ab:

$\vec{n}=\left(\begin{array}{c}2\\ -1\\ 3\end{array}\right)\backslash vec\; n=\backslash begin\{pmatrix\}2\backslash \backslash -1\backslash \backslash 3\backslash end\{pmatrix\}$ und $\vec{m}=\left(\begin{array}{c}1\\ -1\\ -1\end{array}\right)\backslash vec\; m=\backslash begin\{pmatrix\}1\backslash \backslash -1\backslash \backslash -1\backslash end\{pmatrix\}$

Für den Betrag von $\vec{n}\backslash vec\; n$ gilt: $\mathrm{\mid }\vec{n}\mathrm{\mid }=\sqrt{2^2+(-1{)}^2+3^2}=\sqrt{4+1+9}=\sqrt{14}|\backslash vec\; n|=\backslash sqrt\{2^2+(-1)^2+3^2\}\; =\backslash sqrt\{4+1+9\}=\; \backslash sqrt\{14\}$

Für den Betrag von $\vec{m}\backslash vec\; m$ gilt: $\mathrm{\mid }\vec{m}\mathrm{\mid }=\sqrt{1^2+(-1{)}^2+(-1{)}^2}=\sqrt{1+1+1}=\sqrt{3}|\backslash vec\; m|=\backslash sqrt\{1^2+(-1)^2+(-1)^2\}\; =\backslash sqrt\{1+1+1\}=\backslash sqrt\{3\}$

Setze in die oben genannte Formel ein:

Du hast die Gleichung $\cos \alpha =0\backslash cos\backslash ;\backslash alpha=0$ erhalten. Durch Anwendung der Umkehrfunktion des Kosinus kannst du den Winkel $\alpha \backslash alpha$ berechnen. Benutze auf dem Taschenrechner die Funktion ${\cos }^{-1}(x)\backslash cos^\{-1\}(x)$.

$\alpha =\arccos \left(0\right)={90}^{\circ }\backslash alpha=\backslash arccos\backslash left(0\backslash right)=90^\backslash circ$

Antwort: Der Schnittwinkel $\alpha \backslash alpha$ zwischen den beiden Ebenen beträgt ${90}^{\circ }90^\backslash circ$.