Gerade - Gerade

Den Schnittwinkel $\alpha$ zwischen den sich schneidenden Geraden $g1$ und $g2$ kannst du mit folgender Formel berechnen:

$\cos\,\alpha=\dfrac{\left|\vec{u}\cdot\vec{v}\right|}{\left|\vec{u}\right|\cdot \left|\vec{v}\right|}$

Bei $u$ und $v$ handelt es sich um die Richtungsvektoren der Geraden.

Beispiel

Winkel zwischen den Geraden $g1:\vec{x}=\begin{pmatrix}3\\1\\2 \end{pmatrix}+s\begin{pmatrix}1\\2\\1 \end{pmatrix}$ und $g2:$ $\vec{x}=\begin{pmatrix}2\\4\\1\end{pmatrix} + s\begin{pmatrix}2\\3\\1 \end{pmatrix}$ berechnen.

$\begin{array}[t]{rll} \cos\alpha&=&\dfrac{\left|\begin{pmatrix}1\\2\\1\end{pmatrix}\circ\begin{pmatrix}2\\3\\1\end{pmatrix}\right|}{\left|\begin{pmatrix}1\\2\\1\end{pmatrix}\right|\cdot \left|\begin{pmatrix}2\\3\\1\end{pmatrix}\right|} \\[5pt] &=&\dfrac{\left|2+6+1\right|}{\sqrt{1+4+1}\cdot\sqrt{4+9+1}}\\[5pt] &=&\dfrac{9}{\sqrt{6}\cdot\sqrt{14}}\\[5pt] &=&\dfrac{9}{\sqrt{84}}\quad \scriptsize \mid \cos^{-1}\; \\[5pt] \alpha&=&\cos^{-1}\left(\dfrac{9}{\sqrt{84}}\right) \\[5pt] &=& 10,89° \end{array}$

$\alpha\approx43,3°$

Vollständige Lösung anzeigen

$\begin{array}[t]{rl} \cos \alpha = &\frac{{\left| {\left( {\begin{array}{*{20}r} 4 \\ 1 \\ -1 \\ \end{array}} \right) \circ \left( {\begin{array}{*{20}r} 1 \\ 2 \\ 1 \\ \end{array}} \right)} \right|}}{{\left| {\left( {\begin{array}{*{20}r} 4 \\ 1 \\ -1 \\ \end{array}} \right)} \right| \cdot \left| {\left( {\begin{array}{*{20}r} 1 \\ 2 \\ 1 \\ \end{array}} \right)} \right|}} \\[5pt] =&\dfrac{\left|4+2-1\right|}{\sqrt{16+1+1}\cdot\sqrt{1+4+1}} \\[5pt] =&\dfrac{5}{\sqrt{18}\cdot\sqrt{6}} \\[5pt] \Rightarrow&\alpha\approx61,2° \end{array}$

$\begin{array}[t]{rl} \cos \alpha =& \frac{{\left| {\left( {\begin{array}{*{20}r} 2 \\ 5 \\ 3 \\ \end{array}} \right) \circ \left( {\begin{array}{*{20}r} 3 \\ 2 \\ 0 \\ \end{array}} \right)} \right|}}{{\left| {\left( {\begin{array}{*{20}r} 2 \\ 5 \\ 3 \\ \end{array}} \right)} \right| \cdot \left| {\left( {\begin{array}{*{20}r} 3 \\ 2 \\ 0 \\ \end{array}} \right)} \right|}} \\[5pt] =&\dfrac{\left|6+10\right|}{\sqrt{4+25+9}\cdot\sqrt{9+4}} \\[5pt] =&\dfrac{16}{\sqrt{38}\cdot\sqrt{13}} \\[5pt] \Rightarrow&\alpha\approx44° \end{array}$

$\begin{array}[t]{rl} \cos \alpha = &\frac{{\left| {\left( {\begin{array}{*{20}r} 2 \\ 3 \\ -2 \\ \end{array}} \right) \circ \left( {\begin{array}{*{20}r} 2 \\ -1 \\ 3 \\ \end{array}} \right)} \right|}}{{\left| {\left( {\begin{array}{*{20}r} 2 \\ 3 \\ -2 \\ \end{array}} \right)} \right| \cdot \left| {\left( {\begin{array}{*{20}r} 2 \\ -1 \\ 3 \\ \end{array}} \right)} \right|}} \\[5pt] =&\dfrac{\left|4-3-6\right|}{\sqrt{4+9+4}\cdot\sqrt{4+1+9}} \\[5pt] =&\dfrac{5}{\sqrt{17}\cdot\sqrt{14}} \\[5pt] \Rightarrow&\alpha\approx71,1° \end{array}$

$\begin{array}[t]{rl} \cos \alpha =& \frac{{\left| {\left( {\begin{array}{*{20}r} r \\ 1 \\ 0 \\ \end{array}} \right) \circ \left( {\begin{array}{*{20}r} 1 \\ 0 \\ 0 \\ \end{array}} \right)} \right|}}{{\left| {\left( {\begin{array}{*{20}r} r \\ 1 \\ 0 \\ \end{array}} \right)} \right| \cdot \left| {\left( {\begin{array}{*{20}r} 1 \\ 0 \\ 0 \\ \end{array}} \right)} \right|}} \\[5pt] =&\dfrac{\left|r\right|}{\sqrt{r^2+1+0}\cdot\sqrt{1+0+0}} \\[5pt] =&\dfrac{\left|r\right|}{\sqrt{r^2+1}} \end{array}$

$\alpha=45° \quad \Rightarrow \quad \cos\;\alpha=\dfrac{1}{\sqrt{2}}$
(vgl. Kosinuswerte)

$\begin{array}[t]{rll} \dfrac{1}{\sqrt{2}}=&\dfrac{\left|r\right|}{\sqrt{r^2+1}}&\quad\scriptsize\mid (\;\;\;)^2 \\[5pt] \dfrac{1}{2}=& \dfrac{r^2}{r^2+1} & \quad\scriptsize\mid \cdot(r^2+1) \\[5pt] \dfrac{r^2+1}{2}=& r^2 & \quad\scriptsize\mid \cdot 2 \\[5pt] r^2+1=& 2r^2 \\[5pt] 1=& r^2 \\[5pt] r_{1,2}=&\pm1 \end{array}$

$\begin{array}[t]{rl} \cos \alpha = &\frac{{\left| {\left( {\begin{array}{*{20}r} 2 \\ r \\ 0 \\ \end{array}} \right) \circ \left( {\begin{array}{*{20}r} 0 \\ 1 \\ 1 \\ \end{array}} \right)} \right|}}{{\left| {\left( {\begin{array}{*{20}r} 2 \\ r \\ 0 \\ \end{array}} \right)} \right| \cdot \left| {\left( {\begin{array}{*{20}r} 0 \\ 1 \\ 1 \\ \end{array}} \right)} \right|}} \\[5pt] =&\dfrac{\left|r\right|}{\sqrt{4+r^2}\cdot\sqrt{1+1}} \\[5pt] =&\dfrac{\left|r\right|}{\sqrt{8+2r^2}} \end{array}$

$\alpha=60° \quad \Rightarrow \quad \cos\;\alpha=\dfrac{1}{2}$
(vgl. Kosinuswerte)

$\begin{array}[t]{rll} \dfrac{1}{2}=&\dfrac{\left|r\right|}{\sqrt{8+2r^2}}&\quad\scriptsize\mid (\;\;\;)^2 \\[5pt] \dfrac{1}{4}=& \dfrac{r^2}{8+2r^2} & \quad\scriptsize\mid \cdot(8+2r^2) \\[5pt] \dfrac{8+2r^2}{4}=& r^2 & \quad\scriptsize\mid \cdot 4 \\[5pt] 8+2r^2=& 4r^2 \\[5pt] 8=& 2r^2 & \quad\scriptsize\mid :2 \\[5pt] 4=& r^2 \\[5pt] r_{1,2}=&\pm2 \end{array}$

$\begin{array}[t]{rl} \cos \alpha = &\frac{{\left| {\left( {\begin{array}{*{20}r} 1 \\ 0 \\ 1 \\ \end{array}} \right) \circ \left( {\begin{array}{*{20}r} 1 \\ 1 \\ r \\ \end{array}} \right)} \right|}}{{\left| {\left( {\begin{array}{*{20}r} 1 \\ 0 \\ 1 \\ \end{array}} \right)} \right| \cdot \left| {\left( {\begin{array}{*{20}r} 1 \\ 1 \\ r \\ \end{array}} \right)} \right|}} \\[5pt] =&\dfrac{\left|1+r\right|}{\sqrt{1+1}\cdot\sqrt{1+1+r^2}} \\[5pt] =&\dfrac{\left|1+r\right|}{\sqrt{4+2r^2}} \end{array}$

$\alpha=30° \quad \Rightarrow \quad \cos\;\alpha=\dfrac{\sqrt{3}}{2}$
(vgl. Kosinuswerte)

$\begin{array}[t]{rl} r=&2 \end{array}$

Vollständige Lösung anzeigen

1. Schritt: Verbindungsvektoren aufstellen

$\overrightarrow{AB}=\left(\begin{array}{r} 0-1\\ 2-(-2)\\ 1-4\\ \end{array}\right)=\left(\begin{array}{r} -1\\ 4\\ -3\\ \end{array}\right)$

$\overrightarrow{AC}=\left(\begin{array}{r} 3-1\\ 3-(-2)\\ 7-4\\ \end{array}\right)=\left(\begin{array}{r} 2\\ 5\\ 3\\ \end{array}\right)$

$\overrightarrow{BC}=\left(\begin{array}{r} 3-0\\ 3-2\\ 7-1\\ \end{array}\right)=\left(\begin{array}{r} 3\\ 1\\ 6\\ \end{array}\right)$

2. Schritt: Schnittwinkel berechnen

$\blacktriangleright$ $\overrightarrow{AB}$ und $\overrightarrow{AC}$

$\begin{array}[t]{rl} \cos\alpha=&\dfrac{\left|\left(\begin{array}{r} -1\\ 4\\ -3\\ \end{array}\right)\circ\left(\begin{array}{r} 2\\ 5\\ 3\\ \end{array}\right)\right|}{\left|\left(\begin{array}{r} -1\\ 4\\ -3\\ \end{array}\right)\right|\cdot\left|\left(\begin{array}{r} 2\\ 5\\ 3\\ \end{array}\right)\right|} \\[5pt] =&\dfrac{-2+20-9}{\sqrt{1+16+9}\cdot\sqrt{4+25+9}} \\[5pt] =&\dfrac{9}{\sqrt{988}} \\[5pt] \approx&0,29 \end{array}$

$\cos^{-1}\left(0,29\right)\approx73,14°$

$\blacktriangleright$ $\overrightarrow{AB}$ und $\overrightarrow{BC}$

$\begin{array}[t]{rl} \cos\beta=&\dfrac{\left|\left(\begin{array}{r} -1\\ 4\\ -3\\ \end{array}\right)\circ\left(\begin{array}{r} 3\\ 1\\ 6\\ \end{array}\right)\right|}{\left|\left(\begin{array}{r} -1\\ 4\\ -3\\ \end{array}\right)\right|\cdot\left|\left(\begin{array}{r} 3\\ 1\\ 6\\ \end{array}\right)\right|} \\[5pt] =&\dfrac{\left|-3+4-18\right|}{\sqrt{1+16+9}\cdot\sqrt{9+1+36}} \\[5pt] =&\dfrac{17}{\sqrt{1196}} \\[5pt] \approx&0,49 \end{array}$

$\cos^{-1}\left(0,49\right)\approx60,66°$

$\blacktriangleright$ $\overrightarrow{BC}$ und $\overrightarrow{AC}$

$\begin{array}[t]{rl} \cos\gamma=&\dfrac{\left|\left(\begin{array}{r} 3\\ 1\\ 6\\ \end{array}\right)\circ\left(\begin{array}{r} 2\\ 5\\ 3\\ \end{array}\right)\right|}{\left|\left(\begin{array}{r} 3\\ 1\\ 6\\ \end{array}\right)\right|\cdot\left|\left(\begin{array}{r} 2\\ 5\\ 3\\ \end{array}\right)\right|} \\[5pt] =&\dfrac{6+5+18}{\sqrt{9+1+36}\cdot\sqrt{4+25+9}} \\[5pt] =&\dfrac{29}{\sqrt{1748}} \\[5pt] \approx&0,69 \end{array}$

$\cos^{-1}\left(0,69\right)\approx46,37°$

1. Schritt: Verbindungsvektoren aufstellen

$\overrightarrow{AB}=\left(\begin{array}{r} 1-3\\ 4-0\\ 2-(-1)\\ \end{array}\right)=\left(\begin{array}{r} -2\\ 4\\ 3\\ \end{array}\right)$

$\overrightarrow{AC}=\left(\begin{array}{r} 0-3\\ -2-0\\ 4-(-1)\\ \end{array}\right)=\left(\begin{array}{r} -3\\ -2\\ 5\\ \end{array}\right)$

$\overrightarrow{BC}=\left(\begin{array}{r} 0-1\\ -2-4\\ 4-2\\ \end{array}\right)=\left(\begin{array}{r} -1\\ -6\\ 2\\ \end{array}\right)$

2. Schritt: Schnittwinkel berechnen

$\blacktriangleright$ $\overrightarrow{AB}$ und $\overrightarrow{AC}$

$\begin{array}[t]{rl} \cos\alpha=&\dfrac{\left|\left(\begin{array}{r} -2\\ 4\\ 3\\ \end{array}\right)\circ\left(\begin{array}{r} -3\\ -2\\ 5\\ \end{array}\right)\right|}{\left|\left(\begin{array}{r} -2\\ 4\\ 3\\ \end{array}\right)\right|\cdot\left|\left(\begin{array}{r} -3\\ -2\\ 5\\ \end{array}\right)\right|} \\[5pt] =&\dfrac{6-8+15}{\sqrt{4+16+9}\cdot\sqrt{9+4+25}} \\[5pt] =&\dfrac{13}{\sqrt{1102}} \\[5pt] \approx&0,39 \end{array}$

$\cos^{-1}\left(0,39\right)\approx67,04°$

$\blacktriangleright$ $\overrightarrow{AB}$ und $\overrightarrow{BC}$

$\begin{array}[t]{rl} \cos\beta=&\dfrac{\left|\left(\begin{array}{r} -2\\ 4\\ 3\\ \end{array}\right)\circ\left(\begin{array}{r} -1\\ -6\\ 2\\ \end{array}\right)\right|}{\left|\left(\begin{array}{r} -2\\ 4\\ 3\\ \end{array}\right)\right|\cdot\left|\left(\begin{array}{r} -1\\ -6\\ 2\\ \end{array}\right)\right|} \\[5pt] =&\dfrac{\left|2-24+6\right|}{\sqrt{4+16+9}\cdot\sqrt{1+36+4}} \\[5pt] =&\dfrac{16}{\sqrt{1189}} \\[5pt] \approx&0,46 \end{array}$

$\cos^{-1}\left(0,46\right)\approx62,35°$

$\blacktriangleright$ $\overrightarrow{BC}$ und $\overrightarrow{AC}$

$\begin{array}[t]{rl} \cos\gamma=&\dfrac{\left|\left(\begin{array}{r} -1\\ -6\\ 2\\ \end{array}\right)\circ\left(\begin{array}{r} -3\\ -2\\ 5\\ \end{array}\right)\right|}{\left|\left(\begin{array}{r} -1\\ -6\\ 2\\ \end{array}\right)\right|\cdot\left|\left(\begin{array}{r} -3\\ -2\\ 5\\ \end{array}\right)\right|} \\[5pt] =&\dfrac{3+12+10}{\sqrt{1+36+4}\cdot\sqrt{9+4+25}} \\[5pt] =&\dfrac{25}{\sqrt{1558}} \\[5pt] \approx&0,63 \end{array}$

$\cos^{-1}\left(0,63\right)\approx50,7°$