Analytische Geometrie-Vektorrechung in der Ebene-Skalarprodukt - Fläche - Winkel



Beispiel Nr: 03
$ \text{Gegeben:} \\ \text{Vektoren: } \vec{A} =\left( \begin{array}{c} x_a \\ y_a \\ \end{array} \right) \quad \vec{B} =\left( \begin{array}{c} x_b \\ y_b \\ \end{array} \right) \\ \\ \text{Gesucht:} \\ \text{Länge der Vektoren:} \\ \text{Fläche des Parallelogramms} \\ \text{Skalarprodukt} \\ \\ \\ \textbf{Gegeben:} \\ \text{Vektor: } \vec{A} =\left( \begin{array}{c} \frac{3}{10} \\ 1\frac{1}{5} \\ \end{array} \right) \quad \vec{B} =\left( \begin{array}{c} 2\frac{2}{5} \\ 6 \\ \end{array} \right) \\ \\ \\ \textbf{Rechnung:} \\ \text{Vektoren: } \vec{a} =\left( \begin{array}{c} \frac{3}{10} \\ 1\frac{1}{5} \\ \end{array} \right) \quad \vec{b} =\left( \begin{array}{c} 2\frac{2}{5} \\ 6 \\ \end{array} \right) \\ \bullet \text{Steigung} \\ m_s=\dfrac{y_a}{x_a}=\dfrac{1\frac{1}{5}}{\frac{3}{10}}=4 \\ m_b=\dfrac{y_b}{x_b}=\dfrac{6}{2\frac{2}{5}}=2\frac{1}{2} \\ \bullet \text{Länge der Vektoren:} \\ \left|\vec{a}\right| =\sqrt{x_a^2+y_a^2}=\sqrt{\left(\frac{3}{10}\right)^2+\left(1\frac{1}{5}\right)^2} =1,24 \\ \left|\vec{b}\right| =\sqrt{x_b^2+y_b^2} =\sqrt{\left(2\frac{2}{5}\right)^2+6^2} =6,46 \\ \bullet \text{Skalarprodukt:} \\ \vec{a} \circ \vec{b}==\left( \begin{array}{c} \frac{3}{10} \\ 1\frac{1}{5} \\ \end{array} \right) \circ \left( \begin{array}{c} 2\frac{2}{5} \\ 6 \\ \end{array} \right) =\frac{3}{10} \cdot 2\frac{2}{5} + 1\frac{1}{5} \cdot 6 = 7\frac{23}{25} \\ \bullet \text{Fläche des Parallelogramms aus } \vec{a},\vec{b} \\ A= \left| \begin{array}{cc} \frac{3}{10} & 2\frac{2}{5} \\ 1\frac{1}{5} & 6 \end{array} \right| = \frac{3}{10} \cdot 6 - 1\frac{1}{5} \cdot 2\frac{2}{5} = -1\frac{2}{25} \\ \text{ Fläche des Dreiecks aus } \vec{a},\vec{b}\\ A=\frac{1}{2} \left| \begin{array}{cc} \frac{3}{10} & 2\frac{2}{5} \\ 1\frac{1}{5} & 6 \end{array}\right| =\frac{1}{2}(\frac{3}{10} \cdot 6 - 1\frac{1}{5} \cdot 2\frac{2}{5}) = -\frac{27}{50} \\ \bullet \text{Schnittwinkel:} \\ \cos \alpha= \displaystyle\frac{ \vec{a} \circ \vec{b}}{ \left|\vec{a}\right| \cdot \left|\vec{b}\right|}\\ \cos \alpha= \dfrac{\frac{3}{10} \cdot 2\frac{2}{5} + 1\frac{1}{5} \cdot 6 }{\sqrt{\left(\frac{3}{10}\right)^2+\left(1\frac{1}{5}\right)^2}\cdot\sqrt{\left(2\frac{2}{5}\right)^2+6^2}} \\ \cos \alpha= \left|\displaystyle\frac{7\frac{23}{25}}{1,24 \cdot 6,46} \right| \\ \cos \alpha= \left| 0,991 \right| \\ \alpha=7,77 \\ $