Hausaufgabe
Aufgabe

Berechnen Sie den Winkel zwischen den Vektoren \( \overrightarrow{a} = \left(\begin{array}{c} 2 \\ 0 \\ 2 \end{array}\right) \) und \( \overrightarrow{b} = \left(\begin{array}{c} 3 \\ 1 \\ -2 \end{array}\right) \).

Lösung

\( \cos \varphi = \dfrac{\overrightarrow{a} \cdot \overrightarrow{b}}{\left| \overrightarrow{a} \right| \cdot \left| \overrightarrow{b} \right|} \)

\( \cos \varphi = \dfrac{\left(\begin{array}{c} 2 \\ 0 \\ 2 \end{array}\right) \cdot \left(\begin{array}{c} 3 \\ 1 \\ -2 \end{array}\right)}{\sqrt{2^2 + 0^2 + 2^2} \cdot \sqrt{3^2 + 1^2 + (-2)^2}} \)

\( \cos \varphi = \dfrac{2 \cdot 3 + 0 \cdot 1 + 2 \cdot (-2)}{\sqrt{8} \cdot \sqrt{14}} \)

\( \cos \varphi = \dfrac{2}{2,828 \cdot 3,742} \)

\( \cos \varphi = \dfrac{2}{10,583} \)

\( \cos \varphi = 0,189 \)

\( \varphi = \arccos \left(0,189\right) \)

\( \varphi \approx 79,11° \)

Last modified: Tuesday, 18 May 2021, 9:51 AM