Скаларни производ вектора – примери 2

Текст задатака објашњених у видео лекцији.

Пр. 4) Вектори $\overrightarrow a $ и $\overrightarrow b $ образују угао $\angle \left( {\overrightarrow a ,\overrightarrow b } \right) = \frac{\pi }{6}$. Ако је $\left| {\overrightarrow a } \right| = \sqrt 3 ,\left| {\overrightarrow b } \right| = 1$, израчунати угао између вектора $\overrightarrow p = \overrightarrow a + \overrightarrow b $ и $\overrightarrow q = \overrightarrow a - \overrightarrow b $

Пр.4)

\[\begin{gathered}
\measuredangle \left( {\overrightarrow p ,\overrightarrow q } \right) = ? \hfill \\
\cos \measuredangle \left( {\overrightarrow p ,\overrightarrow q } \right) = \frac{{\overrightarrow p \cdot \overrightarrow q }}{{\left| {\overrightarrow p } \right| \cdot \left| {\overrightarrow q } \right|}} \hfill \\
\overrightarrow p \cdot \overrightarrow q = \left( {\overrightarrow a + \overrightarrow b } \right)\left( {\overrightarrow a - \overrightarrow b } \right) = \overrightarrow a \cdot \overrightarrow a + \overrightarrow b \cdot \overrightarrow a - \overrightarrow a \cdot \overrightarrow b - \overrightarrow b \cdot \overrightarrow b = \hfill \\
= {\left| {\overrightarrow a } \right|^2} - {\left| {\overrightarrow b } \right|^2} = 3 - 1 = 2 \hfill \\
\left| {\overrightarrow p } \right| = ? \hfill \\
{\left| {\overrightarrow p } \right|^2} = \overrightarrow p \cdot \overrightarrow p = \left( {\overrightarrow a + \overrightarrow b } \right)\left( {\overrightarrow a + \overrightarrow b } \right) = \overrightarrow a \cdot \overrightarrow a + \overrightarrow b \cdot \overrightarrow a + \overrightarrow a \cdot \overrightarrow b - \overrightarrow b \cdot \overrightarrow b = \hfill \\
= {\left| {\overrightarrow a } \right|^2} + 2\overrightarrow a \cdot \overrightarrow b + {\left| {\overrightarrow b } \right|^2} = {\left| {\overrightarrow a } \right|^2} + 2\left| {\overrightarrow a } \right| \cdot \left| {\overrightarrow b } \right| \cdot \cos \measuredangle \left( {\overrightarrow a ,\overrightarrow b } \right) + {\left| {\overrightarrow b } \right|^2} = \hfill \\
= 3 + 2 \cdot \frac{3}{2} + 1 = 7 \Rightarrow \left| {\overrightarrow p } \right| = \sqrt 7 \hfill \\
\hfill \\
{\left| {\overrightarrow q } \right|^2} = \overrightarrow q \cdot \overrightarrow q = \left( {\overrightarrow a - \overrightarrow b } \right)\left( {\overrightarrow a - \overrightarrow b } \right) = \overrightarrow a \cdot \overrightarrow a - \overrightarrow b \cdot \overrightarrow a - \overrightarrow a \cdot \overrightarrow b - \overrightarrow b \cdot \overrightarrow b = \hfill \\
= {\left| {\overrightarrow a } \right|^2} - 2\overrightarrow a \cdot \overrightarrow b + {\left| {\overrightarrow b } \right|^2} = 3 - 2 \cdot \frac{3}{2} + 1 = 1 \Rightarrow \left| {\overrightarrow q } \right| = 1 \hfill \\
\hfill \\
\cos \measuredangle \left( {\overrightarrow p ,\overrightarrow q } \right) = \frac{2}{{\sqrt 7 }} = \frac{{2\sqrt 7 }}{7} \hfill \\
\measuredangle \left( {\overrightarrow p ,\overrightarrow q } \right) = \arccos \frac{{2\sqrt 7 }}{7} \hfill \\
\end{gathered} \]