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Скаларни производ вектора - примери 1

Угао између вектора. Решени задаци.

Задаци

Текст задатака објашњених у видео лекцији.

Пр.1)   Дати су вектори $\overrightarrow a = \left( {2,2,4} \right)$ и $\overrightarrow b = \left( { - 2,4,2} \right)$. Одредити угао између ова два вектора $\angle \left( {\overrightarrow a ,\overrightarrow b } \right)$.

Пр.2)   Израчунати угао између дијагонала четвороугла $ABCD$, ако су дата темена тог четворозгла $A\left( { - 3, - 7,0} \right),B\left( {3, - 3,1} \right),C\left( {5,0,2} \right)D\left( { - 1,1,1} \right)$.

Пр.3)   Дати су вектрои: $\overrightarrow a = \left( {1,2} \right)$, $\overrightarrow b = \left( { - 1,3} \right)$, $\overrightarrow c = \left( {0, - 1} \right)$ и $\overrightarrow d = \left( { - 3, - 2} \right)$. Одреди угао између вектора $\overrightarrow a - \overrightarrow b $ и $\overrightarrow c + \overrightarrow d $.

Пр.1) 

\[\begin{gathered}
\overrightarrow a \cdot \overrightarrow b = {a_x}{b_x} \cdot {a_y}{b_y} \cdot {a_z}{b_z} \hfill \\
\overrightarrow a \cdot \overrightarrow b = 2 \cdot \left( { - 2} \right) + 2 \cdot 4 + 4 \cdot 2 \hfill \\
\overrightarrow a \cdot \overrightarrow b = - 4 + 8 + 8 \hfill \\
\overrightarrow a \cdot \overrightarrow b = 12 \hfill \\
\left| {\overrightarrow a } \right| = \sqrt {{a_x}^2 + {a_y}^2 + {a_z}^2} \hfill \\
\left| {\overrightarrow a } \right| = \sqrt {{2^2} + {2^2} + {4^2}} = \sqrt {4 + 4 + 16} = \sqrt {24} = 2\sqrt 6 \hfill \\
\left| {\overrightarrow b } \right| = \sqrt {{{\left( { - 2} \right)}^2} + {4^2} + {2^2}} = \sqrt {4 + 16 + 4} = \sqrt {24} = 2\sqrt 6 \hfill \\
\hfill \\
\cos \measuredangle \left( {\overrightarrow a ,\overrightarrow b } \right) = \frac{{\overrightarrow a \cdot \overrightarrow b }}{{\left| {\overrightarrow a } \right| \cdot \left| {\overrightarrow b } \right|}} \hfill \\
\cos \measuredangle \left( {\overrightarrow a ,\overrightarrow b } \right) = \frac{{12}}{{2\sqrt 6 \cdot 2\sqrt 6 }} = \frac{3}{6} = \frac{1}{2} \hfill \\
\measuredangle \left( {\overrightarrow a ,\overrightarrow b } \right) = {60^ \circ } \hfill \\
\end{gathered} \]

Пр.2)

391 png

\[\begin{gathered}
\overrightarrow {AC} = C - A = \left( {8,7,2} \right) \hfill \\
\overrightarrow {BD} = D - B = \left( { - 4,4,0} \right) \hfill \\
\cos \measuredangle \left( {\overrightarrow {AC} ,\overrightarrow {BD} } \right) = \frac{{\overrightarrow {AC} \cdot \overrightarrow {BD} }}{{\left| {\overrightarrow {AC} } \right| \cdot \left| {\overrightarrow {BD} } \right|}} \hfill \\
\overrightarrow {AC} \cdot \overrightarrow {BD} = 8 \cdot \left( { - 4} \right) + 7 \cdot 4 + 2 \cdot 0 = - 4 \hfill \\
\left| {\overrightarrow {AC} } \right| = \sqrt {{8^2} + {7^2} + {2^2}} = \sqrt {117} = 3\sqrt {13} \hfill \\
\left| {\overrightarrow {BD} } \right| = \sqrt {{{\left( { - 4} \right)}^2} + {4^2} + {0^2}} = \sqrt {32} = 4\sqrt 2 \hfill \\
\cos \measuredangle \left( {\overrightarrow {AC} ,\overrightarrow {BD} } \right) = \frac{{ - 4}}{{3\sqrt {13} \cdot 4\sqrt 2 }} = - \frac{{\sqrt {26} }}{{78}} \hfill \\
\measuredangle \left( {\overrightarrow {AC} ,\overrightarrow {BD} } \right) = \arccos \left( { - \frac{{\sqrt {26} }}{{78}}} \right) \hfill \\
\end{gathered} \]

Пр.3)

\[\begin{gathered}
\overrightarrow m = \overrightarrow a - \overrightarrow b = \left( {1,2} \right) - \left( { - 1,3} \right) = \left( {2, - 1} \right) \hfill \\
\overrightarrow n = \overrightarrow c - \overrightarrow d = \left( {0, - 1} \right) + \left( { - 3, - 2} \right) = \left( { - 3, - 3} \right) \hfill \\
\overrightarrow m \cdot \overrightarrow n = - 6 + 3 = - 3 \hfill \\
\left| {\overrightarrow m } \right| = \sqrt {{2^2} + {{\left( { - 1} \right)}^2}} = \sqrt 5 \hfill \\
\left| {\overrightarrow n } \right| = \sqrt {{{\left( { - 3} \right)}^2} + {{\left( { - 3} \right)}^2}} = 3\sqrt 2 \hfill \\
\measuredangle \left( {\overrightarrow m ,\overrightarrow n } \right) = ? \hfill \\
\cos \measuredangle \left( {\overrightarrow m ,\overrightarrow n } \right) = \frac{{\overrightarrow m \cdot \overrightarrow n }}{{\left| {\overrightarrow m } \right| \cdot \left| {\overrightarrow n } \right|}} = - \frac{{\sqrt {10} }}{{10}} \hfill \\
\measuredangle \left( {\overrightarrow m ,\overrightarrow n } \right) = \arccos \left( { - \frac{{\sqrt {10} }}{{10}}} \right) \hfill \\
\end{gathered} \]