Решебник по геометрии 10 класс Атанасян ФГОС 698

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\[\left| \overrightarrow{a} \right| = 1;\ \ \left| \overrightarrow{b} \right| = \left| \overrightarrow{c} \right| = 2;\]

\[\angle\left( \overrightarrow{a}\overrightarrow{c} \right) = \angle\left( \overrightarrow{b}\overrightarrow{c} \right) = 60{^\circ}.\]

\[\left( \overrightarrow{a} + \overrightarrow{b} \right) \cdot \overrightarrow{c} = \overrightarrow{a} \cdot \overrightarrow{c} + \overrightarrow{b} \cdot \overrightarrow{c}:\]

\[\overrightarrow{a}\overrightarrow{c} = \left| \overrightarrow{a} \right| \cdot \left| \overrightarrow{c} \right| \cdot \cos{\angle\left( \overrightarrow{a}\overrightarrow{c} \right)} =\]

\[= 1 \cdot 2 \cdot \cos{60{^\circ}\ } = 2 \cdot \frac{1}{2} = 1;\]

\[\overrightarrow{b}\overrightarrow{c} = \left| \overrightarrow{b} \right| \cdot \left| \overrightarrow{c} \right| \cdot \cos{\angle\left( \overrightarrow{b}\overrightarrow{c} \right)} =\]

\[= 2 \cdot 2 \cdot \cos{60{^\circ}} = 4 \cdot \frac{1}{2} = 2;\]

\[\left( \overrightarrow{a} + \overrightarrow{b} \right) \cdot \overrightarrow{c} = \overrightarrow{a} \cdot \overrightarrow{c} + \overrightarrow{b} \cdot \overrightarrow{c} =\]

\[= 1 + 2 = 3.\]

\[Ответ:3.\]