Решебник по геометрии 10 класс Атанасян ФГОС 689

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\[\overrightarrow{a}\left\{ - 1;2;3 \right\};\ \ \overrightarrow{b}\left\{ 5;x; - 1 \right\}.\]

\[\overrightarrow{a} \cdot \overrightarrow{b} = - 1 \cdot 5 + 2 \cdot x - 1 \cdot 3 =\]

\[= - 5 + 2x - 3 = 2x - 8.\]

\[\textbf{а)}\ \overrightarrow{a} \cdot \overrightarrow{b} = 3:\]

\[2x - 8 = 3\]

\[2x = 11\]

\[x = 5,5.\]

\[\textbf{б)}\ \overrightarrow{a} \cdot \overrightarrow{b} = - 1\]

\[2x - 8 = - 1\]

\[2x = 7\]

\[x = 3,5.\]

\[\textbf{в)}\ \overrightarrow{a}\bot\overrightarrow{b};\ \ \cos\left( \overrightarrow{a};\overrightarrow{b} \right) = 0:\]

\[\overrightarrow{a} \cdot \overrightarrow{b} = 0\]

\[2x - 8 = 0\]

\[2x = 8\]

\[x = 4.\]