Решебник по геометрии 11 класс. Атанасян ФГОС 686

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\[\overrightarrow{a} = 3\overrightarrow{i} - 5\overrightarrow{j} + \overrightarrow{k}:\]

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

\[\overrightarrow{b} = \overrightarrow{j} - 5\overrightarrow{k}:\]

\[\overrightarrow{b}\left\{ 0;1; - 5 \right\}.\]

\[\overrightarrow{i}\left\{ 1;0;0 \right\};\ \ \overrightarrow{j}\left\{ 0;1;0 \right\};\ \ \overrightarrow{k}\left\{ 0;0;1 \right\}.\]

\[\textbf{а)}\ \overrightarrow{a} \cdot \overrightarrow{b} = 0 - 5 - 5 = - 10.\]

\[\textbf{б)}\ \overrightarrow{a} \cdot \overrightarrow{i} = 3 - 0 + 0 = 3.\]

\[\textbf{в)}\ \overrightarrow{b} \cdot \overrightarrow{j} = 0 + 1 - 0 = 1.\]

\[\textbf{г)}\ \left( \overrightarrow{a} + \overrightarrow{b} \right) \cdot \overrightarrow{k} = \overrightarrow{a} \cdot \overrightarrow{k} + \overrightarrow{b} \cdot \overrightarrow{k} =\]

\[= (0 + 0 + 1) + (0 + 0 - 5) =\]

\[= 1 - 5 = - 4.\]

\[\textbf{д)}\ \left( \overrightarrow{a} - 2\overrightarrow{b} \right)\left( \overrightarrow{k} + \overrightarrow{i} - 2\overrightarrow{j} \right) =\]