Решебник по геометрии 10 класс Атанасян ФГОС 693

\[\boxed{\mathbf{693.}еуроки - ответы\ на\ пятёрку}\]

\[Формула:\]

\[\cos\alpha =\]

\[= \frac{x_{1}x_{2} + y_{1}y_{2} + z_{1}z_{2}}{\sqrt{x_{1}^{2} + y_{1}^{2} + z_{1}^{2}} \cdot \sqrt{x_{2}^{2} + y_{2}^{2} + z_{2}^{2}}}.\]

\[\overrightarrow{a}\left\{ 2;1;2 \right\}.\]

\[1)\ \overrightarrow{i}\left\{ 1;0;0 \right\}:\]

\[\cos{\angle\left( \overrightarrow{a};\overrightarrow{i} \right)} =\]

\[= \frac{2 \cdot 1 + 1 \cdot 0 + 2 \cdot 0}{\sqrt{4 + 1 + 4} \cdot \sqrt{1 + 0 + 0}} =\]

\[= \frac{2}{3 \cdot 1} = \frac{2}{3};\]

\[\angle\left( \overrightarrow{a};\overrightarrow{i} \right) = \arccos\frac{2}{3} \approx 50{^\circ}46'.\]

\[2)\ \overrightarrow{j}\left\{ 0;1;0 \right\}:\]

\[\cos{\angle\left( \overrightarrow{a};\overrightarrow{j} \right)} =\]

\[= \frac{2 \cdot 0 + 1 \cdot 1 + 2 \cdot 0}{\sqrt{9} \cdot \sqrt{1}} = \frac{1}{3};\]

\[\angle\left( \overrightarrow{a};\overrightarrow{j} \right) = \arccos\frac{1}{3} \approx 63{^\circ}26'.\]

\[3)\ \overrightarrow{k}\left\{ 0;0;1 \right\}:\]

\[\cos{\angle\left( \overrightarrow{a};\overrightarrow{k} \right)} =\]

\[= \frac{2 \cdot 0 + 1 \cdot 0 + 2 \cdot 1}{\sqrt{9} \cdot \sqrt{1}} = \frac{2}{3};\]

\[\angle\left( \overrightarrow{a};\overrightarrow{k} \right) = \arccos\frac{2}{3} \approx 50{^\circ}46^{'}.\]