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МатематикаРешебник по геометрии 10 класс Атанасян ФГОС 692
692
\[\boxed{\mathbf{692.}еуроки - ответы\ на\ пятёрку}\]
\[Формула:\]
\[\cos\alpha =\]
\[= \frac{x_{1}x_{2} + y_{1}y_{2} + z_{1}z_{2}}{\sqrt{x_{1}^{2} + y_{1}^{2} + z_{1}^{2}} \cdot \sqrt{x_{2}^{2} + y_{2}^{2} + z_{2}^{2}}}.\]
\[\textbf{а)}\ \overrightarrow{a}\left\{ 2; - 2;0 \right\};\ \ \overrightarrow{b}\left\{ 3;0; - 3 \right\}:\]
\[\cos{\angle\left( \overrightarrow{a};\overrightarrow{b} \right)} =\]
\[= \frac{2 \cdot 3 - 2 \cdot 0 - 0 \cdot 3}{\sqrt{4 + 4 + 0} \cdot \sqrt{9 + 9 + 0}} =\]
\[= \frac{6}{\sqrt{8} \cdot \sqrt{18}} = \frac{6}{\sqrt{144}} = \frac{6}{12} = \frac{1}{2};\]
\[\angle\left( \overrightarrow{a};\overrightarrow{b} \right) = 60{^\circ}.\]
\[\textbf{б)}\ \overrightarrow{a}\left\{ \sqrt{2};\sqrt{2};2 \right\};\ \ \overrightarrow{b}\left\{ - 3; - 3;0 \right\}:\]
\[\cos{\angle\left( \overrightarrow{a};\overrightarrow{b} \right)} =\]
\[= \frac{- \sqrt{2} \cdot 3 - \sqrt{2} \cdot 3 - 2 \cdot 0}{\sqrt{2 + 2 + 4} \cdot \sqrt{9 + 9 + 0}} =\]
\[= - \frac{\sqrt{2} \cdot 6}{\sqrt{144}} = - \frac{\sqrt{2}}{2};\]
\[\angle\left( \overrightarrow{a};\overrightarrow{b} \right) = 180{^\circ} - 45{^\circ} = 135{^\circ}.\]
\[\textbf{в)}\ \overrightarrow{a}\left\{ 0;5;0 \right\};\ \ \overrightarrow{b}\left\{ 0; - \sqrt{3};1 \right\}:\]
\[\cos{\angle\left( \overrightarrow{a};\overrightarrow{b} \right)} =\]
\[= \frac{0 \cdot 0 - 5 \cdot \sqrt{3} - 0 \cdot 1}{\sqrt{0 + 25 + 0} \cdot \sqrt{0 + 3 + 1}} =\]
\[= \frac{- 5\sqrt{3}}{5 \cdot 2} = - \frac{\sqrt{3}}{2};\]
\[\angle\left( \overrightarrow{a};\overrightarrow{b} \right) = 180{^\circ} - 30{^\circ} = 150{^\circ}.\]
\[\textbf{г)}\ \overrightarrow{a}\left\{ - 2,5;2,5;0 \right\};\ \ \overrightarrow{b}\left\{ - 5;5;5\sqrt{2} \right\}:\]
\[\cos{\angle\left( \overrightarrow{a};\overrightarrow{b} \right)} =\]
\[= \frac{2,5 \cdot 5 + 2,5 \cdot 5 - 0 \cdot 5\sqrt{2}}{\sqrt{6,25 + 6,25} \cdot \sqrt{25 + 25 + 50}} =\]
\[= \frac{25}{\sqrt{\frac{25}{2}} \cdot 10} = \frac{\sqrt{2} \cdot 25}{50} = \frac{\sqrt{2}}{2};\]
\[\angle\left( \overrightarrow{a};\overrightarrow{b} \right) = 45{^\circ}.\]
\[\textbf{д)}\ \overrightarrow{a}\left\{ - \sqrt{2}; - \sqrt{2}; - 2 \right\};\ \ \]
\[\overrightarrow{b}\left\{ \frac{\sqrt{2}}{2};\frac{\sqrt{2}}{2}; - 1 \right\}:\]
\[\cos{\angle\left( \overrightarrow{a};\overrightarrow{b} \right)} =\]
\[= \frac{- \sqrt{2} \cdot \frac{\sqrt{2}}{2} - \sqrt{2} \cdot \frac{\sqrt{2}}{2} + 2 \cdot 1}{\sqrt{2 + 2 + 4} \cdot \sqrt{\frac{1}{2} + \frac{1}{2} + 1}} =\]
\[= \frac{0}{\sqrt{16}} = 0;\]
\[\angle\left( \overrightarrow{a};\overrightarrow{b} \right) = 90{^\circ}.\]
