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МатематикаГДЗ по геометрии 11 класс. Атанасян ФГОС 705
705
\[\boxed{\mathbf{705.}еуроки - ответы\ на\ пятёрку}\]
\[Дано:\]
\[\angle\left( \overrightarrow{\text{AB}};\overrightarrow{\text{CD}} \right) = \angle\left( \overrightarrow{a}\overrightarrow{b} \right):\]
\[\cos{\angle\left( \overrightarrow{a}\overrightarrow{b} \right)} = \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{\left| \overrightarrow{a} \right| \cdot \left| \overrightarrow{b} \right|} =\]
\[= \frac{x_{a}x_{b} + y_{a}y_{b} + z_{a}z_{b}}{\sqrt{x_{a}^{2} + y_{a}^{2} + z_{a}^{2}} \cdot \sqrt{x_{b}^{2} + y_{b}^{2} + z_{b}^{2}}}.\]
\[\textbf{а)}\ \overrightarrow{\text{AB}}\left\{ 1;1; - 2 \right\};\ \ \overrightarrow{\text{CD}}\left\{ 1;0; - 1 \right\}:\]
\[\cos{\angle\left( \overrightarrow{a}\overrightarrow{b} \right)} =\]
\[= \frac{1 \cdot 1 + 1 \cdot 0 + 2 \cdot 1}{\sqrt{1 + 1 + 4} \cdot \sqrt{1 + 0 + 1}} =\]
\[= \frac{3}{\sqrt{12}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2};\]
\[\angle\left( \overrightarrow{a}\overrightarrow{b} \right) = 30{^\circ}.\]
\[Угол\ между\ AB\ и\ \text{CD\ }равен\ 30{^\circ}.\]
\[\textbf{б)}\ \overrightarrow{\text{AB}}\left\{ 1;0; - 1 \right\};\ \ \overrightarrow{\text{CD}}\left\{ 0; - 2;2 \right\}:\]
\[\cos{\angle\left( \overrightarrow{a}\overrightarrow{b} \right)} =\]
\[= \frac{1 \cdot 0 - 0 \cdot 2 - 2 \cdot 1}{\sqrt{1 + 0 + 1} \cdot \sqrt{0 + 4 + 4}} =\]
\[= - \frac{2}{\sqrt{16}} = - \frac{2}{4} = - \frac{1}{2};\]
\[\angle\left( \overrightarrow{a}\overrightarrow{b} \right) = 120{^\circ}.\]
\[Угол\ между\ AB\ и\ \text{CD\ }равен\ 60{^\circ}.\]
\[\textbf{в)}\ \overrightarrow{\text{AB}}\left\{ 1;1; - 2 \right\};\ \ \overrightarrow{\text{CD}}\left\{ - 2; - 2;4 \right\}:\]
\[\cos{\angle\left( \overrightarrow{a}\overrightarrow{b} \right)} =\]
\[= \frac{- 1 \cdot 2 - 1 \cdot 2 - 2 \cdot 4}{\sqrt{1 + 1 + 4} \cdot \sqrt{4 + 4 + 16}} =\]
\[= - \frac{12}{\sqrt{144}} = - \frac{12}{12} = - 1;\]
\[\angle\left( \overrightarrow{a}\overrightarrow{b} \right) = 180{^\circ}.\]
\[Угол\ между\ AB\ и\ \text{CD\ }равен\ 0{^\circ}.\]
\[\textbf{г)}\ \overrightarrow{\text{AB}}\left\{ - 1;0;1 \right\};\ \ \overrightarrow{\text{CD}}\left\{ 0;0; - 2 \right\}:\]
\[\cos{\angle\left( \overrightarrow{a}\overrightarrow{b} \right)} =\]
\[= \frac{- 1 \cdot 0 + 0 - 1 \cdot 2}{\sqrt{1 + 0 + 1} \cdot \sqrt{0 + 0 + 4}} =\]
\[= - \frac{2}{\sqrt{8}} = - \frac{2}{2\sqrt{2}} = - \frac{\sqrt{2}}{2};\]
\[\angle\left( \overrightarrow{a}\overrightarrow{b} \right) = 135{^\circ}.\]
\[Угол\ между\ AB\ и\ \text{CD\ }равен\ 45{^\circ}.\]
