Решебник по геометрии 10 класс Атанасян ФГОС 709

\[\boxed{\mathbf{709.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[AB = 1;\]

\[BC = 2;\]

\[BB_{1} = 3.\]

\[Решение.\]

\[A(2;0;2);\ \ B(2;1;0);\ \ C(0;1;0);\ \ \]

\[D(0;0;0);\]

\[A_{1}(2;0;3);\ \ B_{1}(2;1;3);\ \ \]

\[C_{1}(0;1;3);\ \ D_{1}(0;0;3).\]

\[\textbf{а)}\ \overrightarrow{\text{AC}}\left\{ - 2;1;0 \right\};\ \ \overrightarrow{D_{1}B}\left\{ 2;1; - 3 \right\}:\]

\[\cos{\angle\left( \overrightarrow{\text{AC}};\overrightarrow{D_{1}B} \right)} =\]

\[= \frac{| - 4 + 1|}{\sqrt{4 + 1} \cdot \sqrt{4 + 1 + 9}} = \frac{3}{\sqrt{70}}.\]

\[\textbf{б)}\ \overrightarrow{AB_{1}}\left\{ 0;1;3 \right\};\ \ \overrightarrow{BC_{1}}\left\{ - 2;0;3 \right\}:\]

\[\cos{\angle\left( \overrightarrow{AB_{1}};\overrightarrow{BC_{1}} \right)} =\]

\[= \frac{|0 + 0 + 9|}{\sqrt{1 + 9} \cdot \sqrt{4 + 9}} = \frac{9}{\sqrt{130}}\text{.\ }\]

\[\textbf{в)}\ \overrightarrow{A_{1}D}\left\{ - 2;0;3 \right\};\ \ \overrightarrow{AC_{1}}\left\{ - 2;1;3 \right\}:\ \]

\[\cos{\angle\left( \overrightarrow{A_{1}D};\overrightarrow{AC_{1}} \right)} =\]

\[= \frac{|4 - 9|}{\sqrt{4 + 9} \cdot \sqrt{4 + 1 + 9}} =\]

\[= \frac{5}{\sqrt{13} \cdot \sqrt{14}} = \frac{5}{\sqrt{182}}.\]