Решебник по геометрии 11 класс. Атанасян ФГОС 708

\[\boxed{\mathbf{708.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[AB = BC = a;\]

\[AA_{1} = 2a.\]

\[Решение.\]

\[Координаты\ вершин\ \]

\[параллелепипеда:\]

\[A(a;0;0);\ \ B(a;a;0);\ \ \]

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

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

\[C_{1}(0;a;2a);\ \ D_{1}(0;0;2a).\]

\[\textbf{а)}\ \overrightarrow{\text{BD}}\left\{ - a; - a;0 \right);\ \ \]

\[\overrightarrow{CD_{1}}\left\{ 0; - a;2a \right\}:\]

\[\cos{\angle\left( \overrightarrow{\text{BD}};\overrightarrow{CD_{1}} \right)} =\]

\[= \frac{\left| 0 + a^{2} + 0 \right|}{\sqrt{a^{2} + a^{2}} \cdot \sqrt{a^{2} + 4a^{2}}} =\]

\[= \frac{a^{2}}{a\sqrt{2} \cdot a\sqrt{5}} = \frac{1}{\sqrt{10}};\]

\[\angle\left( \overrightarrow{\text{BD}};\overrightarrow{CD_{1}} \right) = \arccos\frac{1}{\sqrt{10}} \approx\]

\[\approx 71{^\circ}34^{'}.\]

\[\textbf{б)}\ \overrightarrow{\text{AC}}\left\{ - a;a;0 \right\};\ \ \overrightarrow{AC_{1}}\left\{ - a;a;2a \right\}:\]

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

\[= \frac{\left| a^{2} + a^{2} + 0 \right|}{\sqrt{a^{2} + a^{2}} \cdot \sqrt{a^{2} + a^{2} + 4a^{2}}} =\]

\[= \frac{2a^{2}}{a\sqrt{2} \cdot a\sqrt{6}} = \frac{2\sqrt{2}}{\sqrt{6}} = \frac{1}{\sqrt{3}};\]

\[\angle\left( \overrightarrow{\text{AC}};\overrightarrow{AC_{1}} \right) = \arccos\frac{1}{\sqrt{3}} \approx\]

\[\approx 54{^\circ}44^{'}.\]