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МатематикаГДЗ по геометрии 11 класс. Атанасян ФГОС 707
707
\[\boxed{\mathbf{707.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]
\[Дано:\]
\[AB = a.\]
\[Решение.\]
\[Координаты\ вершин\ \]
\[параллелепипеда:\]
\[A(a;0;0);\ \ B(0;0;0);\ \ C(0;a;0);\ \ \]
\[D(a;a;0);\]
\[A_{1}(a;0;a);\ \ B_{1}(0;0;a);\ \ \]
\[C_{1}(0;a;a);\ \ D_{1}(a;a;a).\]
\[M\left( a;0;\frac{3a}{4} \right);\ \ N\left( 0;\frac{a}{2};0 \right).\]
\[\textbf{а)}\ \overrightarrow{\text{MN}}\left\{ - a;\frac{a}{2};\frac{3a}{4} \right\};\ \ \overrightarrow{DD_{1}}\left\{ 0;0;a \right\}:\ \]
\[\cos{\angle\left( \overrightarrow{\text{MN}};\overrightarrow{DD_{1}} \right)} =\]
\[= \frac{\frac{3a^{2}}{4}}{\sqrt{a^{2} + \frac{a^{2}}{4} + \frac{9a^{2}}{16}} \cdot a} =\]
\[= \frac{3a^{2}}{4a^{2}\sqrt{1^{\backslash 16} + \frac{1^{\backslash 4}}{4} + \frac{9}{16}}} =\]
\[= \frac{3}{\sqrt{16 \cdot \frac{29}{16}}} = \frac{3}{\sqrt{29}}.\]
\[\textbf{б)}\ \overrightarrow{\text{MN}}\left\{ - a;\frac{a}{2};\frac{3a}{4} \right\};\ \ \overrightarrow{\text{BD}}\left\{ a;a;0 \right\}:\]
\[\cos{\angle\left( \overrightarrow{\text{MN}};\overrightarrow{\text{BD}} \right)} =\]
\[= \frac{\left| - \frac{a^{2}}{2} \right|}{\sqrt{a^{2} + \frac{a^{2}}{4} + \frac{9a^{2}}{16}} \cdot a\sqrt{2}} =\]
\[= \frac{1}{2\sqrt{2} \cdot \sqrt{\frac{29}{16}}} = \sqrt{\frac{2}{29}} = \frac{2}{\sqrt{58}}.\]
\[\textbf{в)}\ \overrightarrow{\text{MN}}\left\{ - a;\frac{a}{2};\frac{3a}{4} \right\};\ \ \]
\[\overrightarrow{B_{1}D}\left\{ a;a; - a \right\}:\ \]
\[\cos{\angle\left( \overrightarrow{\text{MN}};\overrightarrow{B_{1}D} \right)} =\]
\[= \frac{\left| \frac{{a^{2}}^{\backslash 2}}{2} - {a^{2}}^{\backslash 4} + \frac{3a^{2}}{4} \right|}{\sqrt{a^{2} + \frac{a^{2}}{4} + \frac{9a^{2}}{16}} \cdot a\sqrt{3}} =\]
\[= \frac{\frac{a^{2}}{4}}{a^{2}\sqrt{3} \cdot \sqrt{\frac{29}{16}}} = \frac{1}{\sqrt{87}}.\]
\[\textbf{г)}\ \overrightarrow{\text{MN}}\left\{ - \frac{a}{2};a;\ - \frac{3a}{4} \right\};\ \ \]
\[\overrightarrow{A_{1}C}\left\{ - a;a; - a \right\}:\ \]
\[\cos{\angle\left( \overrightarrow{\text{MN}};\overrightarrow{A_{1}C} \right)} =\]
\[= \frac{\left| \frac{{a^{2}}^{\backslash 2}}{2} + {a^{2}}^{\backslash 4} + \frac{3a^{2}}{4} \right|}{a\sqrt{\frac{29}{16}} \cdot a\sqrt{3}} =\]
\[= \frac{a^{2} \cdot 9\sqrt{16}}{a^{2} \cdot 4\sqrt{29} \cdot \sqrt{3}} = \frac{3\sqrt{3}}{\sqrt{29}}.\]
