ГДЗ по геометрии 11 класс. Атанасян ФГОС 753

\[\boxed{\mathbf{753.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[Решение.\]

\[Пусть\ ребро\ куба\ равно\ \text{a.}\]

\[Координаты\ вершин\ куба:\]

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

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

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

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

\[M\left( \frac{a}{2};a;\frac{a}{2} \right).\]

\[\textbf{а)}\ \overrightarrow{A_{1}D}\left\{ - a;0; - a \right\};\ \]

\[\overrightarrow{\text{AM}}\left\{ - \frac{a}{2};a;\frac{a}{2} \right\}:\]

\[\cos{\angle\left( \overrightarrow{A_{1}D};\overrightarrow{\text{AM}} \right)} = \frac{\left| \frac{a^{2}}{2} - \frac{a^{2}}{2} \right|}{\sqrt{2a^{2}} \cdot \sqrt{\frac{a^{2}}{2}}} =\]

\[= 0;\]

\[\angle\left( \overrightarrow{A_{1}D};\overrightarrow{\text{AM}} \right) = 90{^\circ}.\]

\[\textbf{б)}\ \overrightarrow{\text{MD}}\left\{ - \frac{a}{2}; - a; - \frac{a}{2} \right\};\ \]

\[\ \overrightarrow{BB_{1}}\left\{ 0;0;a \right\}:\]

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

\[= \frac{- \frac{a^{2}}{2}}{\sqrt{\frac{a^{2}}{4} + a^{2} + \frac{a^{2}}{4}} \cdot \sqrt{a^{2}}} =\]

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

\[\angle\left( \overrightarrow{\text{MD}};\overrightarrow{BB_{1}} \right) \approx 114{^\circ}.\]