ГДЗ по геометрии 11 класс. Атанасян ФГОС 756

\[\boxed{\mathbf{756.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[D(0;0;0) - начало\ системы\ \]

\[координат.\]

\[Длина\ ребра\ куба = \text{a.}\]

\[Координаты\ нужных\ вершин:\]

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

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

\[D(0;0;0);M\left( a;a;\frac{3a}{5} \right);\]

\[N\left( \frac{3a}{5};0;0 \right).\]

\[\textbf{а)}\ \overrightarrow{\text{NM}}\left\{ \frac{2a}{5};a;\frac{3a}{5} \right\};\ \ \overrightarrow{\text{DA}}\left\{ a;0;0 \right\}:\]

\[\sin{\angle\left( \overrightarrow{\text{NM}};\left( DD_{1}C_{1} \right) \right)} =\]

\[= \left| \cos{\angle\left( \overrightarrow{\text{MN}};\overrightarrow{\text{DA}} \right)} \right| =\]

\[= \frac{\left| \frac{2}{5}a^{2} \right|}{\sqrt{\frac{4}{25}a^{2} + a^{2} + \frac{9}{25}a^{2}} \cdot \sqrt{a^{2}}} =\]

\[= \frac{\frac{2}{5}a^{2}}{a^{2}\sqrt{\frac{38}{25}}} = \frac{2}{\sqrt{38}}.\]

\[\textbf{б)}\ \overrightarrow{\text{NM}}\left\{ \frac{2a}{5};a;\frac{3a}{5} \right\};\ \ \overrightarrow{AA_{1}}\left\{ 0;0;a \right\}:\]

\[\sin{\angle\left( \overrightarrow{\text{NM}};\left( A_{1}B_{1}C_{1} \right) \right)} =\]

\[= \left| \cos{\angle\left( \overrightarrow{\text{NM}};\overrightarrow{AA_{1}} \right)} \right| =\]

\[= \frac{\left| \frac{3a^{2}}{5} \right|}{a\sqrt{\frac{38}{25}} \cdot \sqrt{a^{2}}} = \frac{3a^{2}}{5} \cdot \frac{1}{a^{2}\sqrt{\frac{38}{25}}} =\]

\[= \frac{3}{5} \cdot \frac{5}{\sqrt{38}} = \frac{3}{\sqrt{38}}.\]