Решебник по геометрии 10 класс Атанасян ФГОС 711

\[\boxed{\mathbf{711.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[\angle ABD = \angle ABC = \angle DBC = 90{^\circ};\]

\[AB = BD = 2;\]

\[BC = 1.\]

\[Решение.\]

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

\[D(0;0;2);\ \]

\[M(1;0;1);N\left( 0;\frac{1}{2};0 \right).\]

\[\textbf{а)}\ M - середина\ AD;\]

\[N - середина\ BC;\]

\[\overrightarrow{\text{BC}}\left\{ 0;1;0 \right\};\ \ \overrightarrow{\text{MN}}\left\{ - 1;\frac{1}{2}; - 1 \right\}.\]

\[\overrightarrow{\text{BC}}\bot\left( \text{ABD} \right):\]

\[\sin{\angle\left( \overrightarrow{\text{MN}};\left( \text{ABD} \right) \right)} =\]

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

\[= \frac{\left| 0 + \frac{1}{2} + 0 \right|}{\sqrt{1} \cdot \sqrt{1 + \frac{1}{4} + 1\ }} = \frac{\frac{1}{2}}{\sqrt{\frac{9}{4}}} = \frac{1}{3}.\]

\[\textbf{б)}\ \overrightarrow{\text{BA}}\left\{ 2;0;0 \right\};\ \ \overrightarrow{\text{MN}}\left\{ - 1;\frac{1}{2}; - 1 \right\};\]

\[\overrightarrow{\text{BA}}\bot\left( \text{DBC} \right):\]

\[\sin{\angle\left( \overrightarrow{\text{MN}};\left( \text{DBC} \right) \right)} =\]

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

\[= \frac{| - 2|}{\sqrt{4} \cdot \sqrt{1 + \frac{1}{4} + 1}} = \frac{2}{2\sqrt{\frac{9}{4}}} = \frac{2}{3}.\]

\[\textbf{в)}\ \overrightarrow{\text{BD}}\left\{ 0;0;2 \right\};\ \ \overrightarrow{\text{MN}}\left\{ - 1;\frac{1}{2}; - 1 \right\};\]

\[\overrightarrow{\text{BD}}\bot\left( \text{ABC} \right):\]

\[\sin{\angle\left( \overrightarrow{\text{MN}};\left( \text{ABC} \right) \right)} =\]

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

\[= \frac{| - 2|}{\sqrt{4} \cdot \sqrt{1 + \frac{1}{4} + 1\ }} = \frac{2}{3}.\]