Решебник по геометрии 10 класс Атанасян ФГОС 713

\[\boxed{\mathbf{713.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[MQ = a.\]

\[Доказать:\]

\[\textbf{а)}\ PM_{1}\bot MN_{1}Q_{1};\]

\[\textbf{б)}\ PM_{1}\bot QNP_{1}.\]

\[Доказательство.\]

\[Q(0;0;0);Q_{1}(0;0;a);N(a;a;0);\]

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

\[M(a;0;0);M_{1}(a;0;a);P(0;a;0);\]

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

\[\textbf{а)}\ 1)\ \overrightarrow{PM_{1}}\left\{ a; - a;a \right\};\]

\[\cos{\angle\left( \overrightarrow{PM_{1}};\overrightarrow{MQ_{1}} \right)} =\]

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

\[угол\ между\ прямыми\ равен\ \]

\[90{^\circ};\]

\[PM_{1}\bot MQ_{1}.\]

\[2)\ \overrightarrow{MN_{1}}\left\{ 0;a;a \right\};\]

\[\cos{\angle\left( \overrightarrow{PM_{1}};\overrightarrow{MN_{1}} \right)} =\]

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

\[угол\ между\ прямыми\ равен\ \]

\[90{^\circ};\]

\[PM_{1}\bot NM_{1}.\]

\[Отсюда:\]

\[PM_{1}\bot MN_{1} - лежит\ в\ \]

\[плоскости\ MN_{1}Q_{1};\]

\[MQ_{1} - лежит\ в\ плоскости\ \]

\[MN_{1}Q_{1};\]

\[прямые\ пересекаются\ в\ точке\ \]

\[M;\]

\[PM_{1}\bot\left( MN_{1}Q_{1} \right).\]

\[\textbf{б)}\ 1)\ \overrightarrow{\text{QN}}\left\{ a;a;0 \right\};\ \ \overrightarrow{QP_{1}}\left\{ 0;a;a \right\};\ \ \]

\[\overrightarrow{PM_{1}}\left\{ a; - a;a \right\}:\]

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

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

\[угол\ между\ прямыми\ равен\ \]

\[90{^\circ}.\]

\[2)\cos{\angle\left( \overrightarrow{PM_{1}};\overrightarrow{QP_{1}} \right)} =\]

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

\[угол\ между\ прямыми\ равен\ 90{^\circ}.\]

\[Аналогично,\ PM_{1}\bot\left( \text{QN}P_{1} \right).\]

\[Что\ и\ требовалось\ доказать.\]