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МатематикаРешебник по геометрии 10 класс Атанасян ФГОС 605
605
\[\boxed{\mathbf{605.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]
\[Дано:\]
\[\text{ABCD}A_{1}B_{1}C_{1}D_{1} -\]
\[параллелепипед;\ \ \]
\[AM = MB;\ \ \]
\[A_{1}N = ND_{1}.\]
\[Решение.\]
\[\textbf{а)}\ По\ правилу\ \]
\[параллелограмма:\]
\[\overrightarrow{\text{AC}} = \overrightarrow{\text{AB}} + \overrightarrow{\text{AD}}.\]
\[\textbf{б)}\ \overrightarrow{\text{MB}} + \overrightarrow{\text{BC}} = \overrightarrow{\text{MC}} = - \overrightarrow{\text{MC}};\ \ \text{\ \ }\]
\[\overrightarrow{\text{MB}} = \frac{1}{2}\overrightarrow{\text{AB}};\ \ \ \ \overrightarrow{\text{BC}} = \overrightarrow{\text{AD}}:\]
\[\overrightarrow{\text{CM}} = - \left( \frac{1}{2}\overrightarrow{\text{AB}} + \overrightarrow{\text{AD}} \right) =\]
\[= - \frac{1}{2}\overrightarrow{\text{AB}} - \overrightarrow{\text{AD}}.\]
\[\textbf{в)}\ Отметим\ точку\ E -\]
\[середину\ ребра\ AD:\]
\[\ \overrightarrow{C_{1}N} = \overrightarrow{\text{CE}};\]
\[\overrightarrow{\text{ED}} + \overrightarrow{\text{DC}} = \overrightarrow{\text{EC}} = - \overrightarrow{\text{CE}};\ \ \]
\[\overrightarrow{\text{ED}} = \frac{1}{2}\overrightarrow{\text{AD}};\ \ \ \]
\[\overrightarrow{\text{DC}} = \overrightarrow{\text{AB}}.\]
\[Отсюда:\]
\[\overrightarrow{\text{CE}} = - \left( \frac{1}{2}\overrightarrow{\text{AD}} + \overrightarrow{\text{AB}} \right) =\]
\[= - \frac{1}{2}\overrightarrow{\text{AD}} - \overrightarrow{\text{AB}};\ \]
\[\overrightarrow{C_{1}N} = - \overrightarrow{\text{AB}} - \frac{1}{2}\overrightarrow{\text{AD}}.\]
\[\textbf{г)}\ Векторы\ \overrightarrow{AC_{1}},\ \overrightarrow{\text{AB}}\ и\ \overrightarrow{\text{AD}} - не\ \]
\[компланарны,\ так\ как\ \overrightarrow{AC_{1}}\ не\ \]
\[лежит\ в\ плоскости\ ABD.\]
\[\textbf{д)}\ \overrightarrow{A_{1}N} = \overrightarrow{\text{AE}} = \frac{1}{2}\overrightarrow{\text{AD}} + 0 \bullet \overrightarrow{\text{AB}}\text{.\ }\]
\[\textbf{е)}\ Векторы\ \overrightarrow{\text{AN}},\ \overrightarrow{\text{AB}}\ и\ \overrightarrow{\text{AD}} - не\ \]
\[компланарны,\ так\ как\ \overrightarrow{\text{AN}}\ не\ \]
\[лежит\ в\ плоскости\ ABD.\]
\[\textbf{ж)}\ \overrightarrow{\text{AM}} + \overrightarrow{\text{MD}} = \overrightarrow{\text{AD}};\text{\ \ }\]
\[\overrightarrow{\text{AM}} = \frac{1}{2}\overrightarrow{\text{AB}}:\]
\[\overrightarrow{\text{MD}} + \frac{1}{2}\overrightarrow{\text{AB}} = \overrightarrow{\text{AD}}\ \]
\[\overrightarrow{\text{MD}} = \overrightarrow{\text{AD}} - \frac{1}{2}\overrightarrow{\text{AB}}.\]
