ГДЗ по геометрии 11 класс. Атанасян ФГОС 600

Авторы:
Год:2023
Тип:учебник

600

\[\boxed{\mathbf{600.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\ \ \]

\[ABCD - пирамида;\ \]

\[ABCD - параллелограмм;\]

\[M - точка\ пересечения\ \]

\[диагоналей\ \text{ABCD.}\]

\[Разложить:\ \ \]

\[векторы\ \overrightarrow{\text{OD}}\ и\ \overrightarrow{\text{OM}}\ по\ векторам\ \]

\[\overrightarrow{a} = \overrightarrow{\text{OA}};\ \overrightarrow{b} = \ \overrightarrow{\text{OB}}\ и\ \overrightarrow{c} = \overrightarrow{\text{OC}}.\]

\[Решение.\]

\[1)\ \overrightarrow{\text{OB}} = \overrightarrow{\text{OD}} + \overrightarrow{\text{DB}}:\ \]

\[\overrightarrow{\text{OD}} = \overrightarrow{\text{OB}} - \overrightarrow{\text{DB}} = \overrightarrow{b} - \overrightarrow{\text{DB}}\text{.\ \ }\]

\[\overrightarrow{\text{BC}} + \overrightarrow{\text{BA}} = \overrightarrow{\text{BD}}\ \]

\[(по\ правилу\ параллелограмма):\]

\[\overrightarrow{\text{DB}} = - \overrightarrow{\text{BD}} = - \overrightarrow{\text{BC}} - \overrightarrow{\text{BA}} =\]

\[= \overrightarrow{\text{CB}} + \overrightarrow{\text{AB}}.\]

\[2)\ \overrightarrow{\text{OC}} + \overrightarrow{\text{CB}} = \overrightarrow{\text{OB}}:\]

\[\overrightarrow{\text{CB}} = \overrightarrow{\text{OB}} - \overrightarrow{\text{OC}} = \overrightarrow{b} - \overrightarrow{c}.\]

\[3)\ \overrightarrow{\text{OA}} + \overrightarrow{\text{AB}} = \overrightarrow{\text{OB}}:\]

\[\overrightarrow{\text{AB}} = \overrightarrow{\text{OB}} - \overrightarrow{\text{OA}} = \overrightarrow{b} - \overrightarrow{a}.\]

\[4)\ Получаем:\]

\[\overrightarrow{\text{DB}} = \left( \overrightarrow{b} - \overrightarrow{c} \right) + \left( \overrightarrow{b} - \overrightarrow{a} \right) =\]

\[= 2\overrightarrow{b} - \overrightarrow{c} - \overrightarrow{a};\]

\[\overrightarrow{\text{OD}} = \overrightarrow{b} - \left( 2\overrightarrow{b} - \overrightarrow{c} - \overrightarrow{a} \right) =\]

\[= \overrightarrow{a} - \overrightarrow{b} + \overrightarrow{c}.\]

\[5)\ \overrightarrow{\text{OM}} = \overrightarrow{\text{OA}} + \overrightarrow{\text{AM}} = \overrightarrow{\text{OC}} + \overrightarrow{\text{CM}}:\]

\[2\overrightarrow{\text{OM}} = \overrightarrow{\text{OA}} + \overrightarrow{\text{OC}} =\]

\[= \overrightarrow{a} + \overrightarrow{c}\ \left( \overrightarrow{\text{AM}} = - \overrightarrow{\text{CM}} \right).\]

\[Следовательно:\]

\[\overrightarrow{\text{OM}} = \frac{1}{2}\overrightarrow{a} + 0 \bullet \overrightarrow{b} + \frac{1}{2}\overrightarrow{c}.\]

\[Ответ:\ \ \overrightarrow{\text{OD}} = \overrightarrow{a} - \overrightarrow{b} + \overrightarrow{c};\ \ \]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ }\overrightarrow{\text{OM}} = \frac{1}{2}\overrightarrow{a} + 0 \bullet \overrightarrow{b} + \frac{1}{2}\overrightarrow{c}.\]

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