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МатематикаРешебник по геометрии 11 класс. Атанасян ФГОС 600
600
\[\boxed{\mathbf{600.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]
\[Дано:\ \ \]
\[ABCD - пирамида;\ \]
\[ABCD - параллелограмм;\]
\[M - точка\ пересечения\ \]
\[диагоналей\ \text{ABCD.}\]
\[Разложить:\ \ \]
\[векторы\ \overrightarrow{\text{OD}}\ и\ \overrightarrow{\text{OM}}\ по\ векторам\ \]
\[\overrightarrow{a} = \overrightarrow{\text{OA}};\ \overrightarrow{b} = \ \overrightarrow{\text{OB}}\ и\ \overrightarrow{c} = \overrightarrow{\text{OC}}.\]
\[Решение.\]
\[1)\ \overrightarrow{\text{OB}} = \overrightarrow{\text{OD}} + \overrightarrow{\text{DB}}:\ \]
\[\overrightarrow{\text{OD}} = \overrightarrow{\text{OB}} - \overrightarrow{\text{DB}} = \overrightarrow{b} - \overrightarrow{\text{DB}}\text{.\ \ }\]
\[\overrightarrow{\text{BC}} + \overrightarrow{\text{BA}} = \overrightarrow{\text{BD}}\ \]
\[(по\ правилу\ параллелограмма):\]
\[\overrightarrow{\text{DB}} = - \overrightarrow{\text{BD}} = - \overrightarrow{\text{BC}} - \overrightarrow{\text{BA}} =\]
\[= \overrightarrow{\text{CB}} + \overrightarrow{\text{AB}}.\]
\[2)\ \overrightarrow{\text{OC}} + \overrightarrow{\text{CB}} = \overrightarrow{\text{OB}}:\]
\[\overrightarrow{\text{CB}} = \overrightarrow{\text{OB}} - \overrightarrow{\text{OC}} = \overrightarrow{b} - \overrightarrow{c}.\]
\[3)\ \overrightarrow{\text{OA}} + \overrightarrow{\text{AB}} = \overrightarrow{\text{OB}}:\]
\[\overrightarrow{\text{AB}} = \overrightarrow{\text{OB}} - \overrightarrow{\text{OA}} = \overrightarrow{b} - \overrightarrow{a}.\]
\[4)\ Получаем:\]
\[\overrightarrow{\text{DB}} = \left( \overrightarrow{b} - \overrightarrow{c} \right) + \left( \overrightarrow{b} - \overrightarrow{a} \right) =\]
\[= 2\overrightarrow{b} - \overrightarrow{c} - \overrightarrow{a};\]
\[\overrightarrow{\text{OD}} = \overrightarrow{b} - \left( 2\overrightarrow{b} - \overrightarrow{c} - \overrightarrow{a} \right) =\]
\[= \overrightarrow{a} - \overrightarrow{b} + \overrightarrow{c}.\]
\[5)\ \overrightarrow{\text{OM}} = \overrightarrow{\text{OA}} + \overrightarrow{\text{AM}} = \overrightarrow{\text{OC}} + \overrightarrow{\text{CM}}:\]
\[2\overrightarrow{\text{OM}} = \overrightarrow{\text{OA}} + \overrightarrow{\text{OC}} =\]
\[= \overrightarrow{a} + \overrightarrow{c}\ \left( \overrightarrow{\text{AM}} = - \overrightarrow{\text{CM}} \right).\]
\[Следовательно:\]
\[\overrightarrow{\text{OM}} = \frac{1}{2}\overrightarrow{a} + 0 \bullet \overrightarrow{b} + \frac{1}{2}\overrightarrow{c}.\]
\[Ответ:\ \ \overrightarrow{\text{OD}} = \overrightarrow{a} - \overrightarrow{b} + \overrightarrow{c};\ \ \]
\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ }\overrightarrow{\text{OM}} = \frac{1}{2}\overrightarrow{a} + 0 \bullet \overrightarrow{b} + \frac{1}{2}\overrightarrow{c}.\]
