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МатематикаРешебник по геометрии 10 класс Атанасян ФГОС 606
606
\[\boxed{\mathbf{606.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]
\[Дано:\ \ \]
\[OABC - тетраэдр;\ \]
\[\ M - точка\ пересечения\ \]
\[медиан\ грани\ \text{ABC.}\]
\[Разложить:\ \ \]
\[вектор\ \overrightarrow{\text{OA}}\ по\ векторам\ \]
\[\overrightarrow{\text{OB}},\ \overrightarrow{\text{OC}}\ и\ \overrightarrow{\text{OM}}.\]
\[Решение.\]
\[1)\ \ Пусть\ точка\ K - середина\ \]
\[ребра\ AB:\]
\[2\overrightarrow{\text{OK}} = \overrightarrow{\text{OA}} + \overrightarrow{\text{AK}} = \overrightarrow{\text{OB}} + \overrightarrow{\text{BK}} =\]
\[= \overrightarrow{\text{OA}} + \overrightarrow{\text{OB}}\ \]
\[\left( так\ как\ \overrightarrow{\text{AK}} = - \overrightarrow{\text{BK}} \right).\]
\[Получаем:\]
\[\overrightarrow{\text{OK}} = \frac{1}{2}\left( \overrightarrow{\text{OA}} + \overrightarrow{\text{OB}} \right).\]
\[2)\ M - точка\ пересечения\ \]
\[медиан:\ \]
\[\overrightarrow{\text{CM}} = 2\overrightarrow{\text{MK}}.\]
\[\overrightarrow{\text{OC}} + \overrightarrow{\text{CM}} = \overrightarrow{\text{OM}}\text{\ \ }и\ \ \overrightarrow{\text{OM}} + \overrightarrow{\text{MK}} =\]
\[= \overrightarrow{\text{OK}}:\]
\[\overrightarrow{\text{OM}} - \overrightarrow{\text{OC}} = 2 \bullet \left( \overrightarrow{\text{OK}} - \overrightarrow{\text{OM}} \right) =\]
\[= 2\overrightarrow{\text{OK}} - 2\overrightarrow{\text{OM}}.\]
\[Отсюда:\]
\[3\overrightarrow{\text{OM}} = 2\overrightarrow{\text{OK}} + \overrightarrow{\text{OC}}.\]
\[3)\ Следовательно\]
\[\overrightarrow{\text{OM}} = \frac{2\left( \overrightarrow{\text{OK}} + \overrightarrow{\text{OC}} \right)}{3} =\]
\[= \frac{2 \bullet \frac{1}{2}\left( \overrightarrow{\text{OA}} + \overrightarrow{\text{OB}} \right) + \overrightarrow{\text{OC}}}{3} =\]
\[= \frac{\overrightarrow{\text{OA}} + \overrightarrow{\text{OB}} + \overrightarrow{\text{OC}}}{3};\]
\[\overrightarrow{\text{OA}} = 3\overrightarrow{\text{OM}} - \overrightarrow{\text{OB}} - \overrightarrow{\text{OC}}.\]
\[Ответ:\ \ \overrightarrow{\text{OA}} = 3\overrightarrow{\text{OM}} - \overrightarrow{\text{OB}} - \overrightarrow{\text{OC}}\text{.\ \ }\]
