ГДЗ по геометрии 11 класс. Атанасян ФГОС 606

Авторы:
Год:2023
Тип:учебник

606

\[\boxed{\mathbf{606.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\ \ \]

\[OABC - тетраэдр;\ \]

\[\ M - точка\ пересечения\ \]

\[медиан\ грани\ \text{ABC.}\]

\[Разложить:\ \ \]

\[вектор\ \overrightarrow{\text{OA}}\ по\ векторам\ \]

\[\overrightarrow{\text{OB}},\ \overrightarrow{\text{OC}}\ и\ \overrightarrow{\text{OM}}.\]

\[Решение.\]

\[1)\ \ Пусть\ точка\ K - середина\ \]

\[ребра\ AB:\]

\[2\overrightarrow{\text{OK}} = \overrightarrow{\text{OA}} + \overrightarrow{\text{AK}} = \overrightarrow{\text{OB}} + \overrightarrow{\text{BK}} =\]

\[= \overrightarrow{\text{OA}} + \overrightarrow{\text{OB}}\ \]

\[\left( так\ как\ \overrightarrow{\text{AK}} = - \overrightarrow{\text{BK}} \right).\]

\[Получаем:\]

\[\overrightarrow{\text{OK}} = \frac{1}{2}\left( \overrightarrow{\text{OA}} + \overrightarrow{\text{OB}} \right).\]

\[2)\ M - точка\ пересечения\ \]

\[медиан:\ \]

\[\overrightarrow{\text{CM}} = 2\overrightarrow{\text{MK}}.\]

\[\overrightarrow{\text{OC}} + \overrightarrow{\text{CM}} = \overrightarrow{\text{OM}}\text{\ \ }и\ \ \overrightarrow{\text{OM}} + \overrightarrow{\text{MK}} =\]

\[= \overrightarrow{\text{OK}}:\]

\[\overrightarrow{\text{OM}} - \overrightarrow{\text{OC}} = 2 \bullet \left( \overrightarrow{\text{OK}} - \overrightarrow{\text{OM}} \right) =\]

\[= 2\overrightarrow{\text{OK}} - 2\overrightarrow{\text{OM}}.\]

\[Отсюда:\]

\[3\overrightarrow{\text{OM}} = 2\overrightarrow{\text{OK}} + \overrightarrow{\text{OC}}.\]

\[3)\ Следовательно\]

\[\overrightarrow{\text{OM}} = \frac{2\left( \overrightarrow{\text{OK}} + \overrightarrow{\text{OC}} \right)}{3} =\]

\[= \frac{2 \bullet \frac{1}{2}\left( \overrightarrow{\text{OA}} + \overrightarrow{\text{OB}} \right) + \overrightarrow{\text{OC}}}{3} =\]

\[= \frac{\overrightarrow{\text{OA}} + \overrightarrow{\text{OB}} + \overrightarrow{\text{OC}}}{3};\]

\[\overrightarrow{\text{OA}} = 3\overrightarrow{\text{OM}} - \overrightarrow{\text{OB}} - \overrightarrow{\text{OC}}.\]

\[Ответ:\ \ \overrightarrow{\text{OA}} = 3\overrightarrow{\text{OM}} - \overrightarrow{\text{OB}} - \overrightarrow{\text{OC}}\text{.\ \ }\]


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