Решебник по геометрии 10 класс Атанасян ФГОС 604

\[\boxed{\mathbf{604.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\ \ \]

\[ABCD - тетраэдр;\ \ \]

\[AA_{1} - медиана\ грани\ ABC;\ \]

\[AK\ :KA_{1} = 3\ :7.\]

\[Разложить:\ \ \]

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

\[\overrightarrow{\text{DA}},\overrightarrow{\text{DB}}\ и\ \overrightarrow{\text{DC}}.\]

\[Решение.\]

\[1)\ Пусть\ \overrightarrow{\text{AK}} = n \bullet \overrightarrow{KA_{1}};\ \ \ \ \ \]

\[n = \frac{3}{7}.\]

\[2)\ \overrightarrow{\text{DA}} + \overrightarrow{\text{AK}} = \overrightarrow{\text{DK}};\ \text{\ \ }\]

\[\overrightarrow{\text{DK}} + \overrightarrow{KA_{1}} = \overrightarrow{DA_{1}}:\]

\[\overrightarrow{\text{DK}} - \overrightarrow{\text{DA}} = \overrightarrow{\text{AK}} = n \bullet \overrightarrow{KA_{1}} =\]

\[= n \bullet \left( \overrightarrow{DA_{1}} - \overrightarrow{\text{DK}} \right)\]

\[\overrightarrow{\text{DK}}(1 + n) = \overrightarrow{\text{DA}} + n\overrightarrow{DA_{1}}\]

\[\overrightarrow{\text{DK}} = \frac{\overrightarrow{\text{DA}} + n\overrightarrow{DA_{1}}}{1 + n}.\]

\[3)\ \overrightarrow{CA_{1}} = - \overrightarrow{BA_{1}} \rightarrow \ \overrightarrow{DA_{1}} =\]

\[= \frac{1}{2}\left( \overrightarrow{\text{DC}} + \overrightarrow{\text{DB}} \right):\]

\[\overrightarrow{\text{DK}} = \frac{\overrightarrow{\text{DA}} + \frac{3}{7} \bullet \frac{1}{2} \bullet \left( \overrightarrow{\text{DC}} + \overrightarrow{\text{DB}} \right)}{1 + \frac{3}{7}} =\]

\[= \frac{14\overrightarrow{\text{DA}} + 3\overrightarrow{\text{DC}} + 3\overrightarrow{\text{DB}}}{14 + 6}\]

\[\overrightarrow{\text{DK}} = 0,7\overrightarrow{\text{DA}} + 0,15\overrightarrow{\text{DC}} + 0,15\overrightarrow{\text{DB}}.\]

\[Ответ:\ \ \]

\[\overrightarrow{\text{DK}} = 0,7\overrightarrow{\text{DA}} + 0,15\overrightarrow{\text{DC}} + 0,15\overrightarrow{\text{DB}}.\]