Решебник по геометрии 11 класс. Атанасян ФГОС 628

\[\boxed{\mathbf{628.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\ \ \]

\[ABCD - тетраэдр;\ \ \]

\[K - середина\ медианы\ BB_{1}\ \]

\[грани\ \text{BCD.}\]

\[Разложить:\ \ \]

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

\[\overrightarrow{a} = \overrightarrow{\text{AB}},\ \overrightarrow{b} = \overrightarrow{\text{AC}},\ \overrightarrow{c} = \overrightarrow{\text{AD}}.\]

\[Решение.\]

\[1)\ BB_{1} - медиана:\ \]

\[\overrightarrow{CB_{1}} = - \overrightarrow{DB_{1}};\]

\[\overrightarrow{AB_{1}} = \frac{1}{2}\left( \overrightarrow{\text{AD}} + \overrightarrow{\text{AC}} \right).\]

\[2)\ K - середина\ BB_{1};\ \ \]

\[\overrightarrow{\text{BK}} = - \overrightarrow{B_{1}K}:\]

\[\overrightarrow{\text{AK}} = \frac{1}{2}\left( \overrightarrow{\text{AB}} + \overrightarrow{AB_{1}} \right) =\]

\[= \frac{1}{2}\overrightarrow{\text{AB}} + \frac{1}{2}\overrightarrow{AB_{1}} =\]

\[= \frac{1}{2}\overrightarrow{\text{AB}} + \frac{1}{4}\left( \overrightarrow{\text{AD}} + \overrightarrow{\text{AC}} \right).\]

\[3)\ Таким\ образом:\ \ \]

\[\overrightarrow{\text{AK}} = \frac{1}{2}\overrightarrow{a} + \frac{1}{4}\overrightarrow{b} + \frac{1}{4}\overrightarrow{c}.\]

\[Ответ:\ \ \frac{1}{2}\overrightarrow{a} + \frac{1}{4}\overrightarrow{b} + \frac{1}{4}\overrightarrow{c}.\]