ГДЗ по геометрии 11 класс. Атанасян ФГОС 582

\[\boxed{\mathbf{582.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[ABCD - параллелограмм;\]

\[AE = EB;\]

\[BF = FC;\ \ \]

\[O - произвольная\ точка\ \]

\[пространства.\]

\[Выразить:\ \ \]

\[\textbf{а)}\ вектор\ \overrightarrow{\text{OA}} - \overrightarrow{\text{OC}}\ через\ \overrightarrow{\text{EF}};\ \ \]

\[\textbf{б)}\ вектор\ \overrightarrow{\text{OA}} - \overrightarrow{\text{OE}}\ через\ \overrightarrow{\text{DC}}.\]

\[Решение.\]

\[\textbf{а)}\ \overrightarrow{\text{OC}} = \overrightarrow{\text{OA}} + \overrightarrow{\text{AC}}:\ \]

\[\overrightarrow{\text{OA}} - \overrightarrow{\text{OC}} = \overrightarrow{\text{OA}} - \overrightarrow{\text{OA}} - \overrightarrow{\text{AC}} =\]

\[= - \overrightarrow{\text{AC}}.\]

\[Точки\ E\ и\ F - середины\ сторон\ \]

\[\text{AB\ }и\ BC:\]

\[EF - средняя\ линия;\]

\[\overrightarrow{\text{EF}} = \frac{1}{2}\overrightarrow{\text{AC}};\ \]

\[- \overrightarrow{\text{AC}} = - 2\overrightarrow{\text{EF}}.\]

\[\textbf{б)}\ \overrightarrow{\text{OA}} = \overrightarrow{\text{OE}} + \overrightarrow{\text{EA}}:\]

\[\overrightarrow{\text{OA}} - \overrightarrow{\text{OE}} = \overrightarrow{\text{OE}} - \overrightarrow{\text{EA}} - \overrightarrow{\text{OE}} =\]

\[= - \overrightarrow{\text{EA}};\]

\[AE = EB\ \]

\[\left( так\ как\ E - середина\ \text{AB} \right);\]

\[AB = CD\ \]

\[(по\ свойству\ параллелограмма).\]

\[Получаем:\]

\[EA = \frac{1}{2}DC;\]

\[- \overrightarrow{\text{EA}} = - \frac{1}{2}\overrightarrow{\text{DC}}.\]

\[Ответ:\ \ а) - 2\overrightarrow{\text{EF}};\ \ б) - \frac{1}{2}\overrightarrow{\text{DC}}\]