Решебник по геометрии 11 класс. Атанасян ФГОС 583

\[\boxed{\mathbf{583.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[ABCD - трапеция;\]

\[AM = MB;\]

\[DN = NC;\ \ \]

\[O - произвольная\ точка\ \]

\[пространства.\]

\[Выразить:\]

\[вектор\ \overrightarrow{\text{OM}} - \overrightarrow{\text{ON}}\ через\ векторы\ \]

\[\overrightarrow{\text{AD}}\ и\ \overrightarrow{\text{BC}}.\]

\[Решение.\]

\[1)\ \overrightarrow{\text{ON}} = \overrightarrow{\text{OA}} + \overrightarrow{\text{AD}} + \overrightarrow{\text{DN}};\ \ \ \text{\ \ }\]

\[\overrightarrow{\text{OM}} = \overrightarrow{\text{OA}} + \overrightarrow{\text{AM}}:\]

\[\overrightarrow{\text{OM}} - \overrightarrow{\text{ON}} =\]

\[= \overrightarrow{\text{OA}} + \overrightarrow{\text{AM}} - \overrightarrow{\text{OA}} - \overrightarrow{\text{AD}} - \overrightarrow{\text{DN}} =\]

\[= \overrightarrow{\text{AM}} + \overrightarrow{\text{DA}} + \overrightarrow{\text{ND}} =\]

\[= \overrightarrow{\text{ND}} + \overrightarrow{\text{DA}} + \overrightarrow{\text{AM}} = \overrightarrow{\text{NM}}.\]

\[2)\ \overrightarrow{\text{NM}} = \overrightarrow{\text{ND}} + \overrightarrow{\text{DA}} + \overrightarrow{\text{AM}};\ \ \ \ \]

\[\overrightarrow{\text{NM}} = \overrightarrow{\text{NC}} + \overrightarrow{\text{CB}} + \overrightarrow{\text{BM}};\]

\[\overrightarrow{\text{ND}} = - \overrightarrow{\text{NC}};\ \ \ \ \overrightarrow{\text{AM}} = - \overrightarrow{\text{BM}}:\]

\[3)\ \overrightarrow{\text{DA}} = - \overrightarrow{\text{AD}};\ \ \ \ \overrightarrow{\text{CB}} = - \overrightarrow{\text{BC}}:\]

\[2\overrightarrow{\text{NM}} = - \overrightarrow{\text{AD}} - \overrightarrow{\text{BC}}.\]

\[Отсюда:\]

\[\overrightarrow{\text{OM}} - \overrightarrow{\text{ON}} = \overrightarrow{\text{NM}} =\]

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

\[Ответ:\ - \frac{1}{2}(\overrightarrow{\text{AD}} + \overrightarrow{\text{BC}})\]