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МатематикаГДЗ по геометрии 11 класс. Атанасян ФГОС 593
593
\[\boxed{\mathbf{593.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]
\[Дано:\]
\[ABCD - тетраэдр;\]
\[AE = EC;\ \]
\[BF = FD.\]
\[Доказать:\]
\[2\overrightarrow{\text{FE}} = \overrightarrow{\text{BA}} + \overrightarrow{\text{DC}}.\]
\[Найти:\]
\[\overrightarrow{\text{FE}},\overrightarrow{\text{BA}}\ и\ \overrightarrow{\text{DC}} - компланарны.\]
\[Решение.\]
\[1)\ \overrightarrow{\text{FE}} = \overrightarrow{\text{FD}} + \overrightarrow{\text{DC}} + \overrightarrow{\text{CE}};\ \text{\ \ }\]
\[\overrightarrow{\text{FE}} = \overrightarrow{\text{FB}} + \overrightarrow{\text{BA}} + \overrightarrow{\text{AE}}:\]
\[2\overrightarrow{\text{FE}} =\]
\[= \overrightarrow{\text{FD}} + \overrightarrow{\text{DC}} + \overrightarrow{\text{CE}} + \overrightarrow{\text{FB}} + \overrightarrow{\text{BA}} + \overrightarrow{\text{AE}}.\]
\[2)\ \overrightarrow{\text{FD}} = - \overrightarrow{\text{FB}};\ \ \overrightarrow{\text{CE}} = - \overrightarrow{\text{AE}}\ \]
\[(так\ как\ FD = FB\ и\ CE = AE):\]
\[2\overrightarrow{\text{FE}} = \overrightarrow{\text{BA}} + \overrightarrow{\text{DC}}\text{.\ }\]
\[Что\ и\ требовалось\ доказать.\]
\[3)\ 2\overrightarrow{\text{FE}} = \overrightarrow{\text{BA}} + \overrightarrow{\text{DC}}:\ \]
\[\overrightarrow{\text{FE}} = \frac{1}{2}\overrightarrow{\text{BA}} + \frac{1}{2}\overrightarrow{\text{DC}} - признак\ \]
\[компланарности\ трех\ \]
\[векторов.\]
\[Ответ:да.\]
