Решебник по геометрии 10 класс Атанасян ФГОС 569

\[\boxed{\mathbf{569.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Рисунок\ к\ задаче:157.\]

\[Решение.\]

\[1)\ \text{ABCD}A_{1}B_{1}C_{1}D_{1} -\]

\[параллелепипед:\ \ \ \]

\[противоположные\ грани\ \]

\[равны\ и\ параллельны.\]

\[2)\ \overrightarrow{AB_{1}} = \overrightarrow{AD_{1}} + \overrightarrow{D_{1}B_{1}} =\]

\[= \overrightarrow{BC_{1}} + \overrightarrow{\text{DB}} = \overrightarrow{BC_{1}} - \overrightarrow{\text{BD}}.\]

\[3)\ \overrightarrow{\text{DK}} = \overrightarrow{\text{DA}} + \overrightarrow{\text{AK}} = \overrightarrow{\text{CB}} + \overrightarrow{\text{BM}} =\]

\[= \overrightarrow{\text{CB}} - \overrightarrow{\text{MB}}.\]

\[Ответ:\ \ \overrightarrow{AB_{1}} = \overrightarrow{BC_{1}} - \overrightarrow{\text{BD}};\ \ \]

\[\ \overrightarrow{\text{DK}} = \overrightarrow{\text{CB}} - \overrightarrow{\text{MB}}.\]