Решебник по геометрии 11 класс. Атанасян ФГОС 567

\[\boxed{\mathbf{567.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[Решение.\]

\[\textbf{а)}\ \overrightarrow{BA_{1}} = \overrightarrow{b}:\]

\[\ \overrightarrow{A_{1}B} = - \overrightarrow{b};\]

\[\overrightarrow{a} - \overrightarrow{b} = \overrightarrow{C_{1}D_{1}} + \overrightarrow{A_{1}B} =\]

\[= \overrightarrow{C_{1}D_{1}} + \overrightarrow{D_{1}C} = \overrightarrow{C_{1}C}.\]

\[\textbf{б)}\ \overrightarrow{\text{AD}} = \overrightarrow{c}:\]

\[\overrightarrow{\text{DA}} = - \overrightarrow{c};\]

\[\overrightarrow{a} - \overrightarrow{c} = \overrightarrow{C_{1}D_{1}} + \overrightarrow{\text{DA}} =\]

\[= \overrightarrow{C_{1}D_{1}} + \overrightarrow{D_{1}A_{1}} = \overrightarrow{C_{1}A_{1}}.\]

\[\textbf{в)}\ \overrightarrow{C_{1}D_{1}} = a:\ \]

\[\overrightarrow{D_{1}C_{1}} = - \overrightarrow{a};\]

\[\overrightarrow{b} - \overrightarrow{a} = \overrightarrow{BA_{1}} + \overrightarrow{D_{1}C_{1}} =\]

\[= \overrightarrow{BA_{1}} + \overrightarrow{A_{1}B_{1}} = \overrightarrow{BB_{1}}.\]

\[\textbf{г)}\ \overrightarrow{BA_{1}} = b:\]

\[\overrightarrow{A_{1}B} = - \overrightarrow{b}.\]

\[По\ признаку\ параллелограмма:\]

\[\overrightarrow{c} - \overrightarrow{b} = \overrightarrow{\text{AD}} + \overrightarrow{A_{1}B} =\]

\[= \overrightarrow{A_{1}D_{1}} + \overrightarrow{A_{1}B} = \overrightarrow{A_{1}C}.\]

\[\textbf{д)}\ \overrightarrow{C_{1}D_{1}} = a:\ \]

\[\overrightarrow{D_{1}C_{1}} = - \overrightarrow{a};\]

\[\overrightarrow{c} - \overrightarrow{a} = \overrightarrow{\text{AD}} + \overrightarrow{D_{1}C_{1}} =\]

\[= \overrightarrow{\text{AD}} + \overrightarrow{\text{DC}} = \overrightarrow{\text{AC}}.\]