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МатематикаГДЗ по геометрии 11 класс. Атанасян ФГОС 577
577
\[\boxed{\mathbf{577.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]
\[Дано:\]
\[Найти:\]
\[вектор\ \overrightarrow{x}.\]
\[Решение.\]
\[\textbf{а)}\ AA_{1} \parallel BB_{1}\ и\ \ AA_{1} = BB_{1}:\]
\[\overrightarrow{AA_{1}} + \overrightarrow{B_{1}C} - \overrightarrow{x} = \overrightarrow{\text{BA}}\]
\[\overrightarrow{x} = \overrightarrow{AA_{1}} + \overrightarrow{B_{1}C} - \overrightarrow{\text{BA}} =\]
\[= \overrightarrow{BB_{1}} + \overrightarrow{B_{1}C} + \overrightarrow{\text{AB}} = \overrightarrow{\text{AC}}.\]
\[\textbf{б)}\ \overrightarrow{AC_{1}} - \overrightarrow{BB_{1}} + \overrightarrow{x} = \overrightarrow{\text{AB}}\]
\[\overrightarrow{x} = \overrightarrow{\text{AB}} - \overrightarrow{AC_{1}} + \overrightarrow{BB_{1}} =\]
\[= \overrightarrow{C_{1}A} + \overrightarrow{\text{AB}} + \overrightarrow{BB_{1}} = \overrightarrow{C_{1}B_{1}}.\]
\[\textbf{в)}\ \overrightarrow{AB_{1}} + \overrightarrow{x} = \overrightarrow{\text{AC}} - \overrightarrow{x} + \overrightarrow{BC_{1}}\]
\[2\overrightarrow{x} = \overrightarrow{\text{AC}} + \overrightarrow{BC_{1}} - \overrightarrow{AB_{1}} =\]
\[= \left( \overrightarrow{B_{1}A} + \overrightarrow{\text{AC}} \right) + \overrightarrow{BC_{1}} =\]
\[= \overrightarrow{B_{1}C} + \overrightarrow{BC_{1}}\]
\[\overrightarrow{B_{1}C} = \overrightarrow{B_{1}B} + \overrightarrow{\text{BC}}\text{\ \ }и\ \ \overrightarrow{BC_{1}} =\]
\[= \overrightarrow{BB_{1}} + \overrightarrow{B_{1}C_{1}}\]
\[2\overrightarrow{x} = \overrightarrow{B_{1}B} + \overrightarrow{\text{BC}} + \overrightarrow{BB_{1}} + \overrightarrow{B_{1}C_{1}} =\]
\[= \left( \overrightarrow{B_{1}B} + \overrightarrow{BB_{1}} \right) + \overrightarrow{\text{BC}} + \overrightarrow{B_{1}C_{1}} =\]
\[= \overrightarrow{0} + 2\overrightarrow{\text{BC}} = 2\overrightarrow{\text{BC}}.\]
\[BC \parallel B_{1}C_{1};\ \ BC = B_{1}C_{1}:\ \]
\[\ \overrightarrow{x} = \overrightarrow{\text{BC}}.\]
\[Ответ:\ \ а)\ \overrightarrow{\text{AC}};\ \ б)\ \overrightarrow{C_{1}B_{1}};\ \ в)\ \overrightarrow{\text{BC}}.\]
