576 Дан параллелепипед ABCDA1B1C1D1. Укажите вектор \(\overrightarrow{x}\), начало конец которого являются вершинами параллелепипеда, такой, ч a) \(\overrightarrow{DC} + \overrightarrow{D_1A_1} + \overrightarrow{CD_1} + \overrightarrow{x} + \overrightarrow{A_1C_1} = \overrightarrow{DB}\); 6) \(\overrightarrow{DA} + \overrightarrow{x} + \overrightarrow{D_1B} + \overrightarrow{AD_1} + \overrightarrow{BA} =\)

a) \(\overrightarrow{DC} + \overrightarrow{D_1A_1} + \overrightarrow{CD_1} + \overrightarrow{x} + \overrightarrow{A_1C_1} = \overrightarrow{DB}\) \(\overrightarrow{DC} = \overrightarrow{A_1B_1}\), \(\overrightarrow{D_1A_1} = \overrightarrow{CB}\), \(\overrightarrow{CD_1} = \overrightarrow{BA_1}\), \(\overrightarrow{A_1C_1} = \overrightarrow{AC}\) \(\overrightarrow{A_1B_1} + \overrightarrow{CB} + \overrightarrow{BA_1} + \overrightarrow{x} + \overrightarrow{AC} = \overrightarrow{DB}\) \(\overrightarrow{A_1B_1} + \overrightarrow{CB} + \overrightarrow{BA_1} + \overrightarrow{AC} - \overrightarrow{DB} = - \overrightarrow{x}\) \(\overrightarrow{x} = \overrightarrow{DB} - \overrightarrow{A_1B_1} - \overrightarrow{CB} - \overrightarrow{BA_1} - \overrightarrow{AC}\) \(\overrightarrow{x} = \overrightarrow{DB} - \overrightarrow{DC} - \overrightarrow{D_1A_1} - \overrightarrow{CD_1} - \overrightarrow{A_1C_1}\) б) \(\overrightarrow{DA} + \overrightarrow{x} + \overrightarrow{D_1B} + \overrightarrow{AD_1} + \overrightarrow{BA} = \) \(\overrightarrow{DA} + \overrightarrow{x} + \overrightarrow{D_1B} + \overrightarrow{AD_1} + \overrightarrow{BA} = 0\) \(\overrightarrow{x} = - \overrightarrow{DA} - \overrightarrow{D_1B} - \overrightarrow{AD_1} - \overrightarrow{BA}\) \(\overrightarrow{x} = \overrightarrow{AD} + \overrightarrow{BD_1} + \overrightarrow{D_1A} + \overrightarrow{AB}\)

Ответ: a) \(\overrightarrow{x} = \overrightarrow{DB} - \overrightarrow{DC} - \overrightarrow{D_1A_1} - \overrightarrow{CD_1} - \overrightarrow{A_1C_1}\); 6) \(\overrightarrow{x} = \overrightarrow{AD} + \overrightarrow{BD_1} + \overrightarrow{D_1A} + \overrightarrow{AB}\)