Решебник по геометрии 10 класс Атанасян ФГОС 643

\[\boxed{\mathbf{643.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[Доказательство.\]

\[1)\ Векторы\ \overrightarrow{\text{AB}}\ и\ \overrightarrow{\text{DC}} - не\ \]

\[компланарные.\]

\[Вектор\ \overrightarrow{\text{DC}}\ параллельно\ \]

\[перенесли\ и\ получили\ вектор\ \]

\[\overrightarrow{D_{1}C_{1}}:\]

\[\overrightarrow{BC_{1}} = \overrightarrow{D_{1}C_{1}}.\]

\[По\ правилу\ сложения\ векторов:\]

\[\overrightarrow{\text{AB}} + \overrightarrow{\text{DC}} = \overrightarrow{\text{AB}} + \overrightarrow{BC_{1}} = \overrightarrow{AC_{1}}.\]

\[2)\ Если\ AB\left\{ x_{1};y_{1};z_{1} \right\};\ \ \]

\[BC_{1}\left\{ x_{2};y_{2};z_{2} \right\}:\]

\[AC_{1}\left\{ x_{1} + x_{2};y_{1} + y_{2};z_{1} + z_{2} \right\}.\]

\[Координаты\ векторов:\]

\[\overrightarrow{\text{AB}}\left\{ x_{B} - x_{A};y_{B} - y_{A};z_{B} - z_{A} \right\};\]

\[\overrightarrow{BC_{1}}\left\{ x_{C_{1}} - x_{B};y_{C_{1}} - y_{B};z_{C_{1}} - z_{B} \right\};\]

\[\overrightarrow{AC_{1}}\left\{ x_{C_{1}} - x_{A};y_{C_{1}} - y_{A};z_{C_{1}} - z_{A} \right\}.\]

\[3)\ \overrightarrow{\text{AB}} + \overrightarrow{BC_{1}} = \overrightarrow{AC_{1}}:\]

\[x_{1} = x_{B} - x_{A};\ \ x_{2} = x_{C_{1}} - x_{B};\ \ \]

\[y_{1} = y_{B} - y_{A};\ \ y_{2} = {y_{C}}_{1} - y_{B};\]

\[z_{1} = z_{B} - z_{A};\ \ \ z_{2} = z_{C_{1}} - z_{B}.\]

\[4)\ Сумма\ векторов - это\ \]

\[координаты\ вектора\ \overrightarrow{AC_{1}}:\]

\[x_{1} + x_{2} = x_{B} - x_{A} + x_{C_{1}} - x_{B} =\]

\[= x_{C_{1}} - x_{A};\]

\[y_{1} + y_{2} = y_{B} - y_{A} + y_{C_{1}} - y_{B} =\]

\[= y_{C_{1}} - y_{A};\]

\[z_{1} + z_{2} = z_{B} - z_{A} + z_{C_{1}} - z_{B} =\]

\[= z_{C_{1}} - z_{A}.\]

\[Что\ и\ требовалось\ доказать.\]