Алгебра
Английский
Биология
География
Геометрия
Русский
История
Обществознание
Физика
Химия
Информатика
Литература
МатематикаГДЗ по геометрии 11 класс. Атанасян ФГОС 789
789
\[\boxed{\mathbf{789.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]
\[Дано:\]
\[\text{ABCD}A_{1}B_{1}C_{1}D_{1} -\]
\[параллелепипед;\]
\[Доказать:\ \ \]
\[сумма\ квадратов\ диагоналей\ \]
\[равна\ сумме\ квадратов\ ребер.\]
\[Доказательство.\]
\[1)\ Пусть\ \overrightarrow{\text{AB}} = \overrightarrow{a},\ \ \ \overrightarrow{\text{AD}} = \overrightarrow{b},\ \ \ \]
\[\overrightarrow{AA_{1}} = \overrightarrow{c}:\]
\[\ \overrightarrow{AC_{1}} = \overrightarrow{\text{AB}} + \overrightarrow{\text{BC}} + \overrightarrow{CC_{1}} =\]
\[= \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c};\]
\[\overrightarrow{BD_{1}} = \overrightarrow{\text{BA}} + \overrightarrow{\text{AD}} + \overrightarrow{DD_{1}} =\]
\[= - \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c};\]
\[\overrightarrow{CA_{1}} = \overrightarrow{\text{CD}} + \overrightarrow{\text{DA}} + \overrightarrow{AA_{1}} =\]
\[= - \overrightarrow{a} - \overrightarrow{b} + \overrightarrow{c}\ ;\]
\[\overrightarrow{DB_{1}} = \overrightarrow{\text{DC}} + \overrightarrow{\text{CB}} + \overrightarrow{BB_{1}} =\]
\[= \overrightarrow{a} - \overrightarrow{b} + \overrightarrow{c}.\]
\[2)\ Возведем\ каждый\ вектор\ в\ \]
\[квадрат:\]
\[{\overrightarrow{AC_{1}}}^{2} =\]
\[= {\overrightarrow{a}}^{2} + {\overrightarrow{b}}^{2} + {\overrightarrow{c}}^{2} + 2\overrightarrow{\text{ab}} + 2\overrightarrow{\text{ac}} + 2\overrightarrow{\text{cb}};\]
\[{\overrightarrow{BD_{1}}}^{2} =\]
\[= {\overrightarrow{a}}^{2} + {\overrightarrow{b}}^{2} + {\overrightarrow{c}}^{2} - 2\overrightarrow{\text{ab}} - 2\overrightarrow{\text{ac}} + 2\overrightarrow{\text{cb}};\]
\[{\overrightarrow{CA_{1}}}^{2} =\]
\[= {\overrightarrow{a}}^{2} + {\overrightarrow{b}}^{2} + {\overrightarrow{c}}^{2} + 2\overrightarrow{\text{ab}} - 2\overrightarrow{\text{ac}} - 2\overrightarrow{\text{cb}};\]
\[{\overrightarrow{DB_{1}}}^{2} =\]
\[= {\overrightarrow{a}}^{2} + {\overrightarrow{b}}^{2} + {\overrightarrow{c}}^{2} - 2\overrightarrow{\text{ab}} + 2\overrightarrow{\text{ac}} - 2\overrightarrow{\text{cb}}.\]
\[3)\ Таким\ образом:\]
\[{\overrightarrow{AC_{1}}}^{2} + {\overrightarrow{BD_{1}}}^{2} + {\overrightarrow{CA_{1}}}^{2} + {\overrightarrow{DB_{1}}}^{2} =\]
\[= 4{\overrightarrow{a}}^{2} + 4{\overrightarrow{b}}^{2} + 4{\overrightarrow{c}}^{2} =\]
\[Что\ и\ требовалось\ доказать.\]
