ГДЗ по геометрии 11 класс. Атанасян ФГОС 602

\[\boxed{\mathbf{602.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\ \ \]

\[ABCD - параллелограмм;\]

\[O - не\ принадлежит\ ABCD;\]

\[AM = MB;\]

\[MK = KD.\]

\[Разложить:\ \ \]

\[векторы\ \overrightarrow{\text{OM}}\ и\ \overrightarrow{\text{OK}}\ по\ векторам\ \]

\[\overrightarrow{a} = \overrightarrow{\text{OA}};\ \ \overrightarrow{b} = \ \overrightarrow{\text{OB}};\ \ \overrightarrow{c} = \overrightarrow{\text{OC}}.\]

\[Найти:\]

\[длину\ \overrightarrow{\text{AK}},\ если\ ребро\ куба\ \]

\[равно\ \text{m.}\]

\[Решение.\]

\[1)\ \overrightarrow{\text{OM}} = \overrightarrow{\text{OA}} + \overrightarrow{\text{AM}} = \overrightarrow{\text{OB}} + \overrightarrow{\text{BM}}\text{.\ \ }\]

\[\overrightarrow{\text{AM}} = - \overrightarrow{\text{BM}}:\]

\[2\overrightarrow{\text{OM}} = \overrightarrow{\text{OA}} + \overrightarrow{\text{OB}}.\]

\[Отсюда:\]

\[\overrightarrow{\text{OM}} = \frac{1}{2}\overrightarrow{a} + \frac{1}{2}\overrightarrow{b} + 0 \bullet \overrightarrow{c}.\]

\[2)\ \overrightarrow{\text{OM}} = \overrightarrow{\text{OK}} + \overrightarrow{\text{KM}}\ \ и\ \ \]

\[\overrightarrow{\text{KM}} = \frac{1}{2}\overrightarrow{\text{DM}}:\]

\[\overrightarrow{\text{OK}} + \frac{1}{2}\overrightarrow{\text{DM}} = \overrightarrow{\text{OM}}.\]

\[3)\ \overrightarrow{\text{DM}} + \overrightarrow{\text{MB}} + \overrightarrow{\text{BC}} + \overrightarrow{\text{CD}} = \overrightarrow{0}:\]

\[\overrightarrow{\text{DM}} + \frac{1}{2}\overrightarrow{\text{AB}} + \overrightarrow{\text{BC}} + \overrightarrow{\text{CD}} = \overrightarrow{0}.\]

\[4)\ \overrightarrow{\text{OA}} + \overrightarrow{\text{AB}} = \overrightarrow{\text{OB}}:\]

\[\overrightarrow{\text{AB}} = \overrightarrow{\text{OB}} - \overrightarrow{\text{OA}} = \overrightarrow{b} - \overrightarrow{a}.\]

\[\overrightarrow{\text{OB}} + \overrightarrow{\text{BC}} = \overrightarrow{\text{OC}}:\]

\[\overrightarrow{\text{BC}} = \overrightarrow{\text{OC}} - \overrightarrow{\text{OB}} = \overrightarrow{c} - \overrightarrow{b};\]

\[\overrightarrow{\text{CD}} = - \overrightarrow{\text{AB}} = \overrightarrow{a} - \overrightarrow{b}.\]

\[5)\ \overrightarrow{\text{DM}} + \frac{1}{2}\overrightarrow{a} - \frac{3}{2}\overrightarrow{b} + \overrightarrow{c} + \overrightarrow{a} - \overrightarrow{b} =\]

\[= \overrightarrow{0}\]

\[\overrightarrow{\text{DM}} + \frac{1}{2}\overrightarrow{a} - \frac{3}{2}\overrightarrow{b} + \overrightarrow{c} = \overrightarrow{0}\ \]

\[\overrightarrow{\text{DM}} = \frac{3}{2}\overrightarrow{b} - \frac{1}{2}\overrightarrow{a} - \overrightarrow{c}.\]

\[6)\ Получаем:\]

\[\overrightarrow{\text{OK}} + \frac{1}{2}\left( \frac{3}{2}\overrightarrow{b} - \frac{1}{2}\overrightarrow{a} - \overrightarrow{c} \right) =\]

\[= \frac{1}{2}\overrightarrow{a} + \frac{1}{2}\overrightarrow{b}\]

\[\overrightarrow{\text{OK}} =\]

\[= \frac{1}{2}\overrightarrow{a} + \frac{1}{2}\overrightarrow{b} - \frac{3}{4}\overrightarrow{b} + \frac{1}{4}\overrightarrow{a} + \frac{1}{2}\overrightarrow{c} =\]

\[= \frac{3}{4}\overrightarrow{a} - \frac{1}{4}\overrightarrow{b} + \frac{1}{2}\overrightarrow{c}.\]

\[Ответ:\ \ \overrightarrow{\text{OM}} = \frac{1}{2}\overrightarrow{a} + \frac{1}{2}\overrightarrow{b} + 0 \bullet \overrightarrow{c};\ \ \]

\[\overrightarrow{\text{OK}} = \frac{3}{4}\overrightarrow{a} - \frac{1}{4}\overrightarrow{b} + \frac{1}{2}\overrightarrow{c}.\]