\[\boxed{\mathbf{702.}еуроки - ответы\ на\ пятёрку}\]
\[Дано:\]
\[AB = BC = CA = a;\]
\[AD = DC = DB = AB;\]
\[плоские\ углы = 60{^\circ}.\]
\[Доказать:\]
\[\overrightarrow{\text{MN}} \cdot \overrightarrow{\text{AD}} = \overrightarrow{\text{MN}} \cdot \overrightarrow{\text{BC}} = 0.\]
\[Доказательство.\]
\[1)\ \cos{\angle\left( \overrightarrow{a}\overrightarrow{b} \right)} = \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{\left| \overrightarrow{a} \right| \cdot \left| \overrightarrow{b} \right|};\]
\[\overrightarrow{a} \cdot \overrightarrow{b} = \cos{\angle\left( \overrightarrow{a}\overrightarrow{b} \right)} \cdot \left| \overrightarrow{a} \right| \cdot \left| \overrightarrow{b} \right|.\]
\[2)\ \overrightarrow{\text{MN}} \cdot \overrightarrow{\text{AD}} =\]
\[= \left( \overrightarrow{\text{MA}} + \overrightarrow{\text{AB}} + \overrightarrow{\text{BN}} \right)\left( \overrightarrow{\text{AB}} + \overrightarrow{\text{BD}} \right) =\]
\[= a^{2} \cdot \left( - \frac{1}{4} + 1 - \frac{1}{4} - \frac{1}{4} - \frac{1}{2} + \frac{1}{4} \right) =\]
\[= a^{2} \cdot 0 = 0.\]
\[3)\ Аналогично\ доказываем,\ что\ \]
\[\overrightarrow{\text{MN}} \cdot \overrightarrow{\text{BC}} = 0.\]
\[Что\ и\ требовалось\ доказать.\]