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МатематикаГДЗ по геометрии 11 класс. Атанасян ФГОС 696
696
\[\boxed{\mathbf{696.}еуроки - ответы\ на\ пятёрку}\]
\[Пусть\ ребро\ куба\ равно\ \text{a.}\]
\[\textbf{а)}\ ⊿AA_{1}C_{1} - прямоугольный;\ \ \]
\[AA_{1} = a.\]
\[По\ теореме\ Пифагора:\]
\[A_{1}C_{1} = \sqrt{A_{1}B_{1}^{2} + B_{1}C_{1}^{2}} =\]
\[= \sqrt{a^{2} + a^{2}} = a\sqrt{2}.\]
\[\left| \overrightarrow{AC_{1}} \right| = \sqrt{a^{2} + \left( a\sqrt{2} \right)^{2}} = \sqrt{3a^{2}} =\]
\[= a\sqrt{3}.\]
\[\cos{\angle\left( \overrightarrow{AA_{1}};\ \overrightarrow{AC_{1}} \right)} = \cos{\angle A_{1}AC_{1}} =\]
\[= \frac{AA_{1}}{AC_{1}} = \frac{a}{a\sqrt{3}} = \frac{\sqrt{3}}{3}.\]
\[\textbf{б)}\ Векторы\ \overrightarrow{}\ и\ \overrightarrow{}\ лежат\ в\ \]
\[плоскости\ BB_{1}\text{D.}\]
\[BB_{1}D_{1}D - прямоугольник\ со\ \]
\[сторонами\ \text{a\ }и\ a\sqrt{2}.\]
\[По\ теореме\ косинусов\ в\ \]
\[⊿B_{1}OD_{1}:\]
\[\left| \overrightarrow{OB_{1}} \right| = \left| \overrightarrow{OD_{1}} \right| = \frac{1}{2}\left| \overrightarrow{DB_{1}} \right| =\]
\[= \frac{1}{2}\left| \overrightarrow{AC_{1}} \right| = \frac{a\sqrt{3}}{2};\]
\[\left| \overrightarrow{B_{1}D_{1}} \right| = \left| \overrightarrow{A_{1}C_{1}} \right| = a\sqrt{2};\]
\[\cos{\angle B_{1}OD_{1}} =\]
\[= \frac{\left( \frac{a\sqrt{3}}{2} \right)^{2} + \left( \frac{a\sqrt{3}}{2} \right)^{2} - \left( a\sqrt{2} \right)^{2}}{2 \cdot \frac{a\sqrt{3}}{2} \cdot \frac{a\sqrt{3}}{2}} =\]
\[= \frac{\frac{3a^{2}}{4} + \frac{3a^{2}}{4} - 2a^{2}}{\frac{3a^{2}}{2}} =\]
\[= \frac{\frac{3a^{2}}{2} - \frac{4a^{2}}{2}}{\frac{3a^{2}}{2}} = \frac{- a^{2} \cdot 2}{3a^{2} \cdot 2} = - \frac{1}{3}.\]
\[\textbf{в)}\ \overrightarrow{CC_{1}}\bot\ \overrightarrow{\text{BD}};\ \ \overrightarrow{\text{AC}}\bot\overrightarrow{\text{BD}}:\]
\[BD\bot плоскости\ AC_{1}C;\]
\[BD\bot AC_{1}.\]
\[\cos{\angle\left( \overrightarrow{\text{BD}};\overrightarrow{AC_{1}} \right)} = \cos{90{^\circ}} = 0.\]
