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МатематикаГДЗ по геометрии 11 класс. Атанасян ФГОС 684
684
\[\boxed{\mathbf{684.}еуроки - ответы\ на\ пятёрку}\]
\[\textbf{а)}\ \overrightarrow{\text{AD}} \cdot \overrightarrow{B_{1}C_{1}} =\]
\[= \left| \overrightarrow{\text{AD}} \right| \cdot \left| \overrightarrow{B_{1}C_{1}} \right| \cdot \cos{\angle\left( \overrightarrow{\text{AD}};\overrightarrow{B_{1}C_{1}} \right)};\]
\[\left| \overrightarrow{\text{AD}} \right| = \left| \overrightarrow{B_{1}C_{1}} \right| = a;\ \]
\[\cos{\angle\left( \overrightarrow{\text{AD}};\ \overrightarrow{B_{1}C_{1}} \right)} = 1;\]
\[\overrightarrow{\text{AD}} \cdot \overrightarrow{B_{1}C_{1}} = a^{2}.\]
\[\textbf{б)}\ \overrightarrow{\text{AC}} = - \overrightarrow{C_{1}A_{1}};\ \ \]
\[\cos{\angle\left( \overrightarrow{\text{AC}};\overrightarrow{C_{1}A_{1}} \right)} = \cos{180{^\circ}} =\]
\[= - 1;\]
\[\left| \overrightarrow{\text{AC}} \right| = \left| \overrightarrow{C_{1}A_{1}} \right| = \sqrt{a^{2} + a^{2}} =\]
\[= a\sqrt{2};\]
\[\overrightarrow{\text{AC}} \cdot \overrightarrow{C_{1}A_{1}} = a\sqrt{2} \cdot \sqrt{2}a \cdot ( - 1) =\]
\[= - 2a^{2}.\]
\[\textbf{в)}\ \overrightarrow{\text{DB}}\bot\overrightarrow{\text{AC}} - по\ теореме\ о\ трех\ \]
\[перпендикулярах.\]
\[\cos{\angle\left( \overrightarrow{D_{1}B};\overrightarrow{\text{AC}} \right)} = \cos{90{^\circ}} = 0;\]
\[\overrightarrow{D_{1}B} \cdot \overrightarrow{\text{AC}} = 0.\]
\[\textbf{г)}\ ⊿A_{1}BC_{1} - равносторонний;\]
\[\left| \overrightarrow{BA_{1}} \right| = \left| \overrightarrow{BC_{1}} \right| = \sqrt{a^{2} + a^{2}} =\]
\[= a\sqrt{2};\]
\[\angle A_{1}BC_{1} = 60{^\circ};\]
\[\cos{\angle\left( \overrightarrow{BA_{1}};\overrightarrow{BC_{1}} \right)} = \cos{60{^\circ}} = \frac{1}{2};\]
\[\overrightarrow{BA_{1}} \cdot \overrightarrow{BC_{1}} =\]
\[= \left| \overrightarrow{BA_{1}} \right| \cdot \left| \overrightarrow{BC_{1}} \right| \cdot \cos{\angle\left( \overrightarrow{BA_{1}};\overrightarrow{BC_{1}} \right)} =\]
\[= a\sqrt{2} \cdot a\sqrt{2} \cdot \frac{1}{2} = a^{2}.\]
\[\textbf{д)}\ \overrightarrow{A_{1}O_{1}} = \frac{1}{2}\overrightarrow{A_{1}C_{1}};\ \]
\[\cos{\angle(\overrightarrow{A_{1}O_{1}};\overrightarrow{A_{1}C_{1}})} = \cos{0{^\circ}} = 1;\]
\[\left| \overrightarrow{A_{1}O_{1}} \right| = \frac{1}{2}\left| \overrightarrow{A_{1}C_{1}} \right| = \frac{1}{2}a\sqrt{2} =\]
\[= \frac{a\sqrt{2}}{2};\]
\[\overrightarrow{A_{1}O_{1}} \cdot \overrightarrow{A_{1}C_{1}} = a\sqrt{2} \cdot \frac{a\sqrt{2}}{2} = a^{2}.\]
\[\textbf{е)}\ \overrightarrow{D_{1}O_{1}} = \frac{1}{2}\overrightarrow{D_{1}B_{1}};\]
\[\overrightarrow{B_{1}O_{1}} = \frac{1}{2}\overrightarrow{B_{1}D_{1}} = - \frac{1}{2}\overrightarrow{D_{1}B_{1}} =\]
\[= - \overrightarrow{D_{1}O_{1}};\]
\[\angle\left( D_{1}O_{1};B_{1}O_{1} \right) = 180{^\circ};\ \ \ \]
\[\cos{180{^\circ}} = - 1;\]
\[\left| \overrightarrow{D_{1}O_{1}} \right| = \left| \overrightarrow{B_{1}O_{1}} \right| = \frac{1}{2}\sqrt{a^{2} + a^{2}} =\]
\[= \frac{a\sqrt{2}}{2};\]
\[\overrightarrow{D_{1}O_{1}} \cdot \overrightarrow{B_{1}O_{1}} =\]
\[= \frac{1}{2} \cdot \sqrt{2}a \cdot \frac{a\sqrt{2}}{2} \cdot ( - 1) = - \frac{a^{2}}{2}.\]
\[\textbf{ж)}\ ⊿BB_{1}O_{1} - прямоугольный;\]
\[⊿BA_{1}C_{1} - равносторонний.\]
\[\left| \overrightarrow{BB_{1}} \right| = a;\ \ \left| \overrightarrow{B_{1}O_{1}} \right| =\]
\[= \frac{1}{2}\sqrt{a^{2} + a^{2}} = \frac{a\sqrt{2}}{2}.\]
\[\left| \overrightarrow{BO_{1}} \right| = \sqrt{a^{2} + \left( \frac{a\sqrt{2}}{2} \right)^{2}} = \sqrt{\frac{3}{2}}a;\]
\[\overrightarrow{C_{1}B} = \sqrt{a^{2} + a^{2}} = a\sqrt{2}.\]
\[\angle\left( \overrightarrow{BO_{1}};\overrightarrow{C_{1}B} \right) =\]
\[= 180{^\circ} - \angle\left( \overrightarrow{BO_{1}};\ \overrightarrow{BC_{1}} \right) =\]
\[= 180{^\circ} - \angle O_{1}BC_{1};\]
\[\angle O_{1}BC_{1} = \frac{1}{2}\angle A_{1}BC_{1} = \frac{1}{2} \cdot 60{^\circ} =\]
\[= 30{^\circ};\]
\[\overrightarrow{BO_{1}} \cdot \overrightarrow{C_{1}B} =\]
\[= \left| \overrightarrow{BO_{1}} \right| \cdot \left| \overrightarrow{C_{1}B} \right| \cdot \cos(180{^\circ} - 30{^\circ}) =\]
\[= \sqrt{\frac{3}{2}}a \cdot a\sqrt{2} \cdot \left( - \frac{\sqrt{3}}{2} \right) = - \frac{3}{2}a^{2}.\]
