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МатематикаРешебник по геометрии 10 класс Атанасян ФГОС 754
754
\[\boxed{\mathbf{754.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]
\[Дано:\]
\[\angle BAA_{1} = \angle BAD = \angle DAA_{1} =\]
\[= 60{^\circ};\]
\[AB = AA_{1} = AD = 1.\]
\[Воспользуемся\ формулами\ \]
\[суммы\ и\ разности\ векторов.\]
\[\textbf{а)}\ \overrightarrow{AC_{1}} = \overrightarrow{\text{AC}} + \overrightarrow{CC_{1}} =\]
\[= \overrightarrow{\text{AB}} + \overrightarrow{\text{BD}} + \overrightarrow{CC_{1}} =\]
\[= \overrightarrow{\text{AB}} + \overrightarrow{\text{AD}} + \overrightarrow{AA_{1}}.\]
\[\left| \overrightarrow{AC_{1}} \right|^{2} = \left( \overrightarrow{\text{AB}} + \overrightarrow{\text{AD}} + \overrightarrow{AA_{1}} \right)^{2} =\]
\[+ 2\left| \overrightarrow{\text{AD}} \right| \cdot \left| \overrightarrow{AA_{1}} \right| \cdot \cos{60{^\circ}} =\]
\[= 3 + 6 \cdot \frac{1}{2} = 6.\]
\[\left| \overrightarrow{AC_{1}} \right| = \sqrt{6}.\]
\[\textbf{б)}\ \overrightarrow{BD_{1}} = \overrightarrow{\text{BD}} + \overrightarrow{DD_{1}} =\]
\[= \overrightarrow{\text{BA}} + \overrightarrow{\text{AD}} + \overrightarrow{DD_{1}} =\]
\[= \overrightarrow{\text{AD}} + \overrightarrow{AA_{1}} - \overrightarrow{\text{AB}}.\]
\[\left| \overrightarrow{BD_{1}} \right|^{2} = \left( \overrightarrow{\text{AD}} + \overrightarrow{AA_{1}} - \overrightarrow{\text{AB}} \right)^{2} =\]
\[- 2\left| \overrightarrow{AA_{1}} \right| \cdot \left| \overrightarrow{\text{AB}} \right| \cdot \cos{60{^\circ}} =\]
\[= 3 + 1 - 2 = 2.\]
\[\left| \overrightarrow{BD_{1}} \right| = \sqrt{2}.\]
