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МатематикаГДЗ по геометрии 11 класс. Атанасян ФГОС 700
700
\[\boxed{\mathbf{700.}еуроки - ответы\ на\ пятёрку}\]
\[\angle\left( \overrightarrow{a};\overrightarrow{b} \right) = 120{^\circ};\ \ \angle\left( \overrightarrow{b};\overrightarrow{c} \right) = 90{^\circ};\ \ \]
\[\angle\left( \overrightarrow{a};\overrightarrow{c} \right) = 90{^\circ};\]
\[\left| \overrightarrow{a} \right| = \left| \overrightarrow{c} \right| = \left| \overrightarrow{b} \right| = 1.\]
\[\textbf{а)}\ \left( \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} \right) \cdot 2\overrightarrow{b} =\]
\[= 2\overrightarrow{a}\overrightarrow{b} + 2\overrightarrow{b}\overrightarrow{b} + 2\overrightarrow{c}\overrightarrow{b}.\]
\[\overrightarrow{a} \cdot \overrightarrow{b} = \left| \overrightarrow{a} \right| \cdot \left| \overrightarrow{b} \right| \cdot \cos{120{^\circ}} =\]
\[= 1 \cdot 1 \cdot \left( - \frac{1}{2} \right) = - \frac{1}{2};\ \]
\[\overrightarrow{b} \cdot \overrightarrow{b} = \left| \overrightarrow{b} \right| \cdot \left| \overrightarrow{b} \right| \cdot \cos{0{^\circ}} =\]
\[= 1 \cdot 1 \cdot 1 = 1;\]
\[\overrightarrow{b} \cdot \overrightarrow{c} = \left| \overrightarrow{b} \right| \cdot \left| \overrightarrow{c} \right| \cdot \cos{90{^\circ}} = 0.\]
\[\left( \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} \right) \cdot 2\overrightarrow{b} =\]
\[= 2 \cdot \left( - \frac{1}{2} \right) + 2 \cdot 0 + 2 \cdot 1 =\]
\[= - 1 + 2 = 1.\]
\[= \overrightarrow{a} \cdot \overrightarrow{a} - \overrightarrow{a} \cdot \overrightarrow{b} - \overrightarrow{c} \cdot \overrightarrow{c} + \overrightarrow{b} \cdot \overrightarrow{c} =\]
\[= 1 + \frac{1}{2} - 1 + 0 = \frac{1}{2}.\]
\[\textbf{б)}\ По\ теореме\ косинусов.\]
\[\left| \overrightarrow{a} - \overrightarrow{b} \right| =\]
\[= \sqrt{1 + 1 - 2 \cdot 1 \cdot 1 \cdot \left( - \frac{1}{2} \right)} =\]
\[= \sqrt{3}.\]
\[\left| \overrightarrow{a} + \overrightarrow{b} - \overrightarrow{c} \right|:\]
\[= \sqrt{1 + 1 - 2 \cdot 1 \cdot 1 \cdot \frac{1}{2}} = \sqrt{1} = 1;\]
\[\left| \overrightarrow{c} \right| = 1.\]
\[\left| \overrightarrow{a} + \overrightarrow{b} - \overrightarrow{c} \right| = \sqrt{1^{2} + 1^{2}} = \sqrt{2}.\]
