\[\boxed{\mathbf{1051.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунки\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\left| \overrightarrow{a} \right| = 1;\]
\[\left| \overrightarrow{b} \right| = \left| \overrightarrow{c} \right| = 2;\]
\[\widehat{\overrightarrow{a}\overrightarrow{c}} = \widehat{\overrightarrow{b}\overrightarrow{c}} = 60{^\circ}.\]
\[\mathbf{Найти:}\]
\[\left( \overrightarrow{a} + \overrightarrow{b} \right) \bullet \overrightarrow{c}.\]
\[\mathbf{Решение.}\]
\[1)\ \mathrm{\Delta}ABK\ и\ \mathrm{\Delta}AFK -\]
\[прямоугольные:\]
\[\angle DAB = \angle DAK + \angle CAB = 120{^\circ};\]
\[\angle BAK = 120{^\circ} - 90{^\circ} = 30{^\circ}.\]
\[По\ свойству\ прямоугольного\ \]
\[треугольника:\]
\[BK = \frac{1}{2}AB = \frac{1}{2}.\]
\[2)\ FK = BF - BK = 2 - \frac{1}{2} = 1\frac{1}{2}.\]
\[3)\ \left. \ \frac{AK = \sqrt{AB^{2} - KB^{2}}}{AK = \sqrt{AF^{2} - FK^{2}}} \right| \Longrightarrow\]
\[\Longrightarrow AB^{2} - KB^{2} = AF^{2} - FK^{2}:\]
\[1 - \frac{1}{4} = AF^{2} - \frac{9}{4}\]
\[AF^{2} = 1 - \frac{1}{4} + \frac{9}{4} = 3\]
\[AF = \sqrt{3}.\]
\[Отсюда:\]
\[\left| \overrightarrow{a} + \overrightarrow{b} \right| = \sqrt{3}.\]
\[4)\ \overrightarrow{\text{AF}} = \frac{1}{2}\overrightarrow{\text{AE}} = > \left| \overrightarrow{\text{AE}} \right| -\]
\[биссектрисса:\]
\[угол\ между\ \overrightarrow{c}\ и\ \left( \overrightarrow{a} + \overrightarrow{b} \right)\ \]
\[равен\ 30{^\circ}.\]
\[5)\ \left( \overrightarrow{a} + \overrightarrow{b} \right) \bullet \overrightarrow{c} =\]
\[= \left| \overrightarrow{a} + \overrightarrow{b} \right| \bullet \left| \overrightarrow{c} \right| \bullet \cos{30{^\circ}} =\]
\[= \sqrt{3} \bullet 2 \bullet \frac{\sqrt{3}}{2} = 3.\]
\[\mathbf{Ответ:}3.\]
\[\boxed{\mathbf{1051.еуроки - ответы\ на\ пятёрку}}\]
\[\mathbf{Уравнение\ окружности:}\]
\[\left( x - x_{0} \right)^{2} + \left( y - y_{0} \right)^{2} = r^{2};\]
\[x_{0} = y_{0} = 0.\]
\[\textbf{а)}\ r = 3:\ \ \ \]
\[\ x^{2} + y^{2} = 9.\]
\[\textbf{б)}\ r = \sqrt{2}:\ \ \ \ \]
\[x^{2} + y^{2} = 2.\]
\[\textbf{в)}\ r = \frac{5}{2}:\ \ \ \ \]
\[x^{2} + y^{2} = \frac{25}{4}.\]