ГДЗ по геометрии 7 класс Атанасян Задание 1042

\[\boxed{\mathbf{1042.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[\mathrm{\Delta}ABC - равносторон;\]

\[AB = a;\]

\[BD\bot AC.\]

\[\mathbf{Найти:}\]

\[скалярное\ произведение\ \]

\[векторов.\]

\[\mathbf{Решение.}\]

\[1)\ \mathrm{\Delta}ABC - равносторонний:\ \]

\[\angle A = \angle B = \angle C =\]

\[= 60{^\circ}\ (по\ свойству).\]

\[2)\ BD\bot AC \Longrightarrow \angle ADB = 90{^\circ}.\]

\[\textbf{а)}\ \overrightarrow{\text{AB}} \bullet \overrightarrow{\text{AC}} = a \bullet a \bullet \cos{60{^\circ}} =\]

\[= a^{2} \bullet \frac{1}{2} = \frac{a^{2}}{2};\]

\[\textbf{б)}\ \overrightarrow{\text{AC}} \bullet \overrightarrow{\text{CB}} = a \bullet a \bullet \cos{120{^\circ}} =\]

\[= a^{2} \bullet \left( - \frac{1}{2} \right) = - \frac{a^{2}}{2};\]

\[\textbf{в)}\ \overrightarrow{\text{AC}} \bullet \overrightarrow{\text{BD}} =\]

\[= \left| \overrightarrow{\text{AC}} \right| \bullet \left| \overrightarrow{\text{BD}} \right| \bullet \cos{90{^\circ}} = 0;\]

\[\textbf{г)}\ \overrightarrow{\text{AC}} \bullet \overrightarrow{\text{AC}} = a \bullet a \bullet \cos{0{^\circ}} =\]

\[= a^{2} \bullet 1 = a^{2}.\]