\[\boxed{\mathbf{1042.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC - равносторон;\]
\[AB = a;\]
\[BD\bot AC.\]
\[\mathbf{Найти:}\]
\[скалярное\ произведение\ \]
\[векторов.\]
\[\mathbf{Решение.}\]
\[1)\ \mathrm{\Delta}ABC - равносторонний:\ \]
\[\angle A = \angle B = \angle C =\]
\[= 60{^\circ}\ (по\ свойству).\]
\[2)\ BD\bot AC \Longrightarrow \angle ADB = 90{^\circ}.\]
\[\textbf{а)}\ \overrightarrow{\text{AB}} \bullet \overrightarrow{\text{AC}} = a \bullet a \bullet \cos{60{^\circ}} =\]
\[= a^{2} \bullet \frac{1}{2} = \frac{a^{2}}{2};\]
\[\textbf{б)}\ \overrightarrow{\text{AC}} \bullet \overrightarrow{\text{CB}} = a \bullet a \bullet \cos{120{^\circ}} =\]
\[= a^{2} \bullet \left( - \frac{1}{2} \right) = - \frac{a^{2}}{2};\]
\[\textbf{в)}\ \overrightarrow{\text{AC}} \bullet \overrightarrow{\text{BD}} =\]
\[= \left| \overrightarrow{\text{AC}} \right| \bullet \left| \overrightarrow{\text{BD}} \right| \bullet \cos{90{^\circ}} = 0;\]
\[\textbf{г)}\ \overrightarrow{\text{AC}} \bullet \overrightarrow{\text{AC}} = a \bullet a \bullet \cos{0{^\circ}} =\]
\[= a^{2} \bullet 1 = a^{2}.\]