Решебник по геометрии 7 класс Атанасян ФГОС Задание 1052

 

\[\boxed{\mathbf{1052.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[\mathbf{Дано:}\]

\[\left| \overrightarrow{a} \right| = 5;\left| \overrightarrow{b} \right| = 2;\]

\[\left| \overrightarrow{c} \right| = 4;\overrightarrow{a}\bot\overrightarrow{b};\]

\[\overrightarrow{p} = \overrightarrow{a} - \overrightarrow{b} - \overrightarrow{c};\]

\[\overrightarrow{q} = \overrightarrow{a} - \overrightarrow{b} + \overrightarrow{c}.\]

\[\mathbf{Найти:}\]

\[\overrightarrow{p} \bullet \overrightarrow{q} - ?\]

\[\mathbf{Решение.}\]

\[1)\ \overrightarrow{a}\bot\overrightarrow{b} \Longrightarrow \overrightarrow{a} \bullet \overrightarrow{b} = 0.\]

\[2)\ \overrightarrow{p} \bullet \overrightarrow{q} =\]

\[= \left( \overrightarrow{a} - \overrightarrow{b} - \overrightarrow{c} \right)\left( \overrightarrow{a} - \overrightarrow{b} + \overrightarrow{c} \right)\]

\[\overrightarrow{p} \bullet \overrightarrow{q} = \overrightarrow{a^{2}} + \overrightarrow{b^{2}} - \overrightarrow{c^{2}}.\]

\[3)\ \overrightarrow{p} \bullet \overrightarrow{q} = 5^{2} + 2^{2} - 4^{2} =\]

\[= 25 + 4 - 16 = 13.\]

\[\mathbf{Ответ:}13.\]

\[\boxed{\mathbf{1052.еуроки - ответы\ на\ пятёрку}}\]

\[\textbf{а)}\ A(0;5);\ r = 3:\]

\[x^{2} + (y - 5)^{2} = 9.\]

\[\textbf{б)}\ A( - 1;2);\ r = 2:\]

\[{(x + 1)}^{2} + (y - 2)^{2} = 4.\]

\[\textbf{в)}\ A( - 3; - 7);\ r = \frac{1}{2}:\]

\[{(x + 3)}^{2} + (y + 7)^{2} = \frac{1}{4}.\]

\[\textbf{г)}\ A(4; - 3);\ r = 10:\]

\[{(x - 4)}^{2} + (y + 3)^{2} = 100.\]