ГДЗ по геометрии 9 класс Атанасян Задание 989

\[\boxed{\mathbf{989.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[\textbf{а)}\ \overrightarrow{p} = 7\overrightarrow{a} - 3\overrightarrow{b};\ \ \overrightarrow{a}\left( 1; - 1 \right\};\ \]

\[\overrightarrow{b}\left\{ 5; - 2 \right\}:\]

\[\overrightarrow{p} = 7\left\{ 1; - 1 \right\} - 3\left\{ 5; - 2 \right\} =\]

\[= \left\{ 7 - 15; - 5 + 6 \right\} = \left\{ - 8; - 1 \right\};\]

\[\left| \overrightarrow{p} \right| = \sqrt{64 + 1} = \sqrt{65}.\]

\[\textbf{б)}\ \overrightarrow{p} = 4\overrightarrow{a} - 2\overrightarrow{b};\ \ \overrightarrow{a}\left( 6;3 \right\};\ \overrightarrow{b}\left\{ 5;4 \right\}:\]

\[\overrightarrow{p} = 4\left\{ 6;3 \right\} - 2\left\{ 5;4 \right\} =\]

\[= \left\{ 24 - 10;12 - 8 \right\} = \left\{ 14;4 \right\};\]

\[\left| \overrightarrow{p} \right| = \sqrt{196 + 16} = \sqrt{212} = 2\sqrt{53}.\]

\[\textbf{в)}\ \overrightarrow{p} = 5\overrightarrow{a} - 4\overrightarrow{b};\ \ \overrightarrow{a}\left( \frac{3}{5};\frac{1}{5} \right\};\]

\[\ \overrightarrow{b}\left\{ 6; - 1 \right\}:\]

\[\overrightarrow{p} = 5\left\{ \frac{3}{5};\frac{1}{5} \right\} - 4\left\{ 6; - 1 \right\} =\]

\[= \left\{ 3 - 24;1 + 4 \right\} = \left\{ - 21;5 \right\};\]

\[\left| \overrightarrow{p} \right| = \sqrt{441 + 25} = \sqrt{466}.\]

\[\textbf{г)}\ \overrightarrow{p} = 3\left( - 2\overrightarrow{a} - 4\overrightarrow{b} \right);\ \ \overrightarrow{a}\left( 1;5 \right\};\ \]

\[\overrightarrow{b}\left\{ - 1; - 1 \right\}:\]

\[\overrightarrow{p} = 3\left( - 2\left\{ 1;5 \right\} - 4\left\{ - 1; - 1 \right\} \right) =\]

\[= 3\left\{ - 2 + 4; - 10 + 4 \right\} =\]

\[= 3\left\{ 2; - 6 \right\} = \{ 6; - 18\};\]

\[\left| \overrightarrow{p} \right| = \sqrt{36 + 324} = \sqrt{360} =\]

\[= 6\sqrt{10}.\]