\[\boxed{\mathbf{779.}\mathbf{ОК\ ГДЗ - домашка\ на}\ 5}\]
\[\mathbf{Дано:}\]
\[\overrightarrow{p} = 3\overrightarrow{a};\]
\[\overrightarrow{a} \neq \overrightarrow{0}.\]
\[\mathbf{Найти:}\]
\[1)\ направление\ каждого\ из\]
\[векторов\]
\[\overrightarrow{a};\ - \overrightarrow{a};\frac{1}{2}\overrightarrow{a};\ - 2\overrightarrow{a};6\overrightarrow{a}\ по\]
\[отношению\ к\ вектору\ \overrightarrow{p}.\]
\[2)\ выразить\ длины\ этих\ \]
\[векторов\ через\ |\overrightarrow{p}|.\]
\[\mathbf{Решение.}\]
\[1)\ 3 > 0;\ \ 1 > 0:\]
\[\ \overrightarrow{p} \nearrow \nearrow a;\ \ \left| \overrightarrow{p} \right| = 3 \bullet \left| \overrightarrow{a} \right| \Longrightarrow\]
\[\Longrightarrow \left| \overrightarrow{a} \right| = \frac{\left| \overrightarrow{p} \right|}{3}\text{.\ }\]
\[2) - 1 < 0;\ \ 3 > 0:\]
\[\overrightarrow{p} \nearrow \swarrow - \overrightarrow{a};\ \ \left| - \overrightarrow{a} \right| = \left| \overrightarrow{a} \right| = \frac{\left| \overrightarrow{p} \right|}{3}.\]
\[3)\ \frac{1}{2} > 0;\ \ 3 > 0:\]
\[\overrightarrow{p} \nearrow \nearrow \frac{1}{2}\overrightarrow{a};\ \ \left| \frac{1}{2}\overrightarrow{a} \right| = \frac{\left| \overrightarrow{p} \right|}{6}.\]
\[4) - 2 < 0;\ \ 3 > 0:\]
\[\overrightarrow{p} \nearrow \swarrow - 2\overrightarrow{a};\ \ \left| - 2\overrightarrow{a} \right| = \left| 2\overrightarrow{a} \right| = \frac{2}{3}|\overrightarrow{p}|.\]
\[5)\ 6 > 0;\ \ 3 > 0:\]
\[\overrightarrow{p} \nearrow 6\overrightarrow{a};\ \ \left| 6\overrightarrow{a} \right| = 2\left| \overrightarrow{p} \right|.\]