540. Дан вектора (-4; 2). Найдите координаты и модули векторов За, 1-3- a, a. 2 2

Ответ: \(3\overrightarrow{a} = (-12; 6)\), \(|3\overrightarrow{a}| = 6\sqrt{5}\); \(-\frac{1}{2}\overrightarrow{a} = (2; -1)\), \(|-\frac{1}{2}\overrightarrow{a}| = \sqrt{5}\); \(\frac{3}{2}\overrightarrow{a} = (-6; 3)\), \(|\frac{3}{2}\overrightarrow{a}| = 3\sqrt{5}\)

Дано: \(\overrightarrow{a} = (-4; 2)\)

1) \(3\overrightarrow{a} = 3 \cdot (-4; 2) = (-12; 6)\)

Модуль вектора \(3\overrightarrow{a}\):

\[|3\overrightarrow{a}| = \sqrt{(-12)^2 + 6^2} = \sqrt{144 + 36} = \sqrt{180} = \sqrt{36 \cdot 5} = 6\sqrt{5}\]

2) \(-\frac{1}{2}\overrightarrow{a} = -\frac{1}{2} \cdot (-4; 2) = (2; -1)\)

Модуль вектора \(-\frac{1}{2}\overrightarrow{a}\):

\[|-\frac{1}{2}\overrightarrow{a}| = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}\]

3) \(\frac{3}{2}\overrightarrow{a} = \frac{3}{2} \cdot (-4; 2) = (-6; 3)\)

Модуль вектора \(\frac{3}{2}\overrightarrow{a}\):

\[|\frac{3}{2}\overrightarrow{a}| = \sqrt{(-6)^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45} = \sqrt{9 \cdot 5} = 3\sqrt{5}\]

Ответ: \(3\overrightarrow{a} = (-12; 6)\), \(|3\overrightarrow{a}| = 6\sqrt{5}\); \(-\frac{1}{2}\overrightarrow{a} = (2; -1)\), \(|-\frac{1}{2}\overrightarrow{a}| = \sqrt{5}\); \(\frac{3}{2}\overrightarrow{a} = (-6; 3)\), \(|\frac{3}{2}\overrightarrow{a}| = 3\sqrt{5}\)