Задание 5 (Вариант 2): Найдите косинус угла между векторами \(\overrightarrow{m} = 5\overrightarrow{a} + \overrightarrow{b}\) и \(\overrightarrow{n} = 2\overrightarrow{a} - \overrightarrow{b}\), если \(\overrightarrow{a} \perp \overrightarrow{b}\) и \(|\overrightarrow{a}| = |\overrightarrow{b}| = 1\).

Решение: \(\overrightarrow{m} \cdot \overrightarrow{n} = (5\overrightarrow{a} + \overrightarrow{b}) \cdot (2\overrightarrow{a} - \overrightarrow{b}) = 10(\overrightarrow{a} \cdot \overrightarrow{a}) - 5(\overrightarrow{a} \cdot \overrightarrow{b}) + 2(\overrightarrow{b} \cdot \overrightarrow{a}) - (\overrightarrow{b} \cdot \overrightarrow{b})\) Так как \(\overrightarrow{a} \perp \overrightarrow{b}\), то \(\overrightarrow{a} \cdot \overrightarrow{b} = \overrightarrow{b} \cdot \overrightarrow{a} = 0\) \(\overrightarrow{m} \cdot \overrightarrow{n} = 10|\overrightarrow{a}|^2 - |\overrightarrow{b}|^2 = 10(1)^2 - (1)^2 = 10 - 1 = 9\) \(|\overrightarrow{m}| = \sqrt{(5\overrightarrow{a} + \overrightarrow{b}) \cdot (5\overrightarrow{a} + \overrightarrow{b})} = \sqrt{25|\overrightarrow{a}|^2 + 10(\overrightarrow{a} \cdot \overrightarrow{b}) + |\overrightarrow{b}|^2} = \sqrt{25(1)^2 + (1)^2} = \sqrt{26}\) \(|\overrightarrow{n}| = \sqrt{(2\overrightarrow{a} - \overrightarrow{b}) \cdot (2\overrightarrow{a} - \overrightarrow{b})} = \sqrt{4|\overrightarrow{a}|^2 - 4(\overrightarrow{a} \cdot \overrightarrow{b}) + |\overrightarrow{b}|^2} = \sqrt{4(1)^2 + (1)^2} = \sqrt{5}\) \(\cos(\theta) = \frac{\overrightarrow{m} \cdot \overrightarrow{n}}{|\overrightarrow{m}| \cdot |\overrightarrow{n}|} = \frac{9}{\sqrt{26} \cdot \sqrt{5}} = \frac{9}{\sqrt{130}} = \frac{9\sqrt{130}}{130}\)