Задание 5 (Вариант 1): Найдите косинус угла между векторами \(\overrightarrow{a} = 4\overrightarrow{m} - \overrightarrow{p}\) и \(\overrightarrow{b} = \overrightarrow{m} + 2\overrightarrow{p}\), если \(\overrightarrow{m} \perp \overrightarrow{p}\) и \(|\overrightarrow{m}| = |\overrightarrow{p}| = 1\).

Решение: \(\overrightarrow{a} \cdot \overrightarrow{b} = (4\overrightarrow{m} - \overrightarrow{p}) \cdot (\overrightarrow{m} + 2\overrightarrow{p}) = 4(\overrightarrow{m} \cdot \overrightarrow{m}) + 8(\overrightarrow{m} \cdot \overrightarrow{p}) - (\overrightarrow{p} \cdot \overrightarrow{m}) - 2(\overrightarrow{p} \cdot \overrightarrow{p})\) Так как \(\overrightarrow{m} \perp \overrightarrow{p}\), то \(\overrightarrow{m} \cdot \overrightarrow{p} = \overrightarrow{p} \cdot \overrightarrow{m} = 0\) \(\overrightarrow{a} \cdot \overrightarrow{b} = 4|\overrightarrow{m}|^2 - 2|\overrightarrow{p}|^2 = 4(1)^2 - 2(1)^2 = 4 - 2 = 2\) \(|\overrightarrow{a}| = \sqrt{(4\overrightarrow{m} - \overrightarrow{p}) \cdot (4\overrightarrow{m} - \overrightarrow{p})} = \sqrt{16|\overrightarrow{m}|^2 - 8(\overrightarrow{m} \cdot \overrightarrow{p}) + |\overrightarrow{p}|^2} = \sqrt{16(1)^2 + (1)^2} = \sqrt{17}\) \(|\overrightarrow{b}| = \sqrt{(\overrightarrow{m} + 2\overrightarrow{p}) \cdot (\overrightarrow{m} + 2\overrightarrow{p})} = \sqrt{|\overrightarrow{m}|^2 + 4(\overrightarrow{m} \cdot \overrightarrow{p}) + 4|\overrightarrow{p}|^2} = \sqrt{(1)^2 + 4(1)^2} = \sqrt{5}\) \(\cos(\theta) = \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{a}| \cdot |\overrightarrow{b}|} = \frac{2}{\sqrt{17} \cdot \sqrt{5}} = \frac{2}{\sqrt{85}} = \frac{2\sqrt{85}}{85}\)