Найдите косинус угла между векторами $$\vec{a} = \vec{p}$$ и $$\vec{b} = \vec{q}$$, если $$|\vec{p}| = |\vec{q}| = 1$$ и $$\vec{m} = 4\vec{p} + 2\vec{q}$$, $$\vec{n} = \vec{p} - \vec{q}$$.

Решение: $$\vec{m} \cdot \vec{n} = (4\vec{p} + 2\vec{q})(\vec{p} - \vec{q}) = 4\vec{p}^2 - 4\vec{p}\vec{q} + 2\vec{q}\vec{p} - 2\vec{q}^2 = 4|\vec{p}|^2 - 2\vec{p}\vec{q} - 2|\vec{q}|^2 = 4 \cdot 1 - 2 \vec{p}\vec{q} - 2 \cdot 1 = 2 - 2\vec{p}\vec{q}$$ $$\vec{m} \cdot \vec{n} = 2 - 2|\vec{p}||\vec{q}|\cos(\theta) = 2 - 2 \cos(\theta)$$ $$|\vec{m}| = \sqrt{(4\vec{p} + 2\vec{q})^2} = \sqrt{16\vec{p}^2 + 16\vec{p}\vec{q} + 4\vec{q}^2} = \sqrt{16 + 16\cos(\theta) + 4} = \sqrt{20 + 16\cos(\theta)}$$ $$|\vec{n}| = \sqrt{(\vec{p} - \vec{q})^2} = \sqrt{\vec{p}^2 - 2\vec{p}\vec{q} + \vec{q}^2} = \sqrt{1 - 2\cos(\theta) + 1} = \sqrt{2 - 2\cos(\theta)}$$ $$\cos(\theta) = \frac{\vec{m} \cdot \vec{n}}{|\vec{m}||\vec{n}|} = \frac{2 - 2\cos(\theta)}{\sqrt{20 + 16\cos(\theta)} \sqrt{2 - 2\cos(\theta)}} = \frac{2(1 - \cos(\theta))}{\sqrt{4(5 + 4\cos(\theta))2(1 - \cos(\theta))}} = \frac{2(1 - \cos(\theta))}{2\sqrt{2}\sqrt{(5 + 4\cos(\theta))(1 - \cos(\theta))}} = \frac{\sqrt{1 - \cos(\theta)}}{\sqrt{2(5 + 4\cos(\theta))}}$$ $$\cos^2(\theta) = \frac{1 - \cos(\theta)}{2(5 + 4\cos(\theta))}$$ $$2\cos^2(\theta)(5 + 4\cos(\theta)) = 1 - \cos(\theta)$$ $$10\cos^2(\theta) + 8\cos^3(\theta) = 1 - \cos(\theta)$$ $$8\cos^3(\theta) + 10\cos^2(\theta) + \cos(\theta) - 1 = 0$$