Найдите косинус угла между векторами b = 6m - n и с = m + 3n, ec ли т ⊥ n и |m|=|n| = 1.

$$\overrightarrow{b} = 6\overrightarrow{m} - \overrightarrow{n}$$

$$\overrightarrow{c} = \overrightarrow{m} + 3\overrightarrow{n}$$

Т.к. m ⊥ n, то скалярное произведение $$\overrightarrow{m} \cdot \overrightarrow{n} = 0$$

$$|\overrightarrow{m}| = |\overrightarrow{n}| = 1$$

$$\overrightarrow{b} \cdot \overrightarrow{c} = (6\overrightarrow{m} - \overrightarrow{n})(\overrightarrow{m} + 3\overrightarrow{n}) = 6\overrightarrow{m}^2 + 18(\overrightarrow{m} \cdot \overrightarrow{n}) - (\overrightarrow{n} \cdot \overrightarrow{m}) - 3\overrightarrow{n}^2 = 6\overrightarrow{m}^2 - 3\overrightarrow{n}^2 = 6 \cdot 1^2 - 3 \cdot 1^2 = 6 - 3 = 3$$

$$|\overrightarrow{b}| = \sqrt{(6\overrightarrow{m} - \overrightarrow{n})^2} = \sqrt{36\overrightarrow{m}^2 - 12(\overrightarrow{m} \cdot \overrightarrow{n}) + \overrightarrow{n}^2} = \sqrt{36 \cdot 1^2 + 1^2} = \sqrt{37}$$

$$|\overrightarrow{c}| = \sqrt{(\overrightarrow{m} + 3\overrightarrow{n})^2} = \sqrt{\overrightarrow{m}^2 + 6(\overrightarrow{m} \cdot \overrightarrow{n}) + 9\overrightarrow{n}^2} = \sqrt{1^2 + 9 \cdot 1^2} = \sqrt{10}$$

$$cos\alpha = \frac{\overrightarrow{b} \cdot \overrightarrow{c}}{|\overrightarrow{b}| \cdot |\overrightarrow{c}|} = \frac{3}{\sqrt{37} \cdot \sqrt{10}} = \frac{3}{\sqrt{370}} = \frac{3\sqrt{370}}{370}$$

Ответ:$$\frac{3\sqrt{370}}{370}$$