与えられたベクトルの組について、それぞれのなす角$\theta$を求める。

1. 問題の内容

与えられたベクトルの組について、それぞれのなす角$\theta$を求める。

2. 解き方の手順

ベクトル$\vec{a}$と$\vec{b}$のなす角$\theta$は、次の式で求められる。

$$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$$

ここで、$\vec{a} \cdot \vec{b}$は内積、 $|\vec{a}|$ はベクトル$\vec{a}$の大きさである。

(1) $\vec{a} = (2, 1), \vec{b} = (2, 6)$

$\vec{a} \cdot \vec{b} = (2)(2) + (1)(6) = 4 + 6 = 10$

$|\vec{a}| = \sqrt{2^2 + 1^2} = \sqrt{5}$

$|\vec{b}| = \sqrt{2^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}$

$\cos \theta = \frac{10}{\sqrt{5} \cdot 2\sqrt{10}} = \frac{10}{2\sqrt{50}} = \frac{10}{2 \cdot 5\sqrt{2}} = \frac{1}{\sqrt{2}}$

$\theta = \frac{\pi}{4}$

(2) $\vec{a} = (\sqrt{3}, 1), \vec{b} = (-\sqrt{3}, 1)$

$\vec{a} \cdot \vec{b} = (\sqrt{3})(-\sqrt{3}) + (1)(1) = -3 + 1 = -2$

$|\vec{a}| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2$

$|\vec{b}| = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2$

$\cos \theta = \frac{-2}{2 \cdot 2} = -\frac{1}{2}$

$\theta = \frac{2\pi}{3}$

(3) $\vec{a} = (2, -3), \vec{b} = (-4, 6)$

$\vec{a} \cdot \vec{b} = (2)(-4) + (-3)(6) = -8 - 18 = -26$

$|\vec{a}| = \sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}$

$|\vec{b}| = \sqrt{(-4)^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$

$\cos \theta = \frac{-26}{\sqrt{13} \cdot 2\sqrt{13}} = \frac{-26}{2 \cdot 13} = -1$

$\theta = \pi$

(4) $\vec{a} = (2\sqrt{3}, -2), \vec{b} = (-1, \sqrt{3})$

$\vec{a} \cdot \vec{b} = (2\sqrt{3})(-1) + (-2)(\sqrt{3}) = -2\sqrt{3} - 2\sqrt{3} = -4\sqrt{3}$

$|\vec{a}| = \sqrt{(2\sqrt{3})^2 + (-2)^2} = \sqrt{12 + 4} = \sqrt{16} = 4$

$|\vec{b}| = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2$

$\cos \theta = \frac{-4\sqrt{3}}{4 \cdot 2} = -\frac{\sqrt{3}}{2}$

$\theta = \frac{5\pi}{6}$

(5) $\vec{a} = (3, 3), \vec{b} = (1 + \sqrt{3}, 1 - \sqrt{3})$

$\vec{a} \cdot \vec{b} = (3)(1 + \sqrt{3}) + (3)(1 - \sqrt{3}) = 3 + 3\sqrt{3} + 3 - 3\sqrt{3} = 6$

$|\vec{a}| = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$

$|\vec{b}| = \sqrt{(1 + \sqrt{3})^2 + (1 - \sqrt{3})^2} = \sqrt{1 + 2\sqrt{3} + 3 + 1 - 2\sqrt{3} + 3} = \sqrt{8} = 2\sqrt{2}$

$\cos \theta = \frac{6}{3\sqrt{2} \cdot 2\sqrt{2}} = \frac{6}{12} = \frac{1}{2}$

$\theta = \frac{\pi}{3}$

3. 最終的な答え

(1) $\theta = \frac{\pi}{4}$

(2) $\theta = \frac{2\pi}{3}$

(3) $\theta = \pi$

(4) $\theta = \frac{5\pi}{6}$

(5) $\theta = \frac{\pi}{3}$