第13章 13.9

(1)

$\vec{a} \cdot \vec{b} = 3 \cdot 2 + (-1) \cdot 4 + 2 \cdot (-1) = 6 - 4 - 2 = 0$

(2)

$\vec{a} \cdot \vec{c} = 3 \cdot 1 + (-1) \cdot 2 + 2 \cdot 3 = 3 - 2 + 6 = 7$

(3)

$\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{0}{\sqrt{3^2 + (-1)^2 + 2^2} \cdot \sqrt{2^2 + 4^2 + (-1)^2}} = 0 \qquad \theta = 90°$

(4)

$\cos\theta = \frac{\vec{a} \cdot \vec{c}}{|\vec{a}||\vec{c}|} = \frac{7}{\sqrt{3^2 + (-1)^2 + 2^2} \cdot \sqrt{1^2 + 2^2 + 3^2}} = \frac{7}{\sqrt{14}\sqrt{14}} = \frac{1}{2} \qquad \theta = 60°$

(5)

$\sqrt{6}$のベクトルを$\vec{d} = (x, y, z)$とおくと

$\vec{a} \cdot \vec{d} = 3x + (-1) \cdot y + 2z = 0 \qquad 3x - y + 2z = 0 \qquad ①$

$\vec{b} \cdot \vec{d} = 2x + 4y + (-1) \cdot z = 0 \qquad 2x + 4y - z = 0 \qquad ②$

$x^2 + y^2 + z^2 = 6 \qquad ③$

①＋②×２より $7x + 7y = 0 \qquad x = -y \qquad ④$

①×２－②×３ $-14y + 7z = 0 \qquad z = 2y \qquad ⑤$

④、⑤を③に代入して $y^2 + y^2 + 4y^2 = 6 \qquad y^2 = 1 \qquad y = \pm 1 \qquad ⑥$

④、⑥を③に代入して $y^2 + y^2 + z^2 = 6 \qquad 1 + 1 + z^2 = 6 \qquad z = \pm 2 \qquad ⑦$

⑥、⑦を①に代入して $3x \mp 1 \pm 4 = 0 \qquad 3x \pm 3 = 0 \qquad x = \mp 1$

これから

$(\mp 1, \pm 1, \pm 2)$