第12章 12.5

(1)

$a \cdot b = \sqrt{3} \cdot 1 + 1 \cdot \sqrt{3} = 2\sqrt{3}$

$\cos\theta = \frac{a \cdot b}{|a||b|} = \frac{2\sqrt{3}}{\sqrt{\sqrt{3}^2+1^2}\sqrt{1^2+\sqrt{3}^2}} = \frac{\sqrt{3}}{2} \qquad \theta = \frac{\pi}{6}$

(2)

$a \cdot b = 1 \cdot 1 + 2 \cdot (-3) = -5$

$\cos\theta = \frac{a \cdot b}{|a||b|} = \frac{-5}{\sqrt{1^2+2^2}\sqrt{1^2+(-3)^2}} = -\frac{5}{5\sqrt{2}} = -\frac{1}{\sqrt{2}} \qquad \theta = \frac{3\pi}{4}$

(3)

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

$\cos\theta = \frac{a \cdot b}{|a||b|} = 0 \qquad \theta = \frac{\pi}{2}$