Solution (a) $\mathbf{a} \cdot \mathbf{c} = (1 \times 3) + (-1 \times 2) + (2 \times 1) = 3$
(b) $\mathbf{b} \cdot \mathbf{c} = (-2 \times 3) + (0 \times 2) + (2 \times 1) = -4$
(c) $(\mathbf{a} + \mathbf{b}) = (1, -1, 2) + (-2, 0, 2) = (-1, -1, 4)$ 以便
$(\mathbf{a} + \mathbf{b}) \cdot \mathbf{c} = (-1, -1, 4) \cdot (3, 2, 1) = -3 - 2 + 4 = -1$
(请注意 $(\mathbf{a} + \mathbf{b}) \cdot \mathbf{c} = \mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c}$ )
(d) $\mathbf{a} \cdot (2\mathbf{b} + 3\mathbf{c}) = (1, -1, 2) \cdot [(-4, 0, 4) + (9, 6, 3)]$
$= (1, -1, 2) \cdot (5, 6, 7) = (5 - 6 + 14) = 13$
(请注意 $2(\mathbf{a} \cdot \mathbf{b}) + 3(\mathbf{a} \cdot \mathbf{c}) = 4 + 9 = 13$ )
(e) $(\mathbf{a} \cdot \mathbf{b})\mathbf{c} = [(1, -1, 2) \cdot (-2, 0, 2)](3, 2, 1) = [-2 + 0 + 4](3, 2, 1)$
$= 2(3, 2, 1) = (6, 4, 2)$
(请注意 $\mathbf{a} \cdot \mathbf{b}$ 是一个标量, so $(\mathbf{a} \cdot \mathbf{b})\mathbf{c}$ 是一个与……平行或反平行的向量 $\mathbf{c}$ )

请在 MATLAB 中检验以下命令

a = [1 –1 2]; b = [–2 0 2]; c = [3 2 1]; dot(a,c), dot(b,c), dot(a + b,c), dot(a,2*b + 3*c), dot(a,b)*c

请提供需要翻译的 Markdown 原文，我将为您翻译成简体中文。.

Example 4.18 求两向量之间的夹角 $\mathbf{a} = (1, 2, 3)$ and $\mathbf{b} = (2, 0, 4)$ .

Solution 根据定义

$$\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta = a_1 b_1 + a_2 b_2 + a_3 b_3$$

我们在右端有

$$(1, 2, 3) \cdot (2, 0, 4) = 2 + 0 + 12 = 14$$

Also

出现在右端项中 $$(1, 2, 3) \cdot (2, 0, 4) = 2 + 0 + 12 =$$

$$|(1, 2, 3)| = \sqrt{(1^2 + 2^2 + 3^2)} = \sqrt{14}$$

$$|(2, 0, 4)| = \sqrt{(2^2 + 0^2 + 4^2)}$$

and

Thus, 由标量积的定义,

$$14 = \sqrt{(14)\sqrt{(20)}} \cos \theta$$

$$|(2, 0, 4)| = \sqrt[4]{(2^2 + 0^2 + 4^2)} = \sqrt[4]{20}, \text{ from the definition of the scalar pro} \ 14 = \sqrt[4]{(14)\sqrt[4]{(20)}} \cos \theta$$

giving

$$\theta = \cos^{-1}\sqrt{\frac{7}{10}}$$