同济高数 第8章 第8-2-1题

[AI解答]

[AI解答]

**（1）求 $\boldsymbol{a} \cdot \boldsymbol{b}$ 及 $\boldsymbol{a} \times \boldsymbol{b}$**

已知向量
$$
\boldsymbol{a} = 3\mathbf{i} - \mathbf{j} - 2\mathbf{k}, \quad \boldsymbol{b} = \mathbf{i} + 2\mathbf{j} - \mathbf{k}.
$$

点积为：
$$
\boldsymbol{a} \cdot \boldsymbol{b} = (3)(1) + (-1)(2) + (-2)(-1) = 3 - 2 + 2 = 3.
$$

叉积为：
$$
\boldsymbol{a} \times \boldsymbol{b} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
3 & -1 & -2 \\
1 & 2 & -1
\end{vmatrix}
= \mathbf{i}\begin{vmatrix} -1 & -2 \\ 2 & -1 \end{vmatrix}
- \mathbf{j}\begin{vmatrix} 3 & -2 \\ 1 & -1 \end{vmatrix}
+ \mathbf{k}\begin{vmatrix} 3 & -1 \\ 1 & 2 \end{vmatrix}.
$$

计算各分量：
$$
\mathbf{i}: (-1)(-1) - (-2)(2) = 1 + 4 = 5,
$$
$$
\mathbf{j}: (3)(-1) - (-2)(1) = -3 + 2 = -1 \quad \Rightarrow \quad -\mathbf{j}(-1) = \mathbf{j},
$$
注意符号：实际为 $-\mathbf{j} \cdot (-1) = \mathbf{j}$，即第二分量是 $+1$。
$$
\mathbf{k}: (3)(2) - (-1)(1) = 6 + 1 = 7.
$$

因此：
$$
\boldsymbol{a} \times \boldsymbol{b} = 5\mathbf{i} + \mathbf{j} + 7\mathbf{k}.
$$

---

**（2）求 $(-2\boldsymbol{a}) \cdot 3\boldsymbol{b}$ 及 $\boldsymbol{a} \times 2\boldsymbol{b}$**

由点积的线性性质：
$$
(-2\boldsymbol{a}) \cdot (3\boldsymbol{b}) = (-2)(3)(\boldsymbol{a} \cdot \boldsymbol{b}) = -6 \times 3 = -18.
$$

由叉积的线性性质：
$$
\boldsymbol{a} \times (2\boldsymbol{b}) = 2 (\boldsymbol{a} \times \boldsymbol{b}) = 2(5\mathbf{i} + \mathbf{j} + 7\mathbf{k}) = 10\mathbf{i} + 2\mathbf{j} + 14\mathbf{k}.
$$

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**（3）求 $\boldsymbol{a}, \boldsymbol{b}$ 的夹角的余弦**

先计算模长：
$$
|\boldsymbol{a}| = \sqrt{3^2 + (-1)^2 + (-2)^2} = \sqrt{9 + 1 + 4} = \sqrt{14},
$$
$$
|\boldsymbol{b}| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}.
$$

夹角余弦公式：
$$
\cos\theta = \frac{\boldsymbol{a} \cdot \boldsymbol{b}}{|\boldsymbol{a}|\,|\boldsymbol{b}|} = \frac{3}{\sqrt{14} \cdot \sqrt{6}} = \frac{3}{\sqrt{84}} = \frac{3}{2\sqrt{21}} = \frac{\sqrt{21}}{14}.
$$

---

最终答案：
$$
\boxed{\boldsymbol{a} \cdot \boldsymbol{b} = 3,\quad \boldsymbol{a} \times \boldsymbol{b} = 5\mathbf{i} + \mathbf{j} + 7\mathbf{k}}
$$
$$
\boxed{(-2\boldsymbol{a})\cdot 3\boldsymbol{b} = -18,\quad \boldsymbol{a} \times 2\boldsymbol{b} = 10\mathbf{i} + 2\mathbf{j} + 14\mathbf{k}}
$$
$$
\boxed{\cos\theta = \frac{\sqrt{21}}{14}}
$$

难度：★☆☆☆☆