Solution $$(a) \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & 1 & 0 \ 2 & -1 & 1 \end{vmatrix} = \mathbf{i} \begin{vmatrix} 1 & 0 \ -1 & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 2 & 0 \ 2 & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 2 & 1 \ 2 & -1 \end{vmatrix} = (1, -2, -4)$$

$$(b) (\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & -2 & -4 \ 0 & 1 & 1 \end{vmatrix} = \mathbf{i} \begin{vmatrix} -2 & -4 \ 1 & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & -4 \ 0 & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & -2 \ 0 & 1 \end{vmatrix} = (2, -1, 1)$$

$$(c) \mathbf{a} \cdot \mathbf{c} = (2, 1, 0) \cdot (0, 1, 1) = 1, \mathbf{b} \cdot \mathbf{c} = (2, -1, 1) \cdot (0, 1, 1) = 0 \text{ and hence } (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{b} \cdot \mathbf{c})\mathbf{a} = 1\mathbf{b} - 0\mathbf{a} = (2, -1, 1)$$

(请注意 (b) and (c) 给出相同的结果.)

$$(d) \mathbf{b} \times \mathbf{c} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & -1 & 1 \ 0 & 1 & 1 \end{vmatrix} = \mathbf{i} \begin{vmatrix} -1 & 1 \ 1 & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 2 & 1 \ 0 & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 2 & -1 \ 0 & 1 \end{vmatrix} = (-2, -2, 2)$$

$$(e) \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & 1 & 0 \ -2 & -2 & 2 \end{vmatrix} = \mathbf{i} \begin{vmatrix} 1 & 0 \ -2 & 2 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 2 & 0 \ -2 & 2 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 2 & 1 \ -2 & -2 \end{vmatrix} = (2, -4, -2)$$

(请注意 (b) and (e) 不要给出相同的结果，而叉积是 not associative.)

$$(f) \mathbf{a} \cdot \mathbf{c} = (2, 1, 0) \cdot (0, 1, 1) = 1, \mathbf{a} \cdot \mathbf{b} = (2, 1, 0) \cdot (2, -1, 1) = 3 \text{ and hence } (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c} = 1\mathbf{b} - 3\mathbf{c} = (2, -4, -2)$$

(请注意 (e) and (f) 给出相同的结果.)

请在 MATLAB 中检验命令

a = [2 1 0]; b = [2 -1 1]; c = [0 1 1];
cross(a,b)
cross(cross(a,b),c)

返回……的答案 (a) and (b).

Example 4.25求一个与这两个向量所在平面垂直的单位向量。 $\mathbf{a} = (2, -3, 1)$ and $\mathbf{b} = (1, 2, -4)$ .

Solution两个向量所在平面的垂直向量是向量积

$$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & -3 & 1 \ 1 & 2 & -4 \end{vmatrix} = (10, 9, 7)$$