《2025年教材帮高中数学选择性必修第二册苏教版》

例3-4 已知 $\boldsymbol{a} = (1, 2, 3), \boldsymbol{b} = (3, 0, -1), \boldsymbol{c} = (-\frac{1}{5}, 1, -\frac{3}{5})$，下列等式：
① $(\boldsymbol{a} + \boldsymbol{b}) · \boldsymbol{c} = \boldsymbol{a} · (\boldsymbol{b} + \boldsymbol{c})$；
② $(\boldsymbol{a} + \boldsymbol{b} + \boldsymbol{c})^2 = \boldsymbol{a}^2 + \boldsymbol{b}^2 + \boldsymbol{c}^2$；
③ $(\boldsymbol{a} · \boldsymbol{b}) · \boldsymbol{c} = \boldsymbol{a} · (\boldsymbol{b} · \boldsymbol{c})$.
其中正确的有 .(填序号)

答案： 解析 ①
∵ $(\boldsymbol{a} + \boldsymbol{b}) · \boldsymbol{c} = (4, 2, 2) · (-\frac{1}{5}, 1, -\frac{3}{5}) = -\frac{4}{5} + 2 - \frac{6}{5} = 0$，
$\boldsymbol{a} · (\boldsymbol{b} + \boldsymbol{c}) = (1, 2, 3) · (\frac{14}{5}, 1, \frac{-8}{5}) = \frac{14}{5} + 2 - \frac{24}{5} = 0$，$\therefore (\boldsymbol{a} + \boldsymbol{b}) · \boldsymbol{c} = \boldsymbol{a} · (\boldsymbol{b} + \boldsymbol{c})$；
②
∵ $\boldsymbol{a} · \boldsymbol{b} = 3 + 0 - 3 = 0$，$\boldsymbol{a} · \boldsymbol{c} = -\frac{1}{5} + 2 - \frac{9}{5} = 0$，$\boldsymbol{b} · \boldsymbol{c} = -\frac{3}{5} + 0 + \frac{3}{5} = 0$，
$\therefore (\boldsymbol{a} + \boldsymbol{b} + \boldsymbol{c})^2 = \boldsymbol{a}^2 + \boldsymbol{b}^2 + \boldsymbol{c}^2$；
③
∵ $(\boldsymbol{a} · \boldsymbol{b}) · \boldsymbol{c} = 0 · \boldsymbol{c} = 0$，$\boldsymbol{a} · (\boldsymbol{b} · \boldsymbol{c}) = \boldsymbol{a} · 0 = 0$，$\therefore (\boldsymbol{a} · \boldsymbol{b}) · \boldsymbol{c} = \boldsymbol{a} · (\boldsymbol{b} · \boldsymbol{c})$.
答案 ▶ ①②③

例3-5 [教材改编 P27 T5] 在空间直角坐标系中，已知点 $A(1, 0, 1), B(3, 2, 2)$，点 $D$ 满足 $\overrightarrow{AD} = 2\overrightarrow{AB}$，则点 $D$ 的坐标是 .

答案： 解析 ▶ 设点 $D$ 的坐标为 $(x, y, z)$，则 $\overrightarrow{AD} = (x - 1, y, z - 1), \overrightarrow{AB} = (2, 2, 1)$，
因为 $\overrightarrow{AD} = 2\overrightarrow{AB}$，
$\begin{cases}x - 1 = 4, \\y = 4, \\z - 1 = 2,\end{cases}$ 解得 $\begin{cases}x = 5, \\y = 4, \\z = 3,\end{cases}$
所以 $D(5, 4, 3)$.
答案 ▶ $(5, 4, 3)$