(1) 平面 $E$ 的法向量 $\overset{\large\rightharpoonup}{n} = \overset{\large\rightharpoonup}{u} \times \overset{\large\rightharpoonup}{v} = (2,0,1) \times (0,1,1) = (-1, -2, 2)$。 由於平面過原點，方程式為 $-x - 2y + 2z = 0$，即 $x + 2y - 2z = 0$。 對照 $x + by + cz = d$，得 $b = 2, c = -2, d = 0$。 (2) 設 $\overset{\large\rightharpoonup}{w} = \overset{\large\rightharpoonup}{AB}$。則 $\overset{\large\rightharpoonup}{A'B'}$ 為 $\overset{\large\rightharpoonup}{w}$ 在平面 $E$ 上的正射影向量。 可將 $\overset{\large\rightharpoonup}{w}$ 分解為 $\overset{\large\rightharpoonup}{w} = \overset{\large\rightharpoonup}{A'B'} + \overset{\large\rightharpoonup}{w}_{\perp}$，其中 $\overset{\large\rightharpoonup}{w}_{\perp}$ 垂直於平面 $E$。 因為 $\overset{\large\rightharpoonup}{u}$ 在平面 $E$ 上，所以 $\overset{\large\rightharpoonup}{w}_{\perp} \cdot \overset{\large\rightharpoonup}{u} = 0$。 則 $\overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{u} = (\overset{\large\rightharpoonup}{A'B'} + \overset{\large\rightharpoonup}{w}_{\perp}) \cdot \overset{\large\rightharpoonup}{u} = \overset{\large\rightharpoonup}{A'B'} \cdot \overset{\large\rightharpoonup}{u} + 0 = \overset{\large\rightharpoonup}{A'B'} \cdot \overset{\large\rightharpoonup}{u}$。證畢。 (3) 由 (2) 同理可知 $\overset{\large\rightharpoonup}{A'B'} \cdot \overset{\large\rightharpoonup}{v} = \overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{v} = 2$。 將 $\overset{\large\rightharpoonup}{A'B'} = \alpha \overset{\large\rightharpoonup}{u} + \beta \overset{\large\rightharpoonup}{v}$ 代入內積式： $\begin{cases} (\alpha \overset{\large\rightharpoonup}{u} + \beta \overset{\large\rightharpoonup}{v}) \cdot \overset{\large\rightharpoonup}{u} = 5 \\ (\alpha \overset{\large\rightharpoonup}{u} + \beta \overset{\large\rightharpoonup}{v}) \cdot \overset{\large\rightharpoonup}{v} = 2 \end{cases} \implies \begin{cases} \alpha (\overset{\large\rightharpoonup}{u} \cdot \overset{\large\rightharpoonup}{u}) + \beta (\overset{\large\rightharpoonup}{v} \cdot \overset{\large\rightharpoonup}{u}) = 5 \\ \alpha (\overset{\large\rightharpoonup}{u} \cdot \overset{\large\rightharpoonup}{v}) + \beta (\overset{\large\rightharpoonup}{v} \cdot \overset{\large\rightharpoonup}{v}) = 2 \end{cases}$ 計算內積：$\overset{\large\rightharpoonup}{u} \cdot \overset{\large\rightharpoonup}{u} = 5, \overset{\large\rightharpoonup}{v} \cdot \overset{\large\rightharpoonup}{v} = 2, \overset{\large\rightharpoonup}{u} \cdot \overset{\large\rightharpoonup}{v} = 1$。 $\begin{cases} 5\alpha + \beta = 5 \\ \alpha + 2\beta = 2 \end{cases}$ 由第二式得 $\alpha = 2 - 2\beta$，代入第一式： $5(2 - 2\beta) + \beta = 5 \implies 10 - 9\beta = 5 \implies \beta = 5/9$ 則 $\alpha = 2 - 2(5/9) = 8/9$。 故 $\alpha = 8/9, \beta = 5/9$。