2023-M2-10

答案：(a) $\dfrac{-2}{11}\bv{i}-\dfrac{5}{11}\bv{j}-\dfrac{1}{11}\bv{k}$ (b) $-4\bv{i}-10\bv{j}-2\bv{k}$ (c) (i) $-\bv{i}+2\bv{k}$ (ii) $O$、$B$ 及 $G$ 共線。

依題意略繪下圖。

1. 設 $PR:RQ=1:\lambda$。

   在 $\Delta OPQ$ 中，利用截點公式，可得

   $\begin{array}{rcl}
   \av{OR} & = & \dfrac{\lambda \av{OP}+\av{OQ}}{1+\lambda} \\
   \av{OR} & = & \dfrac{\lambda(-2\bv{i}-\bv{k})+(2\bv{i}-\bv{j}+\bv{k})}{1+\lambda} \\
   \av{OR} & = & \dfrac{2-2\lambda}{1+\lambda}\bv{i}-\dfrac{1}{1+\lambda}\bv{j}+\dfrac{1-\lambda}{1+\lambda}\bv{k}
   \end{array}$

   由於 $OR \perp OQ$，可得

   $\begin{array}{rcl}
   \av{OR} \cdot \av{OQ} & = & 0 \\
   \dfrac{2-2\lambda}{1+\lambda}\bv{i}-\dfrac{1}{1+\lambda}\bv{j}+\dfrac{1-\lambda}{1+\lambda}\bv{k} \cdot (2\bv{i}-\bv{j}+\bv{k}) & = & 0 \\
   \dfrac{2-2\lambda}{1+\lambda}\times 2+\dfrac{1}{1+\lambda}+\dfrac{1-\lambda}{1+\lambda} & = & 0 \\
   4-4\lambda+1+1-\lambda & = & \\
   -5\lambda+6 & = & 0 \\
   \lambda & = & \dfrac{6}{5}
   \end{array}$

   由此，可得

   $\begin{array}{rcl}
   \av{OR} & = & \dfrac{2-2(\frac{6}{5})}{1+\frac{6}{5}}\bv{i}-\dfrac{1}{1+\frac{6}{5}}\bv{j}+\dfrac{1-\frac{6}{5}}{1+\frac{6}{5}}\bv{k} \\
   \av{OR} & = & \dfrac{-2}{11}\bv{i}-\dfrac{5}{11}\bv{j}-\dfrac{1}{11}\bv{k}
   \end{array}$

2. 由於 $OR$ 延伸至 $S$，則 $O$、$R$ 及 $S$ 共線。由此，可得 $\av{OS}=\mu \av{OR}$，其中 $\mu$ 為某實數。

   $\begin{array}{rcl}
   \av{OS} & = & \mu\av{OR} \\
   \av{OS} & = & \dfrac{-2\mu}{11}\bv{i}-\dfrac{5\mu}{11}\bv{j}-\dfrac{\mu}{11}\bv{k}
   \end{array}$

   連結 $PQ$。根據同弓形內的圓周角，可得 $\angle QPS = \angle QOS=90^\circ$。所以，$PQ\perp PS$。由此，可得

   $\begin{array}{rcl}
   \av{PQ}\cdot \av{PS} & = & 0 \\
   (\av{OQ}-\av{OP}) \cdot (\av{OS}-\av{OP}) & = & 0 \\
   (4\bv{i}-\bv{j}+2\bv{k})\cdot(\left(\left(\dfrac{-2\mu}{11}+2\right) \bv{i}-\dfrac{5\mu}{11}\bv{j}+\left(\dfrac{-\mu}{11}+1\right)\bv{k}\right) & = & 0 \\
   4\left(\dfrac{-2\mu}{11}+2\right)+\dfrac{5\mu}{11}+2\left(\dfrac{-\mu}{11}+1\right) & = & 0 \\
   -8\mu+88+5\mu-2\mu+22 & = & 0 \\
   -5\mu +110 & = & 0 \\
   \mu & = & 22
   \end{array}$

   所以，$\av{OS}=-4\bv{i}-10\bv{j}-2\bv{k}$。

3. 1. 所求的向量

      $\begin{array}{cl}
      = & \av{OP} \times \av{OQ} \\
      = & \begin{vmatrix} \bv{i} & \bv{j} & \bv{k} \\ -2 & 0 & -1 \\ 2 & -1 & 1 \end{vmatrix} \\
      = & \begin{vmatrix} 0 & -1 \\ -1 & 1 \end{vmatrix}\bv{i} -\begin{vmatrix} -2 & -1 \\ 2 & 1 \end{vmatrix}\bv{j} +\begin{vmatrix} -2 & 0 \\ 2 & -1 \end{vmatrix}\bv{k} \\
      = & -\bv{i}+2\bv{k}
      \end{array}$

   2. 依題意略繪下圖。
      

      由於 $\angle QOS=90^\circ$，根據半圓內的圓周角的逆定理，$QS$ 為一直徑。由此，可得

      $\begin{array}{rcl}
      \av{OG} & = & \dfrac{1}{2}(\av{OS}+\av{OQ}) \\
      \av{OG} & = & \dfrac{1}{2}(-4\bv{i}-10\bv{j}-2\bv{k}+2\bv{i}-\bv{j} +\bv{k}) \\
      \av{OG} & = & -\bv{i}-\dfrac{11}{2}\bv{j}-\dfrac{1}{2}\bv{k}
      \end{array}$

      由於 $\av{BA}$ 與平面 $\Pi$ 垂直，則 $\av{BA} \text{//} (\av{OP}\times \av{OQ})$。利用投影向量的公式，可得

      $\begin{array}{rcl}
      \av{BA} & = & \dfrac{\av{OA}\cdot(\av{OP}\times \av{OQ})}{|\av{OP}\times \av{OQ}|} \left( \dfrac{\av{OP}\times \av{OQ}}{|\av{OP}\times \av{OQ}|}\right) \\
      \av{BA} & = & \dfrac{(-6\bv{i}-22\bv{j}+2\bv{k})\cdot (-\bv{i}+2\bv{k})}{| -\bv{i}+2\bv{k}| } \left( \dfrac{-\bv{i}+2\bv{k}}{|-\bv{i}+2\bv{k}|}\right) \\
      \av{BA} & = & \dfrac{(-6)(-1)+(-22)(0)+(2)(2)}{\sqrt{(-1)^2+(0)^2+(2)^2}} \left( \dfrac{-\bv{i}+2\bv{k}}{\sqrt{(-1)^2+(0)^2+(2)^2}}\right) \\
      \av{BA} & = & 2(-\bv{i}+2\bv{k}) \\
      \av{BA} & = & -2\bv{i}+4\bv{k}
      \end{array}$

      所以，可得

      $\begin{array}{rcl}
      \av{OB} & = & \av{OA}-\av{BA} \\
      \av{OB} & = & (-6\bv{i}-22\bv{j}+2\bv{k}) -(-2\bv{i}+4\bv{k}) \\
      \av{OB} & = & -4\bv{i}-22\bv{j}-2\bv{k} \\
      \av{OB} & = & 4\left(-\bv{i}-\dfrac{11}{2}\bv{j}-\dfrac{1}{2}\bv{k}\right) \\
      \av{OB} & = & 4 \av{OG}
      \end{array}$

      由此，$O$、$B$ 及 $G$ 共線。