四面体 ABCD において \(\overrightarrow{\text{CA}}\) と \(\overrightarrow{\text{CB}}\) , \(\overrightarrow{\text{DA}}\) と \(\overrightarrow{\text{DB}}\) , \(\overrightarrow{\text{AB}}\) と \(\overrightarrow{\text{CD}}\) はそれぞれ垂直であるとする. 　このとき, 頂点 A , 頂点 B および辺 CD の中点 M の \(3\) 点を通る平面は辺 CD と直交することを示せ.

【 解 答 】

\(\overrightarrow{\text{AB}} = \overrightarrow{b}\) , \(\overrightarrow{\text{AC}} = \overrightarrow{c}\) , \(\overrightarrow{\text{AD}} = \overrightarrow{d}\) とおく.
条件より \[\begin{align} \overrightarrow{\text{CA}} \cdot \overrightarrow{\text{CB}} & = -\overrightarrow{c} \cdot \left( \overrightarrow{b} -\overrightarrow{c} \right) = -\overrightarrow{b} \cdot \overrightarrow{c} +\left| \overrightarrow{c} \right|^2 = 0 \\ \overrightarrow{\text{DA}} \cdot \overrightarrow{\text{DB}} & = -\overrightarrow{d} \cdot \left( \overrightarrow{b} -\overrightarrow{d} \right) = -\overrightarrow{b} \cdot \overrightarrow{d} +\left| \overrightarrow{d} \right|^2 = 0 \\ \overrightarrow{\text{AB}} \cdot \overrightarrow{\text{CD}} & = \overrightarrow{b} \cdot \left( \overrightarrow{d} -\overrightarrow{c} \right) = \overrightarrow{b} \cdot \overrightarrow{d} -\overrightarrow{b} \cdot \overrightarrow{c}= 0 \end{align}\] 以上より \[ \overrightarrow{b} \cdot \overrightarrow{c} = \left| \overrightarrow{c} \right|^2 = \overrightarrow{b} \cdot \overrightarrow{d} = \left| \overrightarrow{d} \right|^2 \] これを用いると, \[\begin{align} \overrightarrow{\text{AM}} \cdot \overrightarrow{\text{CD}} & = \dfrac{1}{2}\overrightarrow{b} \cdot \left( \overrightarrow{d} -\overrightarrow{c} \right) \\ & = \dfrac{1}{2} \left( \overrightarrow{b} \cdot \overrightarrow{d} -\overrightarrow{b} \cdot \overrightarrow{c} \right) = 0 \\ \text{∴} \quad & \overrightarrow{\text{AM}} \perp \overrightarrow{\text{CD}} \end{align}\] これと条件 \(\overrightarrow{\text{AB}} \perp \overrightarrow{\text{CD}}\) から, 題意は示された.