四面体 OABC において, \(\text{OA} = \text{OB} = \text{OC} = 1\) とする. \(\angle \text{AOB} = 60^{\circ}\) , \(\angle \text{BOC} = 45^{\circ}\) , \(\angle \text{COA} = 45^{\circ}\) とし, \(\overrightarrow{a} = \overrightarrow{\text{OA}}\) , \(\overrightarrow{b} = \overrightarrow{\text{OB}}\) , \(\overrightarrow{c} = \overrightarrow{\text{OC}}\) とおく. 点 C から面 OAB に垂線を引き, その交点を H とする.
(1) ベクトル \(\overrightarrow{\text{OH}}\) を \(\overrightarrow{a}\) と \(\overrightarrow{b}\) を用いて表せ.
(2) CH の長さを求めよ.
(3) 四面体 OABC の体積を求めよ.
【 解 答 】
(1)
条件より
\[
\left\{\begin{array}{l} \left| \overrightarrow{a} \right| = \left| \overrightarrow{b} \right| = \left| \overrightarrow{c} \right| = 1 \\ \overrightarrow{a} \cdot \overrightarrow{b} = 1 \cdot 1 \cos 60^{\circ} = \dfrac{1}{2} \\ \overrightarrow{b} \cdot \overrightarrow{c} = \overrightarrow{c} \cdot \overrightarrow{a} = 1 \cdot 1 \cos 45^{\circ} = \dfrac{\sqrt{2}}{2} \end{array}\right. \ .
\]
点 H は面 OAB 上の点なので, 実数 \(x , y\) を用いて
\[
\overrightarrow{\text{OH}} = x \overrightarrow{a} +y \overrightarrow{b} \ .
\]
とおける.
また, CH は平面 OAB と垂直なので, \(\overrightarrow{\text{CH}} \perp \overrightarrow{a}\) , \(\overrightarrow{\text{CH}} \perp \overrightarrow{b}\) より
\[\begin{align}
\overrightarrow{\text{CH}} \cdot \overrightarrow{a} & = \left( x \overrightarrow{a} +y \overrightarrow{b} -\overrightarrow{c} \right) \cdot \overrightarrow{a} \\
& = x +\dfrac{y}{2} -\dfrac{\sqrt{2}}{2} = 0 \quad ... [1] \ , \\
\overrightarrow{\text{CH}} \cdot \overrightarrow{b} & = \left( x \overrightarrow{a} +y \overrightarrow{b} -\overrightarrow{c} \right) \cdot \overrightarrow{b} \\
& = \dfrac{x}{2} +y -\dfrac{\sqrt{2}}{2} = 0 \quad ... [2] \ .
\end{align}\]
[1] [2] をとくと
\[
x = y = \dfrac{\sqrt{2}}{3} \ .
\]
よって
\[
\overrightarrow{\text{OH}} = \underline{\dfrac{\sqrt{2}}{3} \left( \overrightarrow{a} +\overrightarrow{b} \right)} \ .
\]
(2)
\[\begin{align} \left| \overrightarrow{\text{CH}} \right|^2 & = \left| \dfrac{\sqrt{2}}{3} \overrightarrow{a} +\dfrac{\sqrt{2}}{3} \overrightarrow{b} -\overrightarrow{c} \right|^2 \\ & = \left( 2 \cdot \dfrac{2}{9} +1 \right) \cdot 1 +2 \cdot \dfrac{2}{9} \cdot \dfrac{1}{2} -4 \cdot \dfrac{\sqrt{2}}{3} \cdot \dfrac{\sqrt{2}}{2} \\ & = \dfrac{13}{9} +\dfrac{2}{9} -\dfrac{4}{3} \\ & = \dfrac{1}{3} \ . \end{align}\] よって \[ \text{CH} = \underline{\dfrac{\sqrt{3}}{3}} \ . \]
(3)
\[ \triangle \text{OAB} = \dfrac{1}{2} \cdot 1^2 \sin 60^{\circ} = \dfrac{\sqrt{3}}{4} \ . \] なので, 求める体積 \(V\) は \[ V = \dfrac{1}{3} \cdot \dfrac{\sqrt{3}}{4} \cdot \dfrac{\sqrt{3}}{3} = \underline{\dfrac{1}{12}} \ . \]