正四面体 OABC の \(1\) 辺の長さが \(1\) とする. 辺 OA を \(2:1\) に内分する点 P , 辺 OB を \(1 : 2\) に内分する点を Q とし, \(0 \lt t \lt 1\) をみたす \(t\) に対して, 辺 OC を \(t : 1-t\) に内分する点を R とする.
(1) PQ の長さを求めよ.
(2) △PQR の面積が最小になるときの \(t\) の値を求めよ.
【 解 答 】
(1)
\(\overrightarrow{\text{OA}} =\overrightarrow{a}\) , \(\overrightarrow{\text{OB}} =\overrightarrow{b}\) , \(\overrightarrow{\text{OC}} =\overrightarrow{c}\) とおくと \[\begin{align} \left| \overrightarrow{a} \right| & = \left| \overrightarrow{b} \right| = \left| \overrightarrow{c} \right| = 1 , \\ \overrightarrow{a} \cdot \overrightarrow{b} & = \overrightarrow{b} \cdot \overrightarrow{c} = \overrightarrow{c} \cdot \overrightarrow{a} = 1 \cdot 1 \cdot \cos 60^{\circ} = \dfrac{1}{2} , \\ \overrightarrow{\text{OP}} & = \dfrac{2}{3} \overrightarrow{a} , \ \overrightarrow{\text{OQ}} = \dfrac{1}{3} \overrightarrow{b} , \ \overrightarrow{\text{OR}} = t \overrightarrow{c} \end{align}\] したがって \[\begin{align} \left| \overrightarrow{\text{PQ}} \right|^2 & = \left| \dfrac{2}{3} \overrightarrow{a} -\dfrac{1}{3} \overrightarrow{b} \right|^2 \\ & = \dfrac{4 -4 \cdot \frac{1}{2} +1}{9} =\dfrac{1}{3} \end{align}\] よって \[ \text{PQ} = \underline{\dfrac{\sqrt{3}}{3}} \]
(2)
\[\begin{align} \left| \overrightarrow{\text{PR}} \right|^2 & = \left| \dfrac{2}{3} \overrightarrow{a} -t \overrightarrow{c} \right|^2 = \dfrac{4}{9} -\dfrac{2t}{3} +t^2 , \\ \overrightarrow{\text{PQ}} \cdot \overrightarrow{\text{PR}} & = \left( \dfrac{2}{3} \overrightarrow{a} -\dfrac{1}{3} \overrightarrow{b} \right) \cdot \left( \dfrac{2}{3} \overrightarrow{a} -t \overrightarrow{c} \right) \\ & = \dfrac{4}{9} -\dfrac{t}{3} -\dfrac{1}{9} +\dfrac{t}{6} \\ & = \dfrac{1}{3} -\dfrac{t}{6} \end{align}\] したがって, △PQR の面積 \(S\) は \[\begin{align} S & = \dfrac{1}{2} \sqrt{\left| \overrightarrow{\text{PQ}} \right|^2 \left| \overrightarrow{\text{PR}} \right|^2 -\left( \overrightarrow{\text{PQ}} \cdot \overrightarrow{\text{PR}} \right)^2} \\ & = \dfrac{1}{2} \sqrt{\dfrac{1}{3} \left( \dfrac{4}{9} -\dfrac{2t}{3} +t^2 \right) -\left( \dfrac{1}{3} -\dfrac{t}{6} \right)^2} \\ & = \dfrac{1}{2} \sqrt{\dfrac{11 t^2}{36} -\dfrac{t}{9} +\dfrac{1}{27}} \\ & = \dfrac{1}{2} \sqrt{\dfrac{11}{36} \left( t -\dfrac{2}{11} \right)^2 +\dfrac{1}{27} -\dfrac{1}{99}} \end{align}\] よって, \(S\) を最小にする \(t\) の値は \[ t = \underline{\dfrac{2}{11}} \]