\(1\) 辺の長さが \(1\) の正四面体 OABC において, \(3\) 辺 OA , OB , AC 上にそれぞれ点 D , E , F を \(\text{OD} = \dfrac{1}{2}\) , \(\text{OE} = t \ ( 0 \lt t \lt 1 )\) , \(\text{AF} = \dfrac{2}{3}\) となるようにとる. \(\overrightarrow{\text{OA}} = \overrightarrow{a}\) , \(\overrightarrow{\text{OB}} = \overrightarrow{b}\) , \(\overrightarrow{\text{OC}} = \overrightarrow{c}\) とおくとき, 次の問いに答えよ.
(1) \(\overrightarrow{\text{DE}} , \overrightarrow{\text{DF}}\) を \(\overrightarrow{a} , \overrightarrow{b} , \overrightarrow{c} , t\) を用いて表せ.
(2) \(\overrightarrow{\text{DE}} \perp \overrightarrow{\text{DF}}\) のとき, \(t\) の値を求めよ.
(3) \(3\) 点 D , E , F が定める平面が直線 BC と交わる点を G とするとき, 線分 BG の長さを \(t\) を用いて表せ.
【 解 答 】
(1)
条件より \[\begin{align} \overrightarrow{\text{OD}} & = \dfrac{1}{2} \overrightarrow{a} , \ \overrightarrow{\text{OE}} = t \overrightarrow{b} , \\ \overrightarrow{\text{OF}} & = \dfrac{\overrightarrow{a} +2 \overrightarrow{c}}{3} \end{align}\] ゆえに \[\begin{align} \overrightarrow{\text{DE}} & = \overrightarrow{\text{OE}} -\overrightarrow{\text{OD}} = \underline{-\dfrac{1}{2} \overrightarrow{a} +t \overrightarrow{b}} , \\ \overrightarrow{\text{DF}} & = \overrightarrow{\text{OF}} -\overrightarrow{\text{OD}} = \underline{-\dfrac{1}{6} \overrightarrow{a} +\dfrac{2}{3} \overrightarrow{c}} \end{align}\]
(2)
条件より \[\begin{align} \big| \overrightarrow{a} \big| & = \big| \overrightarrow{b} \big| = \big| \overrightarrow{c} \big| = 1 , \\ \overrightarrow{a} \cdot \overrightarrow{b} & = \overrightarrow{b} \cdot \overrightarrow{c} = \cdot \overrightarrow{c} \cdot \overrightarrow{a} = 1 \cdot 1 \cdot \cos \dfrac{\pi}{3} = \dfrac{1}{2} \end{align}\] \(\overrightarrow{\text{DE}} \perp \overrightarrow{\text{DF}}\) なので \[\begin{align} \overrightarrow{\text{DE}} \cdot \overrightarrow{\text{DF}} & = \left( -\dfrac{1}{2} \overrightarrow{a} +t \overrightarrow{b} \right) \cdot \left( -\dfrac{1}{6} \overrightarrow{a} +\dfrac{2}{3} \overrightarrow{c} \right) \\ & = \dfrac{1}{12} \big| \overrightarrow{a} \big|^2 -\dfrac{1}{3} \overrightarrow{a} \cdot \overrightarrow{c} -\dfrac{t}{6} \overrightarrow{a} \cdot \overrightarrow{b} +\dfrac{2t}{3} \overrightarrow{b} \cdot \overrightarrow{c} \\ & = \dfrac{t}{4} -\dfrac{1}{12} = 0 \\ \text{∴} \quad t & = \underline{\dfrac{1}{3}} \end{align}\]
(3)
点 G は平面 DEF 上にあるので, 実数 \(u , v\) を用いて
\[\begin{align}
\overrightarrow{\text{OG}} & = \overrightarrow{\text{OD}} +u \overrightarrow{\text{DE}} +v \overrightarrow{\text{DF}} \\
& = \dfrac{1}{2} \overrightarrow{a} +u \left( -\dfrac{1}{2} \overrightarrow{a} +t \overrightarrow{b} \right) +v \left( -\dfrac{1}{6} \overrightarrow{a} +\dfrac{2}{3} \overrightarrow{c} \right) \\
& = \dfrac{3-3u-v}{6} \overrightarrow{a} +t u \overrightarrow{b} +\dfrac{2v}{3} \overrightarrow{c}
\end{align}\]
と表せる.
さらに, 点 G は辺 BC 上にあることから
\[\begin{align}
\dfrac{3-3u-v}{6} = 0 & , \ ut +\dfrac{2v}{3} = 1 \\
3u+v = 3 & , \ 3tu+2v = 3 \\
\text{∴} \quad u = \dfrac{1}{2-t} & , \ v = \dfrac{3(1-t)}{2-t}
\end{align}\]
したがって,
\[
\overrightarrow{\text{OG}} = \dfrac{t}{2-t} \overrightarrow{b} +\dfrac{2(1-t)}{2-t} \overrightarrow{c}
\]
よって, 点 G は辺 BC を \(2(1-t) : t\) に内分するので
\[
\text{BG} = \underline{\dfrac{2(1-t)}{2-t}}
\]