O を原点とする座標空間に, \(4\) 点 \[ \text{A} \ ( -2 , 1 , 3 ) , \ \text{B} \ ( s , 3 , -1 ) , \ \text{C} \ ( 1 , 3 , 4 ) , \ \text{D} \ ( t , 2t , 2t ) \] がある. ただし, \(s , t\) は実数で \(t \neq 0\) である. A を通り \(\overrightarrow{\text{OC}}\) に平行な直線と, B を通り \(\overrightarrow{\text{OD}}\) に平行な直線が点 P で交わるとする. 次の問いに答えよ.
- (1) \(s\) の値および P の座標を求めよ.
以下では, \(\triangle \text{PAB} \sim \triangle \text{OCD}\) を仮定する.
(2) \(t\) の値を求めよ.
(3) D から平面 PAB に下ろした垂線を DH とするとき, H の座標を求めよ.
【 解 答 】
(1)
A を通り \(\overrightarrow{\text{OC}}\) に平行な直線を \(m\) , B を通り \(\overrightarrow{\text{OD}}\) に平行な直線を \(\ell\) とおく.
\[\begin{align}
m : \ \left( \begin{array}{c} x \\ y \\ z \end{array} \right) & = \left( \begin{array}{c} -2 \\ 1 \\ 3 \end{array} \right) +u \left( \begin{array}{c} 1 \\ 3 \\ 4 \end{array} \right) = \left( \begin{array}{c} u-2 \\ 3u-1 \\ 4y=3 \end{array} \right) \\
\ell : \ \left( \begin{array}{c} x \\ y \\ z \end{array} \right) & = \left( \begin{array}{c} s \\ 3 \\ -1 \end{array} \right) +v \left( \begin{array}{c} t \\ 2t \\ 2t \end{array} \right) = \left( \begin{array}{c} tv +s \\ 2tv +3 \\ 2tv -1 \end{array} \right)
\end{align}\]
P は \(m\) と \(\ell\) の交点なので
\[
\left\{ \begin{array}{ll} u-2 = tv +s & ... [1] \\ 3u+1 = 2tv +3 & ... [2] \\ 4u+3 = 2tv -1 & ... [3] \end{array} \right.
\]
\([3] -[2]\) より
\[\begin{align}
u+2 & = -4 \\
\text{∴} \quad u & = -6
\end{align}\]
[3] より
\[
v = \dfrac{4 (u+1)}{2t} = -\dfrac{10}{t}
\]
[1] に代入して
\[
s = u-2 -tv = -6 -2 +10 = \underline{2}
\]
このとき
\[
\text{P} \ \underline{( -8 , -17 , -21 )}
\]
(2)
\(\angle \text{APB} = \angle \text{COD}\) なので, \(\triangle \text{PAB} \sim \triangle \text{OCD}\) となる条件は \[ \text{PA} : \text{PB} = \text{OC} : \text{OD} \quad ... [4] \] ここで \[\begin{align} \text{PA} & = \sqrt{6^2 +18^2 +24^2} \\ & = 6 \sqrt{1+9+4} = 6 \sqrt{14} , \\ \text{PB} & = \sqrt{10^2 +20^2 +20^2} \\ & = 10 \sqrt{1+4+4} = 30 , \\ \text{OC} & = \sqrt{1^2 +3^2 +3^2} = \sqrt{14} , \\ \text{OD} & = \sqrt{t^2 +(2t)^2 +(2t)^2} = 3t \end{align}\] なので, [4] より \[\begin{align} 6 \sqrt{14} : 30 & = \sqrt{14} : 3t \\ 30 \sqrt{14} & = 18 t \sqrt{14} \\ \text{∴} \quad t & = \underline{\dfrac{5}{3}} \end{align}\]
(3)
\(\overrightarrow{\text{PA}} \, / \hspace{-0.2em} / \left( \begin{array}{c} 1 \\ 3 \\ 4 \end{array} \right)\) , \(\overrightarrow{\text{PB}} \, / \hspace{-0.2em} / \left( \begin{array}{c} 1 \\ 2 \\ 2 \end{array} \right)\) を利用すれば, H は平面 PAB 上にあるので, 実数 \(\alpha , \beta\) を用いて \[\begin{align} \overrightarrow{\text{OH}} & = \left( \begin{array}{c} -8 \\ -17 \\ -21 \end{array} \right) +\alpha \left( \begin{array}{c} 1 \\ 3 \\ 4 \end{array} \right) +\beta \left( \begin{array}{c} 1 \\ 2 \\ 2 \end{array} \right) \\ & = \left( \begin{array}{c} -8 +\alpha +\beta \\ -17 +3 \alpha +2 \beta \\ -21 +4 \alpha +2 \beta \end{array} \right) \end{align}\] と表せる. \[\begin{align} \overrightarrow{\text{DH}} & = \overrightarrow{\text{OH}} -\overrightarrow{\text{OD}} \\ & = \left( \begin{array}{c} -\frac{29}{3} +\alpha +\beta \\ -\frac{61}{3} +3 \alpha +2 \beta \\ -\frac{73}{3} +4 \alpha +2 \beta \end{array} \right) \end{align}\] \(\overrightarrow{\text{DH}} \perp \overrightarrow{\text{OC}}\) より \[\begin{align} \overrightarrow{\text{DH}} \cdot \overrightarrow{\text{OC}} & = \left( -\dfrac{29}{3} +\alpha +\beta \right) +3 \left( -\dfrac{61}{3} +3 \alpha +2 \beta \right) \\ & \qquad +4 \left( -\dfrac{73}{3} +4 \alpha +2 \beta \right) \\ & = -168 +26 \alpha +15 \beta = 0 \quad ... [5] \end{align}\] \(\overrightarrow{\text{DH}} \perp \overrightarrow{\text{OD}}\) より \[\begin{align} \overrightarrow{\text{DH}} \cdot \left( \begin{array}{c} 1 \\ 2 \\ 2 \end{array} \right) & = \left( -\dfrac{29}{3} +\alpha +\beta \right) +2 \left( -\dfrac{61}{3} +3 \alpha +2 \beta \right) \\ & \qquad +2 \left( -\dfrac{73}{3} +4 \alpha +2 \beta \right) \\ & = -99 +15 \alpha +9 \beta = 0 \\ \text{∴} \quad & -33 +5 \alpha +3 \beta = 0 \quad ... [6] \end{align}\] \([6] \times 4 -[5]\) より \[\begin{align} -3 +\alpha & = 0 \\ \text{∴} \quad \alpha & = 3 \end{align}\] [6] より \[ \beta = \dfrac{33 -15}{5} = 6 \] よって \[ \overrightarrow{\text{OH}} = \left( \begin{array}{c} -8 +3 +6 \\ -17 +9 +12 \\ -21 +12 +12 \end{array} \right) = \left( \begin{array}{c} 1 \\ 4 \\ 3 \end{array} \right) \] すなわち \[ \text{H} \ \underline{( 1 , 4 , 3 )} \]