（1）
A B → = ( 1 , 1 , − 1 ) \overrightarrow{\mathrm{AB}}=(1,1,-1) AB = ( 1 , 1 , − 1 ) A P → = ( a , 0 , − 1 ) \overrightarrow{\mathrm{AP}}=(a,0,-1) AP = ( a , 0 , − 1 ) A P → = c A B → \overrightarrow{\mathrm{AP}}=c\overrightarrow{\mathrm{AB}} AP = c AB y y y c = 0 c=0 c = 0 z z z − 1 = 0 -1=0 − 1 = 0 A , B , P \mathrm{A},\mathrm{B},\mathrm{P} A , B , P a a a
法線ベクトルを n ⃗ = ( p , q , r ) \vec n=(p,q,r) n = ( p , q , r ) n ⃗ ⋅ A B → = 0 \vec n\cdot\overrightarrow{\mathrm{AB}}=0 n ⋅ AB = 0 n ⃗ ⋅ A P → = 0 \vec n\cdot\overrightarrow{\mathrm{AP}}=0 n ⋅ AP = 0
p + q − r = 0 , a p − r = 0. p+q-r=0,\qquad ap-r=0. p + q − r = 0 , a p − r = 0. p = 1 p=1 p = 1 r = a r=a r = a q = a − 1 q=a-1 q = a − 1 n ⃗ = ( 1 , a − 1 , a ) \vec n=(1,\ a-1,\ a) n = ( 1 , a − 1 , a ) A ( 0 , 0 , 1 ) \mathrm{A}(0,0,1) A ( 0 , 0 , 1 )
x + ( a − 1 ) y + a ( z − 1 ) = 0 , すなわち x + ( a − 1 ) y + a z = a . \begin{aligned}
&x+(a-1)y+a(z-1)=0,\\
&\text{すなわち}\quad x+(a-1)y+az=a.
\end{aligned} x + ( a − 1 ) y + a ( z − 1 ) = 0 , すなわち x + ( a − 1 ) y + a z = a . 直線 ℓ \ell ℓ s s s ( 0 , s , 2 − s ) (0,\ s,\ 2-s) ( 0 , s , 2 − s )
( a − 1 ) s + a ( 2 − s ) = a ⟺ − s + 2 a = a ⟺ s = a . \begin{aligned}
(a-1)s+a(2-s)=a&\iff -s+2a=a\\
&\iff s=a.
\end{aligned} ( a − 1 ) s + a ( 2 − s ) = a ⟺ − s + 2 a = a ⟺ s = a . よって、Q ( 0 , a , 2 − a ) \mathrm{Q}(0,\ a,\ 2-a) Q ( 0 , a , 2 − a )
（2）
線分 A B \mathrm{AB} AB P Q \mathrm{PQ} PQ
O A → + s A B → = ( s , s , 1 − s ) , O P → + u P Q → = ( a − a u , a u , ( 2 − a ) u ) ( 0 ≦ s ≦ 1 , 0 ≦ u ≦ 1 ) . \begin{aligned}
\overrightarrow{\mathrm{OA}}+s\overrightarrow{\mathrm{AB}}&=(s,\ s,\ 1-s),\\
\overrightarrow{\mathrm{OP}}+u\overrightarrow{\mathrm{PQ}}&=(a-au,\ au,\ (2-a)u)\\
&\qquad(0\leqq s\leqq1,\ 0\leqq u\leqq1).
\end{aligned} OA + s AB OP + u PQ = ( s , s , 1 − s ) , = ( a − a u , a u , ( 2 − a ) u ) ( 0 ≦ s ≦ 1 , 0 ≦ u ≦ 1 ) . 2点が一致する条件は、
s = a − a u , s = a u , 1 − s = ( 2 − a ) u . s=a-au,\qquad s=au,\qquad 1-s=(2-a)u. s = a − a u , s = a u , 1 − s = ( 2 − a ) u . はじめの2式から a ( 1 − 2 u ) = 0 a(1-2u)=0 a ( 1 − 2 u ) = 0
a = 0 a=0 a = 0 P ( 0 , 0 , 0 ) \mathrm{P}(0,0,0) P ( 0 , 0 , 0 ) Q ( 0 , 0 , 2 ) \mathrm{Q}(0,0,2) Q ( 0 , 0 , 2 ) A ( 0 , 0 , 1 ) \mathrm{A}(0,0,1) A ( 0 , 0 , 1 ) P Q \mathrm{PQ} PQ P , A , Q \mathrm{P},\mathrm{A},\mathrm{Q} P , A , Q A P B Q \mathrm{APBQ} APBQ
a ≠ 0 a\ne0 a = 0 u = 1 2 u=\dfrac12 u = 2 1 s = a 2 s=\dfrac a2 s = 2 a 1 − a 2 = 2 − a 2 1-\dfrac a2=\dfrac{2-a}{2} 1 − 2 a = 2 2 − a 0 ≦ a 2 ≦ 1 0\leqq\dfrac a2\leqq1 0 ≦ 2 a ≦ 1 0 < a ≦ 2 0< a\leqq2 0 < a ≦ 2
ただし a = 2 a=2 a = 2 s = 1 s=1 s = 1 B \mathrm{B} B P ( 2 , 0 , 0 ) \mathrm{P}(2,0,0) P ( 2 , 0 , 0 ) Q ( 0 , 2 , 0 ) \mathrm{Q}(0,2,0) Q ( 0 , 2 , 0 ) B ( 1 , 1 , 0 ) \mathrm{B}(1,1,0) B ( 1 , 1 , 0 ) P Q \mathrm{PQ} PQ
0 < a < 2 0< a<2 0 < a < 2 M ( a 2 , a 2 , 1 − a 2 ) \mathrm{M}\left(\dfrac a2,\ \dfrac a2,\ 1-\dfrac a2\right) M ( 2 a , 2 a , 1 − 2 a ) A B \mathrm{AB} AB P Q \mathrm{PQ} PQ P \mathrm{P} P A B \mathrm{AB} AB A B \mathrm{AB} AB P Q \mathrm{PQ} PQ A , P , B , Q \mathrm{A},\mathrm{P},\mathrm{B},\mathrm{Q} A , P , B , Q
以上より、求める条件は 0 < a < 2 0< a<2 0 < a < 2
（3）
A B → \overrightarrow{\mathrm{AB}} AB P Q → \overrightarrow{\mathrm{PQ}} PQ θ \theta θ 0 < θ < π 0<\theta<\pi 0 < θ < π A B \mathrm{AB} AB A B P \mathrm{ABP} ABP A B Q \mathrm{ABQ} ABQ M \mathrm{M} M A B \mathrm{AB} AB P \mathrm{P} P P M sin θ \mathrm{PM}\sin\theta PM sin θ Q \mathrm{Q} Q Q M sin θ \mathrm{QM}\sin\theta QM sin θ S S S
S = 1 2 A B ⋅ P M sin θ + 1 2 A B ⋅ Q M sin θ = 1 2 A B ⋅ P Q sin θ . \begin{aligned}
S&=\frac12\mathrm{AB}\cdot\mathrm{PM}\sin\theta+\frac12\mathrm{AB}\cdot\mathrm{QM}\sin\theta\\
&=\frac12\mathrm{AB}\cdot\mathrm{PQ}\sin\theta.
\end{aligned} S = 2 1 AB ⋅ PM sin θ + 2 1 AB ⋅ QM sin θ = 2 1 AB ⋅ PQ sin θ . sin θ = 1 − cos 2 θ \sin\theta=\sqrt{1-\cos^2\theta} sin θ = 1 − cos 2 θ
S = 1 2 ∣ A B → ∣ 2 ∣ P Q → ∣ 2 − ( A B → ⋅ P Q → ) 2 . S=\frac12\sqrt{|\overrightarrow{\mathrm{AB}}|^2|\overrightarrow{\mathrm{PQ}}|^2-(\overrightarrow{\mathrm{AB}}\cdot\overrightarrow{\mathrm{PQ}})^2}. S = 2 1 ∣ AB ∣ 2 ∣ PQ ∣ 2 − ( AB ⋅ PQ ) 2 . A B → = ( 1 , 1 , − 1 ) \overrightarrow{\mathrm{AB}}=(1,1,-1) AB = ( 1 , 1 , − 1 ) P Q → = ( − a , a , 2 − a ) \overrightarrow{\mathrm{PQ}}=(-a,\ a,\ 2-a) PQ = ( − a , a , 2 − a )
∣ A B → ∣ 2 = 3 , ∣ P Q → ∣ 2 = a 2 + a 2 + ( 2 − a ) 2 = 3 a 2 − 4 a + 4 , A B → ⋅ P Q → = − a + a − ( 2 − a ) = a − 2. \begin{aligned}
|\overrightarrow{\mathrm{AB}}|^2&=3,\\
|\overrightarrow{\mathrm{PQ}}|^2&=a^2+a^2+(2-a)^2=3a^2-4a+4,\\
\overrightarrow{\mathrm{AB}}\cdot\overrightarrow{\mathrm{PQ}}&=-a+a-(2-a)=a-2.
\end{aligned} ∣ AB ∣ 2 ∣ PQ ∣ 2 AB ⋅ PQ = 3 , = a 2 + a 2 + ( 2 − a ) 2 = 3 a 2 − 4 a + 4 , = − a + a − ( 2 − a ) = a − 2. よって、
S = 1 2 3 ( 3 a 2 − 4 a + 4 ) − ( a − 2 ) 2 = 1 2 8 a 2 − 8 a + 8 = 2 a 2 − 2 a + 2 . \begin{aligned}
S&=\frac12\sqrt{3(3a^2-4a+4)-(a-2)^2}\\
&=\frac12\sqrt{8a^2-8a+8}\\
&=\sqrt{2a^2-2a+2}.
\end{aligned} S = 2 1 3 ( 3 a 2 − 4 a + 4 ) − ( a − 2 ) 2 = 2 1 8 a 2 − 8 a + 8 = 2 a 2 − 2 a + 2 . 補足 別の方法と特別な場合で確かめる 三角形 A B P \mathrm{ABP} ABP A B → × A P → = ( − 1 , 1 − a , − a ) \overrightarrow{\mathrm{AB}}\times\overrightarrow{\mathrm{AP}}=(-1,\ 1-a,\ -a) AB × AP = ( − 1 , 1 − a , − a ) 1 2 2 a 2 − 2 a + 2 \dfrac12\sqrt{2a^2-2a+2} 2 1 2 a 2 − 2 a + 2 M \mathrm{M} M P Q \mathrm{PQ} PQ A B Q \mathrm{ABQ} ABQ 2 a 2 − 2 a + 2 \sqrt{2a^2-2a+2} 2 a 2 − 2 a + 2
また、a = 1 a=1 a = 1 P ( 1 , 0 , 0 ) \mathrm{P}(1,0,0) P ( 1 , 0 , 0 ) Q ( 0 , 1 , 1 ) \mathrm{Q}(0,1,1) Q ( 0 , 1 , 1 ) A P → = ( 1 , 0 , − 1 ) \overrightarrow{\mathrm{AP}}=(1,0,-1) AP = ( 1 , 0 , − 1 ) P B → = ( 0 , 1 , 0 ) \overrightarrow{\mathrm{PB}}=(0,1,0) PB = ( 0 , 1 , 0 ) 2 \sqrt2 2 1 1 1 2 \sqrt2 2 a = 1 a=1 a = 1 2 \sqrt2 2
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