[출제 의도] 벡터로 표현된 식을 이용하여 두 벡터의 내적의 최댓값과 최솟값을 구할 수 있는가?
조건 (가)에서
P Q ⃗ ∙ A B ⃗ {\vec{{\mathrm{P}} \mathit{\mathrm{Q}}}} \bullet {\vec{{\mathrm{A}} \mathit{\mathrm{B}}}} P Q ∙ A B = 0 = 0 = 0 P Q ⃗ ∙ A D ⃗ {\vec{{\mathrm{P}} \mathit{\mathrm{Q}}}} \bullet {\vec{{\mathrm{A}} \mathit{\mathrm{D}}}} P Q ∙ A D = 0 = 0 = 0 P Q ⃗ ∙ A B ⃗ {\vec{{\mathrm{P}} \mathit{\mathrm{Q}}}} \bullet {\vec{{\mathrm{A}} \mathit{\mathrm{B}}}} P Q ∙ A B = 0 = 0 = 0 P Q ‾ ⊥ A B ‾ \displaystyle \overline{{\mathrm{P}} \mathit{\mathrm{Q}}} \perp \overline{{\mathrm{A}} \mathit{\mathrm{B}}} P Q ⊥ A B O B ⃗ ∙ O P ⃗ ≥ 0 {\vec{{\mathrm{O}} \mathit{\mathrm{B}}}} \bullet {\vec{{\mathrm{O}} \mathit{\mathrm{P}}}} \geq 0 O B ∙ O P ≥ 0 O B ⃗ ∙ O Q ⃗ ≤ 0 {\vec{{\mathrm{O}} \mathit{\mathrm{B}}}} \bullet {\vec{{\mathrm{O}} \mathit{\mathrm{Q}}}} \leq 0 O B ∙ O Q ≤ 0 P \mathrm{P} P A B \mathrm{AB} AB Q \mathrm{Q} Q C D \mathrm{CD} CD
점 P \mathrm{P} P P ( a , 2 − a ) \mathrm{P} \mathit{\left( a , 2 - a \right)} P ( a , 2 − a ) ( 0 ≤ a ≤ 2 ) \left( 0 \leq a \leq 2 \right) ( 0 ≤ a ≤ 2 ) Q \mathrm{Q} Q Q ( a − 2 , a ) \mathrm{Q} \mathit{\left( a - 2 , a \right)} Q ( a − 2 , a ) O A ⃗ ∙ O P ⃗ ≥ − 2 {\vec{{\mathrm{O}} \mathit{\mathrm{A}}}} \bullet {\vec{{\mathrm{O}} \mathit{\mathrm{P}}}} \geq - 2 O A ∙ O P ≥ − 2 ( 2 , 0 ) ∙ ( a , 2 − a ) = 2 a ≥ − 2 \left( 2 , 0 \right) \bullet \left( a , 2 - a \right) = 2 a \geq - 2 ( 2 , 0 ) ∙ ( a , 2 − a ) = 2 a ≥ − 2 a ≥ − 1 a \geq - 1 a ≥ − 1 ⋯ \cdots ⋯ O B ⃗ ∙ O P ⃗ ≥ 0 {\vec{\mathrm{O} \mathit{\mathrm{B}}}} \bullet {\vec{\mathrm{O} \mathit{\mathrm{P}}}} \geq 0 OB ∙ OP ≥ 0 ( 0 , 2 ) ∙ ( a , 2 − a ) \left( 0 , 2 \right) \bullet \left( a , 2 - a \right) ( 0 , 2 ) ∙ ( a , 2 − a ) = 2 ( 2 − a ) ≥ 0 = 2 \left( 2 - a \right) \geq 0 = 2 ( 2 − a ) ≥ 0 a ≤ 2 a \leq 2 a ≤ 2 ⋯ \cdots ⋯ O A ⃗ ∙ O Q ⃗ ≥ − 2 {\vec{{\mathrm{O}} \mathit{\mathrm{A}}}} \bullet {\vec{{\mathrm{O}} \mathit{\mathrm{Q}}}} \geq - 2 O A ∙ O Q ≥ − 2 ( 2 , 0 ) ∙ ( a − 2 , − a ) \left( 2 , 0 \right) \bullet \left( a - 2 , - a \right) ( 2 , 0 ) ∙ ( a − 2 , − a ) = 2 ( a − 2 ) ≥ − 2 = 2 \left( a - 2 \right) \geq - 2 = 2 ( a − 2 ) ≥ − 2 a ≥ 1 a \geq 1 a ≥ 1 ⋯ \cdots ⋯ O B ⃗ ∙ O Q ⃗ ≤ 0 {\vec{{\mathrm{O}} \mathit{\mathrm{B}}}} \bullet {\vec{{\mathrm{O}} \mathit{\mathrm{Q}}}} \leq 0 O B ∙ O Q ≤ 0 ( 0 , 2 ) ∙ ( a − 2 , − a ) \left( 0 , 2 \right) \bullet \left( a - 2 , - a \right) ( 0 , 2 ) ∙ ( a − 2 , − a ) = − 2 a ≤ 0 = - 2 a \leq 0 = − 2 a ≤ 0 a ≥ 0 a \geq 0 a ≥ 0 ⋯ \cdots ⋯ 1 ≤ a ≤ 2 1 \leq a \leq 2 1 ≤ a ≤ 2 ⋯ \cdots ⋯ R ( 4 , 4 ) \mathrm{R} \left( 4 , 4 \right) R ( 4 , 4 ) R P ⃗ {\vec{\mathrm{R} \mathit{\mathrm{P}}}} RP = O P ⃗ − O R ⃗ = {\vec{\mathrm{OP}}} - {\vec{\mathrm{OR}}} = OP − OR = ( a , 2 − a ) − ( 4 , 4 ) = \left( a , 2 - a \right) - \left( 4 , 4 \right) = ( a , 2 − a ) − ( 4 , 4 ) = ( a − 4 , − a − 2 ) = \left( a - 4 , - a - 2 \right) = ( a − 4 , − a − 2 ) R Q ⃗ \mathrm{\vec{RQ}} RQ = O Q ⃗ − O R ⃗ \mathrm{=} \vec{OQ} - \vec{OR} = O Q − O R = ( a − 2 , − a ) − ( 4 , 4 ) = \left( a - 2 , - a \right) - \left( 4 , 4 \right) = ( a − 2 , − a ) − ( 4 , 4 ) = ( a − 6 , − a − 4 ) = \left( a - 6 , - a - 4 \right) = ( a − 6 , − a − 4 ) R P ⃗ ∙ R Q ⃗ {\vec{\mathrm{R} \mathit{\mathrm{P}}}} \bullet {\vec{\mathrm{R} \mathit{\mathrm{Q}}}} RP ∙ RQ = ( a − 4 , − a − 2 ) ∙ ( a − 6 , − a − 4 ) = \left( a - 4 , - a - 2 \right) \bullet \left( a - 6 , - a - 4 \right) = ( a − 4 , − a − 2 ) ∙ ( a − 6 , − a − 4 ) = ( a − 4 ) ( a − 6 ) + ( a + 2 ) ( a + 4 ) = \left( a - 4 \right) \left( a - 6 \right) + \left( a + 2 \right) \left( a + 4 \right) = ( a − 4 ) ( a − 6 ) + ( a + 2 ) ( a + 4 ) = 2 a 2 − 4 a + 32 = 2 a ^{2} - 4 a + 32 = 2 a 2 − 4 a + 32 = 2 ( a − 1 ) 2 + 30 = 2 \left( a - 1 \right) ^{2} + 30 = 2 ( a − 1 ) 2 + 30 30 ≤ R P ⃗ ∙ R Q ⃗ ≤ 32 30 \leq {\vec{\mathrm{R} \mathit{\mathrm{P}}}} \bullet {\vec{\mathrm{R} \mathit{\mathrm{Q}}}} \leq 32 30 ≤ RP ∙ RQ ≤ 32 P Q ⃗ ∙ A D ⃗ = 0 {\vec{{\mathrm{P}} \mathit{\mathrm{Q}}}} \bullet {\vec{{\mathrm{A}} \mathit{\mathrm{D}}}} = 0 P Q ∙ A D = 0 P Q ‾ ⊥ A D ‾ \displaystyle \overline{{\mathrm{P}} \mathit{\mathrm{Q}}} \perp \overline{{\mathrm{A}} \mathit{\mathrm{D}}} P Q ⊥ A D O B ⃗ ∙ O P ⃗ ≥ 0 {\vec{{\mathrm{O}} \mathit{\mathrm{B}}}} \bullet {\vec{{\mathrm{O}} \mathit{\mathrm{P}}}} \geq 0 O B ∙ O P ≥ 0 O B ⃗ ∙ O Q ⃗ ≤ 0 {\vec{{\mathrm{O}} \mathit{\mathrm{B}}}} \bullet {\vec{{\mathrm{O}} \mathit{\mathrm{Q}}}} \leq 0 O B ∙ O Q ≤ 0 P \mathrm{P} P B C \mathrm{BC} BC Q \mathrm{Q} Q A D \mathrm{AD} AD
점 P \mathrm{P} P P ( a , a + 2 ) \mathrm{P} ( \mathit{a} , a + 2 ) P ( a , a + 2 ) ( − 2 ≤ a ≤ 0 ) ( - 2 \leq a \leq 0 ) ( − 2 ≤ a ≤ 0 ) Q \mathrm{Q} Q Q ( a + 2 , a ) \mathrm{Q} ( \mathit{a} + 2 , a ) Q ( a + 2 , a ) O A ⃗ ∙ O P ⃗ ≥ − 2 \mathrm{\vec{OA}} \bullet \vec{OP} \geq - 2 OA ∙ O P ≥ − 2 ( 2 , 0 ) ∙ ( a , a + 2 ) ( 2 , 0 ) \bullet ( a , a + 2 ) ( 2 , 0 ) ∙ ( a , a + 2 ) = 2 a ≥ − 2 = 2 a \geq - 2 = 2 a ≥ − 2 a ≥ − 1 a \geq - 1 a ≥ − 1 ⋯ \cdots ⋯ O B ⃗ ∙ O P ⃗ ≥ 0 \mathrm{\vec{OB}} \bullet {\vec{OP}} \geq 0 OB ∙ O P ≥ 0 ( 0 , 2 ) ∙ ( a , a + 2 ) ( 0 , 2 ) \bullet ( a , a + 2 ) ( 0 , 2 ) ∙ ( a , a + 2 ) = 2 ( a + 2 ) ≥ 0 = 2 ( a + 2 ) \geq 0 = 2 ( a + 2 ) ≥ 0 a ≥ − 2 a \geq - 2 a ≥ − 2 ⋯ \cdots ⋯ O A ⃗ ∙ O Q ⃗ ≥ − 2 {\vec{{\mathrm{O}} \mathit{\mathrm{A}}}} \bullet {\vec{{\mathrm{O}} \mathit{\mathrm{Q}}}} \geq - 2 O A ∙ O Q ≥ − 2 ( 2 , 0 ) ∙ ( a + 2 , a ) \left( 2 , 0 \right) \bullet \left( a + 2 , a \right) ( 2 , 0 ) ∙ ( a + 2 , a ) = 2 ( a + 2 ) ≥ − 2 = 2 \left( a + 2 \right) \geq - 2 = 2 ( a + 2 ) ≥ − 2 a ≥ − 3 a \geq - 3 a ≥ − 3 ⋯ \cdots ⋯ O B ⃗ ∙ O Q ⃗ ≤ 0 {\vec{{\mathrm{O}} \mathit{\mathrm{B}}}} \bullet {\vec{{\mathrm{O}} \mathit{\mathrm{Q}}}} \leq 0 O B ∙ O Q ≤ 0 ( 0 , 2 ) ∙ ( a + 2 , a ) \left( 0 , 2 \right) \bullet \left( a + 2 , a \right) ( 0 , 2 ) ∙ ( a + 2 , a ) = 2 a ≤ 0 = 2 a \leq 0 = 2 a ≤ 0 a ≤ 0 a \leq 0 a ≤ 0 ⋯ \cdots ⋯ − 1 ≤ a ≤ 0 - 1 \leq a \leq 0 − 1 ≤ a ≤ 0 ⋯ \cdots ⋯ R ( 4 , 4 ) \mathrm{R} \left( 4 , 4 \right) R ( 4 , 4 ) R P ⃗ {\vec{{\mathrm{R}} \mathit{\mathrm{P}}}} R P = O P ⃗ − O R ⃗ = {\vec{{\mathrm{O}} \mathit{\mathrm{P}}}} - {\vec{{\mathrm{O}} \mathit{\mathrm{R}}}} = O P − O R = ( a + 2 , a ) − ( 4 , 4 ) = \left( a + 2 , a \right) - \left( 4 , 4 \right) = ( a + 2 , a ) − ( 4 , 4 ) = ( a − 2 , a − 4 ) = \left( a - 2 , a - 4 \right) = ( a − 2 , a − 4 ) R P ⃗ ∙ R Q ⃗ {\vec{{\mathrm{R}} \mathit{\mathrm{P}}}} \bullet {\vec{{\mathrm{R}} \mathit{\mathrm{Q}}}} R P ∙ R Q = ( a − 4 , a − 2 ) ∙ ( a − 2 , a − 4 ) = \left( a - 4 , a - 2 \right) \bullet \left( a - 2 , a - 4 \right) = ( a − 4 , a − 2 ) ∙ ( a − 2 , a − 4 ) = 2 ( a − 4 ) ( a − 2 ) = 2 \left( a - 4 \right) \left( a - 2 \right) = 2 ( a − 4 ) ( a − 2 ) = 2 ( a 2 − 6 a + 8 ) = 2 \left( a ^{2} - 6 a + 8 \right) = 2 ( a 2 − 6 a + 8 ) = 2 ( a − 3 ) 2 − 2 = 2 \left( a - 3 \right) ^{2} - 2 = 2 ( a − 3 ) 2 − 2 16 ≤ R P ⃗ ∙ R Q ⃗ ≤ 30 16 \leq {\vec{{\mathrm{R}} \mathit{\mathrm{P}}}} \bullet {\vec{{\mathrm{R}} \mathit{\mathrm{Q}}}} \leq 30 16 ≤ R P ∙ R Q ≤ 30 16 ≤ R P ⃗ ∙ R Q ⃗ ≤ 32 16 \leq {\vec{{\mathrm{R}} \mathit{\mathrm{P}}}} \bullet {\vec{{\mathrm{R}} \mathit{\mathrm{Q}}}} \leq 32 16 ≤ R P ∙ R Q ≤ 32 R P ⃗ ∙ R Q ⃗ {\vec{{\mathrm{R}} \mathit{\mathrm{P}}}} \bullet {\vec{{\mathrm{R}} \mathit{\mathrm{Q}}}} R P ∙ R Q M = 32 M = 32 M = 32 m = 16 m = 16 m = 16 M + m = 48 M + m = 48 M + m = 48 P \mathrm{P} P Q \mathrm{Q} Q
선분 P Q \mathrm{PQ} PQ N \mathrm{N} N N P ⃗ + N Q ⃗ = 0 {\vec{{\mathrm{N}} \mathit{\mathrm{P}}}} + {\vec{{\mathrm{N}} \mathit{\mathrm{Q}}}} = 0 N P + N Q = 0 N P ⃗ ∙ N Q ⃗ \mathrm{\vec{NP}} \bullet \vec{NQ} NP ∙ N Q = ∣ N P ⃗ ∣ ∣ N Q ⃗ ∣ cos π = \left| {\vec{\mathrm{N} \mathit{\mathrm{P}}}} \right| \left| {\vec{\mathrm{N} \mathit{\mathrm{Q}}}} \right| \cos \pi = NP NQ cos π = − N P ‾ 2 \displaystyle = - {\overline{\mathrm{N} \mathit{\mathrm{P}}}} ^{2} = − NP 2 R P ⃗ ∙ R Q ⃗ {\vec{{\mathrm{R}} \mathit{\mathrm{P}}}} \bullet {\vec{{\mathrm{R}} \mathit{\mathrm{Q}}}} R P ∙ R Q = ( R N ⃗ + N P ⃗ ) ∙ ( R N ⃗ + N Q ⃗ ) = \left( {\vec{{\mathrm{R}} \mathit{\mathrm{N}}}} + {\vec{{\mathrm{N}} \mathit{\mathrm{P}}}} \right) \bullet \left( {\vec{{\mathrm{R}} \mathit{\mathrm{N}}}} + {\vec{{\mathrm{N}} \mathit{\mathrm{Q}}}} \right) = ( R N + N P ) ∙ ( R N + N Q ) = R N ⃗ ∙ R N ⃗ + R N ⃗ ∙ N Q ⃗ + N P ⃗ ∙ R N ⃗ + N P ⃗ ∙ N Q ⃗ = {\vec{\mathrm{R} \mathit{\mathrm{N}}}} \bullet {\vec{\mathrm{R} \mathit{\mathrm{N}}}} + {\vec{\mathrm{R} \mathit{\mathrm{N}}}} \bullet {\vec{\mathrm{N} \mathit{\mathrm{Q}}}} + {\vec{\mathrm{N} \mathit{\mathrm{P}}}} \bullet {\vec{\mathrm{R} \mathit{\mathrm{N}}}} + {\vec{\mathrm{N} \mathit{\mathrm{P}}}} \bullet {\vec{\mathrm{N} \mathit{\mathrm{Q}}}} = RN ∙ RN + RN ∙ NQ + NP ∙ RN + NP ∙ NQ = ∣ R N ⃗ ∣ 2 + R N ⃗ ∙ ( N Q ⃗ + N P ⃗ ) + N P ⃗ ∙ N Q ⃗ = \left| {\vec{{\mathrm{R}} \mathit{\mathrm{N}}}} \right| ^{2} + {\vec{{\mathrm{R}} \mathit{\mathrm{N}}}} \bullet \left( {\vec{{\mathrm{N}} \mathit{\mathrm{Q}}}} + {\vec{{\mathrm{N}} \mathit{\mathrm{P}}}} \right) + {\vec{{\mathrm{N}} \mathit{\mathrm{P}}}} \bullet {\vec{{\mathrm{N}} \mathit{\mathrm{Q}}}} = R N 2 + R N ∙ ( N Q + N P ) + N P ∙ N Q = R N ‾ 2 + 0 − N P ‾ 2 \displaystyle = {\overline{\mathrm{R} \mathit{\mathrm{N}}}} ^{2} + 0 - {\overline{\mathrm{N} \mathit{\mathrm{P}}}} ^{2} = RN 2 + 0 − NP 2 P Q ‾ = 2 2 \displaystyle \overline{{\mathrm{P}} \mathit{\mathrm{Q}}} = 2 \sqrt{2} P Q = 2 2 N P ‾ = 1 2 P Q ‾ = 2 \displaystyle {\overline{{\mathrm{N}} \mathit{\mathrm{P}}}} = \frac{1}{2} {\overline{{\mathrm{P}} \mathit{\mathrm{Q}}}} = \sqrt{2} N P = 2 1 P Q = 2 R P ⃗ ∙ R Q ⃗ {\vec{{\mathrm{R}} \mathit{\mathrm{P}}}} \bullet {\vec{{\mathrm{R}} \mathit{\mathrm{Q}}}} R P ∙ R Q = R N ‾ 2 − 2 \displaystyle = {\overline{{\mathrm{R}} \mathit{\mathrm{N}}}} ^{2} - 2 = R N 2 − 2 R P ⃗ ∙ R Q ⃗ {\vec{{\mathrm{R}} \mathit{\mathrm{P}}}} \bullet {\vec{{\mathrm{R}} \mathit{\mathrm{Q}}}} R P ∙ R Q R N ‾ \displaystyle \overline{{\mathrm{R}} \mathit{\mathrm{N}}} R N R N ‾ \displaystyle \overline{{\mathrm{R}} \mathit{\mathrm{N}}} R N P \mathrm{P} P Q \mathrm{Q} Q ( 2 , 0 ) \left( 2 , 0 \right) ( 2 , 0 ) ( 0 , − 2 ) \left( 0 , - 2 \right) ( 0 , − 2 ) N ( 1 , − 1 ) {\mathrm{N}} \mathit{\left( 1 , - 1 \right)} N ( 1 , − 1 ) R N ‾ = ( 4 − 1 ) 2 + ( 4 + 1 ) 2 = 34 \displaystyle \overline{{\mathrm{R}} \mathit{\mathrm{N}}} = \sqrt{\left( 4 - 1 \right) ^{2} + \left( 4 + 1 \right) ^{2}} = \sqrt{34} R N = ( 4 − 1 ) 2 + ( 4 + 1 ) 2 = 34 R P ⃗ ∙ R Q ⃗ {\vec{{\mathrm{R}} \mathit{\mathrm{P}}}} \bullet {\vec{{\mathrm{R}} \mathit{\mathrm{Q}}}} R P ∙ R Q M = ( 34 ) 2 − 2 = 32 \displaystyle M = \left( \sqrt{34} \right) ^{2} - 2 = 32 M = ( 34 ) 2 − 2 = 32 R N ‾ \displaystyle \overline{{\mathrm{R}} \mathit{\mathrm{N}}} R N P \mathrm{P} P Q \mathrm{Q} Q ( 0 , 2 ) \left( 0 , 2 \right) ( 0 , 2 ) ( 2 , 0 ) \left( 2 , 0 \right) ( 2 , 0 ) N ( 1 , 1 ) \mathrm{N} \left( 1 , 1 \right) N ( 1 , 1 ) R N ‾ = ( 4 − 1 ) 2 + ( 4 − 1 ) 2 = 18 \displaystyle \overline{{\mathrm{R}} \mathit{\mathrm{N}}} = \sqrt{\left( 4 - 1 \right) ^{2} + \left( 4 - 1 \right) ^{2}} = \sqrt{18} R N = ( 4 − 1 ) 2 + ( 4 − 1 ) 2 = 18 R P ⃗ ∙ R Q ⃗ {\vec{{\mathrm{R}} \mathit{\mathrm{P}}}} \bullet {\vec{{\mathrm{R}} \mathit{\mathrm{Q}}}} R P ∙ R Q m = ( 18 ) 2 − 2 = 16 \displaystyle m = \left( \sqrt{18} \right) ^{2} - 2 = 16 m = ( 18 ) 2 − 2 = 16 M + m = 32 + 16 = 48 M + m = 32 + 16 = 48 M + m = 32 + 16 = 48
