第４問
四面体 OABC について，OA ⊥ BC が成り立つための条件を考えよう。次の問いに答えよ。ただし， OA → = a ⃗ \overrightarrow{\text{OA}}=\vec{a} OA = a OB → = b ⃗ \overrightarrow{\text{OB}}=\vec{b} OB = b ∣ OC ∣ → = c ⃗ \overrightarrow{|\text{OC}|}=\vec{c} ∣ OC ∣ = c
(1)
O( 0 , 0 , 0 ) (0,0,0) ( 0 , 0 , 0 ) ( 1 , 1 , 0 ) (1,1,0) ( 1 , 1 , 0 ) ( 1 , 0 , 1 ) (1,0,1) ( 1 , 0 , 1 ) ( 0 , 1 , 1 ) (0,1,1) ( 0 , 1 , 1 ) a ⃗ ⋅ b ⃗ = ア \vec{a}\cdot\vec{b}=\boxed{\text{ ア }} a ⋅ b = ア
OA → ≠ 0 ⃗ \overrightarrow{\text{OA}}\not=\vec{0} OA = 0 BC → ≠ 0 ⃗ \overrightarrow{\text{BC}}\not=\vec{0} BC = 0 OA → ⋅ BC → = イ \overrightarrow{\text{OA}}\cdot\overrightarrow{\text{BC}}=\boxed{\text{ イ }} OA ⋅ BC = イ
正解と解説
ア \boxed{\text{ ア }} ア イ \boxed{\text{ イ }} イ 1 1 1 0 0 0
a ⃗ ⋅ b ⃗ = 1 ⋅ 1 + 1 ⋅ 0 + 0 ⋅ 1 = 1 \vec{a}\cdot\vec{b}=1\cdot1+1\cdot0+0\cdot1=1 a ⋅ b = 1 ⋅ 1 + 1 ⋅ 0 + 0 ⋅ 1 = 1
また
OA → ⋅ BC → = ( 1 , 1 , 0 ) ⋅ ( 0 − 1 , 1 − 0 , 1 − 1 ) \overrightarrow{\text{OA}}\cdot\overrightarrow{\text{BC}}=(1,1,0)\cdot(0-1,1-0,1-1) OA ⋅ BC = ( 1 , 1 , 0 ) ⋅ ( 0 − 1 , 1 − 0 , 1 − 1 ) = 1 ⋅ ( − 1 ) + 1 ⋅ 1 + 0 ⋅ 0 = 0 =1\cdot(-1)+1\cdot1+0\cdot0=0 = 1 ⋅ ( − 1 ) + 1 ⋅ 1 + 0 ⋅ 0 = 0
(2)
四面体 OABC について，OA ⊥ BC となるための必要十分条件を，次の ⓪ 〜 ③ のうちから一つ選べ。 ウ \boxed{\text{ ウ }} ウ
⓪ a ⃗ ⋅ b ⃗ = b ⃗ ⋅ c ⃗ \vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{c} a ⋅ b = b ⋅ c a ⃗ ⋅ b ⃗ = a ⃗ ⋅ c ⃗ \vec{a}\cdot\vec{b}=\vec{a}\cdot\vec{c} a ⋅ b = a ⋅ c b ⃗ ⋅ c ⃗ = 0 \vec{b}\cdot\vec{c}=0 b ⋅ c = 0 ∣ a ⃗ ∣ 2 = b ⃗ ⋅ c ⃗ |\vec{a}|^2=\vec{b}\cdot\vec{c} ∣ a ∣ 2 = b ⋅ c
正解と解説
ウ \boxed{\text{ ウ }} ウ
OA → ⋅ BC → = 0 \overrightarrow{\text{OA}}\cdot\overrightarrow{\text{BC}}=0 OA ⋅ BC = 0 a ⃗ ⋅ ( c ⃗ − b ⃗ ) = 0 \vec{a}\cdot(\vec{c}-\vec{b})=0 a ⋅ ( c − b ) = 0 a ⃗ ⋅ c ⃗ − a ⃗ ⋅ b ⃗ = 0 \vec{a}\cdot\vec{c}-\vec{a}\cdot\vec{b}=0 a ⋅ c − a ⋅ b = 0 a ⃗ ⋅ b ⃗ = a ⃗ ⋅ c ⃗ \vec{a}\cdot\vec{b}=\vec{a}\cdot\vec{c} a ⋅ b = a ⋅ c
(3)
OA ⊥ BC が常に成り立つ四面体を，次の ⓪ 〜 ⑤ のうちから一つ選べ。 エ \boxed{\text{ エ }} エ
⓪ OA = OB かつ ∠AOB = ∠AOC であるような四面体 OABC
正解と解説
エ \boxed{\text{ エ }} エ
必要十分条件である a ⃗ ⋅ b ⃗ = a ⃗ ⋅ c ⃗ \vec{a}\cdot\vec{b}=\vec{a}\cdot\vec{c} a ⋅ b = a ⋅ c
⓪ OA=OB より ∣ a ⃗ ∣ = ∣ b ⃗ ∣ |\vec{a}|=|\vec{b}| ∣ a ∣ = ∣ b ∣ θ \theta θ
a ⃗ ⋅ b ⃗ = ∣ a ⃗ ∣ ∣ b ⃗ ∣ cos θ \vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos\theta a ⋅ b = ∣ a ∣∣ b ∣ cos θ cos θ = a ⃗ ⋅ b ⃗ ∣ a ⃗ ∣ ∣ b ⃗ ∣ ⋯ \cos\theta=\cfrac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}\cdots cos θ = ∣ a ∣∣ b ∣ a ⋅ b ⋯ a ⃗ ⋅ c ⃗ = ∣ a ⃗ ∣ ∣ c ⃗ ∣ cos θ ⋯ \vec{a}\cdot\vec{c}=|\vec{a}||\vec{c}|\cos\theta\cdots a ⋅ c = ∣ a ∣∣ c ∣ cos θ ⋯
①を②に代入
a ⃗ ⋅ c ⃗ = ∣ a ⃗ ∣ ∣ c ⃗ ∣ a ⃗ ⋅ b ⃗ ∣ a ⃗ ∣ ∣ b ⃗ ∣ \vec{a}\cdot\vec{c}=|\vec{a}||\vec{c}|\cfrac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|} a ⋅ c = ∣ a ∣∣ c ∣ ∣ a ∣∣ b ∣ a ⋅ b a ⃗ ⋅ c ⃗ = ∣ c ⃗ ∣ ∣ b ⃗ ∣ a ⃗ ⋅ b ⃗ \vec{a}\cdot\vec{c}=\cfrac{|\vec{c}|}{|\vec{b}|}\vec{a}\cdot\vec{b} a ⋅ c = ∣ b ∣ ∣ c ∣ a ⋅ b
上の式は，もしOB=OC，つまり ∣ b ⃗ ∣ = ∣ c ⃗ ∣ |\vec{b}|=|\vec{c}| ∣ b ∣ = ∣ c ∣ a ⃗ ⋅ b ⃗ = a ⃗ ⋅ c ⃗ \vec{a}\cdot\vec{b}=\vec{a}\cdot\vec{c} a ⋅ b = a ⋅ c
① 同様に選択肢の条件から
∣ a ⃗ ∣ = ∣ b ⃗ ∣ |\vec{a}|=|\vec{b}| ∣ a ∣ = ∣ b ∣ a ⃗ ⋅ b ⃗ = ∣ a ⃗ ∣ ∣ b ⃗ ∣ cos θ \vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos\theta a ⋅ b = ∣ a ∣∣ b ∣ cos θ cos θ = a ⃗ ⋅ b ⃗ ∣ a ⃗ ∣ ∣ b ⃗ ∣ ⋯ \cos\theta=\cfrac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}\cdots cos θ = ∣ a ∣∣ b ∣ a ⋅ b ⋯ b ⃗ ⋅ c ⃗ = ∣ b ⃗ ∣ ∣ c ⃗ ∣ cos θ ⋯ \vec{b}\cdot\vec{c}=|\vec{b}||\vec{c}|\cos\theta\cdots b ⋅ c = ∣ b ∣∣ c ∣ cos θ ⋯
①を②に代入
b ⃗ ⋅ c ⃗ = ∣ b ⃗ ∣ ∣ c ⃗ ∣ a ⃗ ⋅ b ⃗ ∣ a ⃗ ∣ ∣ b ⃗ ∣ \vec{b}\cdot\vec{c}=|\vec{b}||\vec{c}|\cfrac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|} b ⋅ c = ∣ b ∣∣ c ∣ ∣ a ∣∣ b ∣ a ⋅ b b ⃗ ⋅ c ⃗ = ∣ c ⃗ ∣ ∣ a ⃗ ∣ a ⃗ ⋅ b ⃗ \vec{b}\cdot\vec{c}=\cfrac{|\vec{c}|}{|\vec{a}|}\vec{a}\cdot\vec{b} b ⋅ c = ∣ a ∣ ∣ c ∣ a ⋅ b
これより a ⃗ ⋅ b ⃗ = a ⃗ ⋅ c ⃗ \vec{a}\cdot\vec{b}=\vec{a}\cdot\vec{c} a ⋅ b = a ⋅ c
②
∣ b ⃗ ∣ = ∣ c ⃗ ∣ |\vec{b}|=|\vec{c}| ∣ b ∣ = ∣ c ∣ a ⃗ ⋅ b ⃗ = ∣ a ⃗ ∣ ∣ b ⃗ ∣ cos θ \vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos\theta a ⋅ b = ∣ a ∣∣ b ∣ cos θ cos θ = a ⃗ ⋅ b ⃗ ∣ a ⃗ ∣ ∣ b ⃗ ∣ ⋯ \cos\theta=\cfrac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}\cdots cos θ = ∣ a ∣∣ b ∣ a ⋅ b ⋯ a ⃗ ⋅ c ⃗ = ∣ a ⃗ ∣ ∣ c ⃗ ∣ cos θ ⋯ \vec{a}\cdot\vec{c}=|\vec{a}||\vec{c}|\cos\theta\cdots a ⋅ c = ∣ a ∣∣ c ∣ cos θ ⋯
①を②に代入
a ⃗ ⋅ c ⃗ = ∣ a ⃗ ∣ ∣ c ⃗ ∣ a ⃗ ⋅ b ⃗ ∣ a ⃗ ∣ ∣ b ⃗ ∣ \vec{a}\cdot\vec{c}=|\vec{a}||\vec{c}|\cfrac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|} a ⋅ c = ∣ a ∣∣ c ∣ ∣ a ∣∣ b ∣ a ⋅ b a ⃗ ⋅ c ⃗ = ∣ c ⃗ ∣ ∣ b ⃗ ∣ a ⃗ ⋅ b ⃗ \vec{a}\cdot\vec{c}=\cfrac{|\vec{c}|}{|\vec{b}|}\vec{a}\cdot\vec{b} a ⋅ c = ∣ b ∣ ∣ c ∣ a ⋅ b
∣ b ⃗ ∣ = ∣ c ⃗ ∣ |\vec{b}|=|\vec{c}| ∣ b ∣ = ∣ c ∣
a ⃗ ⋅ c ⃗ = a ⃗ ⋅ b ⃗ \vec{a}\cdot\vec{c}=\vec{a}\cdot\vec{b} a ⋅ c = a ⋅ b
したがって，適する。
③ 上と同様にして
b ⃗ ⋅ c ⃗ = ∣ b ⃗ ∣ ∣ c ⃗ ∣ a ⃗ ⋅ c ⃗ \vec{b}\cdot\vec{c}=\cfrac{|\vec{b}|}{|\vec{c}|}\vec{a}\cdot\vec{c} b ⋅ c = ∣ c ∣ ∣ b ∣ a ⋅ c
したがって，不適。
④ 上と同様にして
b ⃗ ⋅ c ⃗ = ∣ b ⃗ ∣ ∣ a ⃗ ∣ a ⃗ ⋅ c ⃗ \vec{b}\cdot\vec{c}=\cfrac{|\vec{b}|}{|\vec{a}|}\vec{a}\cdot\vec{c} b ⋅ c = ∣ a ∣ ∣ b ∣ a ⋅ c
したがって，不適。
⑤ 上と同様にして
b ⃗ ⋅ c ⃗ = ∣ c ⃗ ∣ ∣ a ⃗ ∣ a ⃗ ⋅ b ⃗ \vec{b}\cdot\vec{c}=\cfrac{|\vec{c}|}{|\vec{a}|}\vec{a}\cdot\vec{b} b ⋅ c = ∣ a ∣ ∣ c ∣ a ⋅ b
したがって，不適。
(4) OC = OB = AB = AC を満たす四面体 OABC について，OA ⊥ BC が成り立つことを下のように証明した。
【証明】
線分 OA の中点を D とする。
BD → = 1 2 ( オ + カ ) \overrightarrow{\text{BD}}=\cfrac{1}{2}\Big(\boxed{\text{ オ }}+\boxed{\text{ カ }}\Big) BD = 2 1 ( オ + カ ) OA → = オ − カ \overrightarrow{\text{OA}}=\boxed{\text{ オ }}-\boxed{\text{ カ }} OA = オ − カ
BD → ⋅ OA → = 1 2 { ∣ オ ∣ 2 − ∣ カ ∣ 2 } \overrightarrow{\text{BD}}\cdot\overrightarrow{\text{OA}}=\cfrac{1}{2}\Big\{|\boxed{\text{ オ }}|^2-|\boxed{\text{ カ }}|^2\Big\} BD ⋅ OA = 2 1 { ∣ オ ∣ 2 − ∣ カ ∣ 2 }
また，∣ オ ∣ = ∣ カ ∣ |\boxed{\text{ オ }}|=|\boxed{\text{ カ }}| ∣ オ ∣ = ∣ カ ∣ OA → ⋅ BD → = 0 \overrightarrow{\text{OA}}\cdot\overrightarrow{\text{BD}}=0 OA ⋅ BD = 0
同様に， キ \boxed{\text{ キ }} キ OA → ⋅ CD → = 0 \overrightarrow{\text{OA}}\cdot\overrightarrow{\text{CD}}=0 OA ⋅ CD = 0
このことから OA → ≠ 0 ⃗ \overrightarrow{\text{OA}}\not=\vec{0} OA = 0 BC → ≠ 0 ⃗ \overrightarrow{\text{BC}}\not=\vec{0} BC = 0 OA → ⋅ BC → = OA → ⋅ ( BD → − CD → ) = 0 \overrightarrow{\text{OA}}\cdot\overrightarrow{\text{BC}}=\overrightarrow{\text{OA}}\cdot(\overrightarrow{\text{BD}}-\overrightarrow{\text{CD}})=0 OA ⋅ BC = OA ⋅ ( BD − CD ) = 0
(4)(i)
オ \boxed{\text{ オ }} オ カ \boxed{\text{ カ }} カ
⓪ BA → \overrightarrow{\text{BA}} BA BC → \overrightarrow{\text{BC}} BC BD → \overrightarrow{\text{BD}} BD BO → \overrightarrow{\text{BO}} BO
正解と解説
オ \boxed{\text{ オ }} オ カ \boxed{\text{ カ }} カ
点Bを始点として考えると，点D は OA の中点だから
BD → = 1 2 ( BA → + BO → ) \overrightarrow{\text{BD}}=\cfrac{1}{2}(\overrightarrow{\text{BA}}+\overrightarrow{\text{BO}}) BD = 2 1 ( BA + BO )
また
OA → = BA → − BO → \overrightarrow{\text{OA}}=\overrightarrow{\text{BA}}-\overrightarrow{\text{BO}} OA = BA − BO
これより
BD → ⋅ OA → = 1 2 ( BA → + BO → ) ⋅ ( BA → − BO → ) \overrightarrow{\text{BD}}\cdot\overrightarrow{\text{OA}}=\cfrac{1}{2}(\overrightarrow{\text{BA}}+\overrightarrow{\text{BO}})\cdot(\overrightarrow{\text{BA}}-\overrightarrow{\text{BO}}) BD ⋅ OA = 2 1 ( BA + BO ) ⋅ ( BA − BO )
= 1 2 ( ∣ BA → ∣ 2 − ∣ BO → ∣ 2 ) =\cfrac{1}{2}(|\overrightarrow{\text{BA}}|^2-|\overrightarrow{\text{BO}}|^2) = 2 1 ( ∣ BA ∣ 2 − ∣ BO ∣ 2 )
∣ BA → ∣ = ∣ BO → ∣ |\overrightarrow{\text{BA}}|=|\overrightarrow{\text{BO}}| ∣ BA ∣ = ∣ BO ∣
OA → ⋅ BD → = 0 \overrightarrow{\text{OA}}\cdot\overrightarrow{\text{BD}}=0 OA ⋅ BD = 0
(4)(ii)
キ \boxed{\text{ キ }} キ
⓪ ∣ CO → ∣ = ∣ CB → ∣ |\overrightarrow{\text{CO}}|=|\overrightarrow{\text{CB}}| ∣ CO ∣ = ∣ CB ∣ ∣ CO → ∣ = ∣ CA → ∣ |\overrightarrow{\text{CO}}|=|\overrightarrow{\text{CA}}| ∣ CO ∣ = ∣ CA ∣ ∣ OB → ∣ = ∣ OC → ∣ |\overrightarrow{\text{OB}}|=|\overrightarrow{\text{OC}}| ∣ OB ∣ = ∣ OC ∣ ∣ AB → ∣ = ∣ AC → ∣ |\overrightarrow{\text{AB}}|=|\overrightarrow{\text{AC}}| ∣ AB ∣ = ∣ AC ∣ ∣ BO → ∣ = ∣ BA → ∣ |\overrightarrow{\text{BO}}|=|\overrightarrow{\text{BA}}| ∣ BO ∣ = ∣ BA ∣
正解と解説
キ \boxed{\text{ キ }} キ
点 C を始点として式を作るとよい。
OA → = CA → − CO → \overrightarrow{\text{OA}}=\overrightarrow{\text{CA}}-\overrightarrow{\text{CO}} OA = CA − CO CD → = 1 2 ( CA → + CO → ) \overrightarrow{\text{CD}}=\cfrac{1}{2}(\overrightarrow{\text{CA}}+\overrightarrow{\text{CO}}) CD = 2 1 ( CA + CO )
これより
OA → ⋅ CD → = 1 2 ( CA → − CO → ) ( CA → + CO → ) \overrightarrow{\text{OA}}\cdot\overrightarrow{\text{CD}}=\cfrac{1}{2}(\overrightarrow{\text{CA}}-\overrightarrow{\text{CO}})(\overrightarrow{\text{CA}}+\overrightarrow{\text{CO}}) OA ⋅ CD = 2 1 ( CA − CO ) ( CA + CO ) = 1 2 ( ∣ CA → ∣ 2 − ∣ CO → ∣ 2 ) =\cfrac{1}{2}(|\overrightarrow{\text{CA}}|^2-|\overrightarrow{\text{CO}}|^2) = 2 1 ( ∣ CA ∣ 2 − ∣ CO ∣ 2 )
∣ CO → ∣ = ∣ CA → ∣ |\overrightarrow{\text{CO}}|=|\overrightarrow{\text{CA}}| ∣ CO ∣ = ∣ CA ∣ OA → ⋅ CD → = 0 \overrightarrow{\text{OA}}\cdot\overrightarrow{\text{CD}}=0 OA ⋅ CD = 0
(5)
(4)の証明は，OC = OB = AB = AC のすべての等号が成り立つことを条件として用いているわけではない。このことに注意して，OA ⊥ BC が成り立つ四面体を，次の ⓪ 〜 ③ のうちから一つ選べ。 ク \boxed{\text{ ク }} ク
⓪ OC = AC かつ OB = AB かつ OB ≠ \not= = ≠ \not= = ≠ \not= = ≠ \not= =
正解と解説
ク \boxed{\text{ ク }} ク
(4)の証明に基づいて△OABと△OACについて展開図を描くと以下のようになる。
① 選択肢の条件に基づいて図を描くと，OAとBCは垂直ではない。したがって，不適。
② 上と同様に不適。
③ 上と同様に不適。
※第 5 問については省略
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