题目
Problem
Relative to the origin O O O A A A B B B C C C
O A → = 4 i − 2 j , O B → = 2 i + 8 j + 4 k , O C → = − 2 i + 6 k . \begin{align*}
\overrightarrow{OA}=&\,4\mathbf{i}-2\mathbf{j},\quad \overrightarrow{OB}=2\mathbf{i}+8\mathbf{j}+4\mathbf{k},\\
\overrightarrow{OC}=&\,-2\mathbf{i}+6\mathbf{k}.
\end{align*} O A = O C = 4 i − 2 j , O B = 2 i + 8 j + 4 k , − 2 i + 6 k . The midpoint of A B AB A B M M M
(a) Find the vectors M B → \overrightarrow{MB} M B M C → \overrightarrow{MC} M C
(2)
(b) Calculate the exact value of the cosine of angle C M B CMB C M B
(3)
(c) Hence or otherwise find the exact area of triangle A B C ABC A B C
(4)
题目中文翻译
相对于原点 O O O A A A B B B C C C
O A → = 4 i − 2 j , O B → = 2 i + 8 j + 4 k , O C → = − 2 i + 6 k . \begin{align*}
\overrightarrow{OA}=&\,4\mathbf{i}-2\mathbf{j},\quad \overrightarrow{OB}=2\mathbf{i}+8\mathbf{j}+4\mathbf{k},\\
\overrightarrow{OC}=&\,-2\mathbf{i}+6\mathbf{k}.
\end{align*} O A = O C = 4 i − 2 j , O B = 2 i + 8 j + 4 k , − 2 i + 6 k . 如图所示，A B AB A B M M M
(a) 求向量 M B → \overrightarrow{MB} M B M C → \overrightarrow{MC} M C
(2)
(b) 计算角 C M B CMB C M B
(3)
(c) 由此或使用其他方法，求三角形 A B C ABC A B C
(4)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。
解答
(a)
解法一
思路
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先求中点的位置向量，再分别用终点减起点。
答题过程
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Since M M M A B AB A B
O M → = 1 2 [ ( 4 , − 2 , 0 ) + ( 2 , 8 , 4 ) ] = ( 3 , 3 , 2 ) . \overrightarrow{OM}
=\frac12\big[(4,-2,0)+(2,8,4)\big]=(3,3,2). O M = 2 1 [ ( 4 , − 2 , 0 ) + ( 2 , 8 , 4 ) ] = ( 3 , 3 , 2 ) . Hence
M B → = ( 2 , 8 , 4 ) − ( 3 , 3 , 2 ) = ( − 1 , 5 , 2 ) \boxed{\overrightarrow{MB}
=(2,8,4)-(3,3,2)=(-1,5,2)} M B = ( 2 , 8 , 4 ) − ( 3 , 3 , 2 ) = ( − 1 , 5 , 2 ) and
M C → = ( − 2 , 0 , 6 ) − ( 3 , 3 , 2 ) = ( − 5 , − 3 , 4 ) . \boxed{\overrightarrow{MC}
=(-2,0,6)-(3,3,2)=(-5,-3,4)}. M C = ( − 2 , 0 , 6 ) − ( 3 , 3 , 2 ) = ( − 5 , − 3 , 4 ) .
(b)
解法一
思路
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用从同一顶点出发的两个向量求点积。点积为负，夹角为钝角，余弦应保留负号。
答题过程
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Let θ = ∠ C M B \theta=\angle CMB θ = ∠ C M B
M B → ⋅ M C → = ( − 1 ) ( − 5 ) + 5 ( − 3 ) + 2 ( 4 ) = − 2. \overrightarrow{MB}\cdot\overrightarrow{MC}
=(-1)(-5)+5(-3)+2(4)=-2. M B ⋅ M C = ( − 1 ) ( − 5 ) + 5 ( − 3 ) + 2 ( 4 ) = − 2. Also,
M B = 1 + 25 + 4 = 30 , M C = 25 + 9 + 16 = 50 . \begin{align*}
MB=&\,\sqrt{1+25+4}=\sqrt{30},\\[4mm]
MC=&\,\sqrt{25+9+16}=\sqrt{50}.
\end{align*} M B = M C = 1 + 25 + 4 = 30 , 25 + 9 + 16 = 50 . Therefore
cos θ = − 2 30 50 = − 1 5 15 . \boxed{\cos\theta
=\frac{-2}{\sqrt{30}\sqrt{50}}
=-\frac1{5\sqrt{15}}}. cos θ = 30 50 − 2 = − 5 15 1 .
解法二
思路
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也可先求三边长度，用余弦定理。角的对边是 B C BC B C B B B
答题过程
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The side lengths in triangle B M C BMC B M C
M B 2 = 30 , M C 2 = 50. MB^2=30,\qquad MC^2=50. M B 2 = 30 , M C 2 = 50. Also B C → = ( − 4 , − 8 , 2 ) \overrightarrow{BC}=(-4,-8,2) B C = ( − 4 , − 8 , 2 ) B C 2 = 16 + 64 + 4 = 84 BC^2=16+64+4=84 B C 2 = 16 + 64 + 4 = 84
By the cosine rule,
cos ∠ C M B = 30 + 50 − 84 2 30 50 = − 1 5 15 . \boxed{\cos\angle CMB
=\frac{30+50-84}{2\sqrt{30}\sqrt{50}}
=-\frac1{5\sqrt{15}}}. cos ∠ C M B = 2 30 50 30 + 50 − 84 = − 5 15 1 .
(c)
解法一
思路
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承接(b)求正弦，再求三角形 C M B CMB C M B
答题过程
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From (b), cos 2 θ = 1 / 375 \cos^2\theta=1/375 cos 2 θ = 1/375 θ \theta θ
sin θ = 1 − 1 375 = 374 375 . \sin\theta=\sqrt{1-\frac1{375}}
=\sqrt{\frac{374}{375}}. sin θ = 1 − 375 1 = 375 374 . The triangles C M A CMA C M A C M B CMB C M B M A = M B MA=MB M A = M B
Area ( A B C ) = 2 Area ( C M B ) = 2 ( 1 2 M B ⋅ M C sin θ ) = 30 50 374 375 = 2 374 . \begin{align*}
\operatorname{Area}(ABC)
=&\,2\operatorname{Area}(CMB)\\
=&\,2\bigg(\frac12MB\cdot MC\sin\theta\bigg)\\
=&\,\sqrt{30}\sqrt{50}\sqrt{\frac{374}{375}}\\
=&\,\boxed{2\sqrt{374}}.
\end{align*} Area ( A B C ) = = = = 2 Area ( C M B ) 2 ( 2 1 M B ⋅ M C sin θ ) 30 50 375 374 2 374 .
解法二
思路
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直接用原三角形也可求面积，先由顶点 A A A
答题过程
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The two sides from A A A
A B → = ( − 2 , 10 , 4 ) , A C → = ( − 6 , 2 , 6 ) . \overrightarrow{AB}=(-2,10,4),\qquad
\overrightarrow{AC}=(-6,2,6). A B = ( − 2 , 10 , 4 ) , A C = ( − 6 , 2 , 6 ) . Thus
A B 2 = 120 , A C 2 = 76 , A B → ⋅ A C → = 12 + 20 + 24 = 56. \begin{align*}
AB^2=&\,120,\qquad AC^2=76,\\[4mm]
\overrightarrow{AB}\cdot\overrightarrow{AC}
=&\,12+20+24=56.
\end{align*} A B 2 = A B ⋅ A C = 120 , A C 2 = 76 , 12 + 20 + 24 = 56. Writing ϕ = ∠ B A C \phi=\angle BAC ϕ = ∠ B A C
cos ϕ = 56 120 76 . \cos\phi=\frac{56}{\sqrt{120}\sqrt{76}}. cos ϕ = 120 76 56 . Since 0 < ϕ < π 0<\phi<\pi 0 < ϕ < π
sin ϕ = 120 ( 76 ) − 56 2 120 76 = 4 374 120 76 . \begin{align*}
\sin\phi
=&\,\frac{\sqrt{120(76)-56^2}}{\sqrt{120}\sqrt{76}}\\[4mm]
=&\,\frac{4\sqrt{374}}{\sqrt{120}\sqrt{76}}.
\end{align*} sin ϕ = = 120 76 120 ( 76 ) − 5 6 2 120 76 4 374 . Therefore
Area ( A B C ) = 1 2 120 76 sin ϕ = 2 374 . \begin{align*}
\operatorname{Area}(ABC)
=&\,\frac12\sqrt{120}\sqrt{76}\sin\phi\\[4mm]
=&\,\boxed{2\sqrt{374}}.
\end{align*} Area ( A B C ) = = 2 1 120 76 sin ϕ 2 374 . 
