题目
Problem
The points A A A B B B C C C
4 i + 3 j + 3 k , 2 i + 2 j + k , 5 i + 2 j + 2 k , 4\mathbf{i}+3\mathbf{j}+3\mathbf{k},\quad2\mathbf{i}+2\mathbf{j}+\mathbf{k},\quad5\mathbf{i}+2\mathbf{j}+2\mathbf{k}, 4 i + 3 j + 3 k , 2 i + 2 j + k , 5 i + 2 j + 2 k , respectively, relative to the origin O O O
(a) Find the equation of the plane A B C ABC A B C a x + b y + c z = d ax+by+cz=d a x + b y + cz = d
(5)
The point D D D 2 i + j + 3 k 2\mathbf{i}+\mathbf{j}+3\mathbf{k} 2 i + j + 3 k
(b) Find the acute angle between the line O D OD O D A B C ABC A B C
(3)
(c) Find the shortest distance between the lines A B AB A B C D CD C D
(5)
题目中文翻译
点 A A A B B B C C C O O O
4 i + 3 j + 3 k , 2 i + 2 j + k , 5 i + 2 j + 2 k . 4\mathbf{i}+3\mathbf{j}+3\mathbf{k},\quad2\mathbf{i}+2\mathbf{j}+\mathbf{k},\quad5\mathbf{i}+2\mathbf{j}+2\mathbf{k}. 4 i + 3 j + 3 k , 2 i + 2 j + k , 5 i + 2 j + 2 k . (a) 求平面 A B C ABC A B C a x + b y + c z = d ax+by+cz=d a x + b y + cz = d
(5)
点 D D D 2 i + j + 3 k 2\mathbf{i}+\mathbf{j}+3\mathbf{k} 2 i + j + 3 k
(b) 求直线 O D OD O D A B C ABC A B C
(3)
(c) 求直线 A B AB A B C D CD C D
(5)
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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取平面内两条不同方向的向量并叉积，所得法向量代入任一点得到平面方程。
答题过程
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Two vectors in the plane are
A B → = ( − 2 , − 1 , − 2 ) , A C → = ( 1 , − 1 , − 1 ) . \overrightarrow{AB}=(-2,-1,-2),
\qquad\overrightarrow{AC}=(1,-1,-1). A B = ( − 2 , − 1 , − 2 ) , A C = ( 1 , − 1 , − 1 ) . Their cross product is
n = A B → × A C → = ( − 1 , − 4 , 3 ) . \mathbf n=\overrightarrow{AB}\times\overrightarrow{AC}
=(-1,-4,3). n = A B × A C = ( − 1 , − 4 , 3 ) . Using A = ( 4 , 3 , 3 ) A=(4,3,3) A = ( 4 , 3 , 3 )
− x − 4 y + 3 z = − 4 − 12 + 9 = − 7. -x-4y+3z=-4-12+9=-7. − x − 4 y + 3 z = − 4 − 12 + 9 = − 7. Hence an equation of the plane is − x − 4 y + 3 z = − 7 \boxed{-x-4y+3z=-7} − x − 4 y + 3 z = − 7
(b)
解法一
思路
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先求直线方向与平面法向量的夹角，再取其余角；也可直接用锐角的正弦公式。
答题过程
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The line O D OD O D d = ( 2 , 1 , 3 ) \mathbf d=(2,1,3) d = ( 2 , 1 , 3 ) n = ( − 1 , − 4 , 3 ) \mathbf n=(-1,-4,3) n = ( − 1 , − 4 , 3 )
d ⋅ n = − 2 − 4 + 9 = 3 , ∣ d ∣ = 14 , ∣ n ∣ = 26 . \begin{align*}
\mathbf d\cdot\mathbf n=&\,-2-4+9=3,\\[4mm]
|\mathbf d|=&\,\sqrt{14},\\[4mm]
|\mathbf n|=&\,\sqrt{26}.
\end{align*} d ⋅ n = ∣ d ∣ = ∣ n ∣ = − 2 − 4 + 9 = 3 , 14 , 26 . If ϕ \phi ϕ
sin ϕ = ∣ d ⋅ n ∣ ∣ d ∣ ∣ n ∣ = 3 14 26 . \sin\phi=\frac{|\mathbf d\cdot\mathbf n|}{|\mathbf d||\mathbf n|}
=\frac3{\sqrt{14}\sqrt{26}}. sin ϕ = ∣ d ∣∣ n ∣ ∣ d ⋅ n ∣ = 14 26 3 . Thus ϕ ≈ 9.0 ∘ \boxed{\phi\approx9.0^\circ} ϕ ≈ 9. 0 ∘ 1 1 1
(c)
解法一
思路
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两条线的公垂线方向由方向向量叉积给出。把连接A和C的向量投影到该方向即可求最短距离。
答题过程
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Direction vectors for A B AB A B C D CD C D
u = ( − 2 , − 1 , − 2 ) , v = ( − 3 , − 1 , 1 ) . \mathbf u=(-2,-1,-2),\qquad\mathbf v=(-3,-1,1). u = ( − 2 , − 1 , − 2 ) , v = ( − 3 , − 1 , 1 ) . A common perpendicular direction is
u × v = ( − 3 , 8 , − 1 ) . \mathbf u\times\mathbf v=(-3,8,-1). u × v = ( − 3 , 8 , − 1 ) . Since A C → = ( 1 , − 1 , − 1 ) \overrightarrow{AC}=(1,-1,-1) A C = ( 1 , − 1 , − 1 )
d = ∣ ( 1 , − 1 , − 1 ) ⋅ ( − 3 , 8 , − 1 ) ∣ 9 + 64 + 1 = ∣ − 3 − 8 + 1 ∣ 74 = 10 74 . d=\frac{|(1,-1,-1)\cdot(-3,8,-1)|}{\sqrt{9+64+1}}
=\frac{|-3-8+1|}{\sqrt{74}}
=\boxed{\frac{10}{\sqrt{74}}}. d = 9 + 64 + 1 ∣ ( 1 , − 1 , − 1 ) ⋅ ( − 3 , 8 , − 1 ) ∣ = 74 ∣ − 3 − 8 + 1∣ = 74 10 .
解法二
思路
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也可设两条线上的最近点，其连接向量同时垂直于两条方向，联立求参数后取长度。
答题过程
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Let U = A + λ ( − 2 , − 1 , − 2 ) U=A+\lambda(-2,-1,-2) U = A + λ ( − 2 , − 1 , − 2 ) V = C + μ ( − 3 , − 1 , 1 ) V=C+\mu(-3,-1,1) V = C + μ ( − 3 , − 1 , 1 )
U V → = ( 1 + 2 λ − 3 μ − 1 + λ − μ − 1 + 2 λ + μ ) . \overrightarrow{UV}=\begin{pmatrix}
1+2\lambda-3\mu\\
-1+\lambda-\mu\\
-1+2\lambda+\mu
\end{pmatrix}. U V = 1 + 2 λ − 3 μ − 1 + λ − μ − 1 + 2 λ + μ . Perpendicularity to both line directions gives
1 − 9 λ + 5 μ = 0 , − 3 − 5 λ + 11 μ = 0. \begin{align*}
1-9\lambda+5\mu=&\,0,\\[4mm]
-3-5\lambda+11\mu=&\,0.
\end{align*} 1 − 9 λ + 5 μ = − 3 − 5 λ + 11 μ = 0 , 0. Solving gives λ = 13 / 37 \lambda=13/37 λ = 13/37 μ = 16 / 37 \mu=16/37 μ = 16/37
U V → = 1 37 ( 15 , − 40 , 5 ) . \overrightarrow{UV}=\frac1{37}(15,-40,5). U V = 37 1 ( 15 , − 40 , 5 ) . Hence the shortest distance is
d = 225 + 1600 + 25 37 = 10 74 . d=\frac{\sqrt{225+1600+25}}{37}
=\boxed{\frac{10}{\sqrt{74}}}. d = 37 225 + 1600 + 25 = 74 10 . 