题目
Problem
The points A A A B B B C C C
2 i + 2 j + 4 k , 2 i + 4 j − k , − 3 i − 3 j + 4 k , 2\mathbf{i}+2\mathbf{j}+4\mathbf{k},\qquad 2\mathbf{i}+4\mathbf{j}-\mathbf{k},\qquad -3\mathbf{i}-3\mathbf{j}+4\mathbf{k}, 2 i + 2 j + 4 k , 2 i + 4 j − k , − 3 i − 3 j + 4 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 perpendicular distance from D D D A B C ABC A B C
(2)
(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
2 i + 2 j + 4 k , 2 i + 4 j − k , − 3 i − 3 j + 4 k . 2\mathbf{i}+2\mathbf{j}+4\mathbf{k},\qquad 2\mathbf{i}+4\mathbf{j}-\mathbf{k},\qquad -3\mathbf{i}-3\mathbf{j}+4\mathbf{k}. 2 i + 2 j + 4 k , 2 i + 4 j − k , − 3 i − 3 j + 4 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) 求点 D D D A B C ABC A B C
(2)
(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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用平面内两条边的叉积求法向量，再代入A。
答题过程
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A B → = ( 0 , 2 , − 5 ) , A C → = ( − 5 , − 5 , 0 ) . \begin{align*}
\overrightarrow{AB}=&\,(0,2,-5),\\[4mm]
\overrightarrow{AC}=&\,(-5,-5,0).
\end{align*} A B = A C = ( 0 , 2 , − 5 ) , ( − 5 , − 5 , 0 ) . A normal is
A B → × A C → = ( − 25 , 25 , 10 ) = 5 ( − 5 , 5 , 2 ) . \begin{align*}
\overrightarrow{AB}\times\overrightarrow{AC}=&\,(-25,25,10)\\[4mm]
=&\,5(-5,5,2).
\end{align*} A B × A C = = ( − 25 , 25 , 10 ) 5 ( − 5 , 5 , 2 ) . Using A = ( 2 , 2 , 4 ) A=(2,2,4) A = ( 2 , 2 , 4 ) ( − 5 , 5 , 2 ) (-5,5,2) ( − 5 , 5 , 2 )
− 5 x + 5 y + 2 z = − 10 + 10 + 8 , -5x+5y+2z=-10+10+8, − 5 x + 5 y + 2 z = − 10 + 10 + 8 , so
− 5 x + 5 y + 2 z = 8 . \boxed{-5x+5y+2z=8}. − 5 x + 5 y + 2 z = 8 .
(b)
解法一
思路
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把D代入平面左边，减去常数项并取绝对值，再除以法向量的模。
答题过程
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The distance is
d = ∣ − 5 ( 2 ) + 5 ( 1 ) + 2 ( 3 ) − 8 ∣ 25 + 25 + 4 = 7 54 ( ≈ 0.953 ) . \begin{align*}
d=&\,\frac{|-5(2)+5(1)+2(3)-8|}{\sqrt{25+25+4}}\\[4mm]
=&\,\boxed{\frac7{\sqrt{54}}}\quad(\approx0.953).
\end{align*} d = = 25 + 25 + 4 ∣ − 5 ( 2 ) + 5 ( 1 ) + 2 ( 3 ) − 8∣ 54 7 ( ≈ 0.953 ) .
(c)
解法一
思路
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两直线方向的叉积给公垂线方向，把任意一条连接两线的向量投影到该方向即可得到最短距离。
答题过程
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Direction vectors of A B AB A B C D CD C D
u = ( 0 , 2 , − 5 ) , v = D − C = ( 5 , 4 , − 1 ) . \begin{align*}
\mathbf u=&\,(0,2,-5),\\[4mm]
\mathbf v=&\,D-C=(5,4,-1).
\end{align*} u = v = ( 0 , 2 , − 5 ) , D − C = ( 5 , 4 , − 1 ) . Their common perpendicular direction is
w = u × v = ( 18 , − 25 , − 10 ) , ∣ w ∣ = 1049 . \begin{align*}
\mathbf w=&\,\mathbf u\times\mathbf v=(18,-25,-10),\\[4mm]
|\mathbf w|=&\,\sqrt{1049}.
\end{align*} w = ∣ w ∣ = u × v = ( 18 , − 25 , − 10 ) , 1049 . Since A C → = ( − 5 , − 5 , 0 ) \overrightarrow{AC}=(-5,-5,0) A C = ( − 5 , − 5 , 0 ) A B AB A B C D CD C D
d = ∣ A C → ⋅ w ∣ ∣ w ∣ = ∣ − 90 + 125 ∣ 1049 = 35 1049 ( ≈ 1.08 ) . \begin{align*}
d=&\,\frac{|\overrightarrow{AC}\cdot\mathbf w|}{|\mathbf w|}\\[4mm]
=&\,\frac{|-90+125|}{\sqrt{1049}}\\[4mm]
=&\,\boxed{\frac{35}{\sqrt{1049}}}
\quad(\approx1.08).
\end{align*} d = = = ∣ w ∣ ∣ A C ⋅ w ∣ 1049 ∣ − 90 + 125∣ 1049 35 ( ≈ 1.08 ) . 