题目
Problem
The points A A A B B B C C C
i + j , − i + 2 j + 4 k , − 2 i + j + 3 k , \mathbf{i}+\mathbf{j},\quad-\mathbf{i}+2\mathbf{j}+4\mathbf{k},\quad-2\mathbf{i}+\mathbf{j}+3\mathbf{k}, i + j , − i + 2 j + 4 k , − 2 i + j + 3 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)
(b) Find the perpendicular distance from O O O A B C ABC A B C
(2)
(c) Find a vector equation of the common perpendicular to the lines O C OC O C A B AB A B
(8)
题目中文翻译
点 A A A B B B C C C O O O
i + j , − i + 2 j + 4 k , − 2 i + j + 3 k . \mathbf{i}+\mathbf{j},\quad-\mathbf{i}+2\mathbf{j}+4\mathbf{k},\quad-2\mathbf{i}+\mathbf{j}+3\mathbf{k}. i + j , − i + 2 j + 4 k , − 2 i + j + 3 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)
(b) 求原点 O O O A B C ABC A B C
(2)
(c) 求直线 O C OC O C A B AB A B
(8)
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AB与AC叉积求法向量，再代入A求常数项。
答题过程
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A B → = ( − 2 , 1 , 4 ) , A C → = ( − 3 , 0 , 3 ) . \overrightarrow{AB}=(-2,1,4),\qquad
\overrightarrow{AC}=(-3,0,3). A B = ( − 2 , 1 , 4 ) , A C = ( − 3 , 0 , 3 ) . Their cross product is
A B → × A C → = ( 3 , − 6 , 3 ) = 3 ( 1 , − 2 , 1 ) . \overrightarrow{AB}\times\overrightarrow{AC}=(3,-6,3)=3(1,-2,1). A B × A C = ( 3 , − 6 , 3 ) = 3 ( 1 , − 2 , 1 ) . Using A = ( 1 , 1 , 0 ) A=(1,1,0) A = ( 1 , 1 , 0 )
x − 2 y + z = − 1 . \boxed{x-2y+z=-1}. x − 2 y + z = − 1 .
(b)
解法一
思路
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原点到平面距离是常数项绝对值除法向量模。
答题过程
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For the plane x − 2 y + z + 1 = 0 x-2y+z+1=0 x − 2 y + z + 1 = 0 O O O
∣ 1 ∣ 1 + 4 + 1 = 1 6 . \boxed{\frac{|1|}{\sqrt{1+4+1}}=\frac1{\sqrt6}}. 1 + 4 + 1 ∣1∣ = 6 1 .
(c)
解法一
思路
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分别参数化两直线上的点，连接向量与两方向点积都为零，联立解出公垂足，再配共同垂直方向。
答题过程
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Let P P P O C OC O C Q Q Q A B AB A B
P = ( − 2 λ , λ , 3 λ ) , Q = ( 1 − 2 μ , 1 + μ , 4 μ ) . \begin{align*}
P=&\,(-2\lambda,\lambda,3\lambda),\\[4mm]
Q=&\,(1-2\mu,1+\mu,4\mu).
\end{align*} P = Q = ( − 2 λ , λ , 3 λ ) , ( 1 − 2 μ , 1 + μ , 4 μ ) . Then
P Q → = ( 1 − 2 μ + 2 λ , 1 + μ − λ , 4 μ − 3 λ ) . \overrightarrow{PQ}=(1-2\mu+2\lambda,1+\mu-\lambda,4\mu-3\lambda). P Q = ( 1 − 2 μ + 2 λ , 1 + μ − λ , 4 μ − 3 λ ) . The two perpendicularity conditions give
P Q → ⋅ ( − 2 , 1 , 3 ) = 0 , 17 μ − 14 λ = 1 , P Q → ⋅ ( − 2 , 1 , 4 ) = 0 , 21 μ − 17 λ = 1. \begin{align*}
\overrightarrow{PQ}\cdot(-2,1,3)=&\,0,\\[4mm]
17\mu-14\lambda=&\,1,\\[4mm]
\overrightarrow{PQ}\cdot(-2,1,4)=&\,0,\\[4mm]
21\mu-17\lambda=&\,1.
\end{align*} P Q ⋅ ( − 2 , 1 , 3 ) = 17 μ − 14 λ = P Q ⋅ ( − 2 , 1 , 4 ) = 21 μ − 17 λ = 0 , 1 , 0 , 1. Eliminating λ \lambda λ 5 μ = − 3 5\mu=-3 5 μ = − 3 μ = − 3 / 5 \mu=-3/5 μ = − 3/5 λ = − 4 / 5 \lambda=-4/5 λ = − 4/5
P = ( 8 / 5 , − 4 / 5 , − 12 / 5 ) , Q = ( 11 / 5 , 2 / 5 , − 12 / 5 ) . \begin{align*}
P=&\,(8/5,-4/5,-12/5),\\[4mm]
Q=&\,(11/5,2/5,-12/5).
\end{align*} P = Q = ( 8/5 , − 4/5 , − 12/5 ) , ( 11/5 , 2/5 , − 12/5 ) . The common perpendicular direction is
( − 2 , 1 , 3 ) × ( − 2 , 1 , 4 ) = ( 1 , 2 , 0 ) . (-2,1,3)\times(-2,1,4)=(1,2,0). ( − 2 , 1 , 3 ) × ( − 2 , 1 , 4 ) = ( 1 , 2 , 0 ) . Hence a required equation is
r = ( 8 / 5 − 4 / 5 − 12 / 5 ) + t ( 1 2 0 ) , t ∈ R . \boxed{\mathbf r=\begin{pmatrix}8/5\\-4/5\\-12/5\end{pmatrix}
+t\begin{pmatrix}1\\2\\0\end{pmatrix},\quad t\in\mathbb R}. r = 8/5 − 4/5 − 12/5 + t 1 2 0 , t ∈ R . It contains P P P t = 0 t=0 t = 0 Q Q Q t = 3 / 5 t=3/5 t = 3/5
解法二
思路
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也可先求公垂线方向，令PQ与之平行。零的竖直分量给一个条件，另两个分量比例给第二个条件。
答题过程
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A common perpendicular direction is
w = ( − 2 , 1 , 3 ) × ( − 2 , 1 , 4 ) = ( 1 , 2 , 0 ) . \mathbf w=(-2,1,3)\times(-2,1,4)=(1,2,0). w = ( − 2 , 1 , 3 ) × ( − 2 , 1 , 4 ) = ( 1 , 2 , 0 ) . With P = ( − 2 λ , λ , 3 λ ) P=(-2\lambda,\lambda,3\lambda) P = ( − 2 λ , λ , 3 λ ) Q = ( 1 − 2 μ , 1 + μ , 4 μ ) Q=(1-2\mu,1+\mu,4\mu) Q = ( 1 − 2 μ , 1 + μ , 4 μ ) P Q → \overrightarrow{PQ} P Q w \mathbf w w
4 μ − 3 λ = 0 , 1 + μ − λ = 2 ( 1 − 2 μ + 2 λ ) . \begin{align*}
4\mu-3\lambda=&\,0,\\[4mm]
1+\mu-\lambda=&\,2(1-2\mu+2\lambda).
\end{align*} 4 μ − 3 λ = 1 + μ − λ = 0 , 2 ( 1 − 2 μ + 2 λ ) . Thus λ = 4 μ / 3 \lambda=4\mu/3 λ = 4 μ /3 5 μ − 5 λ = 1 5\mu-5\lambda=1 5 μ − 5 λ = 1 μ = − 3 / 5 \mu=-3/5 μ = − 3/5 λ = − 4 / 5 \lambda=-4/5 λ = − 4/5
Using P = ( 8 / 5 , − 4 / 5 , − 12 / 5 ) P=(8/5,-4/5,-12/5) P = ( 8/5 , − 4/5 , − 12/5 )
r = ( 8 / 5 − 4 / 5 − 12 / 5 ) + t ( 1 2 0 ) , t ∈ R . \boxed{\mathbf r=\begin{pmatrix}8/5\\-4/5\\-12/5\end{pmatrix}
+t\begin{pmatrix}1\\2\\0\end{pmatrix},\quad t\in\mathbb R}. r = 8/5 − 4/5 − 12/5 + t 1 2 0 , t ∈ R . 