题目
Problem
The plane Π \Pi Π x + 3 y + 2 z = 1 x+3y+2z=1 x + 3 y + 2 z = 1
(a) Find the perpendicular distance from the origin O O O Π \Pi Π
(2)
Relative to O O O A A A B B B C C C
− j + 2 k , 2 i − k , 2 i − j − k , -\mathbf{j}+2\mathbf{k},\quad2\mathbf{i}-\mathbf{k},\quad2\mathbf{i}-\mathbf{j}-\mathbf{k}, − j + 2 k , 2 i − k , 2 i − j − k , respectively.
(b) Find the acute angle between the planes O A B OAB O A B Π \Pi Π
(4)
(c) Find an equation for the common perpendicular to the lines O C OC O C A B AB A B
(8)
题目中文翻译
平面 Π \Pi Π x + 3 y + 2 z = 1 x+3y+2z=1 x + 3 y + 2 z = 1
(a) 求原点 O O O Π \Pi Π
(2)
点 A A A B B B C C C O O O
− j + 2 k , 2 i − k , 2 i − j − k . -\mathbf{j}+2\mathbf{k},\quad2\mathbf{i}-\mathbf{k},\quad2\mathbf{i}-\mathbf{j}-\mathbf{k}. − j + 2 k , 2 i − k , 2 i − j − k . (b) 求平面 O A B OAB O A B Π \Pi Π
(4)
(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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原点代入平面左式减常数，再除以法向量长度。
答题过程
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The distance from O = ( 0 , 0 , 0 ) O=(0,0,0) O = ( 0 , 0 , 0 ) Π \Pi Π
d = ∣ 0 + 0 + 0 − 1 ∣ 1 2 + 3 2 + 2 2 = 1 14 . d=\frac{|0+0+0-1|}{\sqrt{1^2+3^2+2^2}}
=\boxed{\frac1{\sqrt{14}}}. d = 1 2 + 3 2 + 2 2 ∣0 + 0 + 0 − 1∣ = 14 1 .
(b)
解法一
思路
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平面OAB的两条位置向量叉积给出法向量；用两法向量的锐角求平面夹角。
答题过程
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A normal to O A B OAB O A B
n 1 = ( 0 , − 1 , 2 ) × ( 2 , 0 , − 1 ) = ( 1 , 4 , 2 ) . \mathbf n_1=(0,-1,2)\times(2,0,-1)=(1,4,2). n 1 = ( 0 , − 1 , 2 ) × ( 2 , 0 , − 1 ) = ( 1 , 4 , 2 ) . A normal to Π \Pi Π n 2 = ( 1 , 3 , 2 ) \mathbf n_2=(1,3,2) n 2 = ( 1 , 3 , 2 )
n 1 ⋅ n 2 = 1 + 12 + 4 = 17 , ∣ n 1 ∣ = 21 , ∣ n 2 ∣ = 14 . \begin{align*}
\mathbf n_1\cdot\mathbf n_2=&\,1+12+4=17,\\[4mm]
|\mathbf n_1|=&\,\sqrt{21},\\[4mm]
|\mathbf n_2|=&\,\sqrt{14}.
\end{align*} n 1 ⋅ n 2 = ∣ n 1 ∣ = ∣ n 2 ∣ = 1 + 12 + 4 = 17 , 21 , 14 . The acute angle ϕ \phi ϕ
cos ϕ = 17 21 14 . \cos\phi=\frac{17}{\sqrt{21}\sqrt{14}}. cos ϕ = 21 14 17 . Hence ϕ ≈ 7.5 ∘ \boxed{\phi\approx7.5^\circ} ϕ ≈ 7. 5 ∘ 1 1 1
(c)
解法一
思路
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分别设两条线上的点，用连接向量与两方向同时垂直列方程；求出最近点和公垂线方向。
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 , − λ ) , Q = ( 2 μ , − 1 + μ , 2 − 3 μ ) . \begin{align*}
P=&\,(2\lambda,-\lambda,-\lambda),\\[4mm]
Q=&\,(2\mu,-1+\mu,2-3\mu).
\end{align*} P = Q = ( 2 λ , − λ , − λ ) , ( 2 μ , − 1 + μ , 2 − 3 μ ) . Then
P Q → = ( 2 μ − 2 λ − 1 + μ + λ 2 − 3 μ + λ ) . \overrightarrow{PQ}=\begin{pmatrix}
2\mu-2\lambda\\
-1+\mu+\lambda\\
2-3\mu+\lambda
\end{pmatrix}. P Q = 2 μ − 2 λ − 1 + μ + λ 2 − 3 μ + λ . For a common perpendicular, its dot products with ( 2 , − 1 , − 1 ) (2,-1,-1) ( 2 , − 1 , − 1 ) ( 2 , 1 , − 3 ) (2,1,-3) ( 2 , 1 , − 3 )
6 μ − 6 λ = 1 , 14 μ − 6 λ = 7. \begin{align*}
6\mu-6\lambda=&\,1,\\[4mm]
14\mu-6\lambda=&\,7.
\end{align*} 6 μ − 6 λ = 14 μ − 6 λ = 1 , 7. Subtracting gives 8 μ = 6 8\mu=6 8 μ = 6
μ = 3 4 , λ = 7 12 . \mu=\frac34,\qquad\lambda=\frac7{12}. μ = 4 3 , λ = 12 7 . Hence
Q = 1 4 ( 6 , − 1 , − 1 ) , P Q → = 1 3 ( 1 , 1 , 1 ) . Q=\frac14(6,-1,-1),
\qquad\overrightarrow{PQ}=\frac13(1,1,1). Q = 4 1 ( 6 , − 1 , − 1 ) , P Q = 3 1 ( 1 , 1 , 1 ) . Therefore an equation of the common perpendicular is
r = 1 4 ( 6 i − j − k ) + t ( i + j + k ) , t ∈ R . \boxed{\mathbf r=\frac14(6\mathbf i-\mathbf j-\mathbf k)
+t(\mathbf i+\mathbf j+\mathbf k),\quad t\in\mathbb R}. r = 4 1 ( 6 i − j − k ) + t ( i + j + k ) , t ∈ R . 