题目
Problem
The points A A A B B B C C C
i − 2 k , i + 2 j + 2 k , 2 i − j − k , \mathbf{i}-2\mathbf{k},\quad \mathbf{i}+2\mathbf{j}+2\mathbf{k},\quad 2\mathbf{i}-\mathbf{j}-\mathbf{k}, i − 2 k , i + 2 j + 2 k , 2 i − j − k , respectively.
(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)
A point D D D i + t k \mathbf{i}+t\mathbf{k} i + t k t ≠ − 2 t\neq-2 t = − 2
(b) Find the acute angle between the planes A B C ABC A B C A B D ABD A B D
(4)
(c) Find the values of t t t A B AB A B C D CD C D 2 \sqrt{2} 2
(7)
题目中文翻译
点 A A A B B B C C C
i − 2 k , i + 2 j + 2 k , 2 i − j − k . \mathbf{i}-2\mathbf{k},\quad \mathbf{i}+2\mathbf{j}+2\mathbf{k},\quad 2\mathbf{i}-\mathbf{j}-\mathbf{k}. i − 2 k , i + 2 j + 2 k , 2 i − j − 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 i + t k \mathbf{i}+t\mathbf{k} i + t k t ≠ − 2 t\neq-2 t = − 2
(b) 求平面 A B C ABC A B C A B D ABD A B D
(4)
(c) 求 t t t A B AB A B C D CD C D 2 \sqrt{2} 2
(7)
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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先取两条平面内方向叉积，再代入A确定平面常数。
答题过程
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We have
A B → = ( 0 , 2 , 4 ) , A C → = ( 1 , − 1 , 1 ) . \begin{align*}
\overrightarrow{AB}=&\,(0,2,4),\\[4mm]
\overrightarrow{AC}=&\,(1,-1,1).
\end{align*} A B = A C = ( 0 , 2 , 4 ) , ( 1 , − 1 , 1 ) . Their cross product is ( 6 , 4 , − 2 ) = 2 ( 3 , 2 , − 1 ) (6,4,-2)=2(3,2,-1) ( 6 , 4 , − 2 ) = 2 ( 3 , 2 , − 1 ) A = ( 1 , 0 , − 2 ) A=(1,0,-2) A = ( 1 , 0 , − 2 )
3 x + 2 y − z = 3 + 0 + 2 = 5. 3x+2y-z=3+0+2=5. 3 x + 2 y − z = 3 + 0 + 2 = 5. Thus 3 x + 2 y − z = 5 \boxed{3x+2y-z=5} 3 x + 2 y − z = 5
(b)
解法一
思路
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AD沿竖直方向，与AB叉积后法向量沿x方向；排除t=−2确保平面ABD确实存在。
答题过程
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Since A D → = ( 0 , 0 , t + 2 ) \overrightarrow{AD}=(0,0,t+2) A D = ( 0 , 0 , t + 2 )
A B → × A D → = ( 2 ( t + 2 ) , 0 , 0 ) . \overrightarrow{AB}\times\overrightarrow{AD}=(2(t+2),0,0). A B × A D = ( 2 ( t + 2 ) , 0 , 0 ) . This is non-zero because t ≠ − 2 t\ne-2 t = − 2 ( 1 , 0 , 0 ) (1,0,0) ( 1 , 0 , 0 ) A B D ABD A B D ( 3 , 2 , − 1 ) (3,2,-1) ( 3 , 2 , − 1 ) A B C ABC A B C
The acute angle ϕ \phi ϕ
cos ϕ = ∣ ( 3 , 2 , − 1 ) ⋅ ( 1 , 0 , 0 ) ∣ 14 = 3 14 . \cos\phi=\frac{|(3,2,-1)\cdot(1,0,0)|}{\sqrt{14}}
=\frac3{\sqrt{14}}. cos ϕ = 14 ∣ ( 3 , 2 , − 1 ) ⋅ ( 1 , 0 , 0 ) ∣ = 14 3 . Hence ϕ ≈ 36.7 ∘ \boxed{\phi\approx36.7^\circ} ϕ ≈ 36. 7 ∘ 1 1 1
(c)
解法一
思路
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先求含t的公垂线方向，连接A与C的向量投影到该方向。距离是非负量，分子必须取绝对值。
答题过程
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Direction vectors for A B AB A B C D CD C D
u = ( 0 , 1 , 2 ) , v = ( − 1 , 1 , t + 1 ) . \begin{align*}
\mathbf u=&\,(0,1,2),\\[4mm]
\mathbf v=&\,(-1,1,t+1).
\end{align*} u = v = ( 0 , 1 , 2 ) , ( − 1 , 1 , t + 1 ) . Their cross product is n = ( t − 1 , − 2 , 1 ) \mathbf n=(t-1,-2,1) n = ( t − 1 , − 2 , 1 ) ( t − 1 ) 2 + 5 > 0 (t-1)^2+5>0 ( t − 1 ) 2 + 5 > 0 A C → = ( 1 , − 1 , 1 ) \overrightarrow{AC}=(1,-1,1) A C = ( 1 , − 1 , 1 )
d = ∣ A C → ⋅ n ∣ ∣ n ∣ = ∣ t + 2 ∣ t 2 − 2 t + 6 . \begin{align*}
d=&\,\frac{|\overrightarrow{AC}\cdot\mathbf n|}{|\mathbf n|}\\[4mm]
=&\,\frac{|t+2|}{\sqrt{t^2-2t+6}}.
\end{align*} d = = ∣ n ∣ ∣ A C ⋅ n ∣ t 2 − 2 t + 6 ∣ t + 2∣ . Set d = 2 d=\sqrt2 d = 2
( t + 2 ) 2 = 2 ( t 2 − 2 t + 6 ) , t 2 − 8 t + 8 = 0. \begin{align*}
(t+2)^2=&\,2(t^2-2t+6), \\[4mm]
t^2-8t+8=&\,0.
\end{align*} ( t + 2 ) 2 = t 2 − 8 t + 8 = 2 ( t 2 − 2 t + 6 ) , 0. Thus
t = 4 + 2 2 or 4 − 2 2 . \boxed{t=4+2\sqrt2\text{ or }4-2\sqrt2}. t = 4 + 2 2 or 4 − 2 2 . Both satisfy t ≠ − 2 t\ne-2 t = − 2
