题目
Problem
The position vectors of the points A A A B B B C C C D D D
7 i + 4 j − k , 11 i + 3 j , 2 i + 6 j + 3 k , 2 i + 7 j + λ k , \begin{aligned}
&\,7\mathbf i+4\mathbf j-\mathbf k,\quad11\mathbf i+3\mathbf j,\\
&\,2\mathbf i+6\mathbf j+3\mathbf k,\quad2\mathbf i+7\mathbf j+\lambda\mathbf k,
\end{aligned} 7 i + 4 j − k , 11 i + 3 j , 2 i + 6 j + 3 k , 2 i + 7 j + λ k , respectively.
(a) Given that the shortest distance between the line A B AB A B C D CD C D 3 3 3 λ 2 − 5 λ + 4 = 0 \lambda^2-5\lambda+4=0 λ 2 − 5 λ + 4 = 0
(7)
Let Π 1 \Pi_1 Π 1 A B D ABD A B D λ = 1 \lambda=1 λ = 1
Let Π 2 \Pi_2 Π 2 A B D ABD A B D λ = 4 \lambda=4 λ = 4
(b) (i) Write down an equation of Π 1 \Pi_1 Π 1 r = a + s b + t c \mathbf r=\mathbf a+s\mathbf b+t\mathbf c r = a + s b + t c
(2)
(ii) Find an equation of Π 2 \Pi_2 Π 2 a x + b y + c z = d ax+by+cz=d a x + b y + cz = d
(4)
(c) Find the acute angle between Π 1 \Pi_1 Π 1 Π 2 \Pi_2 Π 2
(5)
题目中文翻译
点 A A A B B B C C C D D D
7 i + 4 j − k , 11 i + 3 j , 2 i + 6 j + 3 k , 2 i + 7 j + λ k . \begin{aligned}
&\,7\mathbf i+4\mathbf j-\mathbf k,\quad11\mathbf i+3\mathbf j,\\
&\,2\mathbf i+6\mathbf j+3\mathbf k,\quad2\mathbf i+7\mathbf j+\lambda\mathbf k.
\end{aligned} 7 i + 4 j − k , 11 i + 3 j , 2 i + 6 j + 3 k , 2 i + 7 j + λ k . (a) 已知直线 A B AB A B C D CD C D 3 3 3 λ 2 − 5 λ + 4 = 0 \lambda^2-5\lambda+4=0 λ 2 − 5 λ + 4 = 0
(7)
设 Π 1 \Pi_1 Π 1 λ = 1 \lambda=1 λ = 1 A B D ABD A B D
设 Π 2 \Pi_2 Π 2 λ = 4 \lambda=4 λ = 4 A B D ABD A B D
(b) (i) 写出 Π 1 \Pi_1 Π 1 r = a + s b + t c \mathbf r=\mathbf a+s\mathbf b+t\mathbf c r = a + s b + t c
(2)
(ii) 求 Π 2 \Pi_2 Π 2 a x + b y + c z = d ax+by+cz=d a x + b y + cz = d
(4)
(c) 求 Π 1 \Pi_1 Π 1 Π 2 \Pi_2 Π 2
(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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异面直线最短距离可用连接向量在共同法向量上的投影。先确认共同法向量永不为零，再对距离等式平方。
答题过程
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Direction vectors for the lines are
u = A B → = ( 4 , − 1 , 1 ) , v = C D → = ( 0 , 1 , λ − 3 ) . \begin{align*}
\mathbf u=&\,\overrightarrow{AB}=(4,-1,1),\\[4mm]
\mathbf v=&\,\overrightarrow{CD}=(0,1,\lambda-3).
\end{align*} u = v = A B = ( 4 , − 1 , 1 ) , C D = ( 0 , 1 , λ − 3 ) . Their cross product is
n = u × v = ( 2 − λ , 12 − 4 λ , 4 ) , \mathbf n=\mathbf u\times\mathbf v=(2-\lambda,12-4\lambda,4), n = u × v = ( 2 − λ , 12 − 4 λ , 4 ) , which is nonzero for every λ \lambda λ A C → = ( − 5 , 2 , 4 ) \overrightarrow{AC}=(-5,2,4) A C = ( − 5 , 2 , 4 )
A C → ⋅ n = − 5 ( 2 − λ ) + 2 ( 12 − 4 λ ) + 16 = 30 − 3 λ , ∣ n ∣ 2 = ( 2 − λ ) 2 + ( 12 − 4 λ ) 2 + 16 = 17 λ 2 − 100 λ + 164. \begin{align*}
\overrightarrow{AC}\cdot\mathbf n
=&\,-5(2-\lambda)+2(12-4\lambda)+16\\[4mm]
=&\,30-3\lambda,\\[4mm]
|\mathbf n|^2=&\,(2-\lambda)^2+(12-4\lambda)^2+16\\[4mm]
=&\,17\lambda^2-100\lambda+164.
\end{align*} A C ⋅ n = = ∣ n ∣ 2 = = − 5 ( 2 − λ ) + 2 ( 12 − 4 λ ) + 16 30 − 3 λ , ( 2 − λ ) 2 + ( 12 − 4 λ ) 2 + 16 17 λ 2 − 100 λ + 164. The shortest distance is 3 3 3
∣ 30 − 3 λ ∣ 17 λ 2 − 100 λ + 164 = 3. \frac{|30-3\lambda|}{\sqrt{17\lambda^2-100\lambda+164}}=3. 17 λ 2 − 100 λ + 164 ∣30 − 3 λ ∣ = 3. Squaring and simplifying,
9 ( λ − 10 ) 2 = 9 ( 17 λ 2 − 100 λ + 164 ) , 16 λ 2 − 80 λ + 64 = 0 , λ 2 − 5 λ + 4 = 0. \begin{align*}
9(\lambda-10)^2=&\,9(17\lambda^2-100\lambda+164),\\[4mm]
16\lambda^2-80\lambda+64=&\,0,\\[4mm]
\lambda^2-5\lambda+4=&\,0.
\end{align*} 9 ( λ − 10 ) 2 = 16 λ 2 − 80 λ + 64 = λ 2 − 5 λ + 4 = 9 ( 17 λ 2 − 100 λ + 164 ) , 0 , 0.
解法二
思路
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另在两条直线上分别设最近点，令连接向量同时垂直于两条方向向量，解出参数后计算长度。
答题过程
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Let
P = A + s ( 4 , − 1 , 1 ) , Q = C + t ( 0 , 1 , λ − 3 ) . \begin{align*}
P=&\,A+s(4,-1,1),\\[4mm]
Q=&\,C+t(0,1,\lambda-3).
\end{align*} P = Q = A + s ( 4 , − 1 , 1 ) , C + t ( 0 , 1 , λ − 3 ) . Then
P Q → = ( − 5 − 4 s , 2 + t + s , 4 + ( λ − 3 ) t − s ) . \overrightarrow{PQ}=(-5-4s,\ 2+t+s,\ 4+(\lambda-3)t-s). P Q = ( − 5 − 4 s , 2 + t + s , 4 + ( λ − 3 ) t − s ) . For a common perpendicular, its dot product with each line direction must vanish:
− 18 − 18 s + ( λ − 4 ) t = 0 , 4 λ − 10 + ( λ 2 − 6 λ + 10 ) t − ( λ − 4 ) s = 0. \begin{align*}
-18-18s+(\lambda-4)t=&\,0,\\[4mm]
4\lambda-10+(\lambda^2-6\lambda+10)t-(\lambda-4)s=&\,0.
\end{align*} − 18 − 18 s + ( λ − 4 ) t = 4 λ − 10 + ( λ 2 − 6 λ + 10 ) t − ( λ − 4 ) s = 0 , 0. Put D = 17 λ 2 − 100 λ + 164 D=17\lambda^2-100\lambda+164 D = 17 λ 2 − 100 λ + 164 s s s
s = − 1 + λ − 4 18 t , D t = 252 − 90 λ , t = 18 ( 14 − 5 λ ) D , s = − 1 + ( λ − 4 ) ( 14 − 5 λ ) D . \begin{align*}
s=&\,-1+\frac{\lambda-4}{18}t,\\[4mm]
Dt=&\,252-90\lambda,\\[4mm]
t=&\,\frac{18(14-5\lambda)}D,\\[4mm]
s=&\,-1+\frac{(\lambda-4)(14-5\lambda)}D.
\end{align*} s = D t = t = s = − 1 + 18 λ − 4 t , 252 − 90 λ , D 18 ( 14 − 5 λ ) , − 1 + D ( λ − 4 ) ( 14 − 5 λ ) . Here D = 17 ( λ − 50 / 17 ) 2 + 288 / 17 > 0 D=17(\lambda-50/17)^2+288/17>0 D = 17 ( λ − 50/17 ) 2 + 288/17 > 0 P Q → \overrightarrow{PQ} P Q
P Q → = 30 − 3 λ D ( 2 − λ , 12 − 4 λ , 4 ) . \overrightarrow{PQ}=\frac{30-3\lambda}D(2-\lambda,12-4\lambda,4). P Q = D 30 − 3 λ ( 2 − λ , 12 − 4 λ , 4 ) . The squared length of the vector in parentheses is D D D ∣ P Q ∣ 2 = ( 30 − 3 λ ) 2 / D |PQ|^2=(30-3\lambda)^2/D ∣ P Q ∣ 2 = ( 30 − 3 λ ) 2 / D 3 3 3
( 30 − 3 λ ) 2 = 9 D , ( λ − 10 ) 2 = 17 λ 2 − 100 λ + 164 , 16 ( λ 2 − 5 λ + 4 ) = 0. \begin{align*}
(30-3\lambda)^2=&\,9D,\\[4mm]
(\lambda-10)^2=&\,17\lambda^2-100\lambda+164,\\[4mm]
16(\lambda^2-5\lambda+4)=&\,0.
\end{align*} ( 30 − 3 λ ) 2 = ( λ − 10 ) 2 = 16 ( λ 2 − 5 λ + 4 ) = 9 D , 17 λ 2 − 100 λ + 164 , 0. Therefore λ 2 − 5 λ + 4 = 0 \boxed{\lambda^2-5\lambda+4=0} λ 2 − 5 λ + 4 = 0
(b)(i)
解法一
思路
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向量式平面方程由平面上一点与两条不平行方向组成；这里取 A A A A B AB A B A D AD A D
答题过程
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For λ = 1 \lambda=1 λ = 1 D = ( 2 , 7 , 1 ) D=(2,7,1) D = ( 2 , 7 , 1 )
A D → = ( − 5 , 3 , 2 ) , A B → = ( 4 , − 1 , 1 ) . \begin{align*}
\overrightarrow{AD}=&\,(-5,3,2),\\[4mm]
\overrightarrow{AB}=&\,(4,-1,1).
\end{align*} A D = A B = ( − 5 , 3 , 2 ) , ( 4 , − 1 , 1 ) . These vectors are not parallel. Hence an equation of Π 1 \Pi_1 Π 1
r = ( 7 4 − 1 ) + s ( 4 − 1 1 ) + t ( − 5 3 2 ) . \begin{align*}
\mathbf r=&\,\begin{pmatrix}7\\4\\-1\end{pmatrix}
+s\begin{pmatrix}4\\-1\\1\end{pmatrix}\\[4mm]
&\,+t\begin{pmatrix}-5\\3\\2\end{pmatrix}.
\end{align*} r = 7 4 − 1 + s 4 − 1 1 + t − 5 3 2 . Here s , t ∈ R s,t\in\mathbb R s , t ∈ R
(b)(ii)
解法一
思路
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此时 D D D 4 4 4 A D AD A D
答题过程
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For λ = 4 \lambda=4 λ = 4 A D → = ( − 5 , 3 , 5 ) \overrightarrow{AD}=(-5,3,5) A D = ( − 5 , 3 , 5 ) Π 2 \Pi_2 Π 2
( 4 , − 1 , 1 ) × ( − 5 , 3 , 5 ) = ( − 5 − 3 , − 5 − 20 , 12 − 5 ) = ( − 8 , − 25 , 7 ) . \begin{align*}
(4,-1,1)\times(-5,3,5)
=&\,(-5-3,\,-5-20,\,12-5)\\[4mm]
=&\,(-8,-25,7).
\end{align*} ( 4 , − 1 , 1 ) × ( − 5 , 3 , 5 ) = = ( − 5 − 3 , − 5 − 20 , 12 − 5 ) ( − 8 , − 25 , 7 ) . Using A = ( 7 , 4 , − 1 ) A=(7,4,-1) A = ( 7 , 4 , − 1 )
− 8 x − 25 y + 7 z = − 8 ( 7 ) − 25 ( 4 ) + 7 ( − 1 ) . -8x-25y+7z=-8(7)-25(4)+7(-1). − 8 x − 25 y + 7 z = − 8 ( 7 ) − 25 ( 4 ) + 7 ( − 1 ) . Thus − 8 x − 25 y + 7 z = − 163 \boxed{-8x-25y+7z=-163} − 8 x − 25 y + 7 z = − 163
(c)
解法一
思路
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两平面的锐角等于法向量夹角的锐角；点积取绝对值，防止法向量选向影响答案。
答题过程
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A normal to Π 1 \Pi_1 Π 1
n 1 = ( 4 , − 1 , 1 ) × ( − 5 , 3 , 2 ) = ( − 5 , − 13 , 7 ) . \mathbf n_1=(4,-1,1)\times(-5,3,2)=(-5,-13,7). n 1 = ( 4 , − 1 , 1 ) × ( − 5 , 3 , 2 ) = ( − 5 , − 13 , 7 ) . From part (b)(ii), take n 2 = ( − 8 , − 25 , 7 ) \mathbf n_2=(-8,-25,7) n 2 = ( − 8 , − 25 , 7 )
n 1 ⋅ n 2 = 40 + 325 + 49 = 414 , ∣ n 1 ∣ 2 = 25 + 169 + 49 = 243 , ∣ n 2 ∣ 2 = 64 + 625 + 49 = 738. \begin{align*}
\mathbf n_1\cdot\mathbf n_2=&\,40+325+49=414,\\[4mm]
|\mathbf n_1|^2=&\,25+169+49=243,\\[4mm]
|\mathbf n_2|^2=&\,64+625+49=738.
\end{align*} n 1 ⋅ n 2 = ∣ n 1 ∣ 2 = ∣ n 2 ∣ 2 = 40 + 325 + 49 = 414 , 25 + 169 + 49 = 243 , 64 + 625 + 49 = 738. Therefore the acute angle ϕ \phi ϕ
cos ϕ = ∣ 414 ∣ 243 738 , \cos\phi=\frac{|414|}{\sqrt{243}\sqrt{738}}, cos ϕ = 243 738 ∣414∣ , giving ϕ = 12.1 ∘ \boxed{\phi=12.1^\circ} ϕ = 12. 1 ∘
