题目
Problem
The lines l 1 l_1 l 1 l 2 l_2 l 2 r = 2 i + k + λ ( i − j + 2 k ) \mathbf r=2\mathbf i+\mathbf k+\lambda(\mathbf i-\mathbf j+2\mathbf k) r = 2 i + k + λ ( i − j + 2 k ) r = 2 j + 6 k + μ ( i + 2 j − 2 k ) \mathbf r=2\mathbf j+6\mathbf k+\mu(\mathbf i+2\mathbf j-2\mathbf k) r = 2 j + 6 k + μ ( i + 2 j − 2 k )
The point P P P l 1 l_1 l 1 Q Q Q l 2 l_2 l 2 P Q PQ P Q l 1 l_1 l 1 l 2 l_2 l 2
(a) Find the length P Q PQ P Q
(5)
The plane Π 1 \Pi_1 Π 1 P Q PQ P Q l 1 l_1 l 1
The plane Π 2 \Pi_2 Π 2 P Q PQ P Q l 2 l_2 l 2
(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
(1)
(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
直线 l 1 l_1 l 1 l 2 l_2 l 2 r = 2 i + k + λ ( i − j + 2 k ) \mathbf r=2\mathbf i+\mathbf k+\lambda(\mathbf i-\mathbf j+2\mathbf k) r = 2 i + k + λ ( i − j + 2 k ) r = 2 j + 6 k + μ ( i + 2 j − 2 k ) \mathbf r=2\mathbf j+6\mathbf k+\mu(\mathbf i+2\mathbf j-2\mathbf k) r = 2 j + 6 k + μ ( i + 2 j − 2 k )
点 P P P l 1 l_1 l 1 Q Q Q l 2 l_2 l 2 P Q PQ P Q l 1 l_1 l 1 l 2 l_2 l 2
(a) 求 P Q PQ P Q
(5)
平面 Π 1 \Pi_1 Π 1 P Q PQ P Q l 1 l_1 l 1
平面 Π 2 \Pi_2 Π 2 P Q PQ P Q l 2 l_2 l 2
(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
(1)
(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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两直线方向叉积给共同垂直方向，连接两线上任意已知点后作投影。
答题过程
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The line directions are u = ( 1 , − 1 , 2 ) \mathbf u=(1,-1,2) u = ( 1 , − 1 , 2 ) v = ( 1 , 2 , − 2 ) \mathbf v=(1,2,-2) v = ( 1 , 2 , − 2 )
w = ( 2 , − 4 , − 3 ) , \mathbf w=(2,-4,-3), w = ( 2 , − 4 , − 3 ) , which is the negative of u × v \mathbf u\times\mathbf v u × v A = ( 2 , 0 , 1 ) A=(2,0,1) A = ( 2 , 0 , 1 ) l 1 l_1 l 1 B = ( 0 , 2 , 6 ) B=(0,2,6) B = ( 0 , 2 , 6 ) l 2 l_2 l 2
A B → = ( − 2 , 2 , 5 ) . \overrightarrow{AB}=(-2,2,5). A B = ( − 2 , 2 , 5 ) . Hence
P Q = ∣ A B → ⋅ w ∣ ∣ w ∣ = ∣ − 4 − 8 − 15 ∣ 4 + 16 + 9 = 27 29 . \begin{align*}
PQ=&\,\frac{|\overrightarrow{AB}\cdot\mathbf w|}{|\mathbf w|}\\[4mm]
=&\,\frac{|-4-8-15|}{\sqrt{4+16+9}}\\[4mm]
=&\,\boxed{\frac{27}{\sqrt{29}}}.
\end{align*} P Q = = = ∣ w ∣ ∣ A B ⋅ w ∣ 4 + 16 + 9 ∣ − 4 − 8 − 15∣ 29 27 .
(b)(i)
解法一
思路
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平面包含第一条直线及公垂线方向，两方向张成所求平面。
答题过程
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Using the point ( 2 , 0 , 1 ) (2,0,1) ( 2 , 0 , 1 ) l 1 l_1 l 1 ( 1 , − 1 , 2 ) (1,-1,2) ( 1 , − 1 , 2 ) ( 2 , − 4 , − 3 ) (2,-4,-3) ( 2 , − 4 , − 3 )
r = ( 2 0 1 ) + s ( 1 − 1 2 ) + t ( 2 − 4 − 3 ) , \begin{align*}
\mathbf r=&\,\begin{pmatrix}2\\0\\1\end{pmatrix}
+s\begin{pmatrix}1\\-1\\2\end{pmatrix}\\[4mm]
&\,+t\begin{pmatrix}2\\-4\\-3\end{pmatrix},
\end{align*} r = 2 0 1 + s 1 − 1 2 + t 2 − 4 − 3 , where s , t ∈ R s,t\in\mathbb R s , t ∈ R
(b)(ii)
解法一
思路
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第二条直线方向与公垂线方向都在第二平面内，叉积给法向量，再代入第二直线上的点。
答题过程
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A normal to Π 2 \Pi_2 Π 2
( 1 , 2 , − 2 ) × ( 2 , − 4 , − 3 ) = ( − 14 , − 1 , − 8 ) . (1,2,-2)\times(2,-4,-3)=(-14,-1,-8). ( 1 , 2 , − 2 ) × ( 2 , − 4 , − 3 ) = ( − 14 , − 1 , − 8 ) . Taking the opposite normal ( 14 , 1 , 8 ) (14,1,8) ( 14 , 1 , 8 ) ( 0 , 2 , 6 ) (0,2,6) ( 0 , 2 , 6 )
Π 2 : 14 x + y + 8 z = 50 . \boxed{\Pi_2:\ 14x+y+8z=50}. Π 2 : 14 x + y + 8 z = 50 .
(c)
解法一
思路
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第一平面的两方向叉积给法向量，与第二平面法向量点积求两平面锐角。
答题过程
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A normal to Π 1 \Pi_1 Π 1
n 1 = ( 1 , − 1 , 2 ) × ( 2 , − 4 , − 3 ) = ( 11 , 7 , − 2 ) . \mathbf n_1=(1,-1,2)\times(2,-4,-3)=(11,7,-2). n 1 = ( 1 , − 1 , 2 ) × ( 2 , − 4 , − 3 ) = ( 11 , 7 , − 2 ) . Using n 2 = ( 14 , 1 , 8 ) \mathbf n_2=(14,1,8) n 2 = ( 14 , 1 , 8 ) ϕ \phi ϕ
cos ϕ = ∣ n 1 ⋅ n 2 ∣ ∣ n 1 ∣ ∣ n 2 ∣ = ∣ 154 + 7 − 16 ∣ 174 261 = 145 174 261 . \begin{align*}
\cos\phi=&\,\frac{|\mathbf n_1\cdot\mathbf n_2|}{|\mathbf n_1||\mathbf n_2|}\\[4mm]
=&\,\frac{|154+7-16|}{\sqrt{174}\sqrt{261}}\\[4mm]
=&\,\frac{145}{\sqrt{174}\sqrt{261}}.
\end{align*} cos ϕ = = = ∣ n 1 ∣∣ n 2 ∣ ∣ n 1 ⋅ n 2 ∣ 174 261 ∣154 + 7 − 16∣ 174 261 145 . Thus ϕ = 47.1 ∘ \boxed{\phi=47.1^\circ} ϕ = 47. 1 ∘
