题目
Problem
The lines l 1 l_1 l 1 l 2 l_2 l 2
r = i + 3 j − 2 k + λ ( 2 i + j + k ) and r = i − 2 j + 9 k + μ ( i − 4 j + 2 k ) \begin{align*}
\mathbf{r}=&\,\mathbf{i}+3\mathbf{j}-2\mathbf{k}+\lambda(2\mathbf{i}+\mathbf{j}+\mathbf{k})\\
\text{and}\quad\mathbf{r}=&\,\mathbf{i}-2\mathbf{j}+9\mathbf{k}+\mu(\mathbf{i}-4\mathbf{j}+2\mathbf{k})
\end{align*} r = and r = i + 3 j − 2 k + λ ( 2 i + j + k ) i − 2 j + 9 k + μ ( i − 4 j + 2 k ) respectively. The plane Π 1 \Pi_1 Π 1 l 1 l_1 l 1 l 2 l_2 l 2
(a) Find the equation of Π 1 \Pi_1 Π 1 a x + b y + c z = d ax+by+cz=d a x + b y + cz = d
(4)
The plane Π 2 \Pi_2 Π 2 l 2 l_2 l 2 ( 2 , − 1 , 7 ) (2,-1,7) ( 2 , − 1 , 7 )
(b) Find the acute angle between Π 1 \Pi_1 Π 1 Π 2 \Pi_2 Π 2
(4)
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
(c) Find a vector equation for P Q PQ P Q
(7)
题目中文翻译
直线 l 1 l_1 l 1 l 2 l_2 l 2
r = i + 3 j − 2 k + λ ( 2 i + j + k ) r = i − 2 j + 9 k + μ ( i − 4 j + 2 k ) . \begin{align*}
\mathbf{r}=&\,\mathbf{i}+3\mathbf{j}-2\mathbf{k}+\lambda(2\mathbf{i}+\mathbf{j}+\mathbf{k})\\
\mathbf{r}=&\,\mathbf{i}-2\mathbf{j}+9\mathbf{k}+\mu(\mathbf{i}-4\mathbf{j}+2\mathbf{k}).
\end{align*} r = r = i + 3 j − 2 k + λ ( 2 i + j + k ) i − 2 j + 9 k + μ ( i − 4 j + 2 k ) . 平面 Π 1 \Pi_1 Π 1 l 1 l_1 l 1 l 2 l_2 l 2
(a) 求 Π 1 \Pi_1 Π 1 a x + b y + c z = d ax+by+cz=d a x + b y + cz = d
(4)
平面 Π 2 \Pi_2 Π 2 l 2 l_2 l 2 ( 2 , − 1 , 7 ) (2,-1,7) ( 2 , − 1 , 7 )
(b) 求 Π 1 \Pi_1 Π 1 Π 2 \Pi_2 Π 2
(4)
l 1 l_1 l 1 P P P l 2 l_2 l 2 Q Q Q P Q PQ P Q l 1 l_1 l 1 l 2 l_2 l 2
(c) 求 P Q PQ P Q
(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
展开
平面内含第一条直线的方向，又平行第二条直线的方向。两方向叉积是平面法向量，再代第一条直线上的已知点。
答题过程
展开
The plane contains direction vectors u = ( 2 , 1 , 1 ) \mathbf u=(2,1,1) u = ( 2 , 1 , 1 ) v = ( 1 , − 4 , 2 ) \mathbf v=(1,-4,2) v = ( 1 , − 4 , 2 )
u × v = ( 6 , − 3 , − 9 ) = − 3 ( − 2 , 1 , 3 ) . \mathbf u\times\mathbf v=(6,-3,-9)=-3(-2,1,3). u × v = ( 6 , − 3 , − 9 ) = − 3 ( − 2 , 1 , 3 ) . Taking n 1 = ( 2 , − 1 , − 3 ) \mathbf n_1=(2,-1,-3) n 1 = ( 2 , − 1 , − 3 ) ( 1 , 3 , − 2 ) (1,3,-2) ( 1 , 3 , − 2 ) l 1 l_1 l 1
2 x − y − 3 z = 2 ( 1 ) − 3 − 3 ( − 2 ) = 5. 2x-y-3z=2(1)-3-3(-2)=5. 2 x − y − 3 z = 2 ( 1 ) − 3 − 3 ( − 2 ) = 5. Thus
Π 1 : 2 x − y − 3 z = 5 . \boxed{\Pi_1:\ 2x-y-3z=5}. Π 1 : 2 x − y − 3 z = 5 .
(b)
解法一
思路
展开
第二平面用直线方向与该线上一点指向额外给定点的向量求法向量。两平面锐角用两法向量点积的绝对值。
答题过程
展开
A point on l 2 l_2 l 2 B = ( 1 , − 2 , 9 ) B=(1,-2,9) B = ( 1 , − 2 , 9 ) D = ( 2 , − 1 , 7 ) D=(2,-1,7) D = ( 2 , − 1 , 7 ) B D → = ( 1 , 1 , − 2 ) \overrightarrow{BD}=(1,1,-2) B D = ( 1 , 1 , − 2 ) Π 2 \Pi_2 Π 2
A normal to Π 2 \Pi_2 Π 2
n 2 = ( 1 , − 4 , 2 ) × ( 1 , 1 , − 2 ) = ( 6 , 4 , 5 ) . \mathbf n_2=(1,-4,2)\times(1,1,-2)=(6,4,5). n 2 = ( 1 , − 4 , 2 ) × ( 1 , 1 , − 2 ) = ( 6 , 4 , 5 ) . If ϕ \phi ϕ
cos ϕ = ∣ ( 2 , − 1 , − 3 ) ⋅ ( 6 , 4 , 5 ) ∣ 14 77 = ∣ 12 − 4 − 15 ∣ 14 77 = 7 1078 . \begin{align*}
\cos\phi=&\,\frac{|(2,-1,-3)\cdot(6,4,5)|}{\sqrt{14}\sqrt{77}}\\[4mm]
=&\,\frac{|12-4-15|}{\sqrt{14}\sqrt{77}}
=\frac7{\sqrt{1078}}.
\end{align*} cos ϕ = = 14 77 ∣ ( 2 , − 1 , − 3 ) ⋅ ( 6 , 4 , 5 ) ∣ 14 77 ∣12 − 4 − 15∣ = 1078 7 . Hence ϕ ≈ 77.7 ∘ \boxed{\phi\approx77.7^\circ} ϕ ≈ 77. 7 ∘
(c)
解法一
思路
展开
分别参数化P与Q，连线向量与两条直线的方向点积均为零，得到两个线性条件。求出点后用共同垂直方向写直线。
答题过程
展开
Write
P = ( 1 + 2 λ , 3 + λ , − 2 + λ ) , Q = ( 1 + μ , − 2 − 4 μ , 9 + 2 μ ) . \begin{align*}
P=&\,(1+2\lambda,3+\lambda,-2+\lambda),\\[4mm]
Q=&\,(1+\mu,-2-4\mu,9+2\mu).
\end{align*} P = Q = ( 1 + 2 λ , 3 + λ , − 2 + λ ) , ( 1 + μ , − 2 − 4 μ , 9 + 2 μ ) . Then
P Q → = ( μ − 2 λ − 5 − 4 μ − λ 11 + 2 μ − λ ) . \overrightarrow{PQ}=\begin{pmatrix}
\mu-2\lambda\\-5-4\mu-\lambda\\11+2\mu-\lambda
\end{pmatrix}. P Q = μ − 2 λ − 5 − 4 μ − λ 11 + 2 μ − λ . Since P Q PQ P Q
P Q → ⋅ ( 2 , 1 , 1 ) = 0 ⟹ 6 − 6 λ = 0 , P Q → ⋅ ( 1 , − 4 , 2 ) = 0 ⟹ 42 + 21 μ = 0. \begin{align*}
\overrightarrow{PQ}\cdot(2,1,1)=&\,0\\[4mm]
\Longrightarrow&\,6-6\lambda=0,\\[4mm]
\overrightarrow{PQ}\cdot(1,-4,2)=&\,0\\[4mm]
\Longrightarrow&\,42+21\mu=0.
\end{align*} P Q ⋅ ( 2 , 1 , 1 ) = ⟹ P Q ⋅ ( 1 , − 4 , 2 ) = ⟹ 0 6 − 6 λ = 0 , 0 42 + 21 μ = 0. Thus λ = 1 \lambda=1 λ = 1 μ = − 2 \mu=-2 μ = − 2
P = ( 3 , 4 , − 1 ) , Q = ( − 1 , 6 , 5 ) , P Q → = ( − 4 , 2 , 6 ) . \begin{align*}
P=&\,(3,4,-1),\\[4mm]
Q=&\,(-1,6,5),\\[4mm]
\overrightarrow{PQ}=&\,(-4,2,6).
\end{align*} P = Q = P Q = ( 3 , 4 , − 1 ) , ( − 1 , 6 , 5 ) , ( − 4 , 2 , 6 ) . A vector equation of the line P Q PQ P Q
r = ( − 1 6 5 ) + t ( − 2 1 3 ) , t ∈ R . \boxed{\mathbf r=\begin{pmatrix}-1\\6\\5\end{pmatrix}
+t\begin{pmatrix}-2\\1\\3\end{pmatrix},\quad t\in\mathbb R}. r = − 1 6 5 + t − 2 1 3 , t ∈ R . This includes Q Q Q t = 0 t=0 t = 0 P P P t = − 2 t=-2 t = − 2
解法二
思路
展开
两直线的公垂线方向已由叉积给出。令PQ与该方向平行，用分量联立求两参数，避免分别写两个点积。
答题过程
展开
From (a), a common perpendicular direction is w = ( − 2 , 1 , 3 ) \mathbf w=(-2,1,3) w = ( − 2 , 1 , 3 ) P P P Q Q Q
P Q → = s w . \overrightarrow{PQ}=s\mathbf w. P Q = s w . Equating components gives
μ − 2 λ = − 2 s , − 5 − 4 μ − λ = s , 11 + 2 μ − λ = 3 s . \begin{align*}
\mu-2\lambda=&\,-2s,\\[4mm]
-5-4\mu-\lambda=&\,s,\\[4mm]
11+2\mu-\lambda=&\,3s.
\end{align*} μ − 2 λ = − 5 − 4 μ − λ = 11 + 2 μ − λ = − 2 s , s , 3 s . Substituting s = − 5 − 4 μ − λ s=-5-4\mu-\lambda s = − 5 − 4 μ − λ
4 λ + 7 μ = − 10 , λ + 7 μ = − 13. \begin{align*}
4\lambda+7\mu=&\,-10,\\[4mm]
\lambda+7\mu=&\,-13.
\end{align*} 4 λ + 7 μ = λ + 7 μ = − 10 , − 13. Subtracting yields λ = 1 \lambda=1 λ = 1 μ = − 2 \mu=-2 μ = − 2 s = 2 s=2 s = 2
Thus P = ( 3 , 4 , − 1 ) P=(3,4,-1) P = ( 3 , 4 , − 1 ) Q = ( − 1 , 6 , 5 ) Q=(-1,6,5) Q = ( − 1 , 6 , 5 ) Q Q Q w \mathbf w w
r = ( − 1 6 5 ) + t ( − 2 1 3 ) , t ∈ R . \boxed{\mathbf r=\begin{pmatrix}-1\\6\\5\end{pmatrix}
+t\begin{pmatrix}-2\\1\\3\end{pmatrix},\quad t\in\mathbb R}. r = − 1 6 5 + t − 2 1 3 , t ∈ R . 