题目
Problem
The lines l 1 l_1 l 1 l 2 l_2 l 2
r = − 2 i + 4 j + 4 k + λ ( 3 i − 4 j + 2 k ) , r = − 3 i − 2 j + 4 k + μ ( 2 i + 4 j + 3 k ) \begin{align*}
\mathbf{r}=&\,-2\mathbf{i}+4\mathbf{j}+4\mathbf{k}+\lambda(3\mathbf{i}-4\mathbf{j}+2\mathbf{k}),\\
\mathbf{r}=&\,-3\mathbf{i}-2\mathbf{j}+4\mathbf{k}+\mu(2\mathbf{i}+4\mathbf{j}+3\mathbf{k})
\end{align*} r = r = − 2 i + 4 j + 4 k + λ ( 3 i − 4 j + 2 k ) , − 3 i − 2 j + 4 k + μ ( 2 i + 4 j + 3 k ) respectively.
(a) Find the shortest distance between l 1 l_1 l 1 l 2 l_2 l 2
(5)
(b) The plane Π 1 \Pi_1 Π 1 l 1 l_1 l 1 P ( 1 , − 1 , 1 ) P(1,-1,1) P ( 1 , − 1 , 1 )
The plane Π 2 \Pi_2 Π 2 l 2 l_2 l 2
Find the acute angle between Π 1 \Pi_1 Π 1 Π 2 \Pi_2 Π 2
(6)
题目中文翻译
直线 l 1 l_1 l 1 l 2 l_2 l 2
r = − 2 i + 4 j + 4 k + λ ( 3 i − 4 j + 2 k ) , r = − 3 i − 2 j + 4 k + μ ( 2 i + 4 j + 3 k ) . \begin{align*}
\mathbf{r}=&\,-2\mathbf{i}+4\mathbf{j}+4\mathbf{k}+\lambda(3\mathbf{i}-4\mathbf{j}+2\mathbf{k}),\\
\mathbf{r}=&\,-3\mathbf{i}-2\mathbf{j}+4\mathbf{k}+\mu(2\mathbf{i}+4\mathbf{j}+3\mathbf{k}).
\end{align*} r = r = − 2 i + 4 j + 4 k + λ ( 3 i − 4 j + 2 k ) , − 3 i − 2 j + 4 k + μ ( 2 i + 4 j + 3 k ) . (a) 求 l 1 l_1 l 1 l 2 l_2 l 2
(5)
(b) 平面 Π 1 \Pi_1 Π 1 l 1 l_1 l 1 P ( 1 , − 1 , 1 ) P(1,-1,1) P ( 1 , − 1 , 1 )
平面 Π 2 \Pi_2 Π 2 l 2 l_2 l 2
求 Π 1 \Pi_1 Π 1 Π 2 \Pi_2 Π 2
(6)
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
先用两条直线的方向向量叉积找到公垂线方向，再把连接两线上已知点的向量投影到这个方向。
答题过程
展开
Take points A = ( − 2 , 4 , 4 ) A=(-2,4,4) A = ( − 2 , 4 , 4 ) l 1 l_1 l 1 B = ( − 3 , − 2 , 4 ) B=(-3,-2,4) B = ( − 3 , − 2 , 4 ) l 2 l_2 l 2
d 1 = ( 3 , − 4 , 2 ) , d 2 = ( 2 , 4 , 3 ) . \mathbf d_1=(3,-4,2),\qquad\mathbf d_2=(2,4,3). d 1 = ( 3 , − 4 , 2 ) , d 2 = ( 2 , 4 , 3 ) . A common perpendicular direction is
d 1 × d 2 = ( − 12 − 8 , 4 − 9 , 12 + 8 ) = ( − 20 , − 5 , 20 ) = 5 ( − 4 , − 1 , 4 ) . \begin{align*}
\mathbf d_1\times\mathbf d_2
=&\,(-12-8,\ 4-9,\ 12+8) \\[4mm]
=&\,(-20,-5,20)=5(-4,-1,4).
\end{align*} d 1 × d 2 = = ( − 12 − 8 , 4 − 9 , 12 + 8 ) ( − 20 , − 5 , 20 ) = 5 ( − 4 , − 1 , 4 ) . Also A B → = ( − 1 , − 6 , 0 ) \overrightarrow{AB}=(-1,-6,0) A B = ( − 1 , − 6 , 0 )
d = ∣ A B → ⋅ ( − 4 , − 1 , 4 ) ∣ 16 + 1 + 16 = ∣ 4 + 6 ∣ 33 = 10 33 . d=\frac{|\overrightarrow{AB}\cdot(-4,-1,4)|}{\sqrt{16+1+16}}
=\frac{|4+6|}{\sqrt{33}}
=\boxed{\frac{10}{\sqrt{33}}}. d = 16 + 1 + 16 ∣ A B ⋅ ( − 4 , − 1 , 4 ) ∣ = 33 ∣4 + 6∣ = 33 10 .
解法二
思路
展开
也可直接求两条直线上的最近点。它们的连接向量要同时垂直于两条线，给出两个参数方程。
答题过程
展开
Let U U U l 1 l_1 l 1 V V V l 2 l_2 l 2 λ , μ \lambda,\mu λ , μ
U V → = ( − 1 + 2 μ − 3 λ − 6 + 4 μ + 4 λ 3 μ − 2 λ ) . \overrightarrow{UV}=\begin{pmatrix}
-1+2\mu-3\lambda\\
-6+4\mu+4\lambda\\
3\mu-2\lambda
\end{pmatrix}. U V = − 1 + 2 μ − 3 λ − 6 + 4 μ + 4 λ 3 μ − 2 λ . At the nearest points, U V → \overrightarrow{UV} U V
21 − 4 μ − 29 λ = 0 , − 26 + 29 μ + 4 λ = 0. \begin{align*}
21-4\mu-29\lambda=&\,0,\\[4mm]
-26+29\mu+4\lambda=&\,0.
\end{align*} 21 − 4 μ − 29 λ = − 26 + 29 μ + 4 λ = 0 , 0. Solving,
λ = 101 165 , μ = 134 165 , \lambda=\frac{101}{165},\qquad\mu=\frac{134}{165}, λ = 165 101 , μ = 165 134 , and hence
U V → = ( − 40 33 , − 10 33 , 40 33 ) . \overrightarrow{UV}=\bigg(-\frac{40}{33},-\frac{10}{33},\frac{40}{33}\bigg). U V = ( − 33 40 , − 33 10 , 33 40 ) . Therefore the shortest distance is
∣ U V → ∣ = 1600 + 100 + 1600 33 = 10 33 . |\overrightarrow{UV}|=\frac{\sqrt{1600+100+1600}}{33}
=\boxed{\frac{10}{\sqrt{33}}}. ∣ U V ∣ = 33 1600 + 100 + 1600 = 33 10 .
(b)
解法一
思路
展开
每个平面都由两条不平行的平面内方向确定。分别叉积求法向量，再用法向量夹角的余弦绝对值求锐角。
答题过程
展开
For Π 1 \Pi_1 Π 1 P ( 1 , − 1 , 1 ) P(1,-1,1) P ( 1 , − 1 , 1 ) A ( − 2 , 4 , 4 ) A(-2,4,4) A ( − 2 , 4 , 4 ) ( − 3 , 5 , 3 ) (-3,5,3) ( − 3 , 5 , 3 )
n 1 = ( − 3 , 5 , 3 ) × ( 3 , − 4 , 2 ) = ( 22 , 15 , − 3 ) . \mathbf n_1=(-3,5,3)\times(3,-4,2)
=(22,15,-3). n 1 = ( − 3 , 5 , 3 ) × ( 3 , − 4 , 2 ) = ( 22 , 15 , − 3 ) . For Π 2 \Pi_2 Π 2 ( − 3 , − 2 , 4 ) (-3,-2,4) ( − 3 , − 2 , 4 ) l 2 l_2 l 2 ( 2 , 4 , 3 ) (2,4,3) ( 2 , 4 , 3 )
n 2 = ( − 3 , − 2 , 4 ) × ( 2 , 4 , 3 ) = ( − 22 , 17 , − 8 ) . \mathbf n_2=(-3,-2,4)\times(2,4,3)
=(-22,17,-8). n 2 = ( − 3 , − 2 , 4 ) × ( 2 , 4 , 3 ) = ( − 22 , 17 , − 8 ) . Their dot product and squared lengths are
n 1 ⋅ n 2 = − 484 + 255 + 24 = − 205 , ∣ n 1 ∣ 2 = 718 , ∣ n 2 ∣ 2 = 837. \begin{align*}
\mathbf n_1\cdot\mathbf n_2=&\,-484+255+24=-205,\\[4mm]
|\mathbf n_1|^2=&\,718,\\[4mm]
|\mathbf n_2|^2=&\,837.
\end{align*} n 1 ⋅ n 2 = ∣ n 1 ∣ 2 = ∣ n 2 ∣ 2 = − 484 + 255 + 24 = − 205 , 718 , 837. The acute angle ϕ \phi ϕ
cos ϕ = 205 718 837 . \cos\phi=\frac{205}{\sqrt{718}\sqrt{837}}. cos ϕ = 718 837 205 . Thus ϕ ≈ 74.7 ∘ \boxed{\phi\approx74.7^\circ} ϕ ≈ 74. 7 ∘ 1 1 1
