题目
Problem
Two lines have equations
r = ( − 1 3 − 4 ) + λ ( 2 3 − 1 ) \mathbf{r}=\begin{pmatrix}-1\\3\\-4\end{pmatrix}+\lambda\begin{pmatrix}2\\3\\-1\end{pmatrix} r = − 1 3 − 4 + λ 2 3 − 1 and
r = ( 2 − 3 − 1 ) + μ ( − 1 − 2 1 ) . \mathbf{r}=\begin{pmatrix}2\\-3\\-1\end{pmatrix}+\mu\begin{pmatrix}-1\\-2\\1\end{pmatrix}. r = 2 − 3 − 1 + μ − 1 − 2 1 . (a) Show that the lines are skew.
(5)
(b) Find the obtuse angle between the directions of the two lines.
(3)
题目中文翻译
两条直线的方程为
r = ( − 1 3 − 4 ) + λ ( 2 3 − 1 ) \mathbf{r}=\begin{pmatrix}-1\\3\\-4\end{pmatrix}+\lambda\begin{pmatrix}2\\3\\-1\end{pmatrix} r = − 1 3 − 4 + λ 2 3 − 1 和
r = ( 2 − 3 − 1 ) + μ ( − 1 − 2 1 ) . \mathbf{r}=\begin{pmatrix}2\\-3\\-1\end{pmatrix}+\mu\begin{pmatrix}-1\\-2\\1\end{pmatrix}. r = 2 − 3 − 1 + μ − 1 − 2 1 . (a) 证明这两条直线为异面直线。
(5)
(b) 求这两条直线的方向之间的钝角。
(3)
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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异面要求既不平行也不相交。先比较方向向量，再用分量方程的不相容证明不相交。
答题过程
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The direction vectors are
d 1 = ( 2 , 3 , − 1 ) , d 2 = ( − 1 , − 2 , 1 ) . \mathbf d_1=(2,3,-1),\qquad \mathbf d_2=(-1,-2,1). d 1 = ( 2 , 3 , − 1 ) , d 2 = ( − 1 , − 2 , 1 ) . They are not parallel, since 2 / ( − 1 ) = − 2 2/(-1)=-2 2/ ( − 1 ) = − 2 3 / ( − 2 ) = − 3 / 2 3/(-2)=-3/2 3/ ( − 2 ) = − 3/2
If the lines intersected, their components would satisfy
− 1 + 2 λ = 2 − μ , 3 + 3 λ = − 3 − 2 μ , − 4 − λ = − 1 + μ . \begin{align*}
-1+2\lambda=&\,2-\mu,\\[4mm]
3+3\lambda=&\,-3-2\mu,\\[4mm]
-4-\lambda=&\,-1+\mu.
\end{align*} − 1 + 2 λ = 3 + 3 λ = − 4 − λ = 2 − μ , − 3 − 2 μ , − 1 + μ . The first two components give
2 λ + μ = 3 , 3 λ + 2 μ = − 6. \begin{align*}
2\lambda+\mu=&\,3,\\[4mm]
3\lambda+2\mu=&\,-6.
\end{align*} 2 λ + μ = 3 λ + 2 μ = 3 , − 6. Twice the first equation minus the second gives λ = 12 \lambda=12 λ = 12 μ = 3 − 24 = − 21 \mu=3-24=-21 μ = 3 − 24 = − 21
The third components would then be − 4 − 12 = − 16 -4-12=-16 − 4 − 12 = − 16 − 1 − 21 = − 22 -1-21=-22 − 1 − 21 = − 22
Since they are neither parallel nor intersecting, the lines are skew \boxed{\text{the lines are skew}} the lines are skew
(b)
解法一
思路
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题目要求钝角，点积的负号要保留，不能使用锐角公式的绝对值。
答题过程
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Using the given direction vectors,
d 1 ⋅ d 2 = 2 ( − 1 ) + 3 ( − 2 ) + ( − 1 ) ( 1 ) = − 9. \mathbf d_1\cdot\mathbf d_2=2(-1)+3(-2)+(-1)(1)=-9. d 1 ⋅ d 2 = 2 ( − 1 ) + 3 ( − 2 ) + ( − 1 ) ( 1 ) = − 9. Their moduli are 14 \sqrt{14} 14 6 \sqrt6 6 ϕ \phi ϕ
cos ϕ = − 9 14 6 . \cos\phi=-\frac9{\sqrt{14}\sqrt6}. cos ϕ = − 14 6 9 . Therefore
ϕ = 169.1 ∘ . \boxed{\phi=169.1^\circ}. ϕ = 169. 1 ∘ . 