题目
Problem
The equations of two straight lines are
r = i + j + 2 a k + λ ( 3 i + 4 j + a k ) \mathbf{r}=\mathbf{i}+\mathbf{j}+2a\mathbf{k}+\lambda(3\mathbf{i}+4\mathbf{j}+a\mathbf{k}) r = i + j + 2 a k + λ ( 3 i + 4 j + a k ) and
r = − 3 i − j + 4 k + μ ( − i + 2 j + 2 k ) , \mathbf{r}=-3\mathbf{i}-\mathbf{j}+4\mathbf{k}+\mu(-\mathbf{i}+2\mathbf{j}+2\mathbf{k}), r = − 3 i − j + 4 k + μ ( − i + 2 j + 2 k ) , where a a a
(a) Given that the acute angle between the directions of these lines is 1 4 π \frac14\pi 4 1 π a a a
(6)
(b) Given instead that the lines intersect, find the value of a a a
(5)
题目中文翻译
两条直线的方程为
r = i + j + 2 a k + λ ( 3 i + 4 j + a k ) \mathbf{r}=\mathbf{i}+\mathbf{j}+2a\mathbf{k}+\lambda(3\mathbf{i}+4\mathbf{j}+a\mathbf{k}) r = i + j + 2 a k + λ ( 3 i + 4 j + a k ) 和
r = − 3 i − j + 4 k + μ ( − i + 2 j + 2 k ) , \mathbf{r}=-3\mathbf{i}-\mathbf{j}+4\mathbf{k}+\mu(-\mathbf{i}+2\mathbf{j}+2\mathbf{k}), r = − 3 i − j + 4 k + μ ( − i + 2 j + 2 k ) , 其中 a a a
(a) 已知这两条直线方向之间的锐角为 1 4 π \frac14\pi 4 1 π a a a
(6)
(b) 改为已知这两条直线相交，求 a a a
(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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直线夹角由方向向量决定。使用点积与模长的公式，先列出关于 a a a
答题过程
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Direction vectors of the two lines are
d 1 = ( 3 4 a ) , d 2 = ( − 1 2 2 ) . \begin{align*}
\mathbf d_1=&\,\begin{pmatrix}3\\4\\a\end{pmatrix},\\[4mm]
\mathbf d_2=&\,\begin{pmatrix}-1\\2\\2\end{pmatrix}.
\end{align*} d 1 = d 2 = 3 4 a , − 1 2 2 . Their scalar product and moduli are
d 1 ⋅ d 2 = 5 + 2 a , ∣ d 1 ∣ = 25 + a 2 , ∣ d 2 ∣ = 3. \begin{align*}
\mathbf d_1\cdot\mathbf d_2=&\,5+2a,\\[4mm]
|\mathbf d_1|=&\,\sqrt{25+a^2},\\[4mm]
|\mathbf d_2|=&\,3.
\end{align*} d 1 ⋅ d 2 = ∣ d 1 ∣ = ∣ d 2 ∣ = 5 + 2 a , 25 + a 2 , 3. Since the acute angle is π / 4 \pi/4 π /4
∣ 5 + 2 a ∣ 3 25 + a 2 = 1 2 . \frac{|5+2a|}{3\sqrt{25+a^2}}=\frac1{\sqrt2}. 3 25 + a 2 ∣5 + 2 a ∣ = 2 1 . Squaring and simplifying,
2 ( 5 + 2 a ) 2 = 9 ( 25 + a 2 ) , 50 + 40 a + 8 a 2 = 225 + 9 a 2 , a 2 − 40 a + 175 = 0 , ( a − 5 ) ( a − 35 ) = 0. \begin{align*}
2(5+2a)^2=&\,9(25+a^2),\\[4mm]
50+40a+8a^2=&\,225+9a^2,\\[4mm]
a^2-40a+175=&\,0,\\[4mm]
(a-5)(a-35)=&\,0.
\end{align*} 2 ( 5 + 2 a ) 2 = 50 + 40 a + 8 a 2 = a 2 − 40 a + 175 = ( a − 5 ) ( a − 35 ) = 9 ( 25 + a 2 ) , 225 + 9 a 2 , 0 , 0. Both values satisfy the angle condition. Thus
a = 5 or a = 35 . \boxed{a=5\quad\text{or}\quad a=35}. a = 5 or a = 35 .
(b)
解法一
思路
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交点在两条直线上，因此三个分量分别相等。先由前两个分量求参数，再由第三个分量求 a a a
答题过程
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At an intersection,
1 + 3 λ = − 3 − μ , 1 + 4 λ = − 1 + 2 μ , 2 a + a λ = 4 + 2 μ . \begin{align*}
1+3\lambda=&\,-3-\mu,\\[4mm]
1+4\lambda=&\,-1+2\mu,\\[4mm]
2a+a\lambda=&\,4+2\mu.
\end{align*} 1 + 3 λ = 1 + 4 λ = 2 a + aλ = − 3 − μ , − 1 + 2 μ , 4 + 2 μ . The first two equations give
3 λ + μ = − 4 , 2 λ − μ = − 1. \begin{align*}
3\lambda+\mu=&\,-4,\\[4mm]
2\lambda-\mu=&\,-1.
\end{align*} 3 λ + μ = 2 λ − μ = − 4 , − 1. Adding gives 5 λ = − 5 5\lambda=-5 5 λ = − 5 λ = − 1 \lambda=-1 λ = − 1 μ = − 1 \mu=-1 μ = − 1
The third equation now gives
2 a − a = 4 − 2 , a = 2 . 2a-a=4-2,
\qquad \boxed{a=2}. 2 a − a = 4 − 2 , a = 2 . Substituting into either line, the point of intersection has position vector
− 2 i − 3 j + 2 k . \boxed{-2\mathbf i-3\mathbf j+2\mathbf k}. − 2 i − 3 j + 2 k . 