CalcGospel 國際數學圖譜
返回

CIE 9709 2025 June Paper 31 Q8

A Level / CIE / P3

CIE 9709 2025 June Paper 31 Paper · Question 8

题目

Problem

With respect to the origin OO, the points AA and BB have position vectors 2i+4k2\mathbf{i}+4\mathbf{k} and 5i+j+6k5\mathbf{i}+\mathbf{j}+6\mathbf{k} respectively. The line l1l_1 passes through the points AA and BB.

(a) Find a vector equation for the line l1l_1.

(2)

The line l2l_2 has equation r=2i+j+5k+μ(i+2j+3k)\mathbf{r}=2\mathbf{i}+\mathbf{j}+5\mathbf{k}+\mu(\mathbf{i}+2\mathbf{j}+3\mathbf{k}).

(b) Show that l1l_1 and l2l_2 do not intersect.

(4)

(c) Find the acute angle between the directions of l1l_1 and l2l_2.

(3)
题目中文翻译

相对于原点 OO，点 AA 和 BB 的位置向量分别为 2i+4k2\mathbf{i}+4\mathbf{k} 和 5i+j+6k5\mathbf{i}+\mathbf{j}+6\mathbf{k}。直线 l1l_1 经过点 AA 和 BB。

(a) 求直线 l1l_1 的一个向量方程。

(2)

直线 l2l_2 的方程为 r=2i+j+5k+μ(i+2j+3k)\mathbf{r}=2\mathbf{i}+\mathbf{j}+5\mathbf{k}+\mu(\mathbf{i}+2\mathbf{j}+3\mathbf{k})。

(b) 证明 l1l_1 与 l2l_2 不相交。

(4)

(c) 求 l1l_1 与 l2l_2 的方向之间的锐角。

(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

展开

两点之差给出方向向量，再配上直线上的一个已知点。

答题过程

展开

A direction vector for l1l_1 is

AB→=(5,1,6)−(2,0,4)=(3,1,2).\overrightarrow{AB}=(5,1,6)-(2,0,4)=(3,1,2).

Hence

r=(204)+λ(312),λ∈R.\boxed{\mathbf r= \begin{pmatrix}2\\0\\4\end{pmatrix} +\lambda\begin{pmatrix}3\\1\\2\end{pmatrix}, \quad\lambda\in\mathbb R}.

(b)

解法一

思路

展开

前两个分量确定候选参数，第三个分量不相等便构成不相交的证据。

答题过程

展开

If the lines intersected, then

2+3λ= 2+μ,λ= 1+2μ,4+2λ= 5+3μ.\begin{align*} 2+3\lambda=&\,2+\mu,\\[4mm] \lambda=&\,1+2\mu,\\[4mm] 4+2\lambda=&\,5+3\mu. \end{align*}

The first two components give

3λ=μ,λ=1+2μ.3\lambda=\mu,\qquad \lambda=1+2\mu.

Thus λ=1+6λ\lambda=1+6\lambda, so

λ=−15,μ=−35.\lambda=-\frac15,\qquad \mu=-\frac35.

But the third components then become

4+2λ=185,5+3μ=165.4+2\lambda=\frac{18}{5}, \qquad 5+3\mu=\frac{16}{5}.

These are unequal. Hence the three component equations have no simultaneous solution, so l1 and l2 do not intersect\boxed{l_1\text{ and }l_2\text{ do not intersect}}.

(c)

解法一

思路

展开

计算两个方向向量的点积与模长，用锐角公式求夹角。

答题过程

展开

The direction vectors are (3,1,2)(3,1,2) and (1,2,3)(1,2,3). Their scalar product is

3(1)+1(2)+2(3)=11,3(1)+1(2)+2(3)=11,

and both have modulus 14\sqrt{14}.

Thus the acute angle ϕ\phi satisfies

cos⁡ϕ=1114,\cos\phi=\frac{11}{14},

giving

ϕ=cos⁡−11114=0.667 radians.\boxed{\phi=\cos^{-1}\frac{11}{14}=0.667\text{ radians}}.