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CIE 9709 2025 November Paper 33 Q9

A Level / CIE / P3

CIE 9709 2025 Nov Paper 33 Paper · Question 9

题目

Problem

The line l1l_1 passes through the point (3,1,−6)(3,1,-6) and is parallel to the vector 2i+j+4k2\mathbf{i}+\mathbf{j}+4\mathbf{k}.

The line l2l_2 passes through the point (−1,3,−6)(-1,3,-6) and is perpendicular to the vector 3i−2j+k3\mathbf{i}-2\mathbf{j}+\mathbf{k}. The direction vector for l2l_2 has no component in the xx-direction.

(a) Write down a vector equation for l1l_1 and find a vector equation for l2l_2.

(3)

(b) Calculate the acute angle between l1l_1 and l2l_2.

(3)

(c) Find the position vector of the point of intersection of l1l_1 and l2l_2.

(3)
题目中文翻译

直线 l1l_1 经过点 (3,1,−6)(3,1,-6)，且平行于向量 2i+j+4k2\mathbf{i}+\mathbf{j}+4\mathbf{k}。

直线 l2l_2 经过点 (−1,3,−6)(-1,3,-6)，且垂直于向量 3i−2j+k3\mathbf{i}-2\mathbf{j}+\mathbf{k}。l2l_2 的方向向量在 xx 方向上的分量为零。

(a) 写出 l1l_1 的一个向量方程，并求 l2l_2 的一个向量方程。

(3)

(b) 计算 l1l_1 与 l2l_2 之间的锐角。

(3)

(c) 求 l1l_1 与 l2l_2 的交点的位置向量。

(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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第一条直线由已知点和方向直接写出。第二条方向没有 xx 分量，再用垂直的点积条件确定另两个分量比例。

答题过程

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For l1l_1,

r=(31−6)+λ(214),λ∈R.\boxed{\mathbf r= \begin{pmatrix}3\\1\\-6\end{pmatrix} +\lambda\begin{pmatrix}2\\1\\4\end{pmatrix}, \quad\lambda\in\mathbb R}.

Let a direction vector of l2l_2 be (0,a,b)(0,a,b). Perpendicularity to (3,−2,1)(3,-2,1) gives

(0,a,b)⋅(3,−2,1)=−2a+b=0.(0,a,b)\cdot(3,-2,1)=-2a+b=0.

Thus b=2ab=2a, and we may choose (0,1,2)(0,1,2). Hence

r=(−13−6)+μ(012),μ∈R.\boxed{\mathbf r= \begin{pmatrix}-1\\3\\-6\end{pmatrix} +\mu\begin{pmatrix}0\\1\\2\end{pmatrix}, \quad\mu\in\mathbb R}.

(b)

解法一

思路

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两直线的锐角用方向向量的点积绝对值求余弦。

答题过程

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With d1=(2,1,4)\mathbf d_1=(2,1,4) and d2=(0,1,2)\mathbf d_2=(0,1,2),

d1⋅d2=0+1+8=9,\mathbf d_1\cdot\mathbf d_2=0+1+8=9,

and ∣d1∣=21|\mathbf d_1|=\sqrt{21}, ∣d2∣=5|\mathbf d_2|=\sqrt5.

Thus the acute angle θ\theta satisfies

cos⁡θ=9215=9105.\cos\theta=\frac9{\sqrt{21}\sqrt5} =\frac9{\sqrt{105}}.

Therefore

θ=0.498 radians.\boxed{\theta=0.498\text{ radians}}.

(c)

解法一

思路

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联立分量求两个参数，并用第三个分量检验确实相交。

答题过程

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At the intersection,

3+2λ= −1,1+λ= 3+μ,−6+4λ= −6+2μ.\begin{align*} 3+2\lambda=&\,-1,\\[4mm] 1+\lambda=&\,3+\mu,\\[4mm] -6+4\lambda=&\,-6+2\mu. \end{align*}

The first component gives 3+2λ=−13+2\lambda=-1, so λ=−2\lambda=-2.

The second then gives 1−2=3+μ1-2=3+\mu, so μ=−4\mu=-4. The third components agree:

−6+4(−2)=−14=−6+2(−4).-6+4(-2)=-14=-6+2(-4).

Therefore the position vector is

−i−j−14k.\boxed{-\mathbf i-\mathbf j-14\mathbf k}.