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CIE 9709 2025 November Paper 35 Q9

A Level / CIE / P3

CIE 9709 2025 Nov Paper 35 Paper · Question 9

题目

Problem

The equations of two lines are given by

l1:r= (2i+j+4k)+λ(i+2j−3k),l2:r= (3i−j+5k)+μ(2i+3j+ak).\begin{align*} l_1:\quad\mathbf{r}=&\,(2\mathbf{i}+\mathbf{j}+4\mathbf{k})+\lambda(\mathbf{i}+2\mathbf{j}-3\mathbf{k}),\\ l_2:\quad\mathbf{r}=&\,(3\mathbf{i}-\mathbf{j}+5\mathbf{k})+\mu(2\mathbf{i}+3\mathbf{j}+a\mathbf{k}). \end{align*}

(a) Find the value of aa for which l1l_1 is perpendicular to l2l_2.

(2)

(b) Find the value of aa for which l1l_1 and l2l_2 intersect.

(4)

(c) Find the values of aa for which the acute angle between l1l_1 and l2l_2 is equal to cos⁡−1(514)\cos^{-1}\big(\frac5{14}\big).

(4)
题目中文翻译

两条直线的方程为

l1:r= (2i+j+4k)+λ(i+2j−3k),l2:r= (3i−j+5k)+μ(2i+3j+ak).\begin{align*} l_1:\quad\mathbf{r}=&\,(2\mathbf{i}+\mathbf{j}+4\mathbf{k})+\lambda(\mathbf{i}+2\mathbf{j}-3\mathbf{k}),\\ l_2:\quad\mathbf{r}=&\,(3\mathbf{i}-\mathbf{j}+5\mathbf{k})+\mu(2\mathbf{i}+3\mathbf{j}+a\mathbf{k}). \end{align*}

(a) 求使 l1l_1 垂直于 l2l_2 的 aa 的值。

(2)

(b) 求使 l1l_1 与 l2l_2 相交的 aa 的值。

(4)

(c) 求使 l1l_1 与 l2l_2 之间的锐角等于 cos⁡−1(514)\cos^{-1}\big(\frac5{14}\big) 的 aa 的值。

(4)
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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两直线垂直时方向向量点积为零。

答题过程

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The direction vectors are d1=(1,2,−3)\mathbf d_1=(1,2,-3) and d2=(2,3,a)\mathbf d_2=(2,3,a).

Perpendicular directions require

d1⋅d2=2+6−3a=0.\mathbf d_1\cdot\mathbf d_2=2+6-3a=0.

Thus a=83\boxed{a=\frac83}.

(b)

解法一

思路

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交点的三个分量必须同时相等。前两个分量不含 aa，可先求参数，再由第三个分量求 aa。

答题过程

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At an intersection,

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

The first two components give

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

Substituting the first into the second,

2+4μ−3μ=−2⟹μ=−4, λ=−7.2+4\mu-3\mu=-2 \quad\Longrightarrow\quad \mu=-4,\ \lambda=-7.

The third component then requires

4−3(−7)=5−4a.4-3(-7)=5-4a.

Thus 25=5−4a25=5-4a, giving a=−5\boxed{a=-5}.

(c)

解法一

思路

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锐角公式要用点积的绝对值，这样平方后得到的两个方向都可保留。求解后再验证目标余弦。

答题过程

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The acute angle θ\theta between the lines satisfies

cos⁡θ=∣8−3a∣1413+a2=514.\cos\theta =\frac{|8-3a|}{\sqrt{14}\sqrt{13+a^2}} =\frac5{14}.

Squaring and multiplying through gives

14(8−3a)2=25(13+a2).14(8-3a)^2=25(13+a^2).

Expanding,

14(64−48a+9a2)=325+25a2.14(64-48a+9a^2)=325+25a^2.

Collecting terms and factorising,

101a2−672a+571=0⟹(a−1)(101a−571)=0.101a^2-672a+571=0 \quad\Longrightarrow\quad (a-1)(101a-571)=0.

Hence the candidates are a=1a=1 and a=571/101a=571/101. The denominator in the cosine expression is positive, and the absolute value makes both sides nonnegative, so squaring introduces no sign restriction.

Both values give ∣d1⋅d2∣/(∣d1∣∣d2∣)=5/14|\mathbf d_1\cdot\mathbf d_2|/(|\mathbf d_1||\mathbf d_2|)=5/14. Therefore

a=1ora=571101.\boxed{a=1\quad\text{or}\quad a=\frac{571}{101}}.