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CIE 9709 2023 Nov Paper 33 Q11

A Level / CIE / P3

CIE 9709 2023 Nov Paper 33 Paper · Question 11

题目

Problem

The line ll has equation r=i−2j−3k+λ(−i+j+2k)\mathbf{r}=\mathbf{i}-2\mathbf{j}-3\mathbf{k}+\lambda(-\mathbf{i}+\mathbf{j}+2\mathbf{k}). The points AA and BB have position vectors −2i+2j−k-2\mathbf{i}+2\mathbf{j}-\mathbf{k} and 3i−j+k3\mathbf{i}-\mathbf{j}+\mathbf{k} respectively.

(a) Find a unit vector in the direction of ll.

(2)

The line mm passes through the points AA and BB.

(b) Find a vector equation for mm.

(2)

(c) Determine whether lines ll and mm are parallel, intersect or are skew.

(5)
题目中文翻译

直线 ll 的方程为 r=i−2j−3k+λ(−i+j+2k)\mathbf{r}=\mathbf{i}-2\mathbf{j}-3\mathbf{k}+\lambda(-\mathbf{i}+\mathbf{j}+2\mathbf{k})。点 AA 和 BB 的位置向量分别为 −2i+2j−k-2\mathbf{i}+2\mathbf{j}-\mathbf{k} 和 3i−j+k3\mathbf{i}-\mathbf{j}+\mathbf{k}。

(a) 求一个沿 ll 方向的单位向量。

(2)

直线 mm 经过点 AA 和 BB。

(b) 求 mm 的一个向量方程。

(2)

(c) 判断直线 ll 和 mm 是平行、相交还是异面。

(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

展开

将方向向量除以模长，就得到同方向的单位向量。

答题过程

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A direction vector of ll is (−1,1,2)(-1,1,2), whose magnitude is

(−1)2+12+22=6.\sqrt{(-1)^2+1^2+2^2}=\sqrt6.

Thus a unit vector in the direction of ll is

16(−i+j+2k).\boxed{\frac1{\sqrt6}(-\mathbf i+\mathbf j+2\mathbf k)}.

(b)

解法一

思路

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直线通过 AA、BB，用 AB→\overrightarrow{AB} 作方向向量，再选其中一个点的位置向量作起点。

答题过程

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The direction vector is

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

Therefore a vector equation for mm is

r=(−22−1)+μ(5−32).\boxed{\mathbf r= \begin{pmatrix}-2\\2\\-1\end{pmatrix} +\mu\begin{pmatrix}5\\-3\\2\end{pmatrix}}.

(c)

解法一

思路

展开

先检查方向向量是否成比例，排除平行。再联立两个分量求参数，检查第三个分量是否一致。

答题过程

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The direction vectors (−1,1,2)(-1,1,2) and (5,−3,2)(5,-3,2) are not proportional: the ratios from the first two components would be −5-5 and −3-3. Thus the lines are not parallel.

At an intersection, the components would satisfy

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

The first two equations give

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

Subtracting gives 2μ=−12\mu=-1, so μ=−1/2\mu=-1/2 and λ=11/2\lambda=11/2.

The third component on ll is then

−3+2λ=−3+11=8,-3+2\lambda=-3+11=8,

whereas the third component on mm is

−1+2μ=−1−1=−2.-1+2\mu=-1-1=-2.

They are unequal, so the lines do not intersect. Since the lines are neither parallel nor intersecting, l and m are skew\boxed{l\text{ and }m\text{ are skew}}.