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CIE 9709 2023 March Paper 32 Q10

A Level / CIE / P3

CIE 9709 2023 March Paper 32 Paper · Question 10

题目

Problem

With respect to the origin OO, the points AA, BB, CC and DD have position vectors given by

OA→=(3−12),OB→=(12−3),\overrightarrow{OA}=\begin{pmatrix}3\\-1\\2\end{pmatrix},\quad\overrightarrow{OB}=\begin{pmatrix}1\\2\\-3\end{pmatrix}, OC→=(1−25)andOD→=(5−611).\overrightarrow{OC}=\begin{pmatrix}1\\-2\\5\end{pmatrix}\quad\text{and}\quad\overrightarrow{OD}=\begin{pmatrix}5\\-6\\11\end{pmatrix}.

(a) Find the obtuse angle between the vectors OA→\overrightarrow{OA} and OB→\overrightarrow{OB}.

(3)

The line ll passes through the points AA and BB.

(b) Find a vector equation for the line ll.

(2)

(c) Find the position vector of the point of intersection of the line ll and the line passing through CC and DD.

(4)
题目中文翻译

相对于原点 OO，点 AA、BB、CC 和 DD 的位置向量分别为

OA→=(3−12),OB→=(12−3),\overrightarrow{OA}=\begin{pmatrix}3\\-1\\2\end{pmatrix},\quad\overrightarrow{OB}=\begin{pmatrix}1\\2\\-3\end{pmatrix}, OC→=(1−25)和OD→=(5−611).\overrightarrow{OC}=\begin{pmatrix}1\\-2\\5\end{pmatrix}\quad\text{和}\quad\overrightarrow{OD}=\begin{pmatrix}5\\-6\\11\end{pmatrix}.

(a) 求向量 OA→\overrightarrow{OA} 与 OB→\overrightarrow{OB} 之间的钝角。

(3)

直线 ll 经过点 AA 和 BB。

(b) 求直线 ll 的一个向量方程。

(2)

(c) 求直线 ll 与经过 CC 和 DD 的直线的交点的位置向量。

(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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角度是位置向量之间的钝角，不取点积的绝对值。两个向量长度相同，可简化分母。

答题过程

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The scalar product is

OA→⋅OB→=3(1)+(−1)(2)+2(−3)=−5.\overrightarrow{OA}\cdot\overrightarrow{OB} =3(1)+(-1)(2)+2(-3)=-5.

Also,

∣OA→∣=∣OB→∣=14.|\overrightarrow{OA}|=|\overrightarrow{OB}|=\sqrt{14}.

Thus

cos⁡θ=−514.\cos\theta=-\frac5{14}.

Therefore the obtuse angle is

θ=cos⁡−1(−514)≈110.9∘.\theta=\cos^{-1}\bigg(-\frac5{14}\bigg) \approx\boxed{110.9^\circ}.

(b)

解法一

思路

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用 AA 作起点、AB→\overrightarrow{AB} 作方向向量，写出直线方程。

答题过程

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The direction vector is

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

Hence a vector equation is

r=(3−12)+s(−23−5).\boxed{\mathbf r= \begin{pmatrix}3\\-1\\2\end{pmatrix} +s\begin{pmatrix}-2\\3\\-5\end{pmatrix}}.

(c)

解法一

思路

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先写出 CDCD 方程，再与(b)直线的分量联立。由两个分量求参数后，应核对第三个分量。

答题过程

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A direction vector for CDCD is

CD→=(5,−6,11)−(1,−2,5)=(4,−4,6).\overrightarrow{CD}=(5,-6,11)-(1,-2,5)=(4,-4,6).

Thus a point on CDCD is (1+4t,−2−4t,5+6t)(1+4t,-2-4t,5+6t). At an intersection with ll,

3−2s= 1+4t,−1+3s= −2−4t,2−5s= 5+6t.\begin{align*} 3-2s=&\,1+4t,\\[4mm] -1+3s=&\,-2-4t,\\[4mm] 2-5s=&\,5+6t. \end{align*}

The first two equations give

s+2t= 1,3s+4t= −1.\begin{align*} s+2t=&\,1,\\[4mm] 3s+4t=&\,-1. \end{align*}

Subtracting twice the first from the second gives s=−3s=-3, and then t=2t=2.

The third components agree: 2−5(−3)=17=5+6(2)2-5(-3)=17=5+6(2). Hence the intersection is (9,−10,17)(9,-10,17), with position vector

9i−10j+17k.\boxed{9\mathbf i-10\mathbf j+17\mathbf k}.