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CIE 9709 2026 June Paper 35 Q11

A Level / CIE / P3

CIE 9709 2026 June Paper 35 Paper · Question 11

题目

Problem

Relative to a fixed point, OO, the points AA, BB, CC and DD have position vectors

OA→= (235),OB→=(314),OC→= (−10−1)andOD→=(−341).\begin{align*} \overrightarrow{OA}=&\,\begin{pmatrix}2\\3\\5\end{pmatrix},\quad \overrightarrow{OB}=\begin{pmatrix}3\\1\\4\end{pmatrix},\\ \overrightarrow{OC}=&\,\begin{pmatrix}-1\\0\\-1\end{pmatrix}\quad\text{and}\quad \overrightarrow{OD}=\begin{pmatrix}-3\\4\\1\end{pmatrix}. \end{align*}

(a) Show that ABCDABCD is a trapezium, and not a parallelogram.

(4)

(b) Show that the distance from AA to the line through CC and DD is 922\frac92\sqrt2.

(5)

(c) Find the exact area of ABCDABCD. Give your answer in simplified form.

(3)
题目中文翻译

相对于定点 OO，点 AA、BB、CC 和 DD 的位置向量为

OA→= (235),OB→=(314),OC→= (−10−1),OD→=(−341).\begin{align*} \overrightarrow{OA}=&\,\begin{pmatrix}2\\3\\5\end{pmatrix},\quad \overrightarrow{OB}=\begin{pmatrix}3\\1\\4\end{pmatrix},\\ \overrightarrow{OC}=&\,\begin{pmatrix}-1\\0\\-1\end{pmatrix},\quad \overrightarrow{OD}=\begin{pmatrix}-3\\4\\1\end{pmatrix}. \end{align*}

(a) 证明 ABCDABCD 是梯形，而不是平行四边形。

(4)

(b) 证明点 AA 到经过 CC 和 DD 的直线的距离为 922\frac92\sqrt2。

(5)

(c) 求 ABCDABCD 的精确面积。答案写成化简后的形式。

(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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先求两组对边向量。找到一组平行边后，用另一组不平行说明不是平行四边形。

答题过程

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The side vectors are

AB→= (1,−2,−1),DC→= (2,−4,−2)=2AB→.\begin{align*} \overrightarrow{AB}=&\,(1,-2,-1),\\[4mm] \overrightarrow{DC}=&\,(2,-4,-2)=2\overrightarrow{AB}. \end{align*}

Thus ABAB and DCDC are parallel. The other pair gives

AD→= (−5,1,−4),BC→= (−4,−1,−5).\begin{align*} \overrightarrow{AD}=&\,(-5,1,-4),\\[4mm] \overrightarrow{BC}=&\,(-4,-1,-5). \end{align*}

These are not scalar multiples. The points are coplanar because AC→=AD→+2AB→\overrightarrow{AC}=\overrightarrow{AD}+2\overrightarrow{AB}, so they form a trapezium with exactly one pair of parallel sides. The nonparallel second pair also proves that it is not a parallelogram.

(b)

解法一

思路

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设垂足为 CDCD 上的一般点。用垂线与直线方向向量的数量积等于零求参数，再计算垂线长度。

答题过程

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Let P=C+μCD→P=C+\mu\overrightarrow{CD}. Then

AP→=(−3−2μ,−3+4μ,−6+2μ).\overrightarrow{AP}=(-3-2\mu,-3+4\mu,-6+2\mu).

At the perpendicular foot,

 −2(−3−2μ)+4(−3+4μ)+2(−6+2μ)=0, 6+4μ−12+16μ−12+4μ=0, −18+24μ=0.\begin{align*} &\,-2(-3-2\mu)+4(-3+4\mu)+2(-6+2\mu)=0,\\[4mm] &\,6+4\mu-12+16\mu-12+4\mu=0,\\[4mm] &\,-18+24\mu=0. \end{align*}

Thus μ=3/4\mu=3/4 and AP→=(−9/2,0,−9/2)\overrightarrow{AP}=(-9/2,0,-9/2). Its length is (−9/2)2+(−9/2)2=922\sqrt{(-9/2)^2+(-9/2)^2}=\frac92\sqrt2, as required.

解法二

思路

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也可在三角形 ACDACD 内，先用数量积求 CC 点的角。ACAC 乘这个角的正弦就是到 CDCD 的垂距。

答题过程

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The vectors from CC are CA→=(3,3,6)\overrightarrow{CA}=(3,3,6) and CD→=(−2,4,2)\overrightarrow{CD}=(-2,4,2). Their dot product is 1818, and their moduli are 54\sqrt{54} and 24\sqrt{24}. Hence

cos⁡∠ACD= 185424=12,sin⁡∠ACD= 1−14=32.\begin{align*} \cos\angle ACD=&\,\frac{18}{\sqrt{54}\sqrt{24}}=\frac12,\\[4mm] \sin\angle ACD=&\,\sqrt{1-\frac14}=\frac{\sqrt3}{2}. \end{align*}

The perpendicular distance is

∣CA∣sin⁡∠ACD=5432=922.|CA|\sin\angle ACD =\sqrt{54}\frac{\sqrt3}{2}=\frac92\sqrt2.

(c)

解法一

思路

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使用梯形面积公式，两条平行边的长度可由向量模长求得，高就是上一问的垂距。

答题过程

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The parallel side lengths are ∣AB∣=6|AB|=\sqrt6 and ∣DC∣=26|DC|=2\sqrt6. Using the height from part (b), the area is

12(6+26)922=2723.\frac12(\sqrt6+2\sqrt6)\frac92\sqrt2 =\frac{27}{2}\sqrt3.