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CIE 9709 2024 March Paper 32 Q9

A Level / CIE / P3

CIE 9709 2024 March Paper 32 Paper · Question 9

题目

Problem

Relative to the origin OO, the position vectors of the points AA, BB and CC are given by

OA→=5i−2j+k,\overrightarrow{OA}=5\mathbf{i}-2\mathbf{j}+\mathbf{k}, OB→=8i+2j−6k\overrightarrow{OB}=8\mathbf{i}+2\mathbf{j}-6\mathbf{k}

and

OC→=3i+4j−7k.\overrightarrow{OC}=3\mathbf{i}+4\mathbf{j}-7\mathbf{k}.

(a) Show that OABCOABC is a rectangle.

(4)

(b) Use a scalar product to find the acute angle between the diagonals of OABCOABC.

(4)
题目中文翻译

相对于原点 OO，点 AA、BB 和 CC 的位置向量分别为

OA→=5i−2j+k,\overrightarrow{OA}=5\mathbf{i}-2\mathbf{j}+\mathbf{k}, OB→=8i+2j−6k\overrightarrow{OB}=8\mathbf{i}+2\mathbf{j}-6\mathbf{k}

和

OC→=3i+4j−7k.\overrightarrow{OC}=3\mathbf{i}+4\mathbf{j}-7\mathbf{k}.

(a) 证明 OABCOABC 是一个矩形。

(4)

(b) 使用数量积求 OABCOABC 的两条对角线之间的锐角。

(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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先证明一对对边向量相等，得到平行四边形；再用相邻边点积为零证明其中一个角是直角。

答题过程

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The side vectors are

AB→= OB→−OA→=(3,4,−7)=OC→,CB→= OB→−OC→=(5,−2,1)=OA→.\begin{align*} \overrightarrow{AB} =&\,\overrightarrow{OB}-\overrightarrow{OA} =(3,4,-7)=\overrightarrow{OC},\\[4mm] \overrightarrow{CB} =&\,\overrightarrow{OB}-\overrightarrow{OC} =(5,-2,1)=\overrightarrow{OA}. \end{align*}

Thus both pairs of opposite sides are parallel and equal, so OABCOABC is a parallelogram.

Moreover,

OA→⋅OC→=5(3)+(−2)(4)+1(−7)=15−8−7=0.\overrightarrow{OA}\cdot\overrightarrow{OC} =5(3)+(-2)(4)+1(-7)=15-8-7=0.

Hence the adjacent sides OAOA and OCOC are perpendicular. A parallelogram with a right angle is a rectangle, so OABC is a rectangle\boxed{OABC\text{ is a rectangle}}.

(b)

解法一

思路

展开

直接求两条对角线的方向向量，用点积公式计算夹角；最后确认所取的是锐角。

答题过程

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The diagonal vectors are

OB→= (8,2,−6),AC→= OC→−OA→= (−2,6,−8).\begin{align*} \overrightarrow{OB}=&\,(8,2,-6),\\[4mm] \overrightarrow{AC}=&\,\overrightarrow{OC}-\overrightarrow{OA}\\[4mm] =&\,(-2,6,-8). \end{align*}

Their scalar product is

OB→⋅AC→=8(−2)+2(6)+(−6)(−8)=44.\overrightarrow{OB}\cdot\overrightarrow{AC} =8(-2)+2(6)+(-6)(-8)=44.

Their lengths are both 104\sqrt{104}. Therefore the acute angle γ\gamma satisfies

cos⁡γ=44104104=1126.\cos\gamma=\frac{44}{\sqrt{104}\sqrt{104}}=\frac{11}{26}.

Thus

γ=cos⁡−11126≈65.0∘.\gamma=\cos^{-1}\frac{11}{26}\approx\boxed{65.0^\circ}.

解法二

思路

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也可以先用边 OCOC 与对角线 OBOB 的点积求半个锐角。矩形对角线与同一组平行边所夹的锐角相等。

答题过程

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Let δ\delta be the angle between OCOC and OBOB. Then

OC→⋅OB→= 3(8)+4(2)+(−7)(−6)=74,∣OC→∣= 74,∣OB→∣= 104.\begin{align*} \overrightarrow{OC}\cdot\overrightarrow{OB} =&\,3(8)+4(2)+(-7)(-6)=74,\\[4mm] |\overrightarrow{OC}|=&\,\sqrt{74},\\[4mm] |\overrightarrow{OB}|=&\,\sqrt{104}. \end{align*}

Hence

cos⁡δ= 7474104,δ= 32.4855…∘.\begin{align*} \cos\delta=&\,\frac{74}{\sqrt{74}\sqrt{104}},\\[4mm] \delta=&\,32.4855\ldots^\circ. \end{align*}

In the rectangle, the two diagonal directions make equal angles on opposite sides of the direction parallel to OCOC. Thus the acute angle between them is 2δ2\delta:

γ=2δ≈65.0∘.\gamma=2\delta\approx\boxed{65.0^\circ}.