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CIE 9709 2022 November Paper 31 Q11

A Level / CIE / P3

CIE 9709 2022 Nov Paper 31 Paper · Question 11

题目

Problem

In the diagram, OABCDOABCD is a solid figure in which OA=OB=4OA=OB=4 units and OD=3OD=3 units. The edge ODOD is vertical, DCDC is parallel to OBOB and DC=1DC=1 unit. The base, OABOAB, is horizontal and angle AOB=90∘AOB=90^\circ. Unit vectors i\mathbf{i}, j\mathbf{j} and k\mathbf{k} are parallel to OAOA, OBOB and ODOD respectively. The midpoint of ABAB is MM and the point NN on BCBC is such that CN=2NBCN=2NB.

(a) Express vectors MD→\overrightarrow{MD} and ON→\overrightarrow{ON} in terms of i\mathbf{i}, j\mathbf{j} and k\mathbf{k}.

(4)

(b) Calculate the angle in degrees between the directions of MD→\overrightarrow{MD} and ON→\overrightarrow{ON}.

(3)

(c) Show that the length of the perpendicular from MM to ONON is 225\sqrt{\dfrac{22}{5}}.

(4)
题目中文翻译

图中,OABCDOABCD 是一个立体图形,其中 OA=OB=4OA=OB=4 个单位长度,OD=3OD=3 个单位长度。棱 ODOD 竖直,DCDC 平行于 OBOB,且 DC=1DC=1 个单位长度。底面 OABOAB 水平,且角 AOB=90∘AOB=90^\circ。单位向量 i\mathbf{i}、j\mathbf{j} 和 k\mathbf{k} 分别平行于 OAOA、OBOB 和 ODOD。MM 为 ABAB 的中点,BCBC 上的点 NN 满足 CN=2NBCN=2NB。

(a) 用 i\mathbf{i}、j\mathbf{j} 和 k\mathbf{k} 表示向量 MD→\overrightarrow{MD} 和 ON→\overrightarrow{ON}。

(4)

(b) 求 MD→\overrightarrow{MD} 与 ON→\overrightarrow{ON} 的方向之间的夹角,答案以度为单位。

(3)

(c) 证明从 MM 到 ONON 的垂线长为 225\sqrt{\dfrac{22}{5}}。

(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

展开

先根据单位向量方向写出关键点坐标,再用中点与内分比例。CN=2NBCN=2NB 表示从 CC 到 BB 走全长的 2/32/3。

答题过程

展开

The position vectors are

OA→=4i,OB→=4j,OD→=3k.\overrightarrow{OA}=4\mathbf i,\quad \overrightarrow{OB}=4\mathbf j,\quad \overrightarrow{OD}=3\mathbf k.

Since DCDC is parallel to OBOB and has length 11,

OC→=j+3k.\overrightarrow{OC}=\mathbf j+3\mathbf k.

As MM is the midpoint of ABAB,

OM→=2i+2j.\overrightarrow{OM}=2\mathbf i+2\mathbf j.

Therefore

MD→= OD→−OM→= −2i−2j+3k.\begin{align*} \overrightarrow{MD} =&\,\overrightarrow{OD}-\overrightarrow{OM}\\[4mm] =&\,\boxed{-2\mathbf i-2\mathbf j+3\mathbf k}. \end{align*}

Also CN=2NBCN=2NB, so

ON→= OC→+23CB→= (j+3k)+23(3j−3k)= 3j+k.\begin{align*} \overrightarrow{ON} =&\,\overrightarrow{OC}+\frac23\overrightarrow{CB}\\[4mm] =&\,(\mathbf j+3\mathbf k) +\frac23(3\mathbf j-3\mathbf k)\\[4mm] =&\,\boxed{3\mathbf j+\mathbf k}. \end{align*}

(b)

解法一

思路

展开

要求的是两个向量方向之间的角,点积为负意味着钝角,不应取绝对值。

答题过程

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Using part (a),

MD→⋅ON→=(−2)(0)+(−2)(3)+3(1)=−3.\overrightarrow{MD}\cdot\overrightarrow{ON} =(-2)(0)+(-2)(3)+3(1)=-3.

Also ∣MD→∣=17|\overrightarrow{MD}|=\sqrt{17} and ∣ON→∣=10|\overrightarrow{ON}|=\sqrt{10}. Hence

θ=cos⁡−1(−31710)≈103.3∘.\theta=\cos^{-1}\bigg(-\frac3{\sqrt{17}\sqrt{10}}\bigg) \approx\boxed{103.3^\circ}.

(c)

解法一

思路

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设垂足 PP 在 ONON 上,用 MP→\overrightarrow{MP} 与 ONON 方向垂直求参数,再计算长度。

答题过程

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Let OP→=λ(0,3,1)\overrightarrow{OP}=\lambda(0,3,1). Since M=(2,2,0)M=(2,2,0),

MP→=(−2,3λ−2,λ).\overrightarrow{MP}=(-2,3\lambda-2,\lambda).

For perpendicularity,

MP→⋅(0,3,1)=0,\overrightarrow{MP}\cdot(0,3,1)=0,

so

3(3λ−2)+λ= 0,λ= 35.\begin{align*} 3(3\lambda-2)+\lambda=&\,0,\\[4mm] \lambda=&\,\frac35. \end{align*}

Thus

MP→=(−2,−15,35).\overrightarrow{MP}=\bigg(-2,-\frac15,\frac35\bigg).

Therefore

MP=4+125+925=225.\boxed{MP=\sqrt{4+\frac1{25}+\frac9{25}} =\sqrt{\frac{22}{5}}}.

解法二

思路

展开

把点到直线上任意点的距离平方配成完全平方,最小值直接给出垂直距离。

答题过程

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For P=(0,3λ,λ)P=(0,3\lambda,\lambda) on ONON,

MP2= (−2)2+(3λ−2)2+λ2= 10λ2−12λ+8= 10(λ−35)2+225.\begin{align*} MP^2 =&\,(-2)^2+(3\lambda-2)^2+\lambda^2\\[4mm] =&\,10\lambda^2-12\lambda+8\\[4mm] =&\,10\bigg(\lambda-\frac35\bigg)^2+\frac{22}{5}. \end{align*}

Its minimum is 22/522/5, at λ=3/5\lambda=3/5. Equivalently, d(MP2)dλ=20λ−12=0\dfrac{\mathrm{d}(MP^2)}{\mathrm{d}\lambda}=20\lambda-12=0 gives the same parameter, with positive second derivative 2020. Therefore the perpendicular distance is

225.\boxed{\sqrt{\frac{22}{5}}}.

解法三

思路

展开

也可先求 OM→\overrightarrow{OM} 沿 ONON 的投影,再用勾股定理得到垂直分量长度。

答题过程

展开

We have OM→=(2,2,0)\overrightarrow{OM}=(2,2,0) and a direction vector for ONON is d=(0,3,1)\mathbf d=(0,3,1).

The projection length is

OM→⋅d∣d∣=610.\frac{\overrightarrow{OM}\cdot\mathbf d}{|\mathbf d|} =\frac6{\sqrt{10}}.

Since OM2=22+22=8OM^2=2^2+2^2=8, the squared perpendicular distance is

8−(610)2=8−185=225.8-\bigg(\frac6{\sqrt{10}}\bigg)^2 =8-\frac{18}{5}=\frac{22}{5}.

Thus the perpendicular length is 22/5\boxed{\sqrt{22/5}}.