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P4 2021 Oct Q7

A Level / Edexcel / P4

IAL 2021 Oct Paper · Question 7

题目

Problem

With respect to a fixed origin OO,

  • the line ll has equation
r=(42−3)+λ(−4−35)\mathbf{r}=\begin{pmatrix}4\\2\\-3\end{pmatrix}+\lambda \begin{pmatrix}-4\\-3\\5\end{pmatrix}

where λ\lambda is a scalar constant

  • the point AA has position vector 9i−3j+2k9\mathbf{i}-3\mathbf{j}+2\mathbf{k}

Given that XX is the point on ll nearest to AA,

(a) find

(i) the coordinates of XX

(ii) the shortest distance from AA to ll.

Give your answer in the form d\sqrt{d}, where dd is an integer.

(7)

The point BB is the image of AA after reflection in ll.

(b) Find the position vector of BB.

(2)
题目中文翻译

相对于固定原点 OO，

  • 直线 ll 的方程为
r=(42−3)+λ(−4−35)\mathbf{r}=\begin{pmatrix}4\\2\\-3\end{pmatrix}+\lambda \begin{pmatrix}-4\\-3\\5\end{pmatrix}

其中 λ\lambda 为标量常数；

  • 点 AA 的位置向量为 9i−3j+2k9\mathbf{i}-3\mathbf{j}+2\mathbf{k}。

已知 XX 是直线 ll 上距离点 AA 最近的点，

(a) 求

(i) 点 XX 的坐标；

(ii) 点 AA 到直线 ll 的最短距离。

将答案写成 d\sqrt{d} 的形式，其中 dd 为整数。

(7)

点 BB 是点 AA 关于直线 ll 的对称点。

(b) 求点 BB 的位置向量。

(2)

解答

(a)

解法一

思路

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最近点 XX 在直线 ll 上，而且 AXAX 垂直于直线的方向向量。先用参数 λ\lambda 写出 XX 和 AX→\overrightarrow{AX}，再令点积为零求出 λ\lambda。得到 XX 后，用 AXAX 的模求最短距离。

答题过程

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A general point XX on ll has position vector

OX→=(4−4λ2−3λ−3+5λ).\overrightarrow{OX} =\begin{pmatrix} 4-4\lambda\\ 2-3\lambda\\ -3+5\lambda \end{pmatrix}.

Since A=(9,−3,2)A=(9,-3,2),

AX→=(−5−4λ5−3λ−5+5λ).\overrightarrow{AX} =\begin{pmatrix} -5-4\lambda\\ 5-3\lambda\\ -5+5\lambda \end{pmatrix}.

At the nearest point, AXAX is perpendicular to the direction vector of ll. Therefore,

(−5−4λ5−3λ−5+5λ)⋅(−4−35)=0.\begin{pmatrix} -5-4\lambda\\ 5-3\lambda\\ -5+5\lambda \end{pmatrix} \boldsymbol{\cdot} \begin{pmatrix} -4\\-3\\5 \end{pmatrix} =0.

Hence,

 (−5−4λ)(−4) +(5−3λ)(−3) +(−5+5λ)(5)=0, 20+16λ−15+9λ −25+25λ=0, 50λ−20=0.\begin{align*} &\,(-5-4\lambda)(-4) \\ &\,\hspace{2pt}+(5-3\lambda)(-3)\\ &\,\hspace{4pt}+(-5+5\lambda)(5)=0,\\ &\,20+16\lambda-15+9\lambda\\ &\,\hspace{2pt}-25+25\lambda=0,\\ &\,50\lambda-20=0. \end{align*}

Thus,

λ=25.\lambda=\frac25.

For part (i),

 X=(4−4(25), 2−3(25), −3+5(25))= (125,45,−1).\begin{align*} &\,X=\bigg( 4-4\bigg(\frac25\bigg), \\ &\,\hspace{2pt}2-3\bigg(\frac25\bigg),\\ &\,\hspace{4pt}-3+5\bigg(\frac25\bigg) \bigg)\\ =&\,\boxed{\bigg(\frac{12}{5},\frac45,-1\bigg)}. \end{align*}

For part (ii),

AX→=(−335195−3).\overrightarrow{AX} =\begin{pmatrix} -\frac{33}{5}\\ \frac{19}{5}\\ -3 \end{pmatrix}.

Therefore, the shortest distance is

 AX= (−335)2 +(195)2 +(−3)2= 1089+361+22525= 67.\begin{align*} &\,AX=\sqrt{ \begin{aligned} &\,\bigg(-\frac{33}{5}\bigg)^2\\ &\,\hspace{2pt}+\bigg(\frac{19}{5}\bigg)^2\\ &\,\hspace{4pt}+(-3)^2 \end{aligned} }\\ =&\,\sqrt{ \frac{1089+361+225}{25} }\\ =&\,\boxed{\sqrt{67}}. \end{align*}

解法二

思路

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直接把 AX2AX^2 写成 λ\lambda 的二次函数。距离最小时，距离平方也最小，因此令 ddλ(AX2)=0\frac{\mathrm{d}}{\mathrm{d}\lambda}(AX^2)=0，求出最近点所对应的参数。

答题过程

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For a general point XX on ll,

 AX2=(−5−4λ)2 +(5−3λ)2 +(−5+5λ)2.\begin{align*} &\,AX^2=(-5-4\lambda)^2\\ &\,\hspace{2pt}+(5-3\lambda)^2\\ &\,\hspace{4pt}+(-5+5\lambda)^2. \end{align*}

At the minimum distance,

ddλ(AX2)=0.\frac{\mathrm{d}}{\mathrm{d}\lambda}(AX^2)=0.

Therefore,

 −8(−5−4λ) −6(5−3λ) +10(−5+5λ)=0, 40+32λ−30+18λ −50+50λ=0.\begin{align*} &\,-8(-5-4\lambda)\\ &\,\hspace{2pt}-6(5-3\lambda)\\ &\,\hspace{4pt}+10(-5+5\lambda)=0,\\ &\,40+32\lambda-30+18\lambda\\ &\,\hspace{2pt}-50+50\lambda=0. \end{align*}

Hence,

100λ−40= 0λ= 25.\begin{align*} 100\lambda-40=&\,0\\ \lambda=&\,\frac25. \end{align*}

Thus, for part (i),

X=(125,45,−1).\boxed{X=\bigg(\frac{12}{5},\frac45,-1\bigg)}.

For part (ii), substituting λ=25\lambda=\frac25 into the expression for AX2AX^2 gives

 AX2= (−5−4(25))2 +(5−3(25))2 +(−5+5(25))2= 67.\begin{align*} &\,AX^2=\\ &\,\hspace{2pt}\bigg(-5-4\bigg(\frac25\bigg)\bigg)^2\\ &\,+\bigg(5-3\bigg(\frac25\bigg)\bigg)^2\\ &\,+\bigg(-5+5\bigg(\frac25\bigg)\bigg)^2\\ =&\,67. \end{align*}

Therefore,

AX=67.\boxed{AX=\sqrt{67}}.

也可以用配方求距离平方的最小值，不必求导。

Completing the square gives

AX2= 50λ2−40λ+75= 50(λ−25)2+67.\begin{align*} AX^2=&\,50\lambda^2-40\lambda+75\\ =&\,50\bigg(\lambda-\frac25\bigg)^2+67. \end{align*}

The square is at least zero, so the minimum is AX2=67AX^2=67 at λ=25\lambda=\frac25. This gives the same point XX and shortest distance 67\sqrt{67}.

解法三

思路

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取直线上参数为零的点 C=(4,2,−3)C=(4,2,-3)。把 CA→\overrightarrow{CA} 投影到直线的方向上，得到 CXCX；再由 CXCX 与方向向量长度的比值求出 λ\lambda。直角三角形 ACXACX 中可用勾股定理求 AXAX。

答题过程

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Let C=(4,2,−3)C=(4,2,-3) and let the direction vector of ll be v=(−4,−3,5)\mathbf{v}=(-4,-3,5). Then

CA→=(5,−5,5).\overrightarrow{CA}=(5,-5,5).

The scalar product and the length of v\mathbf{v} are

 CA→⋅v= 5(−4)+(−5)(−3) +5(5)= 20, ∣v∣=16+9+25= 50.\begin{align*} &\,\overrightarrow{CA}\boldsymbol{\cdot}\mathbf{v}\\ =&\,5(-4)+(-5)(-3)\\ &\,\hspace{2pt}+5(5)\\ =&\,20,\\ &\,|\mathbf{v}|=\sqrt{16+9+25}\\ =&\,\sqrt{50}. \end{align*}

Since AXAX is perpendicular to CXCX, the projection of CA→\overrightarrow{CA} onto the direction of ll gives

CX=2050=22.CX=\frac{20}{\sqrt{50}}=2\sqrt2.

Hence,

λ=CX∣v∣=2250=25.\lambda=\frac{CX}{|\mathbf{v}|}=\frac{2\sqrt2}{\sqrt{50}}=\frac25.

Substituting into the line equation gives

X=(125,45,−1).\boxed{X=\bigg(\frac{12}{5},\frac45,-1\bigg)}.

Also CA2=52+(−5)2+52=75CA^2=5^2+(-5)^2+5^2=75, so Pythagoras gives

AX= CA2−CX2= 75−8= 67.\begin{align*} AX=&\,\sqrt{CA^2-CX^2}\\ =&\,\sqrt{75-8}\\ =&\,\boxed{\sqrt{67}}. \end{align*}

(b)

解法一

思路

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点 BB 是 AA 关于直线 ll 的对称点，因此垂足 XX 是线段 ABAB 的中点。使用 OB→=2OX→−OA→\overrightarrow{OB}=2\overrightarrow{OX}-\overrightarrow{OA} 即可求出 BB 的位置向量。

答题过程

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Since XX is the midpoint of ABAB,

OB→=2OX→−OA→.\overrightarrow{OB} =2\overrightarrow{OX}-\overrightarrow{OA}.

Therefore,

OB→= 2(12545−1)−(9−32)= (−215235−4).\begin{align*} \overrightarrow{OB} =&\,2 \begin{pmatrix} \frac{12}{5}\\ \frac45\\ -1 \end{pmatrix} -\begin{pmatrix} 9\\-3\\2 \end{pmatrix}\\ =&\, \begin{pmatrix} -\frac{21}{5}\\ \frac{23}{5}\\ -4 \end{pmatrix}. \end{align*}

Hence the position vector of BB is

OB→= −215i +235j−4k.\boxed{ \begin{aligned} \overrightarrow{OB} =&\,-\frac{21}{5}\mathbf{i}\\ &\,\hspace{2pt}+\frac{23}{5}\mathbf{j}-4\mathbf{k} \end{aligned} }.