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FP3 2025 June Q8

A Level / Edexcel / FP3

IAL 2025 June Paper · Question 8

题目

Problem

8.

The plane Π1\Pi_1 has equation

r= (42−1) +λ(75−6) +μ(−312)\begin{align*} \mathbf r =&\,\begin{pmatrix}4\\2\\-1\end{pmatrix}\\ &\,\hspace{2pt}+\lambda\begin{pmatrix}7\\5\\-6\end{pmatrix}\\ &\,\hspace{2pt}+\mu\begin{pmatrix}-3\\1\\2\end{pmatrix} \end{align*}

where λ\lambda and μ\mu are scalar parameters.

(a) Determine

(7i+5j−6k)×(−3i+j+2k)\begin{align*} (7\mathbf{i}+5\mathbf{j}-6\mathbf{k}) \times (-3\mathbf{i}+\mathbf{j}+2\mathbf{k}) \end{align*}
(1)

(b) Hence show that the equation of Π1\Pi_1 can be written in the form

r⋅(8i+2j+11k)= p\begin{align*} \mathbf{r}\cdot(8\mathbf{i}+2\mathbf{j}+11\mathbf{k})=&\,p \end{align*}

where pp is a constant to be determined.

(2)

Given that

  • the plane Π2\Pi_2 has equation x−y+z=7x-y+z=7
  • the planes Π1\Pi_1 and Π2\Pi_2 intersect in the line l1l_1

(c) determine an equation for l1l_1 giving your answer in the form (r−a)×b=0(\mathbf{r}-\mathbf{a})\times \mathbf{b}=0 where a\mathbf{a} and b\mathbf{b} are constant vectors.

(3)

Given also that

  • the point A has coordinates (2,1,3)(2,1,3)
  • the point B has coordinates (3,0,2)(3,0,2)
  • the line l2l_2 passes through A and B

(d) determine the shortest distance between l1l_1 and l2l_2

(4)
题目中文翻译

平面 Π1\Pi_1 的方程为

r= (42−1) +λ(75−6) +μ(−312)\begin{align*} \mathbf r =&\,\begin{pmatrix}4\\2\\-1\end{pmatrix}\\ &\,\hspace{2pt}+\lambda\begin{pmatrix}7\\5\\-6\end{pmatrix}\\ &\,\hspace{2pt}+\mu\begin{pmatrix}-3\\1\\2\end{pmatrix} \end{align*}

其中 λ\lambda 和 μ\mu 为标量参数。

(a) 求

(7i+5j−6k)×(−3i+j+2k)\begin{align*} (7\mathbf{i}+5\mathbf{j}-6\mathbf{k}) \times (-3\mathbf{i}+\mathbf{j}+2\mathbf{k}) \end{align*}
(1)

(b) 从而证明 Π1\Pi_1 的方程可写成

r⋅(8i+2j+11k)= p\begin{align*} \mathbf{r}\cdot(8\mathbf{i}+2\mathbf{j}+11\mathbf{k})=&\,p \end{align*}

其中 pp 为待定常数。

(2)

已知

  • 平面 Π2\Pi_2 的方程为 x−y+z=7x-y+z=7
  • 平面 Π1\Pi_1 与 Π2\Pi_2 的交线为 l1l_1

(c) 求 l1l_1 的方程，答案写成 (r−a)×b=0(\mathbf{r}-\mathbf{a})\times \mathbf{b}=0 的形式，其中 a\mathbf{a} 和 b\mathbf{b} 为常向量。

(3)

另外已知

  • 点 A 的坐标为 (2,1,3)(2,1,3)
  • 点 B 的坐标为 (3,0,2)(3,0,2)
  • 直线 l2l_2 经过 A 和 B

(d) 求直线 l1l_1 和 l2l_2 的最短距离。

(4)

解答

(a)

解法一

思路

展开

直接用行列式计算两个向量的叉积。结果是平面 Π1\Pi_1 的一个法向量；(b) 会把它约去公因数 22。

答题过程

展开 (7i+5j−6k)×(−3i+j+2k)= ∣ijk75−6−312∣= (5(2)−(−6)(1))i −(7(2)−(−6)(−3))j +(7(1)−5(−3))k= 16i+4j+22k.\begin{align*} &(7\mathbf i+5\mathbf j-6\mathbf k) \times(-3\mathbf i+\mathbf j+2\mathbf k)\\ =&\, \begin{vmatrix} \mathbf i&\mathbf j&\mathbf k\\ 7&5&-6\\ -3&1&2 \end{vmatrix}\\ =&\,\big(5(2)-(-6)(1)\big)\mathbf i\\ &\,-\big(7(2)-(-6)(-3)\big)\mathbf j\\ &\,+\big(7(1)-5(-3)\big)\mathbf k\\ =&\,\boxed{16\mathbf i+4\mathbf j+22\mathbf k}. \end{align*}

(b)

解法一

思路

展开

承接 (a)，把叉积除以 22 得到题目指定的法向量 (8,2,11)(8,2,11)。平面上已知基点是 (4,2,−1)(4,2,-1)，将它与法向量作点积即可求常数 pp。

答题过程

展开

From part (a), a normal vector to Π1\Pi_1 is

n=(8211).\mathbf n= \begin{pmatrix} 8\\ 2\\ 11 \end{pmatrix}.

The point with position vector

r0=(42−1)\mathbf r_0= \begin{pmatrix} 4\\ 2\\ -1 \end{pmatrix}

lies on Π1\Pi_1. Hence

p= r0⋅n= 4(8)+2(2)+(−1)(11)= 25.\begin{align*} p =&\,\mathbf r_0\mathbin{\boldsymbol\cdot}\mathbf n\\ =&\,4(8)+2(2)+(-1)(11)\\ =&\,25. \end{align*}

Therefore,

r⋅(8i+2j+11k)=25.\boxed{ \mathbf r\mathbin{\boldsymbol\cdot} (8\mathbf i+2\mathbf j+11\mathbf k)=25}.

(c)

解法一

思路

展开

两平面的交线同时垂直于两个法向量，所以其方向向量可由法向量的叉积求得。再令 x=0x=0 联立两个平面方程，找出交线上的一点，最后写成题目要求的叉积形式。

答题过程

展开

The normal vectors of Π1\Pi_1 and Π2\Pi_2 are

n1=(8211)\mathbf n_1= \begin{pmatrix} 8\\ 2\\ 11 \end{pmatrix}

and

n2=(1−11).\mathbf n_2= \begin{pmatrix} 1\\ -1\\ 1 \end{pmatrix}.

A direction vector of their line of intersection is

n1×n2= ∣ijk82111−11∣= (133−10).\begin{align*} \mathbf n_1\times\mathbf n_2 =&\, \begin{vmatrix} \mathbf i&\mathbf j&\mathbf k\\ 8&2&11\\ 1&-1&1 \end{vmatrix}\\ =&\,\begin{pmatrix}13\\3\\-10\end{pmatrix}. \end{align*}

To find a point on the line, set x=0x=0. The plane equations give

2y+11z= 25,−y+z= 7.\begin{align*} 2y+11z=&\,25,\\ -y+z=&\,7. \end{align*}

From the second equation, y=z−7y=z-7. Substituting into the first,

2(z−7)+11z= 25,13z= 39,z= 3,y=−4.\begin{align*} 2(z-7)+11z=&\,25,\\ 13z=&\,39,\\ z=&\,3,\qquad y=-4. \end{align*}

Thus a point on l1l_1 is (0,−4,3)(0,-4,3). Therefore,

 (r−(−4j+3k)) ×(13i+3j−10k)=0.\begin{align*} &\,\big(\mathbf r-(-4\mathbf j+3\mathbf k)\big)\\ &\,\hspace{2pt}\times (13\mathbf i+3\mathbf j-10\mathbf k) =\mathbf0. \end{align*}

解法二

思路

展开

也可以直接联立两个平面的笛卡尔方程。取 y=ty=t，把 x,zx,z 都表示成 tt，便能同时读出交线上的基点与方向向量，再转换成指定的叉积形式。

答题过程

展开

The two plane equations are

8x+2y+11z= 25,x−y+z= 7.\begin{align*} 8x+2y+11z=&\,25,\\ x-y+z=&\,7. \end{align*}

Let y=ty=t. From the second equation,

x=7+t−z.x=7+t-z.

Substituting into the first equation,

8(7+t−z)+2t+11z= 25,3z= −31−10t.\begin{align*} 8(7+t-z)+2t+11z=&\,25,\\ 3z=&\,-31-10t. \end{align*}

Thus

z=−31−10t3z=\frac{-31-10t}{3}

and

x=52+13t3.x=\frac{52+13t}{3}.

Hence

r= (5230−313) +t(1331−103).\begin{align*} \mathbf r =&\, \begin{pmatrix} \frac{52}{3}\\ 0\\ -\frac{31}{3} \end{pmatrix} \\ &\,\hspace{2pt}+t \begin{pmatrix} \frac{13}{3}\\ 1\\ -\frac{10}{3} \end{pmatrix}. \end{align*}

Multiplying the direction vector by 33, an equation in the required form is

 (r−523i+313k) ×(13i+3j−10k)=0.\begin{align*} &\,\Big( \mathbf r-\frac{52}{3}\mathbf i +\frac{31}{3}\mathbf k\Big)\\ &\,\hspace{2pt}\times (13\mathbf i+3\mathbf j-10\mathbf k) =\mathbf0. \end{align*}

(d)

解法一

思路

展开

先求 l2l_2 的方向向量。两条异面直线的公垂线方向是两个方向向量的叉积；任取两条直线上的点，其连线向量在公垂线方向上的投影长度，就是最短距离。

答题过程

展开

A direction vector of l2l_2 is

AB→=(1−1−1).\overrightarrow{AB} = \begin{pmatrix} 1\\ -1\\ -1 \end{pmatrix}.

Let

d1=(133−10)\mathbf d_1= \begin{pmatrix} 13\\ 3\\ -10 \end{pmatrix}

and

d2=(1−1−1).\mathbf d_2= \begin{pmatrix} 1\\ -1\\ -1 \end{pmatrix}.

A vector perpendicular to both lines is

n= d1×d2= (3(−1)−(−10)(−1))i −(13(−1)−(−10)(1))j +(13(−1)−3(1))k= (−133−16).\begin{align*} \mathbf n =&\,\mathbf d_1\times\mathbf d_2\\ =&\,\big(3(-1)-(-10)(-1)\big)\mathbf i\\[2mm] &\,\hspace{2pt}-\big(13(-1)-(-10)(1)\big)\mathbf j\\[2mm] &\,\hspace{4pt}+\big(13(-1)-3(1)\big)\mathbf k\\[2mm] =&\,\begin{pmatrix}-13\\3\\-16\end{pmatrix}. \end{align*}

Take P=(0,−4,3)P=(0,-4,3) on l1l_1 and A=(2,1,3)A=(2,1,3) on l2l_2. Then

AP→=(−2−50).\overrightarrow{AP} = \begin{pmatrix} -2\\ -5\\ 0 \end{pmatrix}.

The shortest distance is

d= ∣AP→⋅n∣∣n∣= ∣(−2)(−13)+(−5)(3)+0(−16)∣(−13)2+32+(−16)2= 11434.\begin{align*} d =&\,\frac{ \big|\overrightarrow{AP} \mathbin{\boldsymbol\cdot}\mathbf n\big|} {|\mathbf n|}\\ =&\,\frac{|(-2)(-13)+(-5)(3)+0(-16)|} {\sqrt{(-13)^2+3^2+(-16)^2}}\\ =&\,\boxed{\frac{11}{\sqrt{434}}}. \end{align*}

解法二

思路

展开

取两条直线上的一般点 X,YX,Y。最短连线 YX→\overrightarrow{YX} 必须同时垂直于两条直线的方向向量，由两个点积为零的条件解出参数，再求该连线的模。

答题过程

展开

Take general points

X=(2+λ1−λ3−λ)on l2X= \begin{pmatrix} 2+\lambda\\ 1-\lambda\\ 3-\lambda \end{pmatrix} \quad\text{on }l_2

and

Y=(13μ−4+3μ3−10μ)on l1.Y= \begin{pmatrix} 13\mu\\ -4+3\mu\\ 3-10\mu \end{pmatrix} \quad\text{on }l_1.

Then

YX→=(2+λ−13μ5−λ−3μ−λ+10μ).\overrightarrow{YX} = \begin{pmatrix} 2+\lambda-13\mu\\ 5-\lambda-3\mu\\ -\lambda+10\mu \end{pmatrix}.

At the shortest distance, YX→\overrightarrow{YX} is perpendicular to both direction vectors. Hence

YX→⋅(133−10)= 0,YX→⋅(1−1−1)= 0.\begin{align*} \overrightarrow{YX} \mathbin{\boldsymbol\cdot} \begin{pmatrix}13\\3\\-10\end{pmatrix} =&\,0,\\ \overrightarrow{YX} \mathbin{\boldsymbol\cdot} \begin{pmatrix}1\\-1\\-1\end{pmatrix} =&\,0. \end{align*}

Expanding,

41+20λ−278μ= 0,−3+3λ−20μ= 0.\begin{align*} 41+20\lambda-278\mu=&\,0,\\ -3+3\lambda-20\mu=&\,0. \end{align*}

The second equation gives λ=1+203μ\lambda=1+\dfrac{20}{3}\mu. Substituting into the first,

41+20(1+203μ)−278μ= 0,183−434μ= 0,μ= 183434.\begin{align*} 41+20\Big(1+\frac{20}{3}\mu\Big)-278\mu=&\,0,\\[2mm] 183-434\mu=&\,0,\\[2mm] \mu=&\,\frac{183}{434}. \end{align*}

Then

λ= 1+203(183434)= 827217.\begin{align*} \lambda=&\,1+\frac{20}{3}\Big(\frac{183}{434}\Big)\\[2mm] =&\,\frac{827}{217}. \end{align*}

Hence

λ=827217andμ=183434.\lambda=\frac{827}{217} \quad\text{and}\quad \mu=\frac{183}{434}.

Therefore,

YX→=(143434−3343488217).\overrightarrow{YX} = \begin{pmatrix} \frac{143}{434}\\ -\frac{33}{434}\\ \frac{88}{217} \end{pmatrix}.

Its magnitude is

∣YX→∣= 14341432+(−33)2+1762= 11434434= 11434.\begin{align*} |\overrightarrow{YX}| =&\,\frac1{434} \sqrt{143^2+(-33)^2+176^2}\\ =&\,\frac{11\sqrt{434}}{434}\\ =&\,\boxed{\frac{11}{\sqrt{434}}}. \end{align*}