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IAL 2024 June FP3 Q9

A Level / Edexcel / FP3

IAL 2024 June Paper · Question 9

题目

Problem

The plane Π1\Pi_1 has vector equation

r=(530)+s(301)+t(1−22)\mathbf r= \begin{pmatrix}5\\3\\0\end{pmatrix} +s\begin{pmatrix}3\\0\\1\end{pmatrix} +t\begin{pmatrix}1\\-2\\2\end{pmatrix}

where ss and tt are scalar parameters.

(a) Determine a Cartesian equation for Π1\Pi_1

The plane Π2\Pi_2 has vector equation

r⋅(5−23)=1\mathbf r\cdot\begin{pmatrix}5\\-2\\3\end{pmatrix}=1

(b) Determine a vector equation for the line of intersection of Π1\Pi_1 and Π2\Pi_2

Give your answer in the form r=a+λb\mathbf r=\mathbf a+\lambda\mathbf b, where a\mathbf a and b\mathbf b are constant vectors and λ\lambda is a scalar parameter.

The plane Π3\Pi_3 has Cartesian equation 4x−3y−z=04x-3y-z=0

(c) Use the answer to part (b) to determine the coordinates of the point of intersection of Π1\Pi_1, Π2\Pi_2 and Π3\Pi_3

(10)
题目中文翻译

平面 Π1\Pi_1 的向量方程为

r=(530)+s(301)+t(1−22)\mathbf r= \begin{pmatrix}5\\3\\0\end{pmatrix} +s\begin{pmatrix}3\\0\\1\end{pmatrix} +t\begin{pmatrix}1\\-2\\2\end{pmatrix}

其中 ss 和 tt 为标量参数。

(a) 求 Π1\Pi_1 的笛卡尔方程

平面 Π2\Pi_2 的向量方程为

r⋅(5−23)=1\mathbf r\cdot\begin{pmatrix}5\\-2\\3\end{pmatrix}=1

(b) 求平面 Π1\Pi_1 与 Π2\Pi_2 的交线的向量方程。

答案写成 r=a+λb\mathbf r=\mathbf a+\lambda\mathbf b 的形式，其中 a\mathbf a 和 b\mathbf b 为常向量，λ\lambda 为标量参数。

平面 Π3\Pi_3 的笛卡尔方程为 4x−3y−z=04x-3y-z=0

(c) 利用 (b) 的答案求 Π1\Pi_1、Π2\Pi_2 和 Π3\Pi_3 的交点坐标。

解答

(a)

解法一：使用方向向量的叉积

思路

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平面的两个方向向量都与法向量垂直，因此先求它们的叉积。再把平面上的已知点代入点法式，便可得到笛卡尔方程。

答题过程

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The two direction vectors of Π1\Pi_1 are

d1=(301),d2=(1−22).\mathbf d_1= \begin{pmatrix}3\\0\\1\end{pmatrix}, \qquad \mathbf d_2= \begin{pmatrix}1\\-2\\2\end{pmatrix}.

A normal vector to Π1\Pi_1 is therefore

n= d1×d2= (301)×(1−22)= (2−5−6).\begin{align*} \mathbf n =&\,\mathbf d_1\times\mathbf d_2 \\ =&\, \begin{pmatrix}3\\0\\1\end{pmatrix} \times \begin{pmatrix}1\\-2\\2\end{pmatrix} \\ =&\, \begin{pmatrix}2\\-5\\-6\end{pmatrix}. \end{align*}

Since (5,3,0)(5,3,0) lies on Π1\Pi_1,

(530)⋅n= 10−15= −5.\begin{align*} \begin{pmatrix}5\\3\\0\end{pmatrix} \mathbin{\cdot}\mathbf n =&\,10-15 \\ =&\,-5. \end{align*}

Hence a Cartesian equation of Π1\Pi_1 is

2x−5y−6z=−5.\boxed{2x-5y-6z=-5}.

解法二：消去参数

思路

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把向量方程拆成三个坐标方程，先利用 yy 和 zz 分别表示 tt 与 ss，再代入 xx 的方程，便可直接消去两个参数。

答题过程

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The vector equation gives

x=5+3s+t,y=3−2t,z=s+2t.x=5+3s+t, \qquad y=3-2t, \qquad z=s+2t.

Thus

t=3−y2,s=z−2t=y+z−3.t=\frac{3-y}{2}, \qquad s=z-2t=y+z-3.

Substituting these expressions into the equation for xx,

x= 5+3(y+z−3)+3−y2= 5y2+3z−52.\begin{align*} x =&\,5+3(y+z-3)+\frac{3-y}{2} \\ =&\,\frac{5y}{2}+3z-\frac52. \end{align*}

Therefore,

2x−5y−6z=−5.\boxed{2x-5y-6z=-5}.

(b)

解法一：联立平面方程

思路

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交线上的点同时满足两个平面方程。联立后消去 zz 和 xx，把 x,zx,z 都写成 yy 的一次式；再令 y=12λy=12\lambda，可使方向向量的分量全部为整数。

答题过程

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The two plane equations are

2x−5y−6z=−5,5x−2y+3z=1.2x-5y-6z=-5, \qquad 5x-2y+3z=1.

Adding the first equation to twice the second equation gives

12x−9y=−3,12x-9y=-3,

so

x=3y−14.x=\frac{3y-1}{4}.

Also, subtracting twice the second equation from five times the first equation gives

−21y−36z=−27,-21y-36z=-27,

and hence

z=9−7y12.z=\frac{9-7y}{12}.

Let y=12λy=12\lambda. Then

x=−14+9λ,z=34−7λ.x=-\frac14+9\lambda, \qquad z=\frac34-7\lambda.

Therefore, the line of intersection is

r=(−14034)+λ(912−7).\boxed{ \mathbf r= \begin{pmatrix}-\frac14\\0\\\frac34\end{pmatrix} +\lambda \begin{pmatrix}9\\12\\-7\end{pmatrix} }.

解法二：求一点并叉乘法向量

思路

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先令 y=0y=0，联立两个平面方程求出交线上的一点。交线方向同时垂直于两个平面的法向量，所以可由两个法向量的叉积得到。

答题过程

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Set y=0y=0. A point on the line must then satisfy

2x−6z=−5,5x+3z=1.2x-6z=-5, \qquad 5x+3z=1.

Adding the first equation to twice the second equation gives

12x=−3,12x=-3,

so

x=−14,z=34.x=-\frac14, \qquad z=\frac34.

Thus one point on the line is

(−14034).\begin{pmatrix}-\frac14\\0\\\frac34\end{pmatrix}.

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

n1=(2−5−6),n2=(5−23).\mathbf n_1= \begin{pmatrix}2\\-5\\-6\end{pmatrix}, \qquad \mathbf n_2= \begin{pmatrix}5\\-2\\3\end{pmatrix}.

Hence a direction vector of their line of intersection is obtained from

n1×n2= (−27−3621)= −3(912−7).\begin{align*} \mathbf n_1\times\mathbf n_2 =&\, \begin{pmatrix}-27\\-36\\21\end{pmatrix} \\ =&\,-3 \begin{pmatrix}9\\12\\-7\end{pmatrix}. \end{align*}

Therefore, an equivalent vector equation is

r=(−14034)+λ(912−7).\boxed{ \mathbf r= \begin{pmatrix}-\frac14\\0\\\frac34\end{pmatrix} +\lambda \begin{pmatrix}9\\12\\-7\end{pmatrix} }.

(c)

解法一

思路

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按照题目要求使用 (b) 的交线：把交线的三个参数式代入 Π3\Pi_3，先求出 λ\lambda，再代回交线得到唯一的公共点。

答题过程

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Using the line obtained in part (b),

x=−14+9λ,y=12λ,z=34−7λ.x=-\frac14+9\lambda, \qquad y=12\lambda, \qquad z=\frac34-7\lambda.

Substitute these expressions into 4x−3y−z=04x-3y-z=0:

4(−14+9λ) −3(12λ)−(34−7λ)=0,7λ−74= 0,λ= 14.\begin{align*} 4\biggl(-\frac14+9\lambda\biggr) &\,-3(12\lambda) -\biggl(\frac34-7\lambda\biggr)=0, \\ 7\lambda-\frac74=&\,0, \\ \lambda=&\,\frac14. \end{align*}

Therefore,

x= −14+9(14)=2,y= 12(14)=3,z= 34−7(14)=−1.\begin{align*} x=&\,-\frac14+9\biggl(\frac14\biggr)=2, \\ y=&\,12\biggl(\frac14\biggr)=3, \\ z=&\,\frac34-7\biggl(\frac14\biggr)=-1. \end{align*}

Hence the point of intersection of the three planes is

(2,3,−1).\boxed{(2,3,-1)}.