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FP3 2024 Jan Q6

A Level / Edexcel / FP3

IAL 2024 Jan Paper · Question 6

题目

Problem

6.

The points AA, BB and CC have coordinates (3,2,2)(3,2,2), (−1,1,3)(-1,1,3) and (−2,4,2)(-2,4,2) respectively.

The plane Π1\Pi_1 contains the points AA, BB and CC

(a) Determine a Cartesian equation of Π1\Pi_1

(4)

Given that

• point DD has coordinates (−1,1,−2)(-1,1,-2)

• line ll passes through DD and is perpendicular to Π1\Pi_1

• plane Π2\Pi_2 has equation r⋅(14i−j−17k)=−66\mathbf r\cdot(14\mathbf i-\mathbf j-17\mathbf k)=-66

• ll meets Π2\Pi_2 at the point EE

(b) show that DE=p22DE=p\sqrt{22} where pp is a rational number to be determined.

(5)

The point FF has coordinates (4,3,q)(4,3,q) where qq is a constant.

Given that AA, BB, CC and FF are the vertices of a tetrahedron of volume 12

(c) determine the possible values of qq

(3)
题目中文翻译

6.

点 AA、BB 和 CC 的坐标分别为 (3,2,2)(3,2,2)、(−1,1,3)(-1,1,3) 和 (−2,4,2)(-2,4,2)。

平面 Π1\Pi_1 包含点 AA、BB 和 CC

(a) 求 Π1\Pi_1 的笛卡尔方程

(4)

已知

• 点 DD 的坐标为 (−1,1,−2)(-1,1,-2)

• 直线 ll 过点 DD 且垂直于 Π1\Pi_1

• 平面 Π2\Pi_2 的方程为 r⋅(14i−j−17k)=−66\mathbf r\cdot(14\mathbf i-\mathbf j-17\mathbf k)=-66

• ll 与 Π2\Pi_2 交于点 EE

(b) 证明 DE=p22DE=p\sqrt{22}，其中 pp 为待定有理数。

(5)

点 FF 的坐标为 (4,3,q)(4,3,q)，其中 qq 为常数。

已知 AA、BB、CC 和 FF 是体积为 12 的四面体的顶点

(c) 求 qq 的可能值

(3)

解答

(a)

解法一：使用叉积求法向量

思路

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先由三个已知点构造平面内的两个方向向量，再取叉积得到法向量。最后将点 AA 代入点法式，求出平面的常数项。

答题过程

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Two direction vectors in Π1\Pi_1 are

AB→=(−4−11).\overrightarrow{AB}= \begin{pmatrix}-4\\-1\\1\end{pmatrix}.

Also,

AC→=(−520).\overrightarrow{AC}= \begin{pmatrix}-5\\2\\0\end{pmatrix}.

A normal vector to the plane is

AB→×AC→= ((−1)(0)−(1)(2)(1)(−5)−(−4)(0)(−4)(2)−(−1)(−5))= (−2−5−13).\begin{align*} \overrightarrow{AB}\times\overrightarrow{AC} =&\, \begin{pmatrix} (-1)(0)-(1)(2)\\ (1)(-5)-(-4)(0)\\ (-4)(2)-(-1)(-5) \end{pmatrix} \\ =&\, \begin{pmatrix}-2\\-5\\-13\end{pmatrix}. \end{align*}

We may therefore use the opposite normal vector

n=(2513).\mathbf n= \begin{pmatrix}2\\5\\13\end{pmatrix}.

Since A=(3,2,2)A=(3,2,2) lies on Π1\Pi_1,

(322)⋅n= 2(3)+5(2)+13(2)= 42.\begin{align*} \begin{pmatrix}3\\2\\2\end{pmatrix} \mathbin{\cdot}\mathbf n =&\,2(3)+5(2)+13(2) \\ =&\,42. \end{align*}

Hence a Cartesian equation of Π1\Pi_1 is

2x+5y+13z=42.\boxed{2x+5y+13z=42}.

解法二：设一般平面方程

思路

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设平面方程为 αx+βy+γz=1\alpha x+\beta y+\gamma z=1，把三个点逐一代入，得到关于三个系数的线性方程组。解出系数后清除分母即可。

答题过程

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Let the plane have equation

αx+βy+γz=1.\alpha x+\beta y+\gamma z=1.

Substituting AA, BB and CC gives

3α+2β+2γ= 1,−α+β+3γ= 1,−2α+4β+2γ= 1.\begin{align*} 3\alpha+2\beta+2\gamma=&\,1, \\ -\alpha+\beta+3\gamma=&\,1, \\ -2\alpha+4\beta+2\gamma=&\,1. \end{align*}

Subtracting the second equation from the first and third equations respectively gives

4α+β−γ=0,−α+3β−γ=0.4\alpha+\beta-\gamma=0, \qquad -\alpha+3\beta-\gamma=0.

Subtracting these two equations,

5α−2β=0,5\alpha-2\beta=0,

so β=5α/2\beta=5\alpha/2. The second of the preceding equations then gives γ=13α/2\gamma=13\alpha/2.

Substituting into −α+β+3γ=1-\alpha+\beta+3\gamma=1,

−α+5α2+39α2= 1,21α= 1.\begin{align*} -\alpha+\frac{5\alpha}{2} +\frac{39\alpha}{2}=&\,1, \\ 21\alpha=&\,1. \end{align*}

Therefore,

α=121,β=542,γ=1342.\alpha=\frac1{21}, \qquad \beta=\frac5{42}, \qquad \gamma=\frac{13}{42}.

The plane equation is

x21+5y42+13z42=1.\frac{x}{21}+\frac{5y}{42}+\frac{13z}{42}=1.

Multiplying by 4242 gives

2x+5y+13z=42.\boxed{2x+5y+13z=42}.

解法三：写参数方程并消去参数

思路

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取 AA 为平面上的基点，以 AB→\overrightarrow{AB}、AC→\overrightarrow{AC} 为两个方向。把平面参数式拆成三个分量，再依次消去两个参数。

答题过程

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Using A=(3,2,2)A=(3,2,2), AB→=(−4,−1,1)\overrightarrow{AB}=(-4,-1,1) and AC→=(−5,2,0)\overrightarrow{AC}=(-5,2,0), a vector equation of the plane is

r=(322)+λ(−4−11)+μ(−520).\mathbf r=\begin{pmatrix}3\\2\\2\end{pmatrix} +\lambda\begin{pmatrix}-4\\-1\\1\end{pmatrix} +\mu\begin{pmatrix}-5\\2\\0\end{pmatrix}.

Thus

x=3−4λ−5μ,y=2−λ+2μ,z=2+λ.x=3-4\lambda-5\mu,\qquad y=2-\lambda+2\mu,\qquad z=2+\lambda.

The third equation gives λ=z−2\lambda=z-2. Substituting into the second gives

y=4−z+2μ,μ=y+z−42.y=4-z+2\mu,\qquad \mu=\frac{y+z-4}{2}.

Now substitute into the equation for xx:

x= 3−4(z−2)−52(y+z−4),2x= 6−8z+16−5y−5z+20,2x+5y+13z= 42.\begin{align*} x=&\,3-4(z-2)-\frac52(y+z-4),\\ 2x=&\,6-8z+16-5y-5z+20,\\ 2x+5y+13z=&\,42. \end{align*}

Hence

2x+5y+13z=42.\boxed{2x+5y+13z=42}.

(b)

解法一：使用直线参数方程

思路

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直线 ll 垂直于 Π1\Pi_1，所以它的方向向量就是 (a) 得到的法向量。以 DD 为已知点写出参数方程，代入 Π2\Pi_2 求交点参数；由于参数乘以方向向量的长度就是 DEDE，无须先写出 EE 的全部坐标。

答题过程

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Since ll is perpendicular to Π1\Pi_1, it has direction vector

n=(2513).\mathbf n= \begin{pmatrix}2\\5\\13\end{pmatrix}.

As ll passes through D=(−1,1,−2)D=(-1,1,-2), its vector equation is

r= (−11−2)+ λ(2513).\begin{align*} \mathbf r =&\,\begin{pmatrix}-1\\1\\-2\end{pmatrix} \\ +&\,\lambda \begin{pmatrix}2\\5\\13\end{pmatrix}. \end{align*}

Thus a point on ll has coordinates

x=2λ−1,y=5λ+1,z=13λ−2.x=2\lambda-1, \qquad y=5\lambda+1, \qquad z=13\lambda-2.

At EE, these coordinates also satisfy Π2\Pi_2. Therefore,

14(2λ−1) −(5λ+1) −17(13λ−2)=−66,−198λ+19= −66,λ= 85198.\begin{align*} 14(2\lambda-1) &\,-(5\lambda+1) \\ &\,-17(13\lambda-2)=-66, \\ -198\lambda+19=&\,-66, \\ \lambda=&\,\frac{85}{198}. \end{align*}

The magnitude of the direction vector is

∣n∣= 22+52+132= 198=322.\begin{align*} |\mathbf n| =&\,\sqrt{2^2+5^2+13^2} \\ =&\,\sqrt{198}=3\sqrt{22}. \end{align*}

Hence

DE= ∣λ∣ ∣n∣= 85198×322= 856622.\begin{align*} DE =&\,|\lambda|\,|\mathbf n| \\ =&\,\frac{85}{198}\times3\sqrt{22} \\ =&\,\frac{85}{66}\sqrt{22}. \end{align*}

Therefore,

p=8566.\boxed{p=\frac{85}{66}}.

解法二：联立直线的笛卡尔方程

思路

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把 ll 写成对称式，以 xx 表示 y,zy,z，再代入 Π2\Pi_2 直接求出交点 EE。最后将 DE→\overrightarrow{DE} 提取成法向量的倍数，计算距离。

答题过程

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A Cartesian form of ll is

x+12=y−15=z+213.\frac{x+1}{2} =\frac{y-1}{5} =\frac{z+2}{13}.

Hence

y=5x+72,z=13x+92.y=\frac{5x+7}{2}, \qquad z=\frac{13x+9}{2}.

Substituting into 14x−y−17z=−6614x-y-17z=-66 gives

14x−5x+72−17(13x+92)= −66,−198x−160= −132,x= −1499.\begin{align*} 14x-\frac{5x+7}{2} -17\biggl(\frac{13x+9}{2}\biggr) =&\,-66, \\ -198x-160=&\,-132, \\ x=&\,-\frac{14}{99}. \end{align*}

Therefore,

y=623198,z=709198,y=\frac{623}{198}, \qquad z=\frac{709}{198},

so

E=(−1499,623198,709198).E=\biggl(-\frac{14}{99}, \frac{623}{198}, \frac{709}{198}\biggr).

It follows that

DE→= (−1499+1623198−1709198+2)= 85198(2513).\begin{align*} \overrightarrow{DE} =&\, \begin{pmatrix} -\frac{14}{99}+1\\ \frac{623}{198}-1\\ \frac{709}{198}+2 \end{pmatrix} \\ =&\,\frac{85}{198} \begin{pmatrix}2\\5\\13\end{pmatrix}. \end{align*}

Thus

DE= 8519822+52+132= 856622.\begin{align*} DE =&\,\frac{85}{198} \sqrt{2^2+5^2+13^2} \\ =&\,\frac{85}{66}\sqrt{22}. \end{align*}

Therefore,

p=8566.\boxed{p=\frac{85}{66}}.

(c)

解法一：使用标量三重积

思路

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四面体体积是由同一点出发的三个棱向量所成平行六面体体积的六分之一。(a) 中的法向量与 AB→×AC→\overrightarrow{AB}\times\overrightarrow{AC} 只差一个负号，因此可直接用它和 AF→\overrightarrow{AF} 作数量积，再利用绝对值同时保留顶点 FF 位于平面两侧的两种可能。

答题过程

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We have

AF→=(11q−2).\overrightarrow{AF}= \begin{pmatrix}1\\1\\q-2\end{pmatrix}.

Also, from part (a),

AB→×AC→=−(2513).\overrightarrow{AB}\times\overrightarrow{AC} =- \begin{pmatrix}2\\5\\13\end{pmatrix}.

The volume of tetrahedron ABCFABCF is therefore

V= 16∣AF→⋅(AB→×AC→)∣= 16∣2+5+13(q−2)∣= 16∣13q−19∣.\begin{align*} V =&\,\frac16 \Bigg| \overrightarrow{AF}\mathbin{\cdot} \bigl( \overrightarrow{AB}\times\overrightarrow{AC} \bigr) \Bigg| \\ =&\,\frac16 \bigl|2+5+13(q-2)\bigr| \\ =&\,\frac16|13q-19|. \end{align*}

Since the volume is 1212,

16∣13q−19∣= 12,∣13q−19∣= 72.\begin{align*} \frac16|13q-19|=&\,12, \\ |13q-19|=&\,72. \end{align*}

Thus

13q−19=72or13q−19=−72.13q-19=72 \qquad\text{or}\qquad 13q-19=-72.

Hence the possible values are

q=7orq=−5313.\boxed{q=7\quad\text{or}\quad q=-\frac{53}{13}}.