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IAL 2025 Jan Q8

A Level / Edexcel / P4

IAL 2025 Jan Paper · Question 8

题目

Problem

Relative to a fixed origin OO, the line ll has equation

r=(2−13)+λ(42−1)\mathbf{r}= \begin{pmatrix} 2\\ -1\\ 3 \end{pmatrix} +\lambda \begin{pmatrix} 4\\ 2\\ -1 \end{pmatrix}

where λ\lambda is a scalar parameter.

The point AA and the point BB lie on line ll.

Given that

  • AA has coordinates (−2,a,4)(-2,a,4)
  • BB has coordinates (b,3,1)(b,3,1)

(a) find the value of the constant aa and the value of the constant bb.

(2)

(b) Hence find vector AB→\overrightarrow{AB}.

(2)

The point CC has coordinates (4,7,−2)(4,7,-2).

(c) Find the size of angle CABCAB, giving your answer in degrees to one decimal place.

(4)

The point DD lies on the line ll so that the area of triangle CADCAD is twice the area of triangle CABCAB.

(d) Find the coordinates of the two possible positions of DD.

(4)
题目中文翻译

相对于固定原点 OO，直线 ll 的方程为

r=(2−13)+λ(42−1)\mathbf{r}= \begin{pmatrix} 2\\ -1\\ 3 \end{pmatrix} +\lambda \begin{pmatrix} 4\\ 2\\ -1 \end{pmatrix}

其中 λ\lambda 是标量参数。

点 AA 与点 BB 都在线 ll 上。

已知

  • AA 的坐标为 (−2,a,4)(-2,a,4)；
  • BB 的坐标为 (b,3,1)(b,3,1)。

(a) 求常数 aa 与常数 bb 的值。

(b) 进而求向量 AB→\overrightarrow{AB}。

点 CC 的坐标为 (4,7,−2)(4,7,-2)。

(c) 求角 CABCAB 的大小，答案用度数表示并保留到 1 位小数。

点 DD 在线 ll 上，并且三角形 CADCAD 的面积是三角形 CABCAB 面积的两倍。

(d) 求点 DD 的两个可能位置的坐标。

解答

(a)

The line is

r=(2−13)+λ(42−1)\mathbf r= \begin{pmatrix} 2\\ -1\\ 3 \end{pmatrix} +\lambda \begin{pmatrix} 4\\ 2\\ -1 \end{pmatrix}

So a general point on the line has coordinates

(2+4λ, −1+2λ, 3−λ)(2+4\lambda,\ -1+2\lambda,\ 3-\lambda)

For

A=(−2,a,4)A=(-2,a,4)

use the xx coordinate:

2+4λ=−22+4\lambda=-2

so

λ=−1\lambda=-1

Then

a=−1+2(−1)=−3a=-1+2(-1)=-3

For

B=(b,3,1)B=(b,3,1)

use the yy coordinate:

−1+2λ=3-1+2\lambda=3

so

λ=2\lambda=2

Then

b=2+4(2)=10b=2+4(2)=10

Therefore

a=−3,b=10\boxed{a=-3,\qquad b=10}

(b)

Using

A=(−2,−3,4)A=(-2,-3,4)

and

B=(10,3,1)B=(10,3,1)

we get

AB→=B−A\overrightarrow{AB}=B-A

So

AB→=(10−(−2)3−(−3)1−4)=(126−3)\overrightarrow{AB} = \begin{pmatrix} 10-(-2)\\ 3-(-3)\\ 1-4 \end{pmatrix} = \begin{pmatrix} 12\\ 6\\ -3 \end{pmatrix}

Therefore

AB→=(126−3)\boxed{\overrightarrow{AB}= \begin{pmatrix} 12\\ 6\\ -3 \end{pmatrix}}

(c)

We need angle CABCAB, so use vectors

AC→\overrightarrow{AC}

and

AB→\overrightarrow{AB}

Now

C=(4,7,−2)C=(4,7,-2)

so

AC→=C−A=(4−(−2)7−(−3)−2−4)=(610−6)\overrightarrow{AC}=C-A = \begin{pmatrix} 4-(-2)\\ 7-(-3)\\ -2-4 \end{pmatrix} = \begin{pmatrix} 6\\ 10\\ -6 \end{pmatrix}

Also,

AB→=(126−3)\overrightarrow{AB} = \begin{pmatrix} 12\\ 6\\ -3 \end{pmatrix}

The scalar product is

AC→⋅AB→=6(12)+10(6)+(−6)(−3)\overrightarrow{AC}\cdot\overrightarrow{AB} =6(12)+10(6)+(-6)(-3)

So

AC→⋅AB→=150\overrightarrow{AC}\cdot\overrightarrow{AB}=150

Also,

∣AC→∣=62+102+(−6)2=172|\overrightarrow{AC}| =\sqrt{6^2+10^2+(-6)^2} =\sqrt{172}

and

∣AB→∣=122+62+(−3)2=189|\overrightarrow{AB}| =\sqrt{12^2+6^2+(-3)^2} =\sqrt{189}

Therefore

cos⁡∠CAB=150172189\cos\angle CAB =\frac{150}{\sqrt{172}\sqrt{189}}

Hence

∠CAB=33.7∘\angle CAB=33.7^\circ

to one decimal place.

So

33.7∘\boxed{33.7^\circ}

(d)

Since AA, BB and DD lie on the same line, the height from CC to the line is the same for triangles CABCAB and CADCAD.

So the area ratio is the same as the base length ratio:

area of CADarea of CAB=ADAB\frac{\text{area of }CAD}{\text{area of }CAB} =\frac{AD}{AB}

We are told

area of CAD=2⋅area of CAB\text{area of }CAD=2\cdot \text{area of }CAB

Therefore

AD=2ABAD=2AB

On the line, AA corresponds to

λ=−1\lambda=-1

and BB corresponds to

λ=2\lambda=2

So the parameter difference from AA to BB is

33

For AD=2ABAD=2AB, the parameter difference from AA to DD must be

66

Thus

λ=−1+6=5\lambda=-1+6=5

or

λ=−1−6=−7\lambda=-1-6=-7

When λ=5\lambda=5,

D=(2+4(5), −1+2(5), 3−5)D=(2+4(5),\ -1+2(5),\ 3-5)

so

D=(22,9,−2)D=(22,9,-2)

When λ=−7\lambda=-7,

D=(2+4(−7), −1+2(−7), 3−(−7))D=(2+4(-7),\ -1+2(-7),\ 3-(-7))

so

D=(−26,−15,10)D=(-26,-15,10)

Therefore the two possible positions of DD are

(22,9,−2) and (−26,−15,10)\boxed{(22,9,-2)\ \text{and}\ (-26,-15,10)}