AQA Further AS Paper 1 2020 June — Question 13 9 marks

Exam BoardAQA
ModuleFurther AS Paper 1 (Further AS Paper 1)
Year2020
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypeLine intersection with line
DifficultyStandard +0.8 This is a Further Maths question requiring conversion between line forms, understanding parallel/intersecting conditions, and solving simultaneous equations with parameters. Part (a) requires direction vector comparison (routine for Further Maths), while part (b) requires setting up and solving a system where lines intersect, involving parameter elimination and algebraic manipulation. The multi-step nature and parameter handling elevate this above average difficulty, but it follows standard Further Maths techniques without requiring novel geometric insight.
Spec4.04a Line equations: 2D and 3D, cartesian and vector forms4.04e Line intersections: parallel, skew, or intersecting4.04f Line-plane intersection: find point
Line \(l_1\) has equation $$\frac{x - 2}{3} = \frac{1 - 2y}{4} = -z$$ and line \(l_2\) has equation $$\mathbf{r} = \begin{bmatrix} -7 \\ 4 \\ -2 \end{bmatrix} + \mu \begin{bmatrix} 12 \\ a + 3 \\ 2b \end{bmatrix}$$
  1. In the case when \(l_1\) and \(l_2\) are parallel, show that \(a = -11\) and find the value of \(b\). [4 marks]
  2. In a different case, the lines \(l_1\) and \(l_2\) intersect at exactly one point, and the value of \(b\) is 3 Find the value of \(a\). [5 marks]
Question 13:
AnswerMarks Guidance
13(a)Obtains the correct direction vector for line . 1.1b
3
𝑙𝑙1 οΏ½βˆ’2οΏ½
βˆ’1
12 3
οΏ½π‘Žπ‘Ž+3οΏ½ = π‘π‘οΏ½βˆ’2οΏ½
2𝑏𝑏 βˆ’1
12= 3𝑝𝑝
𝑝𝑝 = an4d
π‘Žπ‘Ž+3 = βˆ’2𝑝𝑝 and 2𝑏𝑏 = βˆ’π‘π‘
π‘Žπ‘Ž+3 = βˆ’8 and 2𝑏𝑏 = βˆ’ 4
𝑙𝑙1
Selects a method to find and by equating
π‘Žπ‘Ž 𝑏𝑏
(multiples of) their direction vector and .
12
AnswerMarks Guidance
𝑙𝑙1 οΏ½π‘Žπ‘Ž+3οΏ½3.1a M1
2𝑏𝑏
Equates all components of their vectors and finds
AnswerMarks Guidance
the multiplier β€˜ ’.1.1a M1
𝑝𝑝
AnswerMarks Guidance
Shows correctly that and obtains .2.1 R1
AnswerMarks
13(b)Selects a method to find the intersection of and
by equating vector equations of the two lines or by
Cartesia𝑙𝑙1 𝑙𝑙2
substituting components of in the n
equation of .
AnswerMarks Guidance
𝑙𝑙23.1a M1
π‘₯π‘₯βˆ’2 π‘¦π‘¦βˆ’ 2 π‘§π‘§βˆ’0
= =
3 βˆ’2 βˆ’1
21 3
π‘Ÿπ‘Ÿ = οΏ½ 2 οΏ½+ πœ†πœ†οΏ½βˆ’2οΏ½
0 βˆ’1
βˆ’7+12πœ‡πœ‡ 2+3πœ†πœ†
οΏ½4+πœ‡πœ‡(π‘Žπ‘Ž+3)οΏ½ = οΏ½0.5βˆ’2πœ†πœ†οΏ½
βˆ’2+6πœ‡πœ‡ βˆ’πœ†πœ†
βˆ’7+12πœ‡πœ‡ = 2+3 πœ†πœ†
4+πœ‡πœ‡π‘Žπ‘Ž+3πœ‡πœ‡ = 0.5βˆ’2πœ†πœ†
βˆ’2+6πœ‡πœ‡ = βˆ’πœ†πœ†
βˆ’7+12πœ‡πœ‡ = 2+3(2βˆ’6πœ‡πœ‡)
1
πœ‡πœ‡ = 2
1
πœ†πœ† = 2βˆ’6Γ—2 = βˆ’1
1 1 1
4+2π‘Žπ‘Ž+3Γ—2 = 2βˆ’2Γ—βˆ’1
𝑙𝑙1
Forms an equation in , or writes at least two
simultaneous equations in and another parameter.
Allow one arithmetic eπœ‡πœ‡rror.
AnswerMarks Guidance
πœ‡πœ‡1.1a M1
Obtains the correct value of .1.1b A1
πœ‡πœ‡
Forms an equation in by substituting their value of
into an appropriate equation, from the intersection
of the two lines. π‘Žπ‘Ž
AnswerMarks Guidance
πœ‡πœ‡1.1a M1
Obtains the correct value of .
AnswerMarks Guidance
π‘Žπ‘Ž1.1b A1
Total9 π‘Žπ‘Ž = βˆ’6
QMarking instructions AO 
Question 13:
--- 13(a) ---
13(a) | Obtains the correct direction vector for line . | 1.1b | B1 | direction of is
3
𝑙𝑙1 οΏ½βˆ’2οΏ½
βˆ’1
12 3
οΏ½π‘Žπ‘Ž+3οΏ½ = π‘π‘οΏ½βˆ’2οΏ½
2𝑏𝑏 βˆ’1
12= 3𝑝𝑝
𝑝𝑝 = an4d
π‘Žπ‘Ž+3 = βˆ’2𝑝𝑝 and 2𝑏𝑏 = βˆ’π‘π‘
π‘Žπ‘Ž+3 = βˆ’8 and 2𝑏𝑏 = βˆ’ 4
𝑙𝑙1
Selects a method to find and by equating
π‘Žπ‘Ž 𝑏𝑏
(multiples of) their direction vector and .
12
𝑙𝑙1 οΏ½π‘Žπ‘Ž+3οΏ½ | 3.1a | M1
2𝑏𝑏
Equates all components of their vectors and finds
the multiplier β€˜ ’. | 1.1a | M1
𝑝𝑝
Shows correctly that and obtains . | 2.1 | R1
--- 13(b) ---
13(b) | Selects a method to find the intersection of and
by equating vector equations of the two lines or by
Cartesia𝑙𝑙1 𝑙𝑙2
substituting components of in the n
equation of .
𝑙𝑙2 | 3.1a | M1 | 1
π‘₯π‘₯βˆ’2 π‘¦π‘¦βˆ’ 2 π‘§π‘§βˆ’0
= =
3 βˆ’2 βˆ’1
21 3
π‘Ÿπ‘Ÿ = οΏ½ 2 οΏ½+ πœ†πœ†οΏ½βˆ’2οΏ½
0 βˆ’1
βˆ’7+12πœ‡πœ‡ 2+3πœ†πœ†
οΏ½4+πœ‡πœ‡(π‘Žπ‘Ž+3)οΏ½ = οΏ½0.5βˆ’2πœ†πœ†οΏ½
βˆ’2+6πœ‡πœ‡ βˆ’πœ†πœ†
βˆ’7+12πœ‡πœ‡ = 2+3 πœ†πœ†
4+πœ‡πœ‡π‘Žπ‘Ž+3πœ‡πœ‡ = 0.5βˆ’2πœ†πœ†
βˆ’2+6πœ‡πœ‡ = βˆ’πœ†πœ†
βˆ’7+12πœ‡πœ‡ = 2+3(2βˆ’6πœ‡πœ‡)
1
πœ‡πœ‡ = 2
1
πœ†πœ† = 2βˆ’6Γ—2 = βˆ’1
1 1 1
4+2π‘Žπ‘Ž+3Γ—2 = 2βˆ’2Γ—βˆ’1
𝑙𝑙1
Forms an equation in , or writes at least two
simultaneous equations in and another parameter.
Allow one arithmetic eπœ‡πœ‡rror.
πœ‡πœ‡ | 1.1a | M1
Obtains the correct value of . | 1.1b | A1
πœ‡πœ‡
Forms an equation in by substituting their value of
into an appropriate equation, from the intersection
of the two lines. π‘Žπ‘Ž
πœ‡πœ‡ | 1.1a | M1
Obtains the correct value of .
π‘Žπ‘Ž | 1.1b | A1
Total | 9 | π‘Žπ‘Ž = βˆ’6
Q | Marking instructions | AO | Marks | Typical solution
Line $l_1$ has equation
$$\frac{x - 2}{3} = \frac{1 - 2y}{4} = -z$$

and line $l_2$ has equation
$$\mathbf{r} = \begin{bmatrix} -7 \\ 4 \\ -2 \end{bmatrix} + \mu \begin{bmatrix} 12 \\ a + 3 \\ 2b \end{bmatrix}$$

\begin{enumerate}[label=(\alph*)]
\item In the case when $l_1$ and $l_2$ are parallel, show that $a = -11$ and find the value of $b$.
[4 marks]

\item In a different case, the lines $l_1$ and $l_2$ intersect at exactly one point, and the value of $b$ is 3

Find the value of $a$.
[5 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further AS Paper 1 2020 Q13 [9]}}
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