AQA Further Paper 2 2023 June — Question 11 9 marks

Exam BoardAQA
ModuleFurther Paper 2 (Further Paper 2)
Year2023
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypePlane containing line and point/vector
DifficultyStandard +0.8 This is a multi-part 3D coordinate geometry question requiring finding Cartesian equations from points, checking perpendicularity via direction vectors, and using coplanarity conditions. Part (b)(ii) requires understanding that coplanar lines means a point on one line and both direction vectors are coplanar, leading to a scalar triple product calculation or equivalent system-solving. While systematic, it demands solid vector manipulation and geometric insight beyond routine exercises.
Spec4.04a Line equations: 2D and 3D, cartesian and vector forms4.04e Line intersections: parallel, skew, or intersecting
The line \(l_1\) passes through the points \(A(6, 2, 7)\) and \(B(4, -3, 7)\)
  1. Find a Cartesian equation of \(l_1\) [2 marks]
  2. The line \(l_2\) has vector equation \(\mathbf{r} = \begin{pmatrix} 8 \\ 9 \\ c \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}\) where \(c\) is a constant.
    1. Explain how you know that the lines \(l_1\) and \(l_2\) are not perpendicular. [2 marks]
    2. The lines \(l_1\) and \(l_2\) both lie in the same plane. Find the value of \(c\) [5 marks]
Question 11:
AnswerMarks
11(a)Obtains a direction vector of l
1
AnswerMarks Guidance
PI1.1a M1
   
r= 2 +λ 5
   
 7  0
 
x=6+2λ,y =2+5λ,z =7
x−6 y−2
= ,z =7
2 5
Obtains a correct Cartesian
equation of l
AnswerMarks Guidance
11.1b A1
Subtotal2
QMarking Instructions AO
AnswerMarks
11(b)(i)Obtains correct scalar product of
their direction vector of l and
1
the direction vector of l
AnswerMarks Guidance
21.1b B1
=2×1+5×1+0=7
The scalar product is non-zero, so the
lines are not perpendicular.
Explains that the lines are not
perpendicular because this
AnswerMarks Guidance
scalar product is non-zero.2.4 E1
Subtotal2
QMarking Instructions AO
AnswerMarks
11(b)(ii)Obtains a vector perpendicular
to both lines
Or
Selects a method to obtain the
point of intersection of the two
AnswerMarks Guidance
lines.3.1a M1
i j k −10
 
n= 1 1 2 = 4
 
 
2 5 0  3 
Equation of plane is
−10
 
r• 4 =d
 
 
 3 
−10 6
   
d = 4 • 2 =−31
   
   
 3  7
8 −10
   
9 • 4 = −31
   
   
c  3 
−80+36+3c = −31
13
c =
3
Uses scalar product of their
normal vector and the position
vector of a point on l or l to
1 2
obtain constant term in equation
of plane.
PI
Or
Forms two simultaneous
AnswerMarks Guidance
equations in λ and μ only.1.1a M1
Obtains correct equation of
plane.
Or
Obtains correct simultaneous
AnswerMarks Guidance
equations.1.1b A1
Forms and solves equation in c
using their equation of the plane
or the solutions to their
AnswerMarks Guidance
simultaneous equations.1.1a M1
Obtains correct value of c1.1b A1
Subtotal5
Question total9
QMarking Instructions AO 
Question 11:
--- 11(a) ---
11(a) | Obtains a direction vector of l
1
PI | 1.1a | M1 | 6 2
   
r= 2 +λ 5
   
 7  0
 
x=6+2λ,y =2+5λ,z =7
x−6 y−2
= ,z =7
2 5
Obtains a correct Cartesian
equation of l
1 | 1.1b | A1
Subtotal | 2
Q | Marking Instructions | AO | Marks | Typical solution
--- 11(b)(i) ---
11(b)(i) | Obtains correct scalar product of
their direction vector of l and
1
the direction vector of l
2 | 1.1b | B1 | Scalar product of direction vectors
=2×1+5×1+0=7
The scalar product is non-zero, so the
lines are not perpendicular.
Explains that the lines are not
perpendicular because this
scalar product is non-zero. | 2.4 | E1
Subtotal | 2
Q | Marking Instructions | AO | Marks | Typical solution
--- 11(b)(ii) ---
11(b)(ii) | Obtains a vector perpendicular
to both lines
Or
Selects a method to obtain the
point of intersection of the two
lines. | 3.1a | M1 | Normal to plane
i j k −10
 
n= 1 1 2 = 4
 
 
2 5 0  3 
Equation of plane is
−10
 
r• 4 =d
 
 
 3 
−10 6
   
d = 4 • 2 =−31
   
   
 3  7
8 −10
   
9 • 4 = −31
   
   
c  3 
−80+36+3c = −31
13
c =
3
Uses scalar product of their
normal vector and the position
vector of a point on l or l to
1 2
obtain constant term in equation
of plane.
PI
Or
Forms two simultaneous
equations in λ and μ only. | 1.1a | M1
Obtains correct equation of
plane.
Or
Obtains correct simultaneous
equations. | 1.1b | A1
Forms and solves equation in c
using their equation of the plane
or the solutions to their
simultaneous equations. | 1.1a | M1
Obtains correct value of c | 1.1b | A1
Subtotal | 5
Question total | 9
Q | Marking Instructions | AO | Marks | Typical Solution
The line $l_1$ passes through the points $A(6, 2, 7)$ and $B(4, -3, 7)$

\begin{enumerate}[label=(\alph*)]
\item Find a Cartesian equation of $l_1$
[2 marks]

\item The line $l_2$ has vector equation $\mathbf{r} = \begin{pmatrix} 8 \\ 9 \\ c \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$ where $c$ is a constant.

\begin{enumerate}[label=(\roman*)]
\item Explain how you know that the lines $l_1$ and $l_2$ are not perpendicular.
[2 marks]

\item The lines $l_1$ and $l_2$ both lie in the same plane.

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

\hfill \mbox{\textit{AQA Further Paper 2 2023 Q11 [9]}}
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