AQA Further Paper 1 Specimen — Question 9 13 marks

Exam BoardAQA
ModuleFurther Paper 1 (Further Paper 1)
SessionSpecimen
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypeAngle between line and plane
DifficultyChallenging +1.3 This is a multi-part Further Maths question on 3D coordinate geometry requiring understanding of line-plane relationships, but each part follows standard techniques: checking containment conditions, finding when direction vector isn't parallel to normal, and using the angle formula sin θ = |d·n|/(|d||n|). The main challenge is careful algebraic manipulation across multiple parts rather than novel geometric insight.
Spec4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04d Angles: between planes and between line and plane
A line has Cartesian equations \(x - p = \frac{y + 2}{q} = 3 - z\) and a plane has equation \(\mathbf{r} \cdot \begin{bmatrix} 1 \\ -1 \\ -2 \end{bmatrix} = -3\)
  1. In the case where the plane fully contains the line, find the values of \(p\) and \(q\). [3 marks]
  2. In the case where the line intersects the plane at a single point, find the range of values of \(p\) and \(q\). [3 marks]
  3. In the case where the angle \(\theta\) between the line and the plane satisfies \(\sin \theta = \frac{1}{\sqrt{6}}\) and the line intersects the plane at \(z = 0\)
    1. Find the value of \(q\). [4 marks]
    2. Find the value of \(p\). [3 marks]
Question 9:
AnswerMarks Guidance
9(a)Uses an appropriate method for
ensuring the line lies in the planeAO3.1a M1
y+2
λ= x− p= =3−z
q
,
then
x=λ+ p, y=qλ−2, z=3−λ
sub into equation of plane
(λ+ p)−(qλ−2 )−2 ( 3−λ)=−3
λ(3−q)+(p−1)=0
this is true for all λ
therefore p = 1 and q = 3
ALT
vector equation of line is
 p   1 
   
r= −2 +λ q
   
   
 3  −1
 p 
 
therefore −2 lies on the plane
 
 
 3 
 p   1 
   
−2 . −1 +3=0
   
   −2 
 3   
 1  1 
And  q  is perpendicular to  −1 
   
   
−1 −2
 1   1 
   
q . −1 =0
therefore    
  −1     −2 
⇒q=3 and p=1
AnswerMarks Guidance
Obtains equation(s) in p and qAO1.1a M1
Deduces the values of p and qAO2.2a A1
(b)States that to have a solution the
coefficient of λ in equation from
(a) cannot be 0
AnswerMarks Guidance
OR dot product must ≠ 0AO2.4 R1
   
q . −1 ≠0⇒q≠3
   
   
 −1 −2
p can take any value
AnswerMarks Guidance
Deduces the range of values for qAO2.2a R1
Deduces correct range of values
AnswerMarks Guidance
for pAO2.2a R1
ALT
vector equation of line is
 p   1 
   
r= −2 +λ q
   
   
 3  −1
 p 
 
therefore −2 lies on the plane
 
 
 3 
 p   1 
   
−2 . −1 +3=0
   
   −2 
 3   
 1  1 
And  q  is perpendicular to  −1 
   
   
−1 −2
 1   1 
   
q . −1 =0
therefore    
  −1     −2 
⇒q=3 and p=1
AnswerMarks Guidance
QMarking Instructions AO
(c)(i)Finds the correct scalar product
of the normal to the plane and
AnswerMarks Guidance
the direction vectorAO1.1b B1
   
n= −1 d= q
   
   
 −2 –1
n.d= 3−q
Let α be angle between the line
and the normal to the plane
1 1
sinθ= ⇒cosα=
6 6
 1 
q−3= 6 q2+2×  
 6
( 3−q)2 = q2+2
7
⇒6q = 7 givingq =
6
Correctly deduces the value of
AnswerMarks Guidance
cos αAO2.2a R1
Forms an equation connecting
all relevant parts using
AnswerMarks Guidance
n.d= n dcosθAO3.1a M1
Obtains correct value for qAO1.1b A1
(c)(ii)Uses ‘their’ expressions for x
and y and ‘their’ value for q and
the equation of the plane to form
AnswerMarks Guidance
an equation to find pAO3.1a M1
x− p= =3−z
7
6
z=0⇒x= p+3,y=1.5
p+3  1 
   
1.5 . −1 =−3
   
   
 0  −2
⇒ p+3−1.5=−3
⇒ p=−4.5
Uses z = 0 to deduce
expressions for x and y in terms
AnswerMarks Guidance
of p and qAO2.2a R1
Obtains the correct value of p
AnswerMarks Guidance
CAOAO1.1b A1
Total13
QMarking Instructions AO 
Question 9:
--- 9(a) ---
9(a) | Uses an appropriate method for
ensuring the line lies in the plane | AO3.1a | M1 | Let
y+2
λ= x− p= =3−z
q
,
then
x=λ+ p, y=qλ−2, z=3−λ
sub into equation of plane
(λ+ p)−(qλ−2 )−2 ( 3−λ)=−3
λ(3−q)+(p−1)=0
this is true for all λ
therefore p = 1 and q = 3
ALT
vector equation of line is
 p   1 
   
r= −2 +λ q
   
   
 3  −1
 p 
 
therefore −2 lies on the plane
 
 
 3 
 p   1 
   
−2 . −1 +3=0
   
   −2 
 3   
 1  1 
And  q  is perpendicular to  −1 
   
   
−1 −2
 1   1 
   
q . −1 =0
therefore    
  −1     −2 
⇒q=3 and p=1
Obtains equation(s) in p and q | AO1.1a | M1
Deduces the values of p and q | AO2.2a | A1
(b) | States that to have a solution the
coefficient of λ in equation from
(a) cannot be 0
OR dot product must ≠ 0 | AO2.4 | R1 |  1   1 
   
q . −1 ≠0⇒q≠3
   
   
 −1 −2
p can take any value
Deduces the range of values for q | AO2.2a | R1
Deduces correct range of values
for p | AO2.2a | R1
ALT
vector equation of line is
 p   1 
   
r= −2 +λ q
   
   
 3  −1
 p 
 
therefore −2 lies on the plane
 
 
 3 
 p   1 
   
−2 . −1 +3=0
   
   −2 
 3   
 1  1 
And  q  is perpendicular to  −1 
   
   
−1 −2
 1   1 
   
q . −1 =0
therefore    
  −1     −2 
⇒q=3 and p=1
Q | Marking Instructions | AO | Marks | Typical Solution
(c)(i) | Finds the correct scalar product
of the normal to the plane and
the direction vector | AO1.1b | B1 |  1   1
   
n= −1 d= q
   
   
 −2 –1
n.d= 3−q
Let α be angle between the line
and the normal to the plane
1 1
sinθ= ⇒cosα=
6 6
 1 
q−3= 6 q2+2×  
 6
( 3−q)2 = q2+2
7
⇒6q = 7 givingq =
6
Correctly deduces the value of
cos α | AO2.2a | R1
Forms an equation connecting
all relevant parts using
n.d= n dcosθ | AO3.1a | M1
Obtains correct value for q | AO1.1b | A1
(c)(ii) | Uses ‘their’ expressions for x
and y and ‘their’ value for q and
the equation of the plane to form
an equation to find p | AO3.1a | M1 | y+2
x− p= =3−z
7
6
z=0⇒x= p+3,y=1.5
p+3  1 
   
1.5 . −1 =−3
   
   
 0  −2
⇒ p+3−1.5=−3
⇒ p=−4.5
Uses z = 0 to deduce
expressions for x and y in terms
of p and q | AO2.2a | R1
Obtains the correct value of p
CAO | AO1.1b | A1
Total | 13
Q | Marking Instructions | AO | Marks | Typical Solution
A line has Cartesian equations $x - p = \frac{y + 2}{q} = 3 - z$ and a plane has equation $\mathbf{r} \cdot \begin{bmatrix} 1 \\ -1 \\ -2 \end{bmatrix} = -3$

\begin{enumerate}[label=(\alph*)]
\item In the case where the plane fully contains the line, find the values of $p$ and $q$.
[3 marks]

\item In the case where the line intersects the plane at a single point, find the range of values of $p$ and $q$.
[3 marks]

\item In the case where the angle $\theta$ between the line and the plane satisfies $\sin \theta = \frac{1}{\sqrt{6}}$ and the line intersects the plane at $z = 0$
\begin{enumerate}[label=(\roman*)]
\item Find the value of $q$.
[4 marks]

\item Find the value of $p$.
[3 marks]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 1  Q9 [13]}}
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