AQA Further Paper 1 2024 June — Question 5 5 marks

Exam BoardAQA
ModuleFurther Paper 1 (Further Paper 1)
Year2024
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypeCartesian equation of a plane
DifficultyStandard +0.3 This is a standard Further Maths vectors question requiring the cross product of two vectors in the plane to find the normal, then using point-normal form for the Cartesian equation. Both parts are routine applications of well-practiced techniques with no conceptual challenges or novel problem-solving required. Slightly above average difficulty only because it's Further Maths content, but this is textbook-standard for that level.
Spec4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms
The points \(A\), \(B\) and \(C\) have coordinates \(A(5, 3, 4)\), \(B(8, -1, 9)\) and \(C(12, 5, 10)\) The points \(A\), \(B\) and \(C\) lie in the plane \(\Pi\)
  1. Find a vector that is normal to the plane \(\Pi\) [3 marks]
  2. Find a Cartesian equation of the plane \(\Pi\) [2 marks]
Question 5:
AnswerMarks
5(a)Obtains two of the three vectors
connecting A, B and C
 3   7   4 
 – 4 ,  2 ,  6
5 6 1
AnswerMarks Guidance
Condone one incorrect element.1.1a M1
A B × A C = – 4  2 = 1 7 1
5 6 2
Forms the vector product of two
AnswerMarks Guidance
vectors.1.1a M1
 – 2 
Obtains k 1
AnswerMarks Guidance
21.1b A1
Subtotal3
QMarking instructions AO
AnswerMarks
5(b)Obtains their –2x + y + 2z = d
where d is a number
or
Evaluates the scalar product of
their normal vector and a point
AnswerMarks Guidance
in the plane.1.1a M1
 5   – 2 
d = 3 • 1 = 1
4 2
– 2 x + y + 2 z = 1
AnswerMarks Guidance
Obtains –2x + y + 2z = 1 OE1.1b A1
Subtotal2
Question total5
QMarking instructions AO 
Question 5:
--- 5(a) ---
5(a) | Obtains two of the three vectors
connecting A, B and C
 3   7   4 
 – 4 ,  2 ,  6
5 6 1
Condone one incorrect element. | 1.1a | M1 |  3   7   – 2 
A B × A C = – 4  2 = 1 7 1
5 6 2
Forms the vector product of two
vectors. | 1.1a | M1
 – 2 
Obtains k 1
2 | 1.1b | A1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 5(b) ---
5(b) | Obtains their –2x + y + 2z = d
where d is a number
or
Evaluates the scalar product of
their normal vector and a point
in the plane. | 1.1a | M1 | – 2 x + y + 2 z = d
 5   – 2 
d = 3 • 1 = 1
4 2
– 2 x + y + 2 z = 1
Obtains –2x + y + 2z = 1 OE | 1.1b | A1
Subtotal | 2
Question total | 5
Q | Marking instructions | AO | Marks | Typical solution
The points $A$, $B$ and $C$ have coordinates $A(5, 3, 4)$, $B(8, -1, 9)$ and $C(12, 5, 10)$

The points $A$, $B$ and $C$ lie in the plane $\Pi$

\begin{enumerate}[label=(\alph*)]
\item Find a vector that is normal to the plane $\Pi$ [3 marks]
\item Find a Cartesian equation of the plane $\Pi$ [2 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 1 2024 Q5 [5]}}
This paper (18 questions)
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Q1 1 Q2 1 Q3 1 Q4 1 Q5 5 Q6 4 Q7 5 Q8 4 Q9 8 Q10 6 Q11 5 Q12 10 Q13 9 Q14 7 Q15 5 Q16 9 Q17 7 Q18 12