AQA Further Paper 2 2024 June — Question 8 4 marks

Exam BoardAQA
ModuleFurther Paper 2 (Further Paper 2)
Year2024
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVector Product and Surfaces
TypeVector product properties and identities
DifficultyStandard +0.8 This is a Further Maths vector product question requiring systematic application of cross product properties (distributivity, anticommutativity, and a×a=0). While the algebraic manipulation is straightforward once you know the rules, students must carefully track signs and coefficients across multiple terms. The 4-mark allocation reflects moderate complexity—harder than routine A-level questions but not requiring deep geometric insight or proof.
Spec4.04g Vector product: a x b perpendicular vector
The vectors \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\) are such that \(\mathbf{a} \times \mathbf{b} = \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}\) and \(\mathbf{a} \times \mathbf{c} = \begin{bmatrix} 0 \\ 0 \\ 3 \end{bmatrix}\) Work out \((\mathbf{a} - 4\mathbf{b} + 3\mathbf{c}) \times (2\mathbf{a})\) [4 marks]
Question 8:
AnswerMarks
8Uses distributive
property of
cross-product.
Condone one
AnswerMarks Guidance
numerical error.3.1a M1
= 0 + 8a × b – 6a × c
 16 
 
= 8
 
 –18
Uses a × a = 0
PI by only b × a
and c × a terms
AnswerMarks Guidance
remaining1.1a M1
Uses
anti-commutative
property of
cross-product at
AnswerMarks Guidance
least once.1.1a M1
 16 
 
Obtains 8
 
 –18
AnswerMarks Guidance
1.1b A1
Question total4
QMarking
instructionsAO Marks 
Question 8:
8 | Uses distributive
property of
cross-product.
Condone one
numerical error. | 3.1a | M1 | (a – 4b + 3c) × (2a) = a × 2a – 4b × 2a + 3c × 2a
= 0 + 8a × b – 6a × c
 16 
 
= 8
 
 –18

Uses a × a = 0
PI by only b × a
and c × a terms
remaining | 1.1a | M1
Uses
anti-commutative
property of
cross-product at
least once. | 1.1a | M1
 16 
 
Obtains 8
 
 –18
 | 1.1b | A1
Question total | 4
Q | Marking
instructions | AO | Marks | Typical solution
The vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ are such that $\mathbf{a} \times \mathbf{b} = \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}$ and $\mathbf{a} \times \mathbf{c} = \begin{bmatrix} 0 \\ 0 \\ 3 \end{bmatrix}$

Work out $(\mathbf{a} - 4\mathbf{b} + 3\mathbf{c}) \times (2\mathbf{a})$
[4 marks]

\hfill \mbox{\textit{AQA Further Paper 2 2024 Q8 [4]}}
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