AQA Further Paper 2 Specimen — Question 14 9 marks

Exam BoardAQA
ModuleFurther Paper 2 (Further Paper 2)
SessionSpecimen
Marks9
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Mark schemeDownload PDF ↗
TopicVector Product and Surfaces
TypeVector product properties and identities
DifficultyChallenging +1.2 This is a Further Maths vector product question requiring expansion of cross products, application of standard properties (distributivity, anticommutativity, a×a=0), and use of the perpendicularity condition. While it involves multiple steps and careful algebraic manipulation, it follows a predictable pattern for this topic with no novel geometric insight required. The 9 marks reflect the need to explicitly state properties and show full working, but the conceptual demand is moderate for Further Maths students who have learned vector products.
Spec1.01a Proof: structure of mathematical proof and logical steps4.04g Vector product: a x b perpendicular vector
Given that the vectors a and b are perpendicular, prove that \(|(\mathbf{a} + 5\mathbf{b}) \times (\mathbf{a} - 4\mathbf{b})| = k|\mathbf{a}||\mathbf{b}|\), where \(k\) is an integer to be found. Explicitly state any properties of the vector product that you use within your proof. [9 marks]
Question 14:
AnswerMarks Guidance
14Uses vector product and expands
brackets correctlyAO1.1a M1
 a×a4a×b5b×a20b×b
 04a×b5b×a0
since a is parallel to a and b is parallel to b
then a×a=0 and b×b=0
 4a×b5a×b
since a×b=b×a
 9a×b
9a×b
9a bsin 90
9a b
Uses the correct notation and
correct order with the vector
AnswerMarks Guidance
product.AO2.5 B1
Reduces the number of terms in
‘their’ expression by using
AnswerMarks Guidance
aa bb 0AO1.1a M1
and explains their reasoning
(must have clear statement that
AnswerMarks Guidance
aa0)AO2.4 E1
Uses ab ba to collect
AnswerMarks Guidance
‘their’ terms togetherAO1.1a M1
and explains their reasoning
(must have clear statement that
AnswerMarks Guidance
ab baOE)AO2.4 E1
Recalls correctly the formula for
the modulus of the vector product
(may see a  b sin or may see
AnswerMarks Guidance
a bsin 90 )AO1.2 B1
Obtains ab  a b since
AnswerMarks Guidance
vectors a and b are perpendicularAO1.1b A1
Completes a fully correct proof
giving an answer of 9a b
AnswerMarks Guidance
CAOAO2.2a R1
Total9
QMarking Instructions AO 
Question 14:
14 | Uses vector product and expands
brackets correctly | AO1.1a | M1 | a+5b×a4b
 a×a4a×b5b×a20b×b
 04a×b5b×a0
since a is parallel to a and b is parallel to b
then a×a=0 and b×b=0
 4a×b5a×b
since a×b=b×a
 9a×b
9a×b
9a bsin 90
9a b
Uses the correct notation and
correct order with the vector
product. | AO2.5 | B1
Reduces the number of terms in
‘their’ expression by using
aa bb 0 | AO1.1a | M1
and explains their reasoning
(must have clear statement that
aa0) | AO2.4 | E1
Uses ab ba to collect
‘their’ terms together | AO1.1a | M1
and explains their reasoning
(must have clear statement that
ab baOE) | AO2.4 | E1
Recalls correctly the formula for
the modulus of the vector product
(may see a  b sin or may see

a bsin 90 ) | AO1.2 | B1
Obtains ab  a b since
vectors a and b are perpendicular | AO1.1b | A1
Completes a fully correct proof
giving an answer of 9a b
CAO | AO2.2a | R1
Total | 9
Q | Marking Instructions | AO | Marks | Typical Solution
Given that the vectors a and b are perpendicular, prove that

$|(\mathbf{a} + 5\mathbf{b}) \times (\mathbf{a} - 4\mathbf{b})| = k|\mathbf{a}||\mathbf{b}|$, where $k$ is an integer to be found.

Explicitly state any properties of the vector product that you use within your proof.
[9 marks]

\hfill \mbox{\textit{AQA Further Paper 2  Q14 [9]}}
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