OCR Further Additional Pure AS 2022 June — Question 1 6 marks

Exam BoardOCR
ModuleFurther Additional Pure AS (Further Additional Pure AS)
Year2022
SessionJune
Marks6
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Mark schemeDownload PDF ↗
TopicVector Product and Surfaces
TypeVector product properties and identities
DifficultyStandard +0.8 This is a Further Maths question requiring vector product calculations and verification of a non-trivial vector identity. Part (a) is routine (computing a×b and finding magnitude), but part (b) requires either systematic expansion using the vector triple product formula (a×(b×c) = b(a·c) - c(a·b)) or direct computation of three cross products—both demanding careful algebraic manipulation. The identity itself is not standard textbook material, making this moderately challenging even for Further Maths students.
Spec4.04g Vector product: a x b perpendicular vector4.05a Roots and coefficients: symmetric functions
1 The points \(A , B\) and \(C\) have position vectors \(\mathbf { a } = \left( \begin{array} { l } 1 \\ 1 \\ 3 \end{array} \right) , \mathbf { b } = \left( \begin{array} { r } 2 \\ 3 \\ - 1 \end{array} \right)\) and \(\mathbf { c } = \left( \begin{array} { r } - 5 \\ 1 \\ 2 \end{array} \right)\) respectively, relative to the origin \(O\).
  1. Calculate, in its simplest exact form, the area of triangle \(O A B\).
  2. Show that \(\mathbf { a } \times ( \mathbf { b } \times \mathbf { c } ) + \mathbf { b } \times ( \mathbf { c } \times \mathbf { a } ) + \mathbf { c } \times ( \mathbf { a } \times \mathbf { b } ) = \mathbf { 0 }\).
Question 1:
AnswerMarks Guidance
1a a  b = −10i + 7j + k
Area OAB = 12a  b
= 1 150 or 52 6
AnswerMarks
2B1
M1
A1
AnswerMarks
[3]1.1
1.2
AnswerMarks
1.1Use of area formula
FT their vector product
AnswerMarks
ba  (b  c) + b  (c  a) + c  (a  b)
 1   7   2   1   − 5   − 1 0 
= 1  1 + 3  1 7 + 1  7
3 1 7 − 1 − 6 2 1
 1 4   − 1   − 1 3   0 
= 4 + 1 1 + − 1 5 = 0
AnswerMarks
− 6 3 1 − 2 5 0M1
M1
A1
AnswerMarks
[3]1.1
1.1
AnswerMarks
1.1Attempt at any three vector
products (FT their ab from above,
and at least one other correct)
Three further vector product
attempts (at least one FT correct)
AG Correct final answer from
fully correct and visible working
 x ( x − 1 ) (9 x 2 − 1 5 x + 3 1 ) = 0
AnswerMarks
giving x = 1, y = 13 and z = − 12 47M1
A1
AnswerMarks
[6]1.1
1.1Solving (possibly BC)
All three coordinates of A correctSC1 to replace the final
three marks if all three
coordinates of A correct
without working
Question 1:
1 | a | a  b = −10i + 7j + k
Area OAB = 12 | a  b |
= 1 150 or 52 6
2 | B1
M1
A1
[3] | 1.1
1.2
1.1 | Use of area formula
FT their vector product
b | a  (b  c) + b  (c  a) + c  (a  b)
 1   7   2   1   − 5   − 1 0 
= 1  1 + 3  1 7 + 1  7
3 1 7 − 1 − 6 2 1
 1 4   − 1   − 1 3   0 
= 4 + 1 1 + − 1 5 = 0
− 6 3 1 − 2 5 0 | M1
M1
A1
[3] | 1.1
1.1
1.1 | Attempt at any three vector
products (FT their ab from above,
and at least one other correct)
Three further vector product
attempts (at least one FT correct)
AG Correct final answer from
fully correct and visible working
 x ( x − 1 ) (9 x 2 − 1 5 x + 3 1 ) = 0
giving x = 1, y = 13 and z = − 12 47 | M1
A1
[6] | 1.1
1.1 | Solving (possibly BC)
All three coordinates of A correct | SC1 to replace the final
three marks if all three
coordinates of A correct
without working
1 The points $A , B$ and $C$ have position vectors $\mathbf { a } = \left( \begin{array} { l } 1 \\ 1 \\ 3 \end{array} \right) , \mathbf { b } = \left( \begin{array} { r } 2 \\ 3 \\ - 1 \end{array} \right)$ and $\mathbf { c } = \left( \begin{array} { r } - 5 \\ 1 \\ 2 \end{array} \right)$ respectively, relative to the origin $O$.
\begin{enumerate}[label=(\alph*)]
\item Calculate, in its simplest exact form, the area of triangle $O A B$.
\item Show that $\mathbf { a } \times ( \mathbf { b } \times \mathbf { c } ) + \mathbf { b } \times ( \mathbf { c } \times \mathbf { a } ) + \mathbf { c } \times ( \mathbf { a } \times \mathbf { b } ) = \mathbf { 0 }$.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Additional Pure AS 2022 Q1 [6]}}
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