OCR Further Additional Pure 2019 June — Question 6 12 marks

Exam BoardOCR
ModuleFurther Additional Pure (Further Additional Pure)
Year2019
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVector Product and Surfaces
TypeVector product properties and identities
DifficultyChallenging +1.2 This is a Further Maths question on vector products requiring multiple calculations and understanding of properties. Part (a) involves standard but multi-step computations (scalar triple product, double cross products). Parts (b-d) test conceptual understanding of associativity and geometric interpretation, which elevates it above routine. However, the calculations are methodical and the geometric reasoning is a standard Further Maths topic. Slightly above average difficulty for A-level overall, but typical for Further Additional Pure material.
Spec8.04a Vector product: definition, magnitude/direction, component form8.04b Vector product properties: anti-commutative and distributive8.04c Areas using vector product: triangles and parallelograms
6
  1. For the vectors \(\mathbf { p } = \left( \begin{array} { l } 1 \\ 2 \\ 3 \end{array} \right) , \mathbf { q } = \left( \begin{array} { r } 3 \\ 1 \\ - 1 \end{array} \right)\) and \(\mathbf { r } = \left( \begin{array} { r } 2 \\ - 4 \\ 5 \end{array} \right)\), calculate
    It is given that \(\mathbf { a } , \mathbf { b }\) and \(\mathbf { c }\) are three-dimensional column vectors with real components.
  2. Explain geometrically why the vector \(\mathbf { a } \times ( \mathbf { b } \times \mathbf { c } )\) must be expressible in the form \(\lambda \mathbf { b } + \mu \mathbf { c }\), where \(\lambda\) and \(\mu\) are scalar constants. It is given that the following relationship holds for \(\mathbf { a } , \mathbf { b }\) and \(\mathbf { c }\). \(\mathbf { a } \times ( \mathbf { b } \times \mathbf { c } ) = ( \mathbf { a } \cdot \mathbf { c } ) \mathbf { b } - ( \mathbf { a } \cdot \mathbf { b } ) \mathbf { c }\)
  3. Find an expression for ( \(\mathbf { a } \times \mathbf { b ) } \times \mathbf { c }\) in the form of (*).
Question 6:
AnswerMarks Guidance
6(a) 1  1  1 3 2
   
p • q × r = 2• −17 or 2 1 −4 = –75
   
3 −14 3 −1 5
1  1   23
     
p × (q × r) = 2×−17 =  17
     
3 −14 −19
−5  2 30
     
(p × q) × r =  10×−4 = 15
     
AnswerMarks
−5  5  0 1.1a
1.1
1.1a
1.1
1.1a
AnswerMarks
1.1M1
A1
M1
A1
M1
A1
AnswerMarks
[6]One vector product and a scalar product
Or via the determinant method
Two vector products, in correct order
Two vector products, in correct order
AnswerMarks Guidance
(b)No, since p × (q × r) ≠ (p × q) × r 2.2a
[1]Correct answer with reason
(c)b × c = n is normal/perpendicular to the plane containing b, c (or B, C)
Then a × n is perpendicular to this normal and hence in the plane of b, c
AnswerMarks
so that a × (b × c) is of the form λ b + µ c for scalar constants λ and µ .3.1a
3.2aM1
A1
AnswerMarks
[2]Consideration of fact that the vector product gives a vector
perpendicular to the two original vectors
Fully explained (“planes” not essential; directions suffice)
AnswerMarks
(d)(a × b) × c = – c × (a × b)
= – {(c • b) a – (c • a) b}
AnswerMarks
= (c • a) b – (c • b) a3.1a
2.1
AnswerMarks
2.2aM1
M1
AnswerMarks
A1Anti-commutative property used
Attempted use of given form, with a → c, b → a, c → b
Correct, in required form (order of dot products may be
reversed, of course)
Alt.
AnswerMarks
NB (a × b) × c = – c × (a × b) = – c × – (b × a) = c × (b × a)M1
= (c • a) b – (c • b) aA1
[3]
Question 6:
6 | (a) | 1  1  1 3 2
   
p • q × r = 2• −17 or 2 1 −4 = –75
   
3 −14 3 −1 5
1  1   23
     
p × (q × r) = 2×−17 =  17
     
3 −14 −19
−5  2 30
     
(p × q) × r =  10×−4 = 15
     
−5  5  0  | 1.1a
1.1
1.1a
1.1
1.1a
1.1 | M1
A1
M1
A1
M1
A1
[6] | One vector product and a scalar product
Or via the determinant method
Two vector products, in correct order
Two vector products, in correct order
(b) | No, since p × (q × r) ≠ (p × q) × r | 2.2a | B1
[1] | Correct answer with reason
(c) | b × c = n is normal/perpendicular to the plane containing b, c (or B, C)
Then a × n is perpendicular to this normal and hence in the plane of b, c
so that a × (b × c) is of the form λ b + µ c for scalar constants λ and µ . | 3.1a
3.2a | M1
A1
[2] | Consideration of fact that the vector product gives a vector
perpendicular to the two original vectors
Fully explained (“planes” not essential; directions suffice)
(d) | (a × b) × c = – c × (a × b)
= – {(c • b) a – (c • a) b}
= (c • a) b – (c • b) a | 3.1a
2.1
2.2a | M1
M1
A1 | Anti-commutative property used
Attempted use of given form, with a → c, b → a, c → b
Correct, in required form (order of dot products may be
reversed, of course)
Alt.
NB (a × b) × c = – c × (a × b) = – c × – (b × a) = c × (b × a) | M1
= (c • a) b – (c • b) a | A1
[3]
6
\begin{enumerate}[label=(\alph*)]
\item For the vectors $\mathbf { p } = \left( \begin{array} { l } 1 \\ 2 \\ 3 \end{array} \right) , \mathbf { q } = \left( \begin{array} { r } 3 \\ 1 \\ - 1 \end{array} \right)$ and $\mathbf { r } = \left( \begin{array} { r } 2 \\ - 4 \\ 5 \end{array} \right)$, calculate

\begin{itemize}
  \item $\mathbf { p } \cdot \mathbf { q } \times \mathbf { r }$,
  \item $\mathbf { p } \times ( \mathbf { q } \times \mathbf { r } )$,
  \item $( \mathbf { p } \times \mathbf { q } ) \times \mathbf { r }$.
\item State whether the vector product is associative for three-dimensional column vectors with real components. Justify your answer.
\end{itemize}

It is given that $\mathbf { a } , \mathbf { b }$ and $\mathbf { c }$ are three-dimensional column vectors with real components.
\item Explain geometrically why the vector $\mathbf { a } \times ( \mathbf { b } \times \mathbf { c } )$ must be expressible in the form $\lambda \mathbf { b } + \mu \mathbf { c }$, where $\lambda$ and $\mu$ are scalar constants.

It is given that the following relationship holds for $\mathbf { a } , \mathbf { b }$ and $\mathbf { c }$.\\
$\mathbf { a } \times ( \mathbf { b } \times \mathbf { c } ) = ( \mathbf { a } \cdot \mathbf { c } ) \mathbf { b } - ( \mathbf { a } \cdot \mathbf { b } ) \mathbf { c }$
\item Find an expression for ( $\mathbf { a } \times \mathbf { b ) } \times \mathbf { c }$ in the form of (*).
\end{enumerate}

\hfill \mbox{\textit{OCR Further Additional Pure 2019 Q6 [12]}}
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