OCR Further Additional Pure AS 2019 June — Question 5 8 marks

Exam BoardOCR
ModuleFurther Additional Pure AS (Further Additional Pure AS)
Year2019
SessionJune
Marks8
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Mark schemeDownload PDF ↗
TopicVector Product and Surfaces
TypeSurface area of polyhedron
DifficultyChallenging +1.2 This is a straightforward application of the vector product to find areas of triangular faces. Students must compute four face areas using |a×b|/2, requiring careful arithmetic with the cross product formula and simplification of surds. While tedious with multiple calculations, it follows a standard algorithmic procedure with no conceptual surprises—typical for Further Maths vector geometry but more computational work than average A-level questions.
Spec1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors
5 The tetrahedron \(T\), shown below, has vertices at \(O ( 0,0,0 ) , A ( 1,2,2 ) , B ( 2,1,2 )\) and \(C ( 2,2,1 )\). \includegraphics[max width=\textwidth, alt={}, center]{59fa1650-a296-471e-93b9-0988177cd89d-3_360_464_319_555} Diagram not drawn to scale Show that the surface area of \(T\) is \(\frac { 1 } { 2 } \sqrt { 3 } ( 1 + \sqrt { 51 } )\).
Question 5:
AnswerMarks Guidance
5Area OAB = 1 a  b
2
1 2  2
     
where a  b = 2  1   2
     
2 2 3
= 1 17
2
Area OAC = Area OBC = 1 17 similarly
2
AnswerMarks Guidance
Area ABC = 1(b – a)  (c – a) e.g.
2
 1  1 1
     
where (b – a)  (c – a) =  1   0  1
     
 0 1 1
= 1 3
2
Surface area of T is then 3  1 17+ 1 3
2 2
   
= 1 3 3 171 = 1 3 511
AnswerMarks
2 2M1
B1
A1
B1
M1
A1
M1
A1
AnswerMarks
[8]3.1a
1.1
1.1
1.1
1.1a
1.1
3.1a
AnswerMarks
1.1Use of vector product oe for one simple  area
A correct, relevant vector product calculation
First correct simple  area (exact answer justified)
1 2 2 2 2 3
           
a  c = 2  2   3, b  c = 1  2   2
           
2 1 2 2 1  2
Final, complicated  area attempted
 0
 
c – b =  1
 
1
(First B1 can be earned for this area if not otherwise)
Must follow from correct vector product
Sum of four calculated  areas
AG fully legitimately obtained
Note that OAB, OAC, OBC are congruent isosceles s with sides 3, 3, 2 , while ABC is an equilateral  of side 2
Question 5:
5 | Area OAB = 1 | a  b |
2
1 2  2
     
where a  b = 2  1   2
     
2 2 3
= 1 17
2
Area OAC = Area OBC = 1 17 similarly
2
Area ABC = 1 | (b – a)  (c – a) | e.g.
2
 1  1 1
     
where (b – a)  (c – a) =  1   0  1
     
 0 1 1
= 1 3
2
Surface area of T is then 3  1 17+ 1 3
2 2
   
= 1 3 3 171 = 1 3 511
2 2 | M1
B1
A1
B1
M1
A1
M1
A1
[8] | 3.1a
1.1
1.1
1.1
1.1a
1.1
3.1a
1.1 | Use of vector product oe for one simple  area
A correct, relevant vector product calculation
First correct simple  area (exact answer justified)
1 2 2 2 2 3
           
a  c = 2  2   3, b  c = 1  2   2
           
2 1 2 2 1  2
Final, complicated  area attempted
 0
 
c – b =  1
 
1
(First B1 can be earned for this area if not otherwise)
Must follow from correct vector product
Sum of four calculated  areas
AG fully legitimately obtained
Note that OAB, OAC, OBC are congruent isosceles s with sides 3, 3, 2 , while ABC is an equilateral  of side 2
5 The tetrahedron $T$, shown below, has vertices at $O ( 0,0,0 ) , A ( 1,2,2 ) , B ( 2,1,2 )$ and $C ( 2,2,1 )$.\\
\includegraphics[max width=\textwidth, alt={}, center]{59fa1650-a296-471e-93b9-0988177cd89d-3_360_464_319_555}

Diagram not drawn to scale

Show that the surface area of $T$ is $\frac { 1 } { 2 } \sqrt { 3 } ( 1 + \sqrt { 51 } )$.

\hfill \mbox{\textit{OCR Further Additional Pure AS 2019 Q5 [8]}}
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