OCR Further Additional Pure AS 2018 June — Question 1 8 marks

Exam BoardOCR
ModuleFurther Additional Pure AS (Further Additional Pure AS)
Year2018
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVector Product and Surfaces
TypeArea of triangle using vector product
DifficultyStandard +0.3 This is a straightforward application of the vector product formula for triangle area (½|AB × AC|), followed by using the area-base-height relationship. While it's a Further Maths topic, the question requires only direct application of standard formulas with no conceptual challenges or novel insights, making it slightly easier than average overall.
Spec4.04g Vector product: a x b perpendicular vector4.04h Shortest distances: between parallel lines and between skew lines
1 The points \(A , B\) and \(C\) have position vectors \(6 \mathbf { i } + 2 \mathbf { j } + 4 \mathbf { k } , 13 \mathbf { i } + 2 \mathbf { j } + 5 \mathbf { k }\) and \(16 \mathbf { i } + 6 \mathbf { j } + 3 \mathbf { k }\) respectively.
  1. Using the vector product, calculate the area of triangle \(A B C\).
  2. Hence find, in simplest surd form, the perpendicular distance from \(C\) to the line through \(A\) and \(B\).
Question 1:
AnswerMarks Guidance
1(i) 7 10  3 
     
b – a = 0, c – a =  4, c – b =  4 
     
1 1 2
4
 
(b – a)  (c – a) =  17
 
 28
AnswerMarks
Area ABC = 1their (b – a)  (c – a)
2
33
=
AnswerMarks
2B1
M1
A1
M1
AnswerMarks
A11.1a
1.1a
1.1
1.2
AnswerMarks
1.1Any two of these
Attempt at a suitable vector product
FT correct
FT in surd, rational or decimal form
[5]
AnswerMarks
(ii)AB = 50 or 5 2
Area ABC = 1 (AB) d = 33
2 2
33 33 33 50 33 2
 d = , , ,
AnswerMarks
5 2 50 50 10B1
M1
AnswerMarks
A11.1
1.1a
AnswerMarks
1.1FT their (b – a) from (i)
Equating 1 baseht. (correct
2
setting) to (i)’s answer
AnswerMarks
cao any exact surd form3.3 2 or 21.78 also ok
[3]
Question 1:
1 | (i) | 7 10  3 
     
b – a = 0, c – a =  4, c – b =  4 
     
1 1 2
4
 
(b – a)  (c – a) =  17
 
 28
Area ABC = 1 | their (b – a)  (c – a) |
2
33
=
2 | B1
M1
A1
M1
A1 | 1.1a
1.1a
1.1
1.2
1.1 | Any two of these
Attempt at a suitable vector product
FT correct
FT in surd, rational or decimal form
[5]
(ii) | AB = 50 or 5 2
Area ABC = 1 (AB) d = 33
2 2
33 33 33 50 33 2
 d = , , ,
5 2 50 50 10 | B1
M1
A1 | 1.1
1.1a
1.1 | FT their (b – a) from (i)
Equating 1 baseht. (correct
2
setting) to (i)’s answer
cao any exact surd form | 3.3 2 or 21.78 also ok
[3]
1 The points $A , B$ and $C$ have position vectors $6 \mathbf { i } + 2 \mathbf { j } + 4 \mathbf { k } , 13 \mathbf { i } + 2 \mathbf { j } + 5 \mathbf { k }$ and $16 \mathbf { i } + 6 \mathbf { j } + 3 \mathbf { k }$ respectively.\\
(i) Using the vector product, calculate the area of triangle $A B C$.\\
(ii) Hence find, in simplest surd form, the perpendicular distance from $C$ to the line through $A$ and $B$.

\hfill \mbox{\textit{OCR Further Additional Pure AS 2018 Q1 [8]}}
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