OCR MEI Further Pure Core 2019 June — Question 12 9 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2019
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Cross Product & Distances
TypeArea of triangle using cross product
DifficultyChallenging +1.2 This is a Further Maths question requiring multiple steps: finding intersection points of three lines in 3D (solving systems of equations), then applying the cross product formula for triangle area. While it involves several techniques, the approach is methodical and follows standard procedures taught in Further Pure. The computational work is substantial but straightforward, making it moderately harder than average A-level questions but not requiring novel insight.
Spec4.04c Scalar product: calculate and use for angles4.04h Shortest distances: between parallel lines and between skew lines
12 Three intersecting lines \(L _ { 1 } , L _ { 2 }\) and \(L _ { 3 }\) have equations \(L _ { 1 } : \frac { x } { 2 } = \frac { y } { 3 } = \frac { z } { 1 } , \quad L _ { 2 } : \frac { x } { 1 } = \frac { y } { 2 } = \frac { z } { - 4 } \quad\) and \(\quad L _ { 3 } : \frac { x - 1 } { 1 } = \frac { y - 2 } { 1 } = \frac { z + 4 } { 5 }\).
Find the area of the triangle enclosed by these lines.
Question 12:
AnswerMarks Guidance
12L and L :  = 4 + 5, 8 + 10 = 1 + 
1 3M1 1.1b
  = 1,  = 1, so meet at (2, 3, 1)B1 2.2a
L and L meet at (1, 2, 4)
AnswerMarks Guidance
2 3B1 2.2a
L and L meet at origin
AnswerMarks Guidance
1 2L and L meet at origin
1 2B1 2.2a
(4)
(2i3jk).(i2j4k)
cos
AnswerMarks Guidance
223212 1222(4)2M1 3.1a
M1
4
AnswerMarks Guidance
14 21A1 1.1b
or cos1 ,cos1
AnswerMarks Guidance
14 27 21 27sides 21, 14, 27
 = 76.5 (1.335) = 76.5 (1.335) A1
(3)
AnswerMarks Guidance
Area = ½ 14 21 sin 76.5M1 1.1a
= 8.34 [units2]= 8.34 [units2] A1
(2)
AnswerMarks Guidance
Alternative solutionusing cross product
(2i3jk)(i2j4k)M1 (2i3jk)(ij5k)
14i9jk14i9jk A1
(2)
AnswerMarks Guidance
Area = ½  (142 + 92 + 12)M1 or ½ base  height
= ½ 278 = 8.34 [units2]= ½ 278 = 8.34 [units2] A2
(3)
[9]
3.1a
or cosine rule
A1
1.1b
Question 12:
12 | L and L :  = 4 + 5, 8 + 10 = 1 + 
1 3 | M1 | 1.1b | attempt to solve (any pair)
  = 1,  = 1, so meet at (2, 3, 1) | B1 | 2.2a | by solving or inspection
L and L meet at (1, 2, 4)
2 3 | B1 | 2.2a | by solving or inspection
L and L meet at origin
1 2 | L and L meet at origin
1 2 | B1 | 2.2a | 2.2a | by solving or inspection | by solving or inspection | soi | soi
(4)
(2i3jk).(i2j4k)
cos
223212 1222(4)2 | M1 | 3.1a | or cosine rule
M1
4

14 21 | A1 | 1.1b | 10 17
or cos1 ,cos1
14 27 21 27 | sides 21, 14, 27
 = 76.5 (1.335) |  = 76.5 (1.335) | A1 | 1.1b | 59.044.4 | 59.044.4 | any correct angle
(3)
Area = ½ 14 21 sin 76.5 | M1 | 1.1a | or ½ 14 27 sin 59.0 | or ½ 21 27 sin 44.4
= 8.34 [units2] | = 8.34 [units2] | A1 | 3.2a | 3.2a | art 8.3 or 278/2 | art 8.3 or 278/2
(2)
Alternative solution | using cross product
(2i3jk)(i2j4k) | M1 | (2i3jk)(ij5k) | (i2j4k)(ij5k)
14i9jk | 14i9jk | A1 | 14i9jk | 14i9jk | 14i9jk | 14i9jk
(2)
Area = ½  (142 + 92 + 12) | M1 | or ½ base  height | ½ 14 287/14, etc
= ½ 278 = 8.34 [units2] | = ½ 278 = 8.34 [units2] | A2 | art 8.34 | art 8.34
(3)
[9]
3.1a
or cosine rule
A1
1.1b
12 Three intersecting lines $L _ { 1 } , L _ { 2 }$ and $L _ { 3 }$ have equations\\
$L _ { 1 } : \frac { x } { 2 } = \frac { y } { 3 } = \frac { z } { 1 } , \quad L _ { 2 } : \frac { x } { 1 } = \frac { y } { 2 } = \frac { z } { - 4 } \quad$ and $\quad L _ { 3 } : \frac { x - 1 } { 1 } = \frac { y - 2 } { 1 } = \frac { z + 4 } { 5 }$.\\
Find the area of the triangle enclosed by these lines.

\hfill \mbox{\textit{OCR MEI Further Pure Core 2019 Q12 [9]}}
This paper (17 questions)
View full paper
Q1 4 Q2 3 Q3 7 Q4 3 Q5 5 Q6 4 Q7 8 Q8 8 Q9 7 Q10 8 Q11 12 Q12 9 Q13 11 Q14 13 Q15 8 Q16 12 Q17 22