OCR MEI Further Pure Core 2023 June — Question 2 5 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2023
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypeAngle between line and plane
DifficultyModerate -0.3 This is a straightforward application of the standard formula for angle between a line and plane. Students need to identify the direction vector (3i+2j+k) and plane normal (-i+3j+2k), then use sin θ = |d·n|/(|d||n|). While it requires careful calculation with vectors, it's a direct textbook method with no problem-solving insight needed, making it slightly easier than average.
Spec4.04d Angles: between planes and between line and plane

2 In this question you must show detailed reasoning.
Find the angle between the vector \(3 i + 2 j + \mathbf { k }\) and the plane \(- x + 3 y + 2 z = 8\).

Question 2:
AnswerMarks
2DR
Normal to plane is i + 3j + 2k
Angle between 3i + 2j + k and i + 3j + 2k is 
3×(−1)+2×3+1×2 5
cos𝜃 = =
√3(cid:2870)+2(cid:2870)+1(cid:2870)(cid:3493)(−1)(cid:2870)+3(cid:2870)+2(cid:2870) √14√14
  = 69.07…
90− their 𝜃
AnswerMarks
so angle with plane is 20.9°B1
M1
A1
M1
AnswerMarks
A11.1a
1.1
1.1
1.1
AnswerMarks
2.2asoi
use of scalar product formula with their normal seen
(cid:2873)
or 1.2… rads or arccos
(cid:2869)(cid:2872)
(cid:3095)
or − their 𝜃
(cid:2870)
or 0.365 rads
Alternative method 1
AnswerMarks Guidance
Normal to plane is i + 3j + 2kB1 soi
Angle between 3i + 2j + k and the plane is 
3×(−1)+2×3+1×2 5
sin𝜃 = =
AnswerMarks Guidance
√3(cid:2870)+2(cid:2870)+1(cid:2870)(cid:3493)(−1)(cid:2870)+3(cid:2870)+2(cid:2870) √14√14M2
 = 20.9°A2 or 0.365 rads
Alternative method 2
AnswerMarks Guidance
Normal to plane is i + 3j + 2kB1 soi
Angle between 3i + 2j + k and i + 3j + 2k is 
3 −1 1
(cid:3629)(cid:3437)2(cid:3441)×(cid:3437) 3 (cid:3441)(cid:3629) (cid:3629)(cid:3437)−7(cid:3441)(cid:3629)
1 2 11
sin𝜃 = =
AnswerMarks Guidance
√3(cid:2870)+2(cid:2870)+1(cid:2870)(cid:3493)(−1)(cid:2870)+3(cid:2870)+2(cid:2870) √14√14M1 M1
  = 69.07…A1 √(cid:2869)(cid:2875)(cid:2869)
or 1.2… rads or arcsin
(cid:2869)(cid:2872)
90− their 𝜃
AnswerMarks
so angle with plane is 20.9°M1
A1(cid:3095)
or − their 𝜃
(cid:2870)
or 0.365 rads
[5]
90− their 𝜃
so angle with plane is 20.9°
M1
A1
Question 2:
2 | DR
Normal to plane is i + 3j + 2k
Angle between 3i + 2j + k and i + 3j + 2k is 
3×(−1)+2×3+1×2 5
cos𝜃 = =
√3(cid:2870)+2(cid:2870)+1(cid:2870)(cid:3493)(−1)(cid:2870)+3(cid:2870)+2(cid:2870) √14√14
  = 69.07…
90− their 𝜃
so angle with plane is 20.9° | B1
M1
A1
M1
A1 | 1.1a
1.1
1.1
1.1
2.2a | soi
use of scalar product formula with their normal seen
(cid:2873)
or 1.2… rads or arccos
(cid:2869)(cid:2872)
(cid:3095)
or − their 𝜃
(cid:2870)
or 0.365 rads
Alternative method 1
Normal to plane is i + 3j + 2k | B1 | soi
Angle between 3i + 2j + k and the plane is 
3×(−1)+2×3+1×2 5
sin𝜃 = =
√3(cid:2870)+2(cid:2870)+1(cid:2870)(cid:3493)(−1)(cid:2870)+3(cid:2870)+2(cid:2870) √14√14 | M2
 = 20.9° | A2 | or 0.365 rads
Alternative method 2
Normal to plane is i + 3j + 2k | B1 | soi
Angle between 3i + 2j + k and i + 3j + 2k is 
3 −1 1
(cid:3629)(cid:3437)2(cid:3441)×(cid:3437) 3 (cid:3441)(cid:3629) (cid:3629)(cid:3437)−7(cid:3441)(cid:3629)
1 2 11
sin𝜃 = =
√3(cid:2870)+2(cid:2870)+1(cid:2870)(cid:3493)(−1)(cid:2870)+3(cid:2870)+2(cid:2870) √14√14 | M1 | M1 | complete method with vector product seen | complete method with vector product seen
  = 69.07… | A1 | √(cid:2869)(cid:2875)(cid:2869)
or 1.2… rads or arcsin
(cid:2869)(cid:2872)
90− their 𝜃
so angle with plane is 20.9° | M1
A1 | (cid:3095)
or − their 𝜃
(cid:2870)
or 0.365 rads
[5]
90− their 𝜃
so angle with plane is 20.9°
M1
A1
2 In this question you must show detailed reasoning.\\
Find the angle between the vector $3 i + 2 j + \mathbf { k }$ and the plane $- x + 3 y + 2 z = 8$.

\hfill \mbox{\textit{OCR MEI Further Pure Core 2023 Q2 [5]}}