OCR MEI Further Pure Core 2019 June — Question 2 3 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2019
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypeParallel and perpendicular planes
DifficultyModerate -0.5 This is a straightforward application of the perpendicularity condition for planes: normal vectors must be perpendicular, so their dot product equals zero. It requires only one calculation step (dot product = 0) and solving a simple linear equation, making it easier than average despite being from Further Maths.
Spec4.04b Plane equations: cartesian and vector forms

2 The plane \(x + 2 y + c z = 4\) is perpendicular to the plane \(2 x - c y + 6 z = 9\), where \(c\) is a constant. Find the value of \(c\).

Question 2:
AnswerMarks
2(i + 2j + ck).(2i  cj + 6k) = 0
 2  2c + 6c = 0
AnswerMarks
 c =  ½M1
A1
A1
AnswerMarks
[3]1.1a
1.1b
AnswerMarks
1.1bscalar product = 0
Question 2:
2 | (i + 2j + ck).(2i  cj + 6k) = 0
 2  2c + 6c = 0
 c =  ½ | M1
A1
A1
[3] | 1.1a
1.1b
1.1b | scalar product = 0
2 The plane $x + 2 y + c z = 4$ is perpendicular to the plane $2 x - c y + 6 z = 9$, where $c$ is a constant. Find the value of $c$.

\hfill \mbox{\textit{OCR MEI Further Pure Core 2019 Q2 [3]}}