OCR MEI Further Pure Core Specimen — Question 8 5 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
SessionSpecimen
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypeCartesian equation of a plane
DifficultyStandard +0.3 This is a standard Further Maths question requiring finding two direction vectors, computing their cross product for the normal vector, then substituting into the plane equation. While it involves multiple steps and vector manipulation, it follows a well-established algorithmic procedure taught explicitly in Further Pure courses with no novel insight required.
Spec4.04b Plane equations: cartesian and vector forms
Find the cartesian equation of the plane which contains the three points \((1, 0, -1)\), \((2, 2, 1)\) and \((1, 1, 2)\). [5]
Question 8:
AnswerMarks
8Need a vector perpendicular to the plane
(cid:167)1(cid:183) (cid:167)0(cid:183) (cid:167) 4 (cid:183)
(cid:168) (cid:184) (cid:168) (cid:184) (cid:168) (cid:184)
2 × 1 (cid:32) (cid:16)3
(cid:168) (cid:184) (cid:168) (cid:184) (cid:168) (cid:184)
(cid:168) (cid:184) (cid:168) (cid:184) (cid:168) (cid:184)
(cid:169)2(cid:185) (cid:169)3(cid:185) (cid:169) 1 (cid:185)
Plane has equation 4x(cid:16)3y(cid:14)z(cid:32)d
Passes through (1,0,(cid:16)1) so 4(cid:16)1(cid:32)d
AnswerMarks
Equation of plane is 4x(cid:16)3y(cid:14)z(cid:32)3.M1
B1
M1
A1
A1
AnswerMarks
[5]2.4
3.1a
m
1.1
1.1
i
AnswerMarks
3.2aOr similar language e.g. need
vector perpendicular to [2
calculated vectors] or
Vector product gives vector
perpendicular to plane or
Fnind normal vector using
vector product
eUse vector product with any
two vectors in the plane.
AnswerMarks Guidance
82 1
9i1 1
00 2
9ii3 1
10 5
10i5 2
10ii0 c
20 0
11i0 e
11 0
11ii3 4
12ip
21 0
12ii2 1
S
AnswerMarks Guidance
12iii3 2
13i2 0
13ii1 0
13iiiA1 0
13iiiB0 4
13iv2 0
14iA1 1
14iB1 1
14ii2 3
14iii2 2
14iv1 1 
Question 8:
8 | Need a vector perpendicular to the plane
(cid:167)1(cid:183) (cid:167)0(cid:183) (cid:167) 4 (cid:183)
(cid:168) (cid:184) (cid:168) (cid:184) (cid:168) (cid:184)
2 × 1 (cid:32) (cid:16)3
(cid:168) (cid:184) (cid:168) (cid:184) (cid:168) (cid:184)
(cid:168) (cid:184) (cid:168) (cid:184) (cid:168) (cid:184)
(cid:169)2(cid:185) (cid:169)3(cid:185) (cid:169) 1 (cid:185)
Plane has equation 4x(cid:16)3y(cid:14)z(cid:32)d
Passes through (1,0,(cid:16)1) so 4(cid:16)1(cid:32)d
Equation of plane is 4x(cid:16)3y(cid:14)z(cid:32)3. | M1
B1
M1
A1
A1
[5] | 2.4
3.1a
m
1.1
1.1
i
3.2a | Or similar language e.g. need
vector perpendicular to [2
calculated vectors] or
Vector product gives vector
perpendicular to plane or
Fnind normal vector using
vector product
eUse vector product with any
two vectors in the plane.
8 | 2 | 1 | 2 | 0 | 5
9i | 1 | 1 | m
0 | 0 | 2
9ii | 3 | 1 | i
1 | 0 | 5
10i | 5 | 2 | 0 | 0 | 7
10ii | 0 | c
2 | 0 | 0 | 2
11i | 0 | e
1 | 1 | 0 | 2
11ii | 3 | 4 | 0 | 0 | 7
12i | p
2 | 1 | 0 | 0 | 3
12ii | 2 | 1 | 0 | 0 | 3
S
12iii | 3 | 2 | 2 | 0 | 7
13i | 2 | 0 | 0 | 0 | 2
13ii | 1 | 0 | 2 | 0 | 3
13iiiA | 1 | 0 | 0 | 0 | 1
13iiiB | 0 | 4 | 0 | 0 | 4
13iv | 2 | 0 | 1 | 0 | 3
14iA | 1 | 1 | 0 | 0 | 2
14iB | 1 | 1 | 0 | 0 | 2
14ii | 2 | 3 | 1 | 0 | 6
14iii | 2 | 2 | 1 | 0 | 5
14iv | 1 | 1 | 1 | 0 | 3
Find the cartesian equation of the plane which contains the three points $(1, 0, -1)$, $(2, 2, 1)$ and $(1, 1, 2)$. [5]

\hfill \mbox{\textit{OCR MEI Further Pure Core  Q8 [5]}}
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