OCR MEI Further Extra Pure Specimen — Question 4 16 marks

Exam BoardOCR MEI
ModuleFurther Extra Pure (Further Extra Pure)
SessionSpecimen
Marks16
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypeLine of intersection of planes
DifficultyChallenging +1.8 This Further Pure question requires multivariable calculus (partial derivatives, tangent planes) which is inherently advanced. Parts (i)-(ii) involve routine but careful differentiation using product/chain rules. Part (iii) requires conceptual insight to interpret the partial derivative relation geometrically. Parts (iv)-(v) are standard applications but the multi-step nature, verification requirements, and need to connect algebraic relations to geometric meaning place this significantly above average A-level difficulty, though the techniques themselves are systematic once understood.
Spec4.04b Plane equations: cartesian and vector forms8.05a 3D surfaces: z = f(x,y) and implicit form, partial derivatives8.05d Partial differentiation: first and second order, mixed derivatives8.05g Tangent planes: equation at a given point on surface
A surface \(S\) has equation \(g(x, y, z) = 0\), where \(g(x, y, z) = (y - 2x)(y + z)^2 - 18\).
  1. Show that \(\frac{\partial g}{\partial y} = (y + z)(-4x + 3y + z)\). [2]
  2. Show that \(\frac{\partial g}{\partial x} + 2\frac{\partial g}{\partial y} - 2\frac{\partial g}{\partial z} = 0\). [4]
  3. Hence identify a vector which lies in the tangent plane of every point on \(S\), explaining your reasoning. [3]
  4. Find the cartesian equation of the tangent plane to the surface \(S\) at the point P\((1, 4, -7)\). [3]
The tangent plane to the surface \(S\) at the point Q\((0, 2, 1)\) has equation \(6x - 7y - 4z = -18\).
  1. Find a vector equation for the line of intersection of the tangent planes at P and Q. [4]
Question 4:
AnswerMarks Guidance
4(i) g
(y2x)2(yz)(yz)2
y
(yz)(2y4xyz)
AnswerMarks
(yz)(4x3yz) AGM1
A1
AnswerMarks
[2]1.1a
1.1Attempt at
differentiation,
treating x and z as
constants
AnswerMarks Guidance
4(ii) g
2(yz)2
x
g
2(y2x)(yz)
z
g g g
2 2
x y z
2(yz)22(yz)(4x3yz)4(y2x)(yz)
2(yz)(yz4x3yz2y4x)
AnswerMarks
0 AGB1
B1
M1
A1
AnswerMarks
[4]1.1a
1.1
1.1
AnswerMarks Guidance
2.1Some simplification
4(iii) g
 
x
 
g
is normal vector for tangent plane
 
y
 
g
 
 z 
g
 
x
   1 
g  
. 2 0for every point on S
   
y
   2  
g
 
 z 
 1 
 
2 lies in tangent plane of every point on S
 
 
2
AnswerMarks
E1
E1
B1
AnswerMarks
[3]3.1a
2.4
2.2a
AnswerMarks Guidance
4(iv) g
 
x
  18
g  
At P,  3
   
y
   12  
g
 
 z 
Use of scalar product with correct vector.
AnswerMarks
6x y4z18M1
M1
A1
AnswerMarks
[3]1.1
1.1
AnswerMarks
1.1At least one correct,
or clear attempt to
use formulae from
(i), (ii) or redone here
oe
AnswerMarks Guidance
4(v) 8y8z0
(-3, 0, 0)
 1 
 
2 in both planes, so is direction vector for line of intersection
 
 
2
3  1 
   
r 0 t 2
   
   
AnswerMarks
 0   2M1
A1
M1
A1
AnswerMarks
[4]3.1a
1.1
3.1a
AnswerMarks
2.2aDealing with planes
as simultaneous
equations to find
equation of a
common point
Finding a common
point
Question 4:
4 | (i) | g
(y2x)2(yz)(yz)2
y
(yz)(2y4xyz)
(yz)(4x3yz) AG | M1
A1
[2] | 1.1a
1.1 | Attempt at
differentiation,
treating x and z as
constants
4 | (ii) | g
2(yz)2
x
g
2(y2x)(yz)
z
g g g
2 2
x y z
2(yz)22(yz)(4x3yz)4(y2x)(yz)
2(yz)(yz4x3yz2y4x)
0 AG | B1
B1
M1
A1
[4] | 1.1a
1.1
1.1
2.1 | Some simplification
4 | (iii) | g
 
x
 
g
is normal vector for tangent plane
 
y
 
g
 
 z 
g
 
x
   1 
g  
. 2 0for every point on S
   
y
   2  
g
 
 z 
 1 
 
2 lies in tangent plane of every point on S
 
 
2
 | E1
E1
B1
[3] | 3.1a
2.4
2.2a
4 | (iv) | g
 
x
  18
g  
At P,  3
   
y
   12  
g
 
 z 
Use of scalar product with correct vector.
6x y4z18 | M1
M1
A1
[3] | 1.1
1.1
1.1 | At least one correct,
or clear attempt to
use formulae from
(i), (ii) or redone here
oe
4 | (v) | 8y8z0
(-3, 0, 0)
 1 
 
2 in both planes, so is direction vector for line of intersection
 
 
2

3  1 
   
r 0 t 2
   
   
 0   2 | M1
A1
M1
A1
[4] | 3.1a
1.1
3.1a
2.2a | Dealing with planes
as simultaneous
equations to find
equation of a
common point
Finding a common
point
A surface $S$ has equation $g(x, y, z) = 0$, where $g(x, y, z) = (y - 2x)(y + z)^2 - 18$.

\begin{enumerate}[label=(\roman*)]
\item Show that $\frac{\partial g}{\partial y} = (y + z)(-4x + 3y + z)$. [2]

\item Show that $\frac{\partial g}{\partial x} + 2\frac{\partial g}{\partial y} - 2\frac{\partial g}{\partial z} = 0$. [4]

\item Hence identify a vector which lies in the tangent plane of every point on $S$, explaining your reasoning. [3]

\item Find the cartesian equation of the tangent plane to the surface $S$ at the point P$(1, 4, -7)$. [3]
\end{enumerate}

The tangent plane to the surface $S$ at the point Q$(0, 2, 1)$ has equation $6x - 7y - 4z = -18$.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{4}
\item Find a vector equation for the line of intersection of the tangent planes at P and Q. [4]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Extra Pure  Q4 [16]}}
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