OCR MEI Further Extra Pure 2023 June — Question 3 8 marks

Exam BoardOCR MEI
ModuleFurther Extra Pure (Further Extra Pure)
Year2023
SessionJune
Marks8
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Mark schemeDownload PDF ↗
TopicInvariant lines and eigenvalues and vectors
TypeFind line of invariant points
DifficultyChallenging +1.8 This is a Further Maths question requiring multiple sophisticated techniques: computing a gradient vector for the normal, finding a tangent plane equation, then solving simultaneous equations for line-plane intersection. While each step is methodical, the multi-stage nature and 3D coordinate geometry with partial derivatives places it well above average difficulty.
Spec4.04f Line-plane intersection: find point8.05g Tangent planes: equation at a given point on surface
3 A surface, \(S\), is defined by \(g ( x , y , z ) = 0\) where \(g ( x , y , z ) = 2 x ^ { 3 } - x ^ { 2 } y + 2 x y ^ { 2 } + 27 z\). The normal to \(S\) at the point \(\left( 1,1 , - \frac { 1 } { 9 } \right)\) and the tangent plane to \(S\) at the point \(( 3,3 , - 3 )\) intersect at \(P\). Determine the position vector of P .
Question 3:
AnswerMarks
3g
=6x2 −2xy+2y2
x
g
=−x2 +4xy
y
 6 x 2 − 2 x y + 2 y 2 
 g = − x 2 + 4 x y
2 7
x = 1 , y = 1 
 6  1 2 − 2  1  1 + 2  1 2    6  
 g = − 1 2 + 4  1  1 = 3
2 7 2 7
1 2    
So equation of normal is ( ) 1 1  r = +
AnswerMarks
19 9 −B1*
B1*
M1
M1
AnswerMarks
dep*A13.1a
3.1a
1.2
1.1
AnswerMarks
1.1Can be embedded in g.
Can be embedded in  g .
Attempt to form g using their
partial derivatives, either in
general form or using numerical
values for either of the given
points, with a consistent z
component. This mark can be
awarded for the 3 components
clearly and consistently shown,
even if not put into vector form.
Using their g correctly to find a
direction vector for normal line to
 2 
( )
S at 1 , 1 , − 19 . Condone " = 1 "
9
or with any equivalent direction
AnswerMarks
vector.Could be rearranged to z = f(x, y):
f −6x2 +2xy−2y2
=
x 27
f x2 −4xy
=
y 27
 − 6 x 2 + 2 x y − 2 y 2 
2 7
x 2 − 4 x y
 g =
2− 7
1
 − 29191 
x = 1 , y = 1   g = −
−
29191  −  6  
eg  − or 3 
− 2 7
x=3, y=3
632 −233+232 54
   
g=  −32 +433  =  27 
   
 27  27
2  3  2
     
So equation of plane is r. 1 = 3 . 1 (=6)
     
     
AnswerMarks
1 −3 1M1
dep*A11.1
1.1Using their g correctly to find a
normal vector to the tangent plane
to S at (3, 3, –3) (or using its
components in the tangent plane
2
 
formula). Condone "= 1 "
 
 
1
oe eg 2x + y + z = 6 or
AnswerMarks
54x + 27y + 27z = 162 − 2 
x = 3 , y = 3   g = − 1
− 1
 − 2   3   − 2 
r . − 1 = 3− . − 1 = − 6
− 1 3 − 1
  
1 2 2
       26 2
 1  + 1 . 1 = +14=6=
    9 9
−1 9  1
 9 
so position vector of P is
13
 
1 2  9  13
  1   + 2 1  =11= 1 11 
9   9  9 
AnswerMarks Guidance
  −1 9   9   1 9 7  17A1
[8]2.2a Or from subbing x = 1 + 2 etc
into 2x + y + z = 6.
Condone presentation of final
answer as coordinates but not
with 1/9 outside.
 1 3 
Do not ISW if 1 1 seen as final
1 7
AnswerMarks
answer.29191     −  
1 2  
2 6 1 4
1 1   + − . = −
9 9 
19 1 − −
6 2  =  = −
s o p o s i t i o n v e c t o r o f P i s
29191 1 391    −   
1 1 3  
1
1 2 191 1 1 − − = =
9
19 79 1 7 − −
[5]
Question 3:
3 | g
=6x2 −2xy+2y2
x
g
=−x2 +4xy
y
 6 x 2 − 2 x y + 2 y 2 
 g = − x 2 + 4 x y
2 7
x = 1 , y = 1 
 6  1 2 − 2  1  1 + 2  1 2    6  
 g = − 1 2 + 4  1  1 = 3
2 7 2 7
1 2    
So equation of normal is ( ) 1 1  r = +
19 9 − | B1*
B1*
M1
M1
dep*A1 | 3.1a
3.1a
1.2
1.1
1.1 | Can be embedded in g.
Can be embedded in  g .
Attempt to form g using their
partial derivatives, either in
general form or using numerical
values for either of the given
points, with a consistent z
component. This mark can be
awarded for the 3 components
clearly and consistently shown,
even if not put into vector form.
Using their g correctly to find a
direction vector for normal line to
 2 
( )
S at 1 , 1 , − 19 . Condone " = 1 "
9
or with any equivalent direction
vector. | Could be rearranged to z = f(x, y):
f −6x2 +2xy−2y2
=
x 27
f x2 −4xy
=
y 27
 − 6 x 2 + 2 x y − 2 y 2 
2 7
x 2 − 4 x y
 g =
2− 7
1
 − 29191 
x = 1 , y = 1   g = −
−
29191  −  6  
eg  − or 3 
− 2 7
x=3, y=3
632 −233+232 54
   
g=  −32 +433  =  27 
   
 27  27
2  3  2
     
So equation of plane is r. 1 = 3 . 1 (=6)
     
     
1 −3 1 | M1
dep*A1 | 1.1
1.1 | Using their g correctly to find a
normal vector to the tangent plane
to S at (3, 3, –3) (or using its
components in the tangent plane
2
 
formula). Condone "= 1 "
 
 
1
oe eg 2x + y + z = 6 or
54x + 27y + 27z = 162 |  − 2 
x = 3 , y = 3   g = − 1
− 1
 − 2   3   − 2 
r . − 1 = 3− . − 1 = − 6
− 1 3 − 1
  
1 2 2
       26 2
 1  + 1 . 1 = +14=6=
    9 9
−1 9  1
 9 
so position vector of P is
13
 
1 2  9  13
  1   + 2 1  =11= 1 11 
9   9  9 
  −1 9   9   1 9 7  17 | A1
[8] | 2.2a | Or from subbing x = 1 + 2 etc
into 2x + y + z = 6.
Condone presentation of final
answer as coordinates but not
with 1/9 outside.
 1 3 
Do not ISW if 1 1 seen as final
1 7
answer. | 29191     −  
1 2  
2 6 1 4
1 1   + − . = −
9 9 
19 1 − −
6 2  =  = −
s o p o s i t i o n v e c t o r o f P i s
29191 1 391    −   
1 1 3  
1
1 2 191 1 1 − − = =
9
19 79 1 7 − −
[5]
3 A surface, $S$, is defined by $g ( x , y , z ) = 0$ where $g ( x , y , z ) = 2 x ^ { 3 } - x ^ { 2 } y + 2 x y ^ { 2 } + 27 z$. The normal to $S$ at the point $\left( 1,1 , - \frac { 1 } { 9 } \right)$ and the tangent plane to $S$ at the point $( 3,3 , - 3 )$ intersect at $P$.

Determine the position vector of P .

\hfill \mbox{\textit{OCR MEI Further Extra Pure 2023 Q3 [8]}}
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