Question 12:
--- 12(a) ---
12(a) | Calculates A using the
determinants of three 2x2
matrices | 1.1a | M1 | −2 𝑝𝑝 4 𝑝𝑝 4 −2
|𝐀𝐀|=1� �−5� �+3� �
5 −11 8 −11 8 5
|𝐀𝐀 |==3 (2 2 −+3 5 𝑝𝑝) −=5 ( −(4 4 −+8 𝑝𝑝)) +3(20+16)
A 5 0 5 p 3 5 1 0 p
Cofactors:
22−5p 44+8p 36 
 
70 −35 35
 
 5p+6 12− p −22
22−5p 70 5p+6
1  
A-1 = 44+8p −35 12− p
350+35p  
 36 35 −22 
p≠−10
Obtains correct A | 1.1b | A1
Obtains matrix of
minors/cofactors with four
elements correct | 1.1a | M1
Obtains fully correct matrix
of cofactors | 1.1b | A1
Transposes their matrix of
cofactors (with at most one
further error) & divides by
their determinant | 1.1b | A1F
Obtains fully correct answer
including p≠−10 | 2.1 | R1
Total | 6
Q | Marking Instructions | AO | Marks | Typical solution
--- 12(b)(i) ---
12(b)(i) | Uses their to form a
product to fin−d1 the
coordinates𝐀𝐀 of the point of
intersection.
Must include
5
� 24 �
or −30
Eliminates one variable to
form two simultaneous
equations in two variables | 3.1a | M1 | x 22−5p 70 5p+6 5 
  1   
y = 44+8p −35 12− p 24
  35(10+ p)  
z  36 35 −22 −30
110−25p+1680−150p−180
1  
= 220+40p−840−360+30p
35(10+ p) 
 180+840+660 
1610−175p
1  
= −980+70p
35(10+ p) 
 1680 
 46−5p 
1  
= −28+2p
10+ p  
 48 
46−5p −28+2p 48
x= ; y= ; z=
10+ p 10+ p 10+ p
Point of intersection is:
46−5𝑝𝑝 −28+2𝑝𝑝 48
� , , �
10+𝑝𝑝 10+𝑝𝑝 10+𝑝𝑝
One component correct
from their , can be
unsimplified− 1
or 𝐀𝐀
Obtains one correct value
for x, y or z , can be
unsimplified | 1.1b | A1F
Two components correct
A-1
from their , can be
unsimplified
or
Obtains a second correct
value for x, y or z , can be
unsimplified | 1.1b | A1F
All three correct, like terms
collected, but can be
unsimplified
Condone any form of the
answer | 2.1 | R1
Total | 4
Q | Marking Instructions | AO | Marks | Typical solution
--- 12(b)(ii) ---
12(b)(ii) | Obtains the scalar product of
normal vectors of two planes | 3.1a | M1 | x +5y +3z = 5
4x −2y +2z = 24
8x +5y −11z = −30
1 4
�5�∙�−2� = 4−10+6 = 0
3 2
So the planes represented by the first
two equations are perpendicular
1 4 16 8
�5�×�−2� = � 10 � = 2� 5 �
3 2 −22 −11
As the vector perpendicular to the first
two planes is a multiple of the normal
vector of the third plane, the three
planes are mutually perpendicular
Calculates that the scalar
product is 0 and interprets this
as meaning the two planes are
perpendicular. | 3.2a | A1
Obtains vector product of the
same two normal vectors
or
Obtains the scalar products of
the other two pairs of normal
vector combinations | 3.1a | M1
Clearly shows the vector
product is a multiple of third
plane’s normal vector
and interprets this as meaning
that the three planes are
mutually perpendicular
or
States that all three scalar
products are zero
and interprets this as meaning
that the three planes are
mutually perpendicular | 3.2a | R1
Total | 4
Question total | 14
Q | Marking Instructions | AO | Marks | Typical solution