Question 10:
--- 10(a) ---
10(a) | Obtains correct normal vector to
the plane | 2.2a | B1 | 4
 
Normal to plane n= p
 
 5

Let AB then
 2 
�  ����⃗ = 𝐜𝐜
c= −5
 
  1  
c𝐧𝐧=·𝐜𝐜3=01 a3n−d 5 n 𝑝𝑝 = 41+ p2
15
As α is acute, sinα=cosθ=
75
13−5p 15
Hence =
30 41+ p2 75
2 2 2
82+2𝑝𝑝 = 5 (13−5𝑝𝑝)
2
or
623𝑝𝑝 −3250𝑝𝑝+4143= 0
1381
𝑝𝑝 = 3 𝑝𝑝 =
623
𝑝𝑝 ∈ℤ
Obtains correct expression for
 
AB or BA | 1.2 | B1
Obtains their correct scalar (or
vector) product | 1.1b | B1F
Uses scalar (or vector) product
to obtain an equation in p | 3.1a | M1
Forms an equation in p , by
squaring and removing any
rational functions
Eg 82+2p2 =52( 13−5p )2 | 1.1a | M1
Solves quadratic and selects
correct answer, discarding the
other root
Condone lack of modulus sign in
the working | 2.1 | R1
Total | 6 | ∴ 𝑝𝑝 = 3
Q | Marking instructions | AO | Marks | Typical solution
--- 10(b) ---
10(b) | Recognises need to divide
constant term of the plane
equation by n | 1.1a | M1 | n =5 2
Distance is
9 5 7 2
+ = =1.98 cm
5 2 5 2 5
Finds correct distance for their
p
, exact or decimal at least 2sf
(condone 2)
Condone missing units | 1.1b | A1F
Total | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 10(c) ---
10(c) | Obtains correct equation of the
line through A & A’ for their
p
Condone lack of | 3.1a | B1F | 7 4
At 𝐫𝐫 = � 2 �+𝜇𝜇�3�
−3 5
Π1
4(7+4𝜇𝜇)+3(2+3𝜇𝜇)
+5(−3+5𝜇𝜇) = 9
−1
𝜇𝜇 =
At image point 5
−2
𝜇𝜇 =
5
27 4
� , ,−5�
5 5
𝐫𝐫 =
Forms an equation to find the
value of for their line
µ
Π
Condone use of
2 | 3.1a | M1
Doubles their value of and
µ
uses it to find image point for
their line
Π
Condone use of
2 | 3.2a | M1
Obtains correct coordinates for
their
p
Do not accept position vector
Π
Do not condone use of
2 | 1.1b | A1F
Total | 4
Question total | 12
Q | Marking instructions | AO | Marks | Typical solution