Question 9:
--- 9(a) ---
9(a) | Obtains two vectors in the
plane of the triangle | 1.1b | B1 | −3 −2 18
 
     
AB×AC = −9 × −3 = −6
     
  0    −2   −9 
 
1 1
Area= AB×AC = 182+(−6)2+(−9)2
2 2
21
=
2
Selects a method to find the
area of a triangle for
example by taking the
vector product of their two
vectors | 3.1a | M1
Uses a correct formula for
the area of a triangle | 1.2 | M1
Obtains the correct area
with no incorrect working | 1.1b | A1
Subtotal | 4
Q | Marking instructions | AO | Marks | Typical solution
--- 9(b) ---
9(b) | Forms the scalar product of
their normal and a position
vector of a point in Π | 1.1a | M1 |  0   6 
   
d = −2 • −2 =4
   
  0     −3 
 6 
 
r• −2 =4
 
 −3
 
Obtains a correct equation
of Π | 1.1b | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 9(c) ---
9(c) | Selects a method to find the
parallel plane that P lies in
by using the scalar product
of their normal vector and P
to obtain their “–8”
or
Uses a correct formula to
find the required distance
or
Substitutes equation of
perpendicular line from P to
Π into equation of Π | 3.1a | M1 | 1  6 
   
D = 4 • −2 =−8
2    
 2  −3
  
n = 62 +(−2 )2 +(−3 )2 =7
4 −8
d = & d =
1 7 2 7
4 −8 12
Distance of P from Π = − =
7 7 7
Uses the magnitude of their
normal vector to divide their
“4” and “–8” in order to find
the distance of P from Π
or
Substitutes correctly into a
formula for the required
distance | 2.2a | M1
Obtains the correct distance
of P from Π | 1.1b | A1
Subtotal | 3
Question total | 9
Q | Marking instructions | AO | Marks | Typical solution