Calculator Question
1.
(a)
 4
Finding correct vectors, AB   
 3
  3
AC   
 1 
A1A1
Substituting correctly in the scalar product AB AC = 4(–3) + 3(1)
= –9
AB  5
(b)
AC  10
A1
AG
N0
(A1)(A1)
Evidence of using scalar product formula
9
e.g. cos BÂC 
= –0.569 (3 s.f.)
5 10
M1
BÂC = 2.47 (radians), 125°
A1
N3
[7]
Non Calculator Questions
1.
METHOD 1
At point of intersection:
5 + 3λ = –2 + 4t
l – 2λ = 2 + t
(M1)
(M1)
Attempting to solve the linear system
λ = –l (or t = 1)
(M1)
(A1)
 2
OP   
 3
(A1)(A1) (C6)
METHOD 2
(changing to Cartesian coordinates)
2x + 3y = 13, x – 4y = –10
Attempt to solve the system
 2
OP   
 3
(A1)(A1) (C6)
Note:
2.
Award (C5) for the point P(2, 3).
Direction vectors are a = i – 3j and b = i – j.
a  b = (1 + 3)
a = 10 , b = 2
cos θ =
cos θ =
(M1)(A1)(A1)
(M1)
ab 
4 
 

ab
10 2 
4
20
(A2)
(A1)
(A1)
(M1)
(A1) (C6)
[6]
3.
(a)

 1    x 

AB    , OR  
  3
 3  3x 

(b)
N2

AB OR  x  3 3  3x 

A1A1
A1

AB OR  0 10 x  9  0
9 3
R is  , 
 10 10 
M1
A1A1
N2
[6]
4.
(a)

 5 
PQ =  
  3
(b)
Using r = a + tb
 x  1  5 
      t  
 y   6    3
A1A1
N2
A2A1A1
N4
[6]
5.
(a)
(b)
(c)
16  9 =
25 = 5
  2  4  6
   2    
 1  3   7 
  2  4
r =    t  
 1  3 
Note:
(M1)(A1) (C2)
(so B is (6, 7) )
(not unique)
(M1)(A1) (C2)
(A2) (C2)
Award (A1) if “ r = ” is omitted, ie not
an equation.
[6]
6.
(a)
(i)
evidence of combining vectors
e.g. AB  OB  OA
  2

AB  
  3
(b)
(M1)
A1
N2
(ii)
 3 

AC  
  2
A1
N1
(i)
AB AC = (–2)(3) + (–3)(–2) = 0
A1
N1
(ii)
scalar product 0 =  perpendicular, θ = 90°
sin θ = 1
(R1)
A1
N2
[6]
7.
(a)
evidence of appropriate approach
(M1)
 3
e.g. AC  AB,  4
4
 1 

BC  1 
 3
(b)
A1
N2
2
N0
2
N0
2
N0
3
METHOD 1
AD  1 
 3
correct approach
(A1)
A1
1 3
e.g. AD  AB, 3  1


  2

BD  
 2
AG
METHOD 2
Recognizing CD  BA
correct approach
(A1)
A1
1 3
e.g. BC  CD, 3  1


BD    2 
 2
(c)
AG
METHOD 1
evidence of scalar product
(M1)
2
4
e.g. BD  AC, 2    4 
   
correct substitution
A1
e.g. (–2)(4) + (2)(4), –8 + 8
BD  AC  0
A1
therefore vectors BD and AC are perpendicular
AG
METHOD 2
attempt to find angle between two vectors
e.g.
ab
ab
correct substitution
e.g.
(M1)
A1
 24  24 , cos  0
8 32
  90 
A1
therefore vectors BD and AC are perpendicular
AG
N0
[7]
8.
(a)
0  5 
p = 2     2  
12    3 
10 
=  
6
(b)
(A1)
(accept any other vector notation, including (10, 6) )
(A1) (N2)
METHOD 1
(i)
equating components
(M1)
0 + 5p = 14 + q , 12  3p = 0 + 3q
(A1)
p = 3, q =1
(ii)
(A1)(A1)(N1)(N1)
The coordinates of P are (15, 3)
(accept x = 15, y = 3 )
(A1)(A1)(N1)(N1)
METHOD 2
(i)
Setting up Cartesian equations
x = 5p
x = 14 + q
y =12  3p
y = 3q
giving
3x  y = 42
3x + 5y = 60
(M1)
(A1)
Solving simultaneously gives x = 15, y = 3
Substituting to find p and q
15   0 
     
 3  12 
p=3
(ii)
 5 
p   ,
  3
15  14 
1
      q   ,
3 0
 3
q=1
From above, P is (15, 3)
(A1)(A1)(N1)(N1)
(accept x = 15, y = 3 seen above) (A1)(A1)(N1)(N1)
[8]
9.
(a)

12  4 
 8
 =  
DE = 
 11  5 
 6
(M1)(A1) (N2)

(b)
 DE =

82  6 2
64  36

= 10
(c)
(M1)
(A1) (N2)
Vector geometry approach
Using DG = 10
(M1)
(x  4)2 + (y  5)2 = 100
(A1)
Using (DG) perpendicular to (DE)
(M1)

 6
  6 
Leading to DG =   , DG =  
  8
 8


(A1)(A1)

Using DG = DO  OG
G (2, 13), G (10, 3)
(O is the origin)
(accept position vectors)
(M1)
(A1)(A1)
Algebraic approach
gradient of DE =
6
8
gradient of DG = 
(A1)
8
6
equation of line DG is y  5 = 
(A1)
4
( x  4)
3
(A1)
Using DG = 10
(M1)
(x  4)2 + (y  5)2 = 100
(A1)
Solving simultaneous equation
(M1)
G ( 2, 13), G (10, 3) (accept position vectors)
Note: Award full marks for an appropriately
labelled diagram (eg showing that DG =10 ,
displacements of 6 and 8), or an accurate
diagram leading to the correct answers.
(A1)(A1)
[12]

10.
(a)
(i)


BC  OC OB
 6i  2 j

(ii)

(A1)(A1) (N2)

OD  OA BC
 2i  0 j ( 2i )

(b)

(A1)(A1) (N2)

BD  OD OB
 3i  3 j


4
(A1)

AC  OC OA
 9i  7 j
(A1)


Let  be the angle between BD and AC


 (3i  3 j )  (9i  7 j ) 
cos θ  
 (3i  3 j )  9i  7 j 


(M1)
numerator = + 27 – 21 (= 6)
(A1)

denominator  18 130  2340
therefore, cos 

(A1)
6
2340
  82.9 (1.45 rad)
(c)
r  i  3 j  t (2i  7 j )
(d)
EITHER
  (1  2t )i  (3  7t ) j 
(A1) (N3)
6
(A1) (N1)
1
4i  2 j  s(i  4 j )  i  3 j  t (2i  7 j ) (may be implied)
(M1)
4  s  1  2t 

2  4s  3  7t 
(A1)
t  7 and/or s  11
(A1)
Position vector of P is 15i  46 j
(A1) (N2)
OR
7 x  2 y  13 or equivalent
(A1)
4 x  y  14 or equivalent
(A1)
x  15 , y  46
(A1)
Position vector of P is 15i  46 j
(A1) (N2)
4
[15]
11.
(a)
(i)
  3  2 
AB  OB  OA      
  1   2
(M1)
  5
=  
 1 
(A1)
(ii)
(N2)
AB  25  1
=
2
(M1)
26 (= 5.10 to 3 sf)
(A1)
(N2)
2
(A1)(A1)
2
Note: An answer of 5.1 is subject to AP.
(b)
AD  OD  OA
d  2 
=     
 23    2 
 d  2

= 
 25 
(c)
(i)
EITHER
BÂD  90  AB  AD = 0 or mention of scalar (dot) product.
(M1)
  5  d  2
 = 0
    
 1   25 
–5d + 10 + 25 = 0
d=7
(A1)
(AG)
OR
1 
5 

25 
Gradient of AD 
d  2 
Gradient of AB  
 25   1 

     = –1
 d  2  5
d=7
(ii)
(d)
7
OD    (correct answer only)
 23 
AD  BC
(A1)
(A1)
(AG)
(A1)
(M1)
3
5
BC   
 25 
(A1)
OC  OB  BC
(M1)
  3  5 
OC      
  1   25 
2
=  
 24 
(A1) (N3)
4
Note: Many other methods, including scale drawing, are acceptable.
(e)
AD  or BC   5 2  25 2  650


Area =
(A1)
26  650 =( 5.099 × 25.5)
= 130
(A1)
2
[15]
12.
(a)
(i)
correct approach
5 1
e.g. OC  OA,     
 2  0
 4
AC   
 2
(ii)
appropriate approach
 4  1
e.g. D – B,      , move 3 to the right and 6 down
  1  5 
 3 
BD   
  6
(iii)
(b)
(i)
A1
AG
N0
(M1)
A1
N2
finding the scalar product
e.g. 4(3) + 2(–6), 12 – 12
A1
valid reasoning
e.g. 4(3) + 2(–6) = 0, scalar product is zero
R1
AC is perpendicular to BD
AG
N0
A1A1
N2
A2
N2
correct “position” vector for u; “direction” vector for v
5
1
 4
  2
e.g. u =  , u   ; v   , v   
 2
 0
 2
 1
 5   4
accept in equation e.g.    t  
 2   2
(ii)
any correct equation in the form r = a + tb, where b = BD
 1   3   x   4    1
e.g. r =    t  ,       t  
 5    6   y    1  2 
(c)
METHOD 1
substitute (3, k) into equation for (AC) or (BD)
e.g. 3 = 1 + 4s, 3 = 1 + 3t
(M1)
value of t or s
1 1
2 1
e.g. s = , , t  , ,
2 2
3 3
A1
substituting
A1
e.g. k = 0 +
1
( 2) ,
2
k=1
AG
N0
METHOD 2
setting up two equations
e.g. 1 + 4s = 4 + 3t, 2s = –1 – 6t; setting vector equations of lines equal
(d)
(M1)
value of t or s
1 1
2 1
e.g. s = , , t  ,
2 2
3 3
A1
substituting
 4  1 3 
e.g. r =      ,
  1 3   6 
k=1
A1
AG
 1 

PD  
  2
(A1)
PD  2 2  12 ( 5 )
(A1)
AC  4 2  2 2 ( 20 )
(A1)
area =
1
 AC  PD
2
 1

   20  5 
 2

N0
M1
=5
A1
N4
[17]
13.
(a)
(i)

 200   600 
AB  


 400   200 
(A1)
 800 


 600 
(A1) (N2)

(ii)
AB  8002  6002  1000 (must be seen)
unit vector 
1  800 


1000  600 
(M1)
(A1)
 0.8 
(AG) (N0)
 
 0.6 
Note: A reverse method is not acceptable in “show that” questions.
(b)
(i)
 0.8 
v  250  
 0.6 
4
(M1)
 200 
(AG) (N0)


 150 
Note: A correct alternative method is using the given vector equation with
t = 4.
(ii)
at 13:00, t = 1
 x   600   200 
 y    200   1 150 
  
 

  400 


 50 
(M1)
(A1) (N1)

(iii)
AB  1 000
Time 
1000
 4 (hours)
250
over town B at 16:00 (4 pm, 4:00 pm)
(Do not accept 16 or 4:00 or 4)
(M1)(A1)
(A1) (N3)
6
(c)
Note: There are a variety of approaches. The table shows some of them, with the mark
allocation. Use discretion, following this allocation as closely as possible.
Time for A to B to C
= 9 hours
Distance from A to B to C
= 2250 km
Fuel used from A to B
= 1800  4  7200 litres
Light goes on after
16000 litres
Light goes on after
16000 litres
Fuel remaining
= 9800 litres
Time for 16 000
litres
16000

1800
8
 8 ( 8.889 )
9
Time remaining is
Distance on 16000 litres
Hours before light
8800
1800
16000

 250
1800
8
 4   4.889 
9
1
= ( 0.111) hour
9
1
Distance   250
9
= 27.8 km
2
 2222 ( 2222 .22) km
9
Distance to C
= 2250 – 2222.22
= 27.8 km
Time remaining is
1
   0.111 hour
9
1
Distance   250
9
= 27.8 km
(A1)
(A1)
(A1)(A
1)
(A1)
(A2)
(N4) 7
[17]