IB Math Standard Level – Vector Practice 0506
Alei - Desert Academy
SL Vector Practice 0506
1.
The points P(−2, 4), Q (3, 1) and R (1, 6) are shown in the diagram below.
(a) Find the vector PQ .
(b) Find a vector equation for the line through R parallel to the line (PQ).
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(Total 6 marks)
2.
The position vector of point A is 2i + 3 j + k and the position vector of point B is 4i − 5 j + 21k.
(a)
(i)
Show that AB = 2i −8 j + 20k.
(ii)
Find the unit vector u in the direction of AB .
(iii)
Show that u is perpendicular to OA .
(6)
Let S be the midpoint of [AB]. The line L1 passes through S and is parallel to OA .
(b) (i)
Find the position vector of S.
(ii) Write down the equation of L1.
(4)
The line L2 has equation r = (5i +10 j +10k) + s (−2i + 5 j − 3k).
(c) Explain why L1 and L2 are not parallel.
(2)
(d)
The lines L1 and L2 intersect at the point P. Find the position vector of P.
(7)
(Total 19 marks)
3.
The line L passes through the points A (3, 2, 1) and B (1, 5, 3).
(a) Find the vector AB .
(b) Write down a vector equation of the line L in the form r = a + tb.
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(Total 6 marks)
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IB Math Standard Level – Vector Practice 0506
4.
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The line L passes through A (0, 3) and B (1, 0). The origin is at O. The point R (x, 3 − 3x) is on L,
and (OR) is perpendicular to L.
(a) Write down the vectors AB and OR .
(b) Use the scalar product to find the coordinates of R.
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(Total 6 marks)
5.
1
0
In this question the vector   represents a displacement of 1 km east, and the vector  
0
1
represents a displacement of 1 km north.
The diagram below shows the positions of towns A, B and C in relation to an airport O, which is at
the point (0, 0). An aircraft flies over the three towns at a constant speed of 250 km h–1.
y
B
O
x
A
C
Town A is 600 km west and 200 km south of the airport.
Town B is 200 km east and 400 km north of the airport.
Town C is 1200 km east and 350 km south of the airport.
(a)
(i)
Find AB .
(ii)
 0. 8 
Show that the vector of length one unit in the direction of AB is   .
 0. 6 
(4)
–1
An aircraft flies over town A at 12:00, heading towards town B at 250 km h .
 p
Let   be the velocity vector of the aircraft. Let t be the number of hours in flight after 12:00. The
q
position of the aircraft can be given by the vector equation
 x    600  p 
   
  t   .
 y    200  q 
(b)
(i)
(ii)
(iii)
 200
 .
Show that the velocity vector is 
 150 
Find the position of the aircraft at 13:00.
At what time is the aircraft flying over town B?
(6)
Over town B the aircraft changes direction so it now flies towards town C. It takes five hours to
travel the 1250 km between B and C. Over town A the pilot noted that she had 17 000 litres of fuel
left. The aircraft uses 1800 litres of fuel per hour when travelling at 250 km h–1. When the fuel gets
below 1000 litres a warning light comes on.
(c) How far from town C will the aircraft be when the warning light comes on?
(7)
(Total 17 marks)
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IB Math Standard Level – Vector Practice 0506
6.
Alei - Desert Academy
A boat B moves with constant velocity along a straight line. Its velocity vector is given by
 4
v =   .
 3
At time t = 0 it is at the point (−2, 1).
(a) Find the magnitude of v.
(b) Find the coordinates of B when t = 2.
(c) Write down a vector equation representing the position of B, giving your answer in the form r
= a + tb.
Working:
Answers:
(a) .....................................................
(b) .....................................................
(c) .....................................................
(Total 6 marks)
7.
Consider the point D with coordinates (4, 5), and the point E, with coordinates (12, 11).
(a)
Find DE .
(2)
(b)
Find DE .
(c)
The point D is the centre of a circle and E is on the circumference as shown in the following
diagram.
(2)
The point G is also on the circumference. DE is perpendicular to DG . Find the possible
coordinates of G.
(8)
(Total 12 marks)
8.
Car 1 moves in a straight line, starting at point A (0, 12). Its
 x  0 
 5 
position p seconds after it starts is given by   =   + p   .
 y  12 
  3
(a) Find the position vector of the car after 2 seconds.
(2)
Car 2 moves in a straight line starting at point B (14, 0). Its
 x  14 
1
position q seconds after it starts is given by   =   + q   .
 y  0 
 3
Cars 1 and 2 collide at point P.
(b) (i)
Find the value of p and the value of q when the collision occurs.
(ii) Find the coordinates of P.
(6)
(Total 8 marks)
9.
 3
  2
Find the cosine of the angle between the two vectors   and   .
 4
 1
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IB Math Standard Level – Vector Practice 0506
Alei - Desert Academy
Working:
Answer:
....……………………………………..........
(Total 6 marks)
10.
The following diagram shows a solid figure ABCDEFGH. Each of the six faces is a parallelogram.
The coordinates of A and B are A (7, –3, –5), B(17, 2, 5).
(a) Find
(i)
AB;
(ii)
AB .
(4)
The following information is given.
  6
  2
 
 
AD =  6  , AD = 9, AE =   4  , AE = 6
 3 
 4 
 
 
(b)
(i)
Calculate AD • AE .
(ii)
Calculate AB • AD .
(iii)
(iv)
Calculate AB • AE .
Hence, write down the size of the angle between any two intersecting edges.
(5)
(c)
Calculate the volume of the solid ABCDEFGH.
(d)
The coordinates of G are (9, 4, 12). Find the coordinates of H.
(e)
The lines (AG) and (HB) intersect at the point P.
2
 
Given that AG =  7  , find the acute angle at P.
17 
 
(2)
(3)
(5)
(Total 19 marks)
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IB Math Standard Level – Vector Practice 0506 - MarkScheme
Alei - Desert Academy 2013-14
SL Vector Practice 0506 MarkScheme
1.
(a)
(b)

 5 
PQ =  
  3
Using r = a + tb
 x  1  5 
      t  
 y   6    3
A1A1N2
A2A1A1N4
[6]
(a)
 1  3
    
AB  OB  OA    5    2  
  3  1  
    
(b)
  2
 
AB   3 
 2 
 
Using r = a + tb
 x   3   2  x  1   2
           
 y    2   t  3  or  y    5   t  3 
 z   1   2   z   3  2 
           

2.


(M1)

A2
N3
A1A1A1N3
[6]
3.
(a)
(i)
Evidence of subtracting all three components in the correct order



eg AB  OB  OA  4i  5 j  21k   2i  3 j  k 
= 2i 8j + 20k

(ii)
 AB  =
u=
(iii)
M1

2 2   8  202 
2
1
468
468  6 13  2 117  21.6

2i  8 j  20k 


2
8
20
 
i
j
k , 0.0925i  0.370 j  0.925k , etc.
468
468
468


If the scalar product is zero, the vectors are perpendicular.
Note: Award R1 for stating the relationship between
the scalar product and perpendicularity, seen
anywhere in the solution.





Finding an appropriate scalar product  u  OA or AB  OA 



 2 
 8 
 20 
  2  
  3  
  1
eg u  OA  
 468 
 468 
 468 
AG
N0
(A1)
A1
N2
R1
M1
 4  24  20 
 

468 



AB  OA  2  2   8  3  20  1

(b)
(i)


u  OA  0 or AB  OA  0
EITHER
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A1
N0
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IB Math Standard Level – Vector Practice 0506 - MarkScheme
 2  4 3  5 1  21 
S
,
,

2
2 
 2
Alei - Desert Academy 2013-14
(M1)(A1)

Therefore, OS = 3i  j + 11k (accept (3, 1, 11))
OR


1 
OS  OA  AB
2
1
= (2i + 3j + k) + (2i + 8j + 20k)
2
A1
N3
(M1)
(A1)

(c)
(d)
OS = 3i  j + 11k
(ii) L1 : r = (3i  j + 11k) + t (2i + 3j + 1k)
Using direction vectors (eg 2i + 3j + 1k and 2i + 5j  3k)
Valid explanation of why L1 is not parallel to L2
eg. Direction vectors are not scalar multiples of each other.
Angle between the direction vectors is not zero or 180.
Finding the angle
d1 • d2  d1d2.
Note: Award R0 for “direction vectors are not equal”.
Setting up any two of the three equations
For each correct equation
eg 3 + 2t = 5  2s, 1 + 3t = 10 + 5s, 11 + t = 10  3s
Attempt to solve these equations
Finding one correct parameter (s = 1, t = 2)
P has position vector 7i + 5j + 13k
Notes: Award (M1)A2 if the same parameter is used
for both lines in the initial correct equations.
Award no further marks.
A1 N3
A1 N1
(M1)
R1 N2
(M1)
A1A1
(M1)
(A1)
A2 N4
[19]
(a)
 1    x 

AB    , OR  
  3
 3  3x 
(b)
AB  OR  x  3 3  3x 

4.




A1A1N2
A1
AB  OR  0 10x  9  0
M1
9 3
R is  , 
 10 10 
A1A1N2
[6]
 200   600 
AB  


 400   200 
 800 


 600 

5.
(a)
(i)
(A1)
(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.
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IB Math Standard Level – Vector Practice 0506 - MarkScheme
(b)
(i)
(ii)
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 0.8 
v  250  
(M1)
 0.6 
 200 

(AG)(N0)

 150 
Note: A correct alternative method is using the given vector equation with t = 4.
at 13:00, t = 1
 x   600   200 
(M1)
 y    200   1 150 
  
 

  400 


 50 
(A1)(N1)

(iii)
AB  1000
1000
 4 (hours)
(M1)(A1)
250
over town B at 16:00 (4 pm, 4:00 pm)
(Do not accept 16 or 4:00 or 4)
(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.
Distance from A to B to
Fuel used from A to B
Time for A to B to C
C
(A1)
= 9 hours
= 1800  4  7200 litres
= 2250 km
Light goes on after
Light goes on after
Fuel remaining
16000 litres
16000 litres
= 9800 litres
(A1)
Time for 16 000
Distance on 16000 litres
Hours before light
litres
(A1)(A
8800
1)
16000
1800

(A1)
16000
1800
8

 250

4

4.889


1800
8
9
 8 ( 8.889)
9
2
Time remaining is
Time remaining is
 2222 ( 2222.22) km
9
1
1
   0.111 hour
= ( 0.111) hour
9
9
1
1
Distance to C
Distance   250
Distance   250
(A2)
(N4) 7
= 2250 – 2222.22
9
9
= 27.8 km
= 27.8 km
= 27.8 km
Time 
[17]
6.
(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]
12  4   8 
 =  
DE = 
 11 5   6 

7.
(a)
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(M1)(A1) (N2)
Page 3 of 6
IB Math Standard Level – Vector Practice 0506 - MarkScheme


(b)
(c)

 64 36
 DE = 8 2  6 2
= 10
Vector geometry approach
Using DG = 10
(x  4)2 + (y  5)2 = 100
Using (DG) perpendicular to (DE)

  6 
 6
Leading to DG =   , DG =  
 8
  8


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(M1)
(A1)(N2)
(M1)
(A1)
(M1)
(A1)(A1)

Using DG = DO  OG
G (2, 13), G (10, 3)
Algebraic approach
6
gradient of DE =
8
8
gradient of DG = 
6
(O is the origin)
(accept position vectors)
equation of line DG is y  5 = 
(M1)
(A1)(A1)
(A1)
(A1)
4
( x  4)
3
Using DG = 10
(x  4)2 + (y  5)2 = 100
Solving simultaneous equation
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)
(M1)
(A1)
(M1)
(A1)(A1)
[12]
8.
(a)
(b)
0  5 
p = 2     2  
12   3 
10 
=  
(accept any other vector notation, including (10, 6) )
6
METHOD 1
(i)
equating components
0 + 5p = 14 + q , 12  3p = 0 + 3q
p = 3, q =1
(ii) The coordinates of P are (15, 3) (accept x = 15, y = 3 )
METHOD 2
(i)
Setting up Cartesian equations
x = 5p
x = 14 + q
y =12  3p
y = 3q
giving 3x + 5y = 60 3x  y = 42
Solving simultaneously gives x = 15, y = 3
Substituting to find p and q
15  0 
 5  15 14  1 
      p   ,       q   ,
 3  12
  3  3   0   3
p=3
q=1
(ii) From above, P is (15, 3) (accept x = 15, y = 3 seen above)
(A1)
(A1)(N2)
(M1)
(A1)
(A1)(A1)(N1)(N1)
(A1)(A1)(N1)(N1)
(M1)
(A1)
(A1)(A1)(N1)(N1)
(A1)(A1)(N1)(N1)
[8]
9.
METHOD 1
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IB Math Standard Level – Vector Practice 0506 - MarkScheme
Using a  b = ab cos  (may be implied)
 3   – 2  3   – 2
          cos
 4  1   4  1 
 3   – 2
Correct value of scalar product       3  – 2  4 1  – 2
 4  1 
 3
 2 
Correct magnitudes    25   5 ,    5
 4
1
cos 
2
Alei - Desert Academy 2013-14
(M1)
(A1)
(A1)
(A1)(A1)
(A1)(C6)
125
METHOD 2
 3
   25
 4
(A1)
 2 
   5
1
(A1)
 5
   34
 3
Using cosine rule
34  25  5  25 5 cos
2
cos  
125
(A1)
(M1)
(A1)
(A1)(C6)
[6]

10.
(a)
(i)


AB  OB  OA
17   7 
   
=  2     3
 5   5
   
(A1)
10 
 
=  5
10 
 
A1
N2

AB  102  5 2  102
= 15
Evidence of correct calculation of scalar product (may be in (i), (ii)
or (iii))
(ii)
(b)
(i)
(ii)






AB AE  0 ((6)(2) + 6(4) + 3(4))
AB AD  0 ((10)(6) + 5(6) + 10(3))
A1
A1
N1
A1
N1
(iii)
AB AE  0 ((10)(2) + 5(4) + 10(4))
A1
N1
(iv)
 
90  or 
 2
A1
N1

(c)
(M1)
A1 N2


Volume =  AB    AD    AE 
= 15  9  6
= 810 (cubic units)
C:\Users\Bob\Documents\Dropbox\Desert\SL\5Vectors\SLVectorsPractice0506.docx on 8/29/14 at 8:45 AM
(A1)
A1
N2
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IB Math Standard Level – Vector Practice 0506 - MarkScheme
(d)
Setting up a valid equation involving H. There are many possibilities.
 x  9    10








 

eg OH  OG  GH , OH  OA  AE  EH ,  y  4     5 
 z 12   10

 

Using equal vectors



Alei - Desert Academy 2013-14
(M1)
(M1)

eg GH   AB , EH  AD
 9  10   1
 7    2    6    1

              
OH   4    5     1 , OH    3     4    6     1
12 10  2 
  5   4   3  2 
     
       
(e)
coordinates of H are ( 1, 1, 2)
18

 
HB   3 
 3
 
A1
N3
A1
Attempting to use formula cos P̂ 




AG HB
(M1)
AG HB
=
2 18  7  3 17 3


108




342
342


2 2  7 2  172 182  32  32
= 0.31578...
(= 1.25 radians)
P̂  71.6
A1
(A1)
A1 N3
[19]
C:\Users\Bob\Documents\Dropbox\Desert\SL\5Vectors\SLVectorsPractice0506.docx on 8/29/14 at 8:45 AM
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