# Chapter5Notes-VectorsinMechanics ```VECTORS
1. Vectors in Mechanics
We have seen already that speed and velocity are very similar concepts but that velocity is a
vector whilst speed is a scalar. Velocity, the vector quantity, contains more information as it also
indicates the direction of motion as well as the actual speed. We say that it has both magnitude
(size) and a direction. Speed contains only information about the magnitude. Our concept of
velocity is, however, still quite limited because we have only looked at how to represent the
direction by considering a change of sign. This has limited our work on velocity to acting along a
line in 1 dimension.
Consider, now, the following situation involving velocity in 2 dimensions:
A bird is flying due south at a speed of 4ms-1, but it is caught in a wind blowing east at 3ms-1.
What is the bird’s actual velocity?
Sketching a picture to show what happens to the birds flight over a 1 second interval. The bird
heads 3m east and 4m south.
N
Using Pythagoras’ Theorem the distance travelled in 1s is
d  32  4 2
d  5m
The speed of the bird over the ground is therefore 5ms-1
3m
θ
4m
The angle θ is given by
tan  
4
3
  tan 1 43
  53.1o
Therefore the bearing is 90&deg; + 53.1&deg; = 143.1&deg;.
So the birds velocity is 5ms-1 on a Bearing of 143.1&deg;.
Now consider the following situations involving displacement and forces:
 If you walk 3m east and then 4m south, how far and in what direction will you be from
 A sledge is being pulled with forces of 3N to the east and 4N to the south. What is the
combined effect of these forces?
It is easy to deduce that the answer to both of these questions will contain a magnitude of 5 at an
angle of 143.1&deg; from North.
2. Vector Notation
We can represent a vector in 3 different ways.
Consider the following situation to describe the velocity of our bird from our first example. It
had a seed of 5ms-1 on a Bearing of 143.1&deg;.

We can represent its velocity as an arrow of length 5 units drawn at the correct angle
N
143.1

We could represent its velocity by splitting it into component parts
v = 3i – 4j
(where i represents the quantity in the horizontal direction and j represents the quantity in the
vertical direction)

We could represent its velocity as a column vector
 3 
v =  
  4
We may refer to the vector by a single letter ‘v’ or by using the labels at the end points
Eg. The vector joining A to B is denoted AB , whereas the distance from A to B is denoted AB (so
the arrow above the letters indicates a vector quantity rather than a scalar. It also indicates the
direction, from A to B).
Vectors that describe the same ‘movement’ or ‘resultant’ are equivalent. Equivalent vectors will
have the same magnitude AND the same direction. Vectors like this have no specific location
and are sometimes called ‘free vectors’. Vectors that are used to describe a position based on a
coordinate grid are called ‘position vectors’. So in the diagram below, AB  PQ .
B
A
Q
P
Note that we can consider vectors in 3d by extending these ideas. It is obviously difficult to draw
vectors in 3d as they would need to be represented by a line pointing in 3d space, but you can
easily add an extra column to the column vector format. Using components we simply introduce
an extra letter k to represent the 3rd direction that we can move in) Eg v = 3i – 4j + 5k
3. Vector Arithmetic
Consider the vectors
AB = -3i + 4j
CD = 5i + 2j
a) Multiplying a vector by a number (scalar)
This changes the magnitude of the vector, but not its direction.
In in the diagram below, The vector AB has been multiplied by two:
PQ = 2 AB
Q
B
A
P
Using unit vector notation
we say that PQ = 2(-3i + 4j)
= -6i + 8j
Using column vector notation
  3
we say that PQ = 2   =
 4 
  6
 
 8 
In general, if AB  k PQ , then AB and PQ are parallel, with AB being k times longer than
PQ .
Multiplying a vector by a negative number changes the magnitude of the vector, and reverses
its direction. So in the diagram below PQ = -2 AB
Q
B
A
P
In general the return journey from B to A is the exact opposite of the journey from A to B, so we
can say that AB   BA .
To add two vectors together we place the beginning of one vector
at the end of the other. The resultant is the vector formed by
adding the two vectors together. In the diagram opposite,
OB  OA  AB
A
O
In the example you can see how we are
adding AB and CD together using a
diagram.
B
D
C
We can also add them together using unit
vector notation:
B
AB + CD = (-3i + 4j) + (5i + 2j)
= 2i + 6j
A
Or using column vector notation:
  3  5   2 
AB + CD =        
 4   2  6
b) Subtracting two vectors
To subtract two vectors we consider adding the ‘negative’ of the second vector to the first.
In the example you can see how to find
AB - CD using a diagram. We are in fact
finding AB + - CD
D
We can also add them together using unit
vector notation:
AB - CD = (-3i + 4j) - (5i + 2j)
= -8i + 2j
Or using column vector notation:
  3  5    8 
AB - CD =        
 4   2  2 
Exercise 5B page 88 q 1 - 11
C
B
A
4. Vector Geometry
Activity 1
Vector Geometry powerpoint
We can use the rules for adding vectors to form a kind of ‘algebra of vectors’, known as vector
geometry.
B
C
Example 1 : The diagram shows a parallelogram OACB with
b
OA  a and OB  b . D is the midpoint of AC.
Express the following vectors in terms of a and b.
O
a) BC
b) AC
c) CA
d) OC
e) AB
D
A
a
f) OD .
a) Since OA and BC are equal and parallel, BC  OA  a
b) Since AC and OB are equal and parallel, AC  OB  b
c) Since CA is the return journey of AC, CA   AC  b
d) To get from O to C, we can go from O to A to C, so OC  OA  AC  a  b
e) Similarly, we have AB  AC  CB  b  a .
f) The journey from A to D is half the distance from A to C, so we have
 OA  12 AC
 a  12 b
S
Example 2 : The trapezium PQRS is such that PQ  2SR . X is
the midpoint of PQ, and Y is the midpoint of QR. If a and b
are defined as in the diagram opposite, express XY in terms of
a and b.
b
R
Y
a
P
X
Q
Now
XY  XQ  QY
 12 PQ  12 QR

 b  12 QP  PS  SR

 b  12 (2b  a  b)
 12 (a  b)
Example 3 : In the triangle PQR, PQ  a and PR  b . The point X is on the line QR such that
QX : XR  2 : 3 . Find PX in terms of a and b.
It helps to draw a diagram.
PX  PQ  QX
Q
 PQ  52 QR
2

 PQ  52 QP  PR
X
a

 a  (a  b)
2
5
3
 15 (3a  2b)
P
b
R
Exercise 5A page 82 q 1 - 11
5. Resolving Vectors
Activity 2
Resolving Vectors powerpoint
Use this activity to introduce the idea of splitting a vector up into its component parts – a
horizontal (i) part and a vertical (j) part
Example 1 : A vector R is of magnitude 16km and on a bearing of 030&deg;. Taking i as a unit vector
due east and j as a unit vector due north, write R in the form ai + bj.
a  16sin 30
8
b  16 cos 30
8 3
R  8i  8 3 j
R
30
16
bj
ai
Example 2 : A vector R is of magnitude 20km and on a bearing of 210&deg;. Taking i as a unit vector
due east and j as a unit vector due north, write R in the form ai + bj.
a  -20sin600
= -10 3
b = -20cos600
= -10
R = -10 3 i – 10j
Example 3 : Two Forces P and Q with magnitude 4 and 5 respectively act as shown on a partcile
Q
5
P
30
o
4
60o
Find the magnitude and direction of the resultant force P + Q
It is not obvious from the diagram
what the precise size and direction
of the resultant force will be.
Q
5
P+Q
60o
P
o
4
30
Instead of solving this problem with a diagram, we resolve both forces into
component parts and use the following facts:
The vertical component of P + Q will be the same as the vertical component of P
plus the vertical component of Q
The horizontal component of P + Q will be the same as the horizontal component of
P plus the horizontal component of Q
Considering the vertical components
↑ 4sin30 + 5sin60 = 6.33
Considering the horizontal components → 4cos30 – 5cos60 = 0.96
So P + Q = 0.96i + 6.33j
The magnitude of the resultant is
The direction is
0.962  6.332 = 6.4
33
tan θ = 06..96
θ = 81.4o relative to the positive i direction
Exercise 5D page 95 q 1 - 12
6. Finding the magnitude and direction of a vector in 2d and in 3d
The magnitude of a vector a is its length or size. We denote the magnitude of vector a as a . We
can find the magnitude of a vector by using pythagoras. This works in both 2d and 3d. A unit
vector is a vector of length 1 unit, that points in the same direction as the original vector. To
change a vector into a unit vector simply divide it by its magnitude.
 1 
 
Example 1 : Find the magnitude of the vector   2  and find a unit vector that points in the
 3 
 
same direction.
Magnitude = 12  (2) 2  3 2
= 14
 1 
 
A unit vector in the same direction as   2 
 3 
 




would be given by 




1 

14 
2 

14 
3 

14 
Example 2 : Find the magnitude of the vector a = 3i + 4j. Hence find a unit vector in the
direction of a.
3i  4 j
4j
3i
Using Pythagoras’ Theorem,
a 2  32  42
 25
a5
The magnitude is 5 units. A unit vector in the direction of a must have length 1, and therefore be
one-fifth of the length of a. The unit vector is therefore 15 (3i  4 j) .
7. Using Vectors in Mechanics
Example 1 : Anne attempts to swim directly across a river. If her swimming speed is 1 ms-1, and
the current is flowing at 2 ms-1, find her actual speed, and the angle which her path
makes with the bank of the river.
-1
2 ms
1 ms-1
1 ms-1

v
2 ms-1
The first diagram shows the two components of Anne’s speed. The second diagram shows how
the vectors are added, by placing the beginning of one vector at the end of the other. The
resultant vector is v.
To find the actual speed, we use Pythagoras’ Theorem.
v 2  12  22
v 5
 2.24 ms-1 (2 d.p.)
To find the angle made with the bank,
1
2
  26.6 (1 d.p.)
tan  
Example 2 : A pilot wishes to fly his plane north-west, but a southerly wind of speed 60 km/h is
blowing. If the plane is capable of 400 km/h in still air, on what bearing does the
plane actually fly, and what will its actual speed be?
400 km/h
45
400 km/h
45
v
60 km/h
Using the cosine rule,
v 2  4002  602  2  400  60  cos135
v  444.46 km/h (2 d.p.)
Using the sine rule,

60 km/h
sin  sin135

400
v
400sin135
sin  
v
  39.5,   140.5
The plane flies at 444.46 km/h, on a bearing of 360 – 39.5 = 320.5&deg;.
Example 3 : The pilot now compensates for the southerly wind so that the plane does actually fly
due north-west. On what bearing must the pilot head the plane, and what is it speed?
We require the resultant vector to be in a north-west direction. So we have
400 km/h


v
45 60 km/h
By the sine rule,
sin  sin 45

60
400
60sin 45
sin  
400
  6.1 (1 d.p.)
  180  45  
 128.9
The plane must aim on a bearing of 180&deg; + 128.9&deg; = 308.9&deg;.
To find the actual speed of the plane, we use the cosine rule.
v 2  4002  602  2  400  60  cos 
v  440.17 km/h (2 d.p.)
The actual speed of the plane is 440.17 km/h.
Exercise 5E page 98 q 1 - 9
We can extend our work on kinematics into two or even three dimensions.
Example 4 : A particle is travelling with uniform velocity. The position vector of the particle is
3i  5 j m at time t  2 , and 7i  8 j m at time t  7 . Find the velocity of the particle.
change in displacement
time
(7i  8 j)  (3i  5 j)

5
 0.8i  2.6 j ms -1
v
The equations of uniform acceleration readily become vector equations, although the only one
needed in unit M1 is
v  u  at
Example 5 : A particle has velocity 3i  2 j ms-1 when t  0 , and velocity 7i  4 j ms-1 when
t  2 . Find the constant acceleration of the particle.
v  u  at
7i  4 j  3i  2 j  2a
a  2i  j ms -2
Here are some more examples which use common vector methods.
Example 6 : A particle is moving with a constant velocity 12i  5 j ms-1. It passes through the
point with position vector 4i  5 j m at t  0 . Find the speed of the particle and its
distance from O when t  3 seconds.
speed  122  52
 13 ms-1
The position vector of the particle at time t  3 is given by
r  4i  5 j  3(12i  5 j)
 40i  20 j
distance  40 2  20 2
 44.72 m (2 d.p.)
Example 7 : At noon a lighthouse keeper observes two ships A and B which have position vectors
4i  3j km and 4i  9 j km respectively, relative to the lighthouse O. (The unit
vectors i and j are directed due east and due north). The ships are moving with
constant velocities 4i  17 j km/h and 12i  5 j km/h respectively.
a) Write down in the form p  qt the position vector of A and the position vector of
B at time t hours after noon.
b) Show that A and B will collide and find the time when this collision will occur
and the position vector of the point of collision.
c) In order to prevent the collision, at 12.15 pm ship A increases its speed and
changes its velocity to 16i  13j km/h. Ship B maintains its original velocity. Find
the distance between A and B at 12.30 pm now.
a) The position vectors are
rA  (4i  3j)  (4i  17 j)t
rB  (4i  9 j)  (12i  5 j)t
b) The ships will collide if they have the same position at the same time. We begin by assuming
this is the case at time t. We can then equate the i and j components of the position vectors.
For the i components,
4  4t  4  12t
16t  8
t  0.5
For the j components,
3  17t  9  5t
12t  6
t  0.5
As both ships have the same position vector at the same time, they do indeed collide. The time
of collision is 12.30 pm, and the position vector can be found by substituting t  0.5 into one
of the position vectors.
rA  (4i  3j)  0.5(4i  17 j)
 2i  11.5 j
Activity 1 : This can be demonstrated on Autograph. Set the axes from –20 to 20 in each
direction. For ship A, type in the equation x  4  4a, y  3  17a (don’t use t as the
variable for time). Under ‘draw option’ select 4 12 point. Type in the equation for
ship B in the same way. Then use the constant controller to manually increment a in
steps of 601 , starting at 0. The position of each ship will be plotted every minute.
c) The position vector of B is unchanged, but the new position vector of A at 12.30 pm is
rA  (4i  3j)  0.25(4i  17 j)  0.25(16i  13 j)
 i  10.5 j
The displacement from ship B to ship A is therefore 3i  j , and the distance is given by
distance  32  12
 3.16 km (2 d.p.)
Example 5 : A cruiser C is sailing due east at a constant speed of 20 km/h and a destroyer D is
sailing due north at a constant speed of 10 km/h. At noon C and D are at points with
position vectors 5i km and 20 j km respectively, relative to a fixed origin O. The
unit vectors i and j are due east and due north respectively.
a) Find the position vector of C relative to D at time t hours after noon.
b) Hence prove that C and D are closest together at 12.36 pm, and find the distance
between them at this time.
a) The position vectors at time t are
rC  5i  20ti
 (20t  5)i
rD  20 j  10tj
 (10t  20) j
The position vector of C relative to D is given by
rC  rD  (20t  5)i  (10t  20) j
b) The distance d between the two vessels is simply the magnitude of this vector.
d 2  (20t  5) 2  (10t  20) 2
 400t 2  200t  25  100t 2  400t  400
 500t 2  600t  425
Differentiating to minimise d 2 ,
d d 2 
dt
2
d d 2 
dt 2
 1000t  600
 1000
The distance is a minimum when
1000t  600  0
t  0.6 hours
 36 minutes
So the ships are closest at 12.36 pm. We know this is a minimum since the second derivative
is positive. The distance is given by
d  500t 2  600t  425
 500  0.62  600  0.6  425
 15.65 km (2 d.p.)
Activity 2 : Again, this can be demonstrated on Autograph.
Hard Extension Problem
Example 8 : At noon the position vectors and the velocity vectors of two ships A and B are
rA  2i  j km
v A  3i  j km/h
rB  i  4 j km
v B  11i  3j km/h
a) Find the distance of closest approach of the ships, and the time at which this
occurs.
b) Find the times at which the ships are 5km apart.
a) The position vectors at time t are
rA  2i  j  (3i  j)t
 (3t  2)i  (t  1) j
rB  i  4 j  (11i  3j)t
 (11t  1)i  (3t  4) j
The position vector of A relative to B is given by
rA  rB  (3t  2)i  (t  1) j  (11t  1)i  (3t  4) j
 (3  8t )i  (5  2t ) j
The distance d between the two ships is given by
d 2  (3  8t ) 2  (5  2t ) 2
 9  48t  64t 2  25  20t  4t 2
 68t 2  68t  34
Differentiating to minimise d 2 ,
d d 2 
dt
2
d d 2 
dt 2
 136t  68
 136
The distance is a minimum when
136t  68  0
t  0.5 hours
So the ships are closest at 12.30 pm. We know this is a minimum since the second derivative
is positive. The distance is given by
d  68t 2  68t  34
 68  0.52  68  0.5  34
 4.12 km (2 d.p.)
b) We require
68t 2  68t  34  52
68t 2  68t  9  0
68  (68) 2  4  68  9
2  68
t  0.157, t  0.843
t
The times are 12.09 pm and 12.51 pm, to the nearest minute.
Activity 3 : Again, this can be demonstrated on Autograph.
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