M1 - VECTORS
VECTORS – THE BASICS
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
B
a
SCALARS – Have magnitude (size) ONLY
VECTORS – Have magnitude AND direction
A
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Vectors are written in the form ⃗⃗⃗⃗⃗
𝐴𝐵 – the arrow signifies the direction (i.e. from A to B).
⃗⃗⃗⃗⃗ is written as |𝐴𝐵
⃗⃗⃗⃗⃗ |.
The magnitude of 𝐴𝐵
Alternatively you can label a vector with a single letter in bold, such as a. When
handwriting this, it is usual to write it underlined rather than in bold, such as a.
Its magnitude would be written as |a| or |a|.
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A unit vector, is a vector with a magnitude of 1. The unit vector in the direction of vector a
a
̂ and it is given by
is usually labelled as 𝒂
|a |
Vectors are equal if they have the same magnitude AND the same direction.
Multiplying a vector by a number (scalar) alters the magnitude but NOT the direction.
A negative vector has the same magnitude as the vector but is in the opposite direction.
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Two vectors are added using the TRIANGLE LAW.
b
Q
b
a
a
R
a+b
c
PQ  QR  PR
P
a+b+c

Adding the vectors PQ and QP gives the zero vector 0
Q
P
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Any vector parallel to the vector a may be written as λa, where λ is a non-zero scalar.
Subtracting a vector is equivalent to ‘adding a negative vector’, so a – b is defined to be
a + (-b).
a
b
-b
a
a-b
M1 - VECTORS
Example 1
The vector a is directed due north and |a| = 24. The vector b is directed due west and|b| = 7.
Find |a + b|
Example 2
An expedition travels 10km on a bearing of 080° and then 8km on a bearing of 045°.
What is the final position in relation to the starting position?
Example 3
A swimmer who can swim at a speed of 0.8ms-1 in still water, wishes to cross a river flowing at a
speed of 0.5ms-1.
(a)
If she aims straight across the river, what will be her actual velocity?
(b)
If she wishes to travel straight across, in what direction should she aim and what will her
actual speed be?
M1 - VECTORS
Example 4
In the diagram, OA = a, OB = b and BC = 1.5a. M is the mid-point of BC , N is the mid-point of AC
and P is the mid-point of OB .
(a)
Find in terms of a and b the vectors…
(i)
(b)
AC
(ii)
OM
(iii)
BN
Prove that PN is parallel to OA .
B
P
b
M
O
C
a
N
A
M1 - VECTORS
Example 5
OACB is a parallelogram. OA is represented by the vector a. OB is represented by the vector b. P
is the point 2/3 the distance from O to C, and M is the mid-point of AC . Show that B, P and M lie
on the same straight line.
B
C
b
O
P
a
M
A
M1 - VECTORS
Example 6
In the diagram OA = a, OB = b, 3 OC = 2 OA and 4 OD = 7 OB .
The line DC meets the line AB at E.
(a)
Write down, in terms of a and b, expressions for
(i)
AB
(ii)
O
DC
Given that DE   DC and EB   AB where λ and µ are constants:
(b)
Use triangle EBD to form an equation relating to a, b, λ and µ.
Hence:
9
(c)
Show that  
.
13
(d)
Find the exact value of µ.
(e)
B
E
C
Express OE in terms of a and b.
F
The line OE produced meets the line AD at F.
Given that OF  k OE where k is a constant and that AF 
(f)
Find the value of k.
D
1
(7b  4a) :
10
A
M1 - VECTORS
COMPONENTS OF A VECTOR
A unit vector is a vector is a vector with magnitude 1.
The unit vector in the x-direction is called i.
The unit vector in the y-direction is called j.
Rather than specify a vector using its magnitude and direction, you can state it in component form
in terms of i and j.
y
⃗⃗⃗⃗⃗ = 𝑂𝐴
⃗⃗⃗⃗⃗ + 𝐴𝑃
⃗⃗⃗⃗⃗
𝑂𝑃
P
2
⃗⃗⃗⃗⃗
𝑂𝐴 = 3𝒊
⃗⃗⃗⃗⃗
𝐴𝑃 = 2𝒋
So, ⃗⃗⃗⃗⃗
𝑂𝑃 = 3𝒊 + 2𝒋 in component form
A
O
x
⃗⃗⃗⃗⃗ = (3) as a column vector
Or, 𝑂𝑃
2
3
When you are given a vector as a magnitude and direction, you can convert it to component form.
This is called resolving the vector into components.
P(x, y)
r
y
⃗⃗⃗⃗⃗ = 𝑟𝑐𝑜𝑠𝜃𝒊 + 𝑟𝑠𝑖𝑛𝜃𝒋 = (𝑟𝑐𝑜𝑠𝜃)
𝑂𝑃
𝑟𝑠𝑖𝑛𝜃
θ
O
The vector ⃗⃗⃗⃗⃗
𝑂𝑃 has magnitude r and direction θ to the
positive x-direction.
From the triangle,
x = rcosθ and y = rsinθ, so
x
When you are given a vector in component form, you can calculate its magnitude and direction.
P(x, y)
r
y
⃗⃗⃗⃗⃗ = 𝑥𝒊 + 𝑦𝒋
The vector 𝑂𝑃
The magnitude is given by 𝑟 = √𝑥 2 + 𝑦 2
𝑦
Its direction is given by θ, where 𝑡𝑎𝑛𝜃 = 𝑥
θ
O
x
M1 - VECTORS
Example 7
⃗⃗⃗⃗⃗ , 𝑂𝑄
⃗⃗⃗⃗⃗⃗ and 𝑂𝑅
⃗⃗⃗⃗⃗ in component form
For the diagrams shown, express the vectors 𝑂𝑃
Example 8
Find the magnitude and direction of the following vectors:
(a)
p = 2i + 5j
(b)
q = 3i – 2j
(c)
r = -i – 2j
M1 - VECTORS
Example 9
Given p = 12i + 5j and q = 3i – 4j, find:
(a)
i.
p–q
ii.
2p + 3q
(b)
a vector parallel to p with magnitude 39
̂
(c)
the unit vector 𝒒
Example 10
Find the magnitude and direction of the resultant of the vectors shown in the diagram
y
5N
4N
20°
74°
3N
x
55°
7N
Example 11
Given that p = mi + nj and q = (2n + 5)i + (1 – m)j, find the values of m and n for which p = q.
M1 - VECTORS
POSITION, DISPLACEMENT AND VELOCITY
The vector ⃗⃗⃗⃗⃗
𝑂𝐴 describes the position of a point, A
relative to an origin O.
⃗⃗⃗⃗⃗
𝑂𝐴 is the POSITION VECTOR of A.
The diagram shows points A and B with position
vectors a and b relative to an origin O.
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ − ⃗⃗⃗⃗⃗
𝐴𝐵 = 𝑂𝐵
𝐴𝑂 = 𝑂𝐵
𝑂𝐴
⃗⃗⃗⃗⃗ = 𝒃 − 𝒂
𝐴𝐵
⃗⃗⃗⃗⃗ | = |𝒃 − 𝒂|
The distance from A to B = |𝐴𝐵
Example 12
Points A and B have position vectors a = 2i + j and b = 5i – 6j respectively. Find the distance AB.
Example 13
A particle leaves point A, with position vector (3i + 7j)m, and travels with constant velocity
(2i – j)ms-1. Find its position, B, after 3 seconds.
M1 - VECTORS
Example 14
At a certain time, particle A is at the point with position vector (i + 4j)m and is moving with
constant velocity (3i + 3j)ms-1. At the same time, particle B is at the point (5i + 2j)m and is moving
with constant velocity (2i + 3.5j)ms-1.
⃗⃗⃗⃗⃗ at time t seconds.
(a)
Find the vector 𝐴𝐵
(b)
Hence show that the particles will collide and find the position vector of the point of
collision.
M1 - VECTORS
Example 15
A particle moving with velocity (i + 2j)ms-1 undergoes a constant acceleration of (2i – j)ms-2 for a
period of 4 seconds. Find its velocity and speed at the end of this period.
Example 16
Take east and north to be the x- and y-directions respectively. A bird, flying with velocity
(3i – 4j)ms-1, accelerates with constant acceleration (i + 2j)ms-2.
(a)
At what subsequent time is the bird flying due east?
(b)
Find the bird’s speed and direction of flight after 5 seconds.
M1 - VECTORS
APPLICATIONS OF VECTORS
[In these questions, i and j are horizontal unit vectors due east and due north respectively and
position vectors are given with respect to a fixed origin.]
Example 17
A ship S is moving along a straight line with constant velocity. At time t hours the position vector
of S is s km. When t = 0, s = 9i – 6j. When t = 4, s = 21i + 10j. Find
(a)
the speed of S,
(4)
(b)
the direction in which S is moving, giving your answer as a bearing.
(2)
(c)
Show that s = (3t + 9) i + (4t – 6) j.
(2)
A lighthouse L is located at the point with position vector (18i + 6j) km. When t = T, the ship S is
10 km from L.
(d)
Find the possible values of T.
(6)
M1 - VECTORS
Example 18
Two ships P and Q are moving with constant velocities. Ship P moves with velocity
(2i – 3j) km h–1 and ship Q moves with velocity (3i + 4j) km h–1.
(a)
Find, to the nearest degree, the bearing on which Q is moving.
(2)
At 2 p.m., ship P is at the point with position vector (i + j) km and ship Q is at the point with
position vector (–2j) km.
At time t hours after 2 p.m., the position vector of P is p km and the position vector of Q is q km.
(b)
Write down expressions, in terms of t, for
PQ .
(i)
p,
(ii)
q,
(iii)
(5)
(c)
Find the time when
(i)
Q is due north of P
(ii)
Q is north-west of P.
(4)