6)_M1_Vectors.pdf

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Introduction

• In this chapter you will learn about Vectors

• You will have seen vectors at GCSE level, this

chapter focuses on using them to solve problems

involving SUVAT equations and forces

• Sometimes using vectors offers an easier

alternative to regular methods

• Vectors are used in video games in the movement of

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Vectors

You can use vectors to describe displacements

A vector has both direction and magnitude

For example:

 An object is moving north at 20ms-1

 A horizontal force of 7N

 An object has moved 5m to the left These are all vectors. A scalar quantity would

be something such as:

A force of 10N

(It is scalar since it has no direction)

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Vectors

A girl walks 2km due east from a fixed point O, to A, and then 3km due south from A to a point B. Describe the displacement of B from

O.

 Start, as always, with a diagram!

 To describe the displacement you need the distance from O as well as the

direction (as a bearing)

 Remember bearings are always measured from north!

“Point B is 3.61km from O on a bearing of 146˚”

6A

2km 3km O A B θ N

Describing the displacement

The distance – use Pythagoras’ Theorem

= + = 2 + 3 = 3.61

The bearing – use Trigonometry to find angle θ

=

= 3 2 = 56.3˚

= 146˚

Sub in a and b Calculate

Sub in opp and adj Use inverse Tan

Bearings are measured from

north. Add the north line and add 90˚

Opp Adj

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Vectors

In an orienteering exercise, a cadet leaves the starting point S and walks 15km on a

bearing of 120˚ to reach A, the first checkpoint.

From A he then walks 9km on a bearing of 240˚ to the second checkpoint, and point B.

From B he then returns directly to S. Describe the displacement of S from B.

 Start with a diagram!

 We need the distance B to S and the bearing from B to S as well. You will need

to use angle rules you have learnt pre A-level. N N S A B 15km 9km 120° 240° You can use interior

angles to find an angle in the triangle

Interior angles add up to 180°

The missing angle next to 240 is 60°

The angle inside the triangle must also be

60°

Finding the distance B to S

=

+

− 2

= 15 + 9 − 2(15 × 9)

60 = 171

= 13.1

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Vectors

In an orienteering exercise, a cadet leaves the starting point S and walks 15km on a

bearing of 120˚ to reach A, the first checkpoint.

From A he then walks 9km on a bearing of 240˚ to the second checkpoint, and point B.

From B he then returns directly to S. Describe the displacement of S from B.

 Start with a diagram!

 We need the distance B to S and the bearing from B to S as well. You will need

to use angle rules you have learnt pre A-level. N N S A B 15km 9km 120° 240° 60° 60° 13.1km N θ

Finding the bearing from B to S

 Show the bearing at B

 It can be split into 2

sections, one of which is 180°

 Find angle θ inside the triangle

9 =

60 13.1

θ = 60

13.1 × 9

= 36.6° =

Sub in values Rearrange Calculate θ

156.6° 180°

You can now use Alternate angles to

find the unknown part of the bearing

Add on 180°

The bearing is 336.6°

S is 13.1km from B on a bearing of 337°

A

a

b

B

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Vectors

You can add and represent vectors using line segments

A vector can be represented as a directed line segment

Two vectors are equal if they have the same magnitude and direction

Two vectors are parallel if they have the same direction

You can add vectors using the triangle law of addition

6B

A C

B

a

3 a

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Vectors

OACB is a parallelogram. The points P, Q, M and N are the midpoints of

the sides.

OA = a

OB = b

Express the following in terms of a

and b.

a) OC b) AB c) QC

d) CN e) QN

6B

M

B

P

O

N

D

Q

A

C

b

a

+

b

-

a

1

_/

2

b

-

1

_/

2

a

1

/

2

b

-

1

/

2

a

What can you deduce about AB and QN,

looking at the vectors?

=

−

₌

1

2 −

1

2 =

1

2 ( − )

QN is a multiple of

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Vectors

In triangle OAB, M is the midpoint of OA and N divides AB in the ratio

1:2.

OM = a

OB = b

Express ON in terms of a and b

6B

A B O M N a b a 1 2

=

+

= 2

=

1

3

Use the ratio. If N divides AB in the ratio 1:2, show this on the diagram

 You can see now that AN is one-third of AB

 We therefore need to know AB

 To get from A to B, use AO + OB

=

+

= −2 +

= −

2

3 +

1

3

Sub in AO and OB AN = 1_/

3AB

= 2 + −

2

3 +

1

3 =

4

3 +

1

3

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Vectors

OABC is a parallelogram. P is the point where OB and AC intersect.

The vectors a and c represent OA and OC respectively.

Prove that the diagonals bisect each other.

 If the diagonals bisect each other, then P must be the midpoint of both AC and OB…

 Try to find a way to represent OP in different ways…

(make sure you don’t ‘accidentally’ assume P is the midpoint – this is

what we need to prove!)

6B

P

O

A _B

C

a

c

One way to get from O to P

 Start with OB

=

+

= λ( + )

OP is parallel to OB so is a multiple of (a + c)

 We don’t know how much for now, so can use λ (lamda) to represent the unknown quantity

c

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Vectors

6B

P O A _B C a c

Another way to get from O to P

 Go from O to A, then A to P

 We will need AC first…

c

= λ( + )

-a

=

−

= ( − )

=

+

=

+ ( − )

AP is parallel to AC so is a multiple of it. Use a different symbol

(usually μ, ‘mew’, for this multiple) Now we have another way to get from O to P

Sub in vectors

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Vectors

6B

P O A _B C a

= λ( + )

=

+ ( − )

= λ( + ) = + ( − )

λ + = + ( − ) λ + λ = + − λ + λ = (1 − ) +

As these represent the same vector, the expressions must be equal!

Multiply out brackets

Factorise the ‘a’ terms on the right side

Now compare sides – there must be the same number of ‘a’s and ‘c’s on each

λ = 1 − λ = λ = 1 − λ

λ = 0.5 μ = 0.5

Sub 2nd _{equation into the first}

They are equal

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Vectors

You can describe vectors using the i, j notation

A unit vector is a vector of length 1. Unit vectors along Cartesian (x, y) axes are usually denoted by i and j

respectively.

You can write any two-dimensional vector in the form ai + bj

Draw a diagram to represent the vector -3i + j

6C

O (0,1)

(1,0)

j

i

A B

C

5i

2j

5i + 2j ₌ ₊

= 5 + 2

-3i j

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Vectors

You can solve problems with vectors written using the i, j notation

When vectors are written in terms of the unit vectors i and j you can add them together by adding the terms in

i and j separately. Subtraction works in a similar way.

Given that:

p = 2i + 3j q = 5i+ j

Find p + q in terms of i and j

6D

+

= (2 + 3 ) + (5 + )

+

= 7 + 4

Add the i terms and j terms

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Vectors

When vectors are written in terms of the unit vectors i and j you can add them together by adding the terms in

i and j separately. Subtraction works in a similar way.

Given that:

a = 5i + 2j b = 3i- 4j

Find 2a – b in terms of i and j

6D

2 + = 2(5 + 2 ) − (3 − 4 )

_{Multiply out}

the bracket

2 + = 10 + 4 − (3 − 4 )

2 + = 10 + 4 − 3 + 4

2 + = 7 + 8

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Vectors

When a vector is given in terms of the unit vectors i and j, you can find

its magnitude using Pythagoras’ Theorem.

The magnitude of vector a is written as |a|

Find the magnitude of the vector: 3i – 7j

6D

3i

-7j

3i - 7j

=

3 + (−7)

= 58

= 7.62

(3sf)

Put in the values from the vectors and calculate

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Vectors

You can also use trigonometry to find an angle between a vector and the

axes

Find the angle between the vector -4i

+ 5j and the positive x-axis

Draw a diagram

6D

-4i

x

θ

5j

y

Opp

Adj

=

= 5 4 = 51.3°

= 128.7°

Sub in values Inverse Tan

The angle we want is between the vector and the positive x-axis

 Subtract θ from 180°

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Vectors

Given that:

a = 3i - j b = i + j

Find µ if a + µb is parallel to 3i + j

 Start by calculating a + µb in terms of a, b and µ

6D

+

= (3 − ) + ( + )

= 3 − +

+

= 3 +

− +

= 3 +

+ −1 +

As the vector must be parallel to 3i + j, the i term must be 3 times the j term!

3 + = 3(−1 + )

3 + = −3 + 3

6 = 2

3 =

Multiply out the brackets

Divide by 2

= 3

Move the iand jterms together

Factorise the terms in i and j

Multiply out the bracket

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= 2(3 + )

Vectors

Given that:

a = 3i - j b = i + j

Find µ if a + µb is parallel to 3i + j

 Start by calculating a + µb in terms of a, b and µ

6D

= 3

To show that this works…

+

+ 3

= (3 − ) + 3( + )

= 3 − + 3 + 3

= 6 + 2

Multiply out the brackets We now know µ

Group terms

Factorise

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Vectors

You can express the velocity of a particle as a vector

The velocity of a particle is a vector in the direction of motion. The magnitude of the vector is its speed.

Velocity is usually represented by v.

A particle is moving with constant velocity given by:

v = (3i + j) ms-1

Find:

a) The speed of the particle b) The distance moved every 4

seconds

6E

Finding the speed

 The speed of the particle is the magnitude of the vector

 Use Pythagoras’ Theorem

3i

j

3i + j

=

3 + 1

= 3.16

Finding the distance travelled every 4 seconds

 Use GCSE relationships

 Distance = Speed x Time

=

×

= 3.16 × 4

= 12.6

Sub in values (use the exactspeed!)

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Vectors

You can solve problems involving velocity, time, mass and forces (as in earlier chapters) by using vector

notation

If a particle starts from the point with position vector r₀ and moves with

constant velocity v, then its displacement from its initial position

at time t is given by:

6F

=

+

Final

position Starting position

Velocity

Time

A particle starts from the point with position vector (3i+ 7j) m and then moves constant velocity (2i – j)

ms-1_{. Find the position vector of the particle 4} seconds later.

(a position vector tells you where a particle is in relation to the origin O)

= (3 + 7 )

= (2 − )

= 4

=

+

= 3 + 7

+ (2 − )(4)

= 3 + 7 + 8 − 4

= 11 + 3

Sub in values

Multiply/remove brackets

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Vectors

notation

If a particle starts from the point with position vector r₀ and moves with

constant velocity v, then its displacement from its initial position

at time t is given by:

6F

=

+

Final

position Starting position

Velocity

Time

A particle moving at a constant velocity, ‘v’, and is at the point with position vector (2i + 4j) m at time t = 0. Five

seconds later the particle is at the point with position vector (12i + 16j) m. Find the velocity of the particle.

= (2 − 4 )

= (12 + 16 )

= 5

=

+

(12 + 16 ) = 2 − 4

+ ( )(5)

Sub in values

12 + 16 = 2 − 4 + 5

10 + 20 = 5

2 + 4 =

The velocity of the particle is (2i + 4j) ms-1

Deal with the brackets!

Subtract 2i and add 4j

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Vectors

You can solve problems involving velocity, time, mass and forces (as in

earlier chapters) by using vector notation

At time t = 0, a particle has position vector 4i + 7jand is moving at a speed of

15ms-1 _{in the direction 3}_i _{– 4}_j_{. Find its}

position vector after 2 seconds.

 You have to be careful here, you have been given the speed of the particle. The direction vector does not necessarily have the same speed

 Find the speed of the direction vector as it is given in the question

 Then ‘multiply up’ to get the required speed (we need 15ms-1_{, not}

5ms-1₎

Multiplying the vectors will allow you to use the correct velocity

6F

=

+

3i

-4j

3i – 4j

= 3 + (−4)

= 5

9i

-12j

9i – 12j

5ms-1 _15ms-1

Multiply all vectors by 3

= 15

= 9 − 12

Calculate

We can use the vectors as the velocity

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Vectors

You can solve problems involving velocity, time, mass and forces (as in

earlier chapters) by using vector notation

At time t = 0, a particle has position vector 4i + 7jand is moving at a speed of

15ms-1 _{in the direction 3}_i _{– 4}_j_{. Find its}

position vector after 2 seconds.

 You have to be careful here, you have been given the speed of the particle. The direction vector does not necessarily have the same speed

 Find the speed of the direction vector as it is given in the question

 Then ‘multiply up’ to get the required speed (we need 15ms-1_{, not}

5ms-1₎

Multiplying the vectors will allow you to use the correct velocity

6F

=

+

= 9 − 12

= (4 + 7 )

= (9 − 12 )

= 2

=

+

= (4 + 7 ) + (9 − 12 )(2)

= 4 + 7 + 18 − 24

= 22 − 17

Sub in values

‘Deal with’ the brackets

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Vectors

notation

You can also solve problems involving acceleration by using:

v = u + at

Where v, u and a are all given in vector form.

Particle P has velocity (-3i + j) ms-1 _at time t = 0. The particle moves along

with constant acceleration a = (2i + 3j) ms-2_{. Find the speed of the}

particle after 3 seconds.

6F

=? = (−3 + ) =? = (2 + 3 ) = 3

= +

= (−3 + ) + (2 + 3 )(3)

= −3 + + 6 + 9

= 3 + 10

= 10.4

Sub in values

Group terms

Remember this is the velocity, not the speed!

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Vectors

notation

A force applied to a particle has both a magnitude and a direction, so force is a vector. The force will cause the particle

to accelerate.

Remember from chapter 3:

F= ma

A constant force, FN, acts on a particle of mass 2kg for 10 seconds. The particle is initially at rest, and 10 seconds later it has a velocity of (10i – 24j) ms-1_{. Find}_F_.

 We need to find a first…

6F

=?

= 0

= (10 − 24 )

=?

= 10

=

+

(10 − 24 ) = 0 + (10)

10 − 24 = 10

− 2.4 =

Sub in values

‘Tidy up’

Divide by 10

=

= (2)( − 2.4 )

= 2 − 4.8

Sub in values

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Vectors

You can use vectors to solve problems about forces

If a particle is resting in equilibrium, then the resultant of all the forces

acting on it is zero.

The forces (2i + 3j), (4i – j), (-3i + 2j) and (ai + bj) are acting on a particle

which is in equilibrium.

Calculate the values of a and b.

 Set the sum of all the vectors equal to 0

6G

2 + 3

+ 4 −

+ −3 + 2

+

= 0

3 +

+

= 0

3 +

= 0

= −3

4 +

= 0

= −4

Group together the

numerical terms

The ‘i’ terms must sum to 0

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Vectors

If several forces are involved in a question a good starting point is to

find the resultant force.

The following forces:

F₁ = (2i + 4j) N

F₂ = (-5i + 4j) N

F₃ = (6i – 5j) N

all act on a particle of mass 3kg. Find the acceleration of the particle.

Start by finding the overall resultant force.

6G

=

+

= 2 + 4

+ −5 + 4

+ (6 − 5 )

= 3 + 3

=

(3 + 3 ) = 3

+ =

The acceleration is (

i

+

j

) ms

-2

Sub in values Group

up

Sub in the resultant force, and the mass

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Vectors

A particle P, of weight 7N is suspended in equilibrium by two light inextensible strings attached to P and to points A and

B as shown in the diagram.

Line AB is horizontal. Find the tensions in the two strings

 Draw a sketch of the forces acting on P

 These can be rearranged into a triangle of forces

(the reason being, if the particle is in equilibrium then the overall force is zero

– ie) The particle ends up where it started)

 You will now need to work out the angles in the triangle…

6G

A B

P

30° _40°

P

T_A T_B

7N

T_A

T_B

7N

These are the forces

acting on P

These are the forces

rearranged as a

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Vectors

A particle P, of weight 7N is suspended in equilibrium by two light inextensible strings attached to P and

to points A and B as shown in the diagram.

 You will now need to work out the angles in the triangle…

 Consider the original diagram, you could work out more angles on it

as shown, some of which correspond to our triangle of

forces…

6G

T_A

T_B

7N

The angle between 7N and T_A is 60°

A B P 30° _40° 7N 50° 60° 60° 50° 70°

The angle between 7N and T_B is 50°

(It is vertically opposite on our triangle of forces) The final angle can be

worked out from the triangle of forces alone Now we can calculate

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Vectors

 To calculate the tensions you can now use the Sine rule

(depending on the information given, you may have to use the Cosine rule

instead!)

6G

T_A

T_B

7N

A B

P

30° _40°

7N 50° 60°

60° 50°

70° 50 =

7 70

= 7

70× 50

= 5.71

Multiply by Sin50

Calculate

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Vectors

 To calculate the tensions you can now use the Sine rule

(depending on the information given, you may have to use the Cosine rule

instead!)

6G

T_A

T_B

7N

A B

P

30° _40°

7N 50° 60°

60° 50°

70° 60 =

7 70

= 7

70× 60

= 6.45

Multiply by Sin60

Calculate

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Vectors

You need to be able to solve worded problems in practical contexts

The mixed exercise in this chapter is very important as it contains questions in context,

the type of which are often on exam papers

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Vectors

A ship S is moving with constant velocity (–2.5i + 6j) kmh–1_{. At time 1200, the position}

vector of S relative to a fixed origin O is (16i + 5j) km.

Find:

(a) the speed of S

(b) the bearing on which S is moving.

The ship is heading directly towards a submerged rock R. A radar tracking station

calculates that, if S continues on the same course with the same speed, it will hit R at

the time 1500.

(c) Find the position vector of R.

6H

-2.5i

6j

The speed of S

= (−2.5) +6

= 6.5 ℎ

Use Pythagoras’

Theorem

Calculate

6.5 kmh-1

N

θ

180°

The bearing on which S is travelling

= 6 2.5

 Find angle θ

= 67.4°

= 337°

Use Tan = Opp_/ Adj

Calculate

Consider the north line and read clockwise… 337°

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Vectors

Find:

(a) the speed of S

(b) the bearing on which S is moving.

The ship is heading directly towards a submerged rock R. A radar tracking station

calculates that, if S continues on the same course with the same speed, it will hit R at

the time 1500.

(c) Find the position vector of R.

6H

6.5 kmh-1

337°

= (16 + 5 )

=

+

= (−2.5 + 6 )

= 3

= (16 + 5 ) + (−2.5 + 6 )(3)

= 16 + 5 − 7.5 + 18

= 8.5 + 23

Sub in values

Group terms

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Vectors

The tracking station warns the ship’s captain of the situation. The captain maintains S on

its course with the same speed until the time is 1400. He then changes course so that

S moves due north at a constant speed of 5 km h–1_{. Assuming that S continues to move}

with this new constant velocity.

Find:

(d) an expression for the position vector of the ship t hours after 1400,

(e) the time when S will be due east of R, (f) the distance of S from R at the time

1600

6H

 Find the position vector of the ship at 1400

= (16 + 5 )

=

+

= (−2.5 + 6 )

= 2

= (16 + 5 ) + (−2.5 + 6 )(2)

= 16 + 5 − 5 + 12

= 11 + 17

Sub in values

Group terms

So at 1400 hours, the ship is at position vector (11i + 17j)

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Vectors

Find:

1600

6H

 At 1400 the ship is at (11i+ 17j)

 Find an expression for its position t hours after 1400

 Use the same formula, with the updated information

= (11 + 17 )

= 5

=

+

= (11 + 17 ) + (5 )( )

= 11 + 17 + 5

= 11 + (17 + 5 )

= 8.5 + 23

Sub in values

Factorise the j

terms

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Vectors

Find:

1600

6H

= 8.5 + 23

= 11 + (17 + 5 )

 Find the time when S will be due east of R

R S

8.5 + 23

11 + (17 + 5 )

If S is due east of R, then their j terms must be equal!

23 = 17 + 5

6 = 5

1.2 =

Subtract 17

Divide by 5

 1.2 hours = 1 hour 12 minutes

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Vectors

Find:

1600

6H

= 8.5 + 23

= 11 + (17 + 5 )

1512

Find the distance of S from R at the time 1600

 Find where S is at 1600 hours…

= 11 + (17 + 5 )

= 11 + (17 + 5(2))

= 11 + 27

Sub in t = 2 (1400 – 1600

hours)

Simplify/calculate

So the position vectors of the rock and the ship at 1600 hours are:

= 8.5 + 23 = 11 + 27

To calculate the vector between them, calculate S - R

11 + 27 − (8.5 + 23 ) = 2.5 + 4

= 2.5 + 4

= 4.72

Calculate

Now use Pythagoras’ Theorem to work out the distance

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Summary

• We have seen how to use vectors in

problems involving forces and SUVAT

equations

• We have also seen how to answer

multi-part worded questions

• It is essential you practice the mixed