vectors

Vectors

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 characters and by engineers in the design of buildings, bridges and other structures

Teachings for Exercise 6A

VectorsYou 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)

6A

Vectors have both direction and magnitude!

http://www.youtube.com/watch?v=GKQdkS0qG3g






displacements

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

𝑐=√𝑎2+𝑏2𝑐=√22+32𝑐=3.61𝑘𝑚The bearing – use Trigonometry to find angle θ

𝑇𝑎𝑛𝜃=𝑂𝑝𝑝𝐴𝑑𝑗

𝑇𝑎𝑛𝜃=32

𝜃=56.3 ˚𝐵𝑒𝑎𝑟𝑖𝑛𝑔=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

56.3˚


displacements

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

NS

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°

60°

60°

Finding the distance B to S

𝑎2=𝑏2+𝑐2−2𝑏𝑐𝐶𝑜𝑠𝐴𝑎2=152+92−2 (15×9)𝐶𝑜𝑠60

𝑎2=171𝑎=13.1𝑘𝑚

Sub in values

a

b

c

Work out

Square root

13.1km


displacements

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

NS

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

𝑆𝑖𝑛𝜃9

=𝑆𝑖𝑛6013.1

𝑆𝑖𝑛θ=𝑆𝑖𝑛 6013.1

×9

𝜃=36.6 °

𝑆𝑖𝑛𝐴𝑎

=𝑆𝑖𝑛𝐵𝑏

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

bB

36.6°

Teachings for Exercise 6B

VectorsYou 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 3a

𝐴𝐶=𝐴𝐵+𝐵𝐶



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

ND

QA C

b

a

a + b b - a 1/2b

-1/2a 1/2b - 1/2a

What can you deduce about AB and QN, looking at the vectors?

�⃗�𝐵=𝒃−𝒂 �⃗�𝑁=12𝒃−

12𝒂

�⃗�𝑁=12(𝒃−𝒂)

QN is a multiple of AB, so they are

parallel!



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

BO

MN

a

b

a1

2

�⃗�𝑁=�⃗�𝐴+ �⃗�𝑁

�⃗�𝐴=2𝑎

�⃗�𝑁=13�⃗�𝐵

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

�⃗�𝐵= �⃗�𝑂+�⃗�𝐵

�⃗�𝐵=−2𝒂+𝒃

�⃗�𝑁=−23𝒂+13𝒃

Sub in AO and OB

AN = 1/3AB

�⃗�𝑁=2𝒂+−23𝒂+13𝒃

�⃗�𝑁=43𝒂+13𝒃

Sub in values

Simplify



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

�⃗�𝐵=𝑎+𝑐

�⃗�𝑃= λ(𝑎+𝑐 )

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

�⃗�𝑃= λ(𝑎+𝑐 )











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

�⃗�𝑃= λ(𝑎+𝑐 )

-a

�⃗�𝐶=𝑐−𝑎

�⃗�𝑃=𝜇(𝑐−𝑎)

�⃗�𝑃=�⃗�𝐴+ �⃗�𝑃

�⃗�𝑃=𝑎+𝜇(𝑐−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

�⃗�𝑃=𝑎+𝜇(𝑐−𝑎)











P

O

A B

C

a

�⃗�𝑃= λ(𝑎+𝑐 )�⃗�𝑃=𝑎+𝜇(𝑐−𝑎)

�⃗�𝑃= λ(𝑎+𝑐 ) �⃗�𝑃=𝑎+𝜇(𝑐−𝑎)

λ (𝑎+𝑐 )=𝑎+𝜇(𝑐−𝑎)

λ 𝑎+ λ𝑐=𝑎+𝜇𝑐−𝜇𝑎λ 𝑎+ λ𝑐=(1−𝜇)𝑎+𝜇𝑐

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−𝜇 λ=𝜇λ=1−λλ=0.5μ=0.5

Sub 2nd equation into the first

They are equal

Rearrange and solve So P is halfway along OB and AC and hence the lines bisect each

other!

Teachings for Exercise 6C

VectorsYou 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

2j5i + 2j �⃗�𝐶= �⃗�𝐵+ �⃗�𝐶

�⃗�𝐶=5 𝒊+2 𝒋

-3i

j-3i + j

Teachings for Exercise 6D

VectorsYou 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 + 3jq = 5i + j

Find p + q in terms of i and j

6D

𝒑+𝒒=¿(2 𝒊+3 𝒋)+(5 𝒊+ 𝒋)

𝒑+𝒒=¿7 𝒊+4 𝒋

Add the i terms and j terms separately


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:a = 5i + 2jb = 3i - 4j

Find 2a – b in terms of i and j

6D

2 2(5 𝒊+2 𝒋 )−(3 𝒊−4 𝒋) Multiply out the bracket

2 10 𝒊+4 𝒋−(3 𝒊−4 𝒋)

2 10 𝒊+4 𝒋−3 𝒊+4 𝒋2 7 𝒊+8 𝒋

Careful with the subtraction here!

Group terms…


notation

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

|𝒗|=√32+(−7 )2

|𝒗|=√58 (3sf)

Put in the values from the vectors and

calculateRound if necessary


notation

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

-4ix

θ5j

y

Opp

Adj

𝑇𝑎𝑛𝜃=𝑂𝑝𝑝𝐴𝑑𝑗

𝑇𝑎𝑛𝜃=54

𝜃=51.3 °

𝐴𝑐𝑡𝑢𝑎𝑙𝑎𝑛𝑔𝑙𝑒=128.7 °

Sub in values

Inverse Tan

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

Subtract θ from 180°

51.3°


notation

Given that:

a = 3i - jb = 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

𝒂+𝜇𝒃=¿(3 𝒊− 𝒋)+𝜇(𝒊+ 𝒋)

¿3 𝒊− 𝒋+𝜇𝒊+𝜇 𝒋

¿3 𝒊+𝜇𝒊− 𝒋+𝜇 𝒋

¿ (3+𝜇) 𝒊+(−1+𝜇) 𝒋

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

3+𝜇=¿3 (−1+𝜇)3+𝜇=¿−3+3𝜇6=¿2𝜇3=¿𝜇

Multiply out the brackets

Divide by 2

𝜇=3

Move the i and j terms together

Factorise the terms in i and j

Multiply out the bracket

Subtract µ, and add 3

¿2(3 𝒊+ 𝒋)


notation

Given that:

a = 3i - jb = 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

𝜇=3

To show that this works…

𝒂+𝜇𝒃𝒂+3𝒃¿ (3 𝒊− 𝒋)+3 (𝒊+ 𝒋)

¿3 𝒊− 𝒋+3 𝒊+3 𝒋¿6 𝒊+2 𝒋

Multiply out the brackets

We now know µ

Group terms

Factorise

You can see that using the value of µ = 3, we get a vector which is parallel

to 3i + j

Teachings for Exercise 6E

VectorsYou 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

j3i + j

|𝑣|=√32+12|𝑣|=3.16𝑚𝑠− 1

Finding the distance travelled every 4 seconds

Use GCSE relationships Distance = Speed x Time

𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒=𝑠𝑝𝑒𝑒𝑑×𝑡𝑖𝑚𝑒𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒=3.16×4𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒=12.6𝑚

Sub in values (use the exact speed!)

Calculate

Calculate

Teachings for Exercise 6F

VectorsYou 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

𝒓=𝒓 0+𝒗 𝑡

Final positio

n

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)

𝒓 0=(3 𝒊+7 𝒋)𝒗=(2 𝒊− 𝒋)𝑡=4


𝒓=(3 𝒊+7 𝒋 )+(2 𝒊− 𝒋 )(4 )

𝒓=3 𝒊+7 𝒋+8 𝒊−4 𝒋𝒓=11𝒊+3 𝒋

Sub in values

Multiply/remove brackets

Simplify




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 positio

n

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.𝒓 0=(2 𝒊−4 𝒋 )𝒓=(12 𝒊+16 𝒋 )𝑡=5


(12 𝒊+16 𝒋)=(2 𝒊−4 𝒋 )+(𝒗 )(5)Sub in values

12 𝒊+16 𝒋=2 𝒊−4 𝐣+5 𝒗10 𝒊+20 𝒋=5 𝒗2 𝒊+4 𝒋=𝒗

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

Deal with the brackets!

Subtract 2i and add 4j

Divide by 5

VectorsYou 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 + 7j and is moving at a speed of

15ms-1 in the direction 3i – 4j. 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

-4j3i – 4j

𝑆𝑝𝑒𝑒𝑑=√32+(−4)2

𝑆𝑝𝑒𝑒𝑑=5𝑚𝑠−1

9i

-12j9i – 12j

5ms-1 15ms-1

Multiply all vectors by 3

𝑆𝑝𝑒𝑒𝑑=15𝑚𝑠−1

𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦=(9 𝒊−12 𝒋 )𝑚𝑠−1

Calculate

We can use the vectors as the

velocity


VectorsYou 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 + 7j and is moving at a speed of

15ms-1 in the direction 3i – 4j. 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



𝒓 0=(4 𝒊+7 𝒋)𝒗=(9 𝒊−12 𝒋) 𝑡=2


𝒓=(4 𝒊+7 𝒋)+(9 𝒊−12 𝒋 )(2)

𝒓=4 𝒊+7 𝒋+18 𝒊−24 𝒋𝒓=22 𝒊−17 𝒋

Sub in values

‘Deal with’ the brackets

Group terms




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

𝑠=?𝒖=(−3 𝒊+ 𝒋)𝒗=? 𝒂=(2 𝒊+3 𝒋 )𝑡=3

𝒗=𝒖+𝒂𝑡

𝒗=(−3 𝒊+ 𝒋)+(2 𝒊+3 𝒋)(3)

𝒗=−3 𝒊+ 𝒋+6 𝒊+9 𝒋

𝒗=3 𝒊+10 𝒋

𝑆𝑝𝑒𝑒𝑑=√32+102𝑆𝑝𝑒𝑒𝑑=10.4𝑚𝑠−1

Sub in values


Group terms

Remember this is the velocity, not the speed!

Calculate!




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

𝑠=?𝒖=0𝒗=(10 𝒊−24 𝒋 )𝒂=?𝑡=10

𝒗=𝒖+𝒂𝑡(10 𝒊−24 𝒋 )=0+𝒂(10)

10 𝒊−24 𝒋=10𝒂𝒊−2.4 𝒋=𝒂

Sub in values

‘Tidy up’

Divide by 10

𝑭=𝑚𝒂𝑭=(2)(𝒊−2.4 𝒋)

𝑭=(2 𝒊−4.8 𝒋 ) 𝑁

Sub in values

Calculate

Teachings for Exercise 6G

VectorsYou 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 𝒊+3 𝒋 )+(4 𝒊− 𝒋 )+ (−3 𝒊+2 𝒋 )+ (𝑎𝒊+𝑏 𝒋 )=0

(3 𝒊+𝟒 𝒋 )+(𝑎 𝒊+𝑏 𝒋 )=0

3+𝑎=0𝑎=−3

4+𝑏=0𝑏=−4

Group together the

numerical terms

The ‘i’ terms must sum to

0

The ‘j’ terms must sum to

0



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

𝑅𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 𝐹𝑜𝑟𝑐𝑒=𝑭 1+𝑭 2+𝑭 3

¿ (2 𝒊+4 𝒋 )+(−5 𝒊+4 𝒋 )+(6 𝒊−5 𝒋)

¿ (3 𝒊+3 𝒋 )

𝑭=𝑚𝒂

(3 𝒊+3 𝒋)=3𝒂

𝒊+ 𝒋=𝒂

The acceleration is (i + j) ms-2

Sub in values

Group up

Sub in the resultant force, and the mass

Divide by 3



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

TA TB

7NTA

TB

7N

These are the forces acting on P

These are the forces rearranged as a

triangle

7N



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

TA

TB

7NThe angle between 7N

and TA is 60°

A B

P

30° 40°

7N

50°60°

60°

50°

70°

The angle between 7N and TB 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 the tensions!







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

TA

TB

7N

A B

P

30° 40°

7N

50°60°

60°

50°

70°

𝑇 𝐴

𝑆𝑖𝑛50= 7𝑆𝑖𝑛70

𝑇 𝐴=7

𝑆𝑖𝑛70×𝑆𝑖𝑛50

𝑇 𝐴=5.71𝑁

Multiply by Sin50

Calculate

𝑇 𝐴=5.71𝑁







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

TA

TB

7N

A B

P

30° 40°

7N

50°60°

60°

50°

70°

𝑇 𝐵

𝑆𝑖𝑛60= 7𝑆𝑖𝑛70

𝑇 𝐵=7

𝑆𝑖𝑛70×𝑆𝑖𝑛 60

𝑇 𝐵=6.45𝑁

Multiply by Sin60

Calculate

𝑇 𝐴=5.71𝑁

𝑇 𝐵=6.45𝑁

Teachings for Exercise 6H(the mixed exercise – essenti al!)

VectorsYou 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

6H



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

𝑆𝑝𝑒𝑒𝑑=√(−2.5)2+62

𝑆𝑝𝑒𝑒𝑑=6.5𝑘𝑚h−1

Use Pythagoras’

Theorem

Calculate

6.5 kmh-1

N

θ180°

The bearing on which S is travelling

𝑇𝑎𝑛𝜃=62.5

Find angle θ

𝜃=67.4 °

𝐵𝑒𝑎𝑟𝑖𝑛𝑔=337 °

Use Tan = Opp/Adj

Calculate

Consider the north line and

read clockwise…

337°

67.4°





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°

𝒓 0=(16 𝒊+5 𝒋)


𝒗=(−2.5 𝒊+6 𝒋)𝒕=3

𝒓=(16 𝒊+5 𝒋)+(−2.5 𝒊+6 𝒋)(3)

𝒓=16 𝒊+5 𝒋−7.5 𝒊+18 𝒋𝒓=8.5 𝒊+23 𝒋

Sub in values


Group terms

𝑹=8.5 𝒊+23 𝒋





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

𝒓 0=(16 𝒊+5 𝒋)


𝒗=(−2.5 𝒊+6 𝒋)𝒕=2

𝒓=(16 𝒊+5 𝒋)+(−2.5 𝒊+6 𝒋)(2)

𝒓=16 𝒊+5 𝒋−5 𝒊+12 𝒋𝒓=11𝒊+17 𝒋

Sub in values


Group terms

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

𝑹=8.5 𝒊+23 𝒋








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

𝒓 0=(11𝒊+17 𝒋) 𝒗=5 𝒋 𝑡=𝑡


𝒓=(11 𝒊+17 𝒋)+(5 𝒋)(𝑡)

𝒓=11𝒊+17 𝒋+5 𝑡 𝒋𝒓=11𝒊+(17+5 𝑡) 𝒋

𝑹=8.5 𝒊+23 𝒋Sub in values


Factorise the j terms

𝒓=11𝒊+(17+5 𝑡) 𝒋








Find:



1600 6H

𝑹=8.5 𝒊+23 𝒋

𝒓=11𝒊+(17+5 𝑡) 𝒋

Find the time when S will be due east of R

R S

8.5 𝒊+23 𝒋 11𝒊+(17+5 𝑡) 𝒋If S is due east of R, then their j terms must be equal!

23=17+5 𝑡6=5 𝑡1.2=𝑡

Subtract 17

Divide by 5

1.2 hours = 1 hour 12 minutes

So S will be due east of R at 1512 hours!

1512








Find:



1600 6H

𝑹=8.5 𝒊+23 𝒋

𝒓=11𝒊+(17+5 𝑡) 𝒋

1512

Find the distance of S from R at the time 1600 Find where S is at 1600 hours…

𝒓=11𝒊+(17+5 𝑡) 𝒋

𝒓=11𝒊+(17+5(2)) 𝒋

𝒓=11𝒊+27 𝒋

Sub in t = 2 (1400 – 1600

hours)

Simplify/calculate

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

𝑹=8.5 𝒊+23 𝒋 𝑺=11𝒊+27 𝒋To calculate the vector between them, calculate S - R

(11𝒊+27 𝒋 )−(8.5 𝒊+23 𝒋)¿2.5 𝒊+4 𝒋

𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒=√2.52+42

𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒=4.72𝑘𝑚Calculate

Now use Pythagoras’ Theorem to work out the distance

4.72𝑘𝑚

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 exercise in this chapter

vectors

Education

distance b

displacement of b

b express

b c work

b p o n d qa c b

seen vectors

regular methods vectors

bearing use trigonometry