P1 Chapter 11 :: Vectors
Transcript of P1 Chapter 11 :: Vectors
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Chapter Overview
1:: Add/scale factors and show vectors are parallel.
4:: Solve both geometric problems.
An orienteer leaves position π and walks 15km at a bearing of 120Β° to
the position A. Determine ππ΄.
2:: Calculate magnitude and direction of a vector.
Vectors used to be a Year 2 A Level topic. However, the more difficult content has been moved to Further Maths. This chapter is now more intended as providing the βfundamentalsβ of vectors (most of which you learned at GCSE), supporting Mechanics and FM.
If π = 3π + 4π, determine |π|.
3:: Understand and use position vectors.
A donut has position vector 4π + 3πβ¦
5:: Understand speed vs velocity
If a ship moves at a velocity of 12π + 5π ππ/β, what is its speed?
Vector Basics
Whereas a coordinate represents a position in space, a vectorrepresents a displacement in space.
A
A vector has 2 properties:β’ Directionβ’ Magnitude (i.e. length)
B
If π and π are points then ππ is the vector between them.
π
π
If two vectors ππ and π π have the same magnitude and direction, theyβre the same vector and are parallel.
π
π
π
π
This might seem obvious, but students sometimes think the vector is different because the movement occurred at a different point in space. Nope!
D Triangle Law for vector addition:
π΄π΅ + π΅πΆ = π΄πΆ
The vector of multiple vectors is known as the resultant vector. (you will encounter this term in Mechanics)
ππ
π + ππ΄
π΅
πΆ
C π΄π΅ = βπ΅π΄ and the two vectors are parallel, equal in magnitude but in opposite directions.
π΄
π΅
π΄
π΅π
βπ
βMultiplyingβ two sets (known again as the cross product) finds each possible combination of members, one from each:
π, π, π Γ π, π = { π, π , π, π , {π, π},π, π , π, π , π, π }
It has the nice property that π π΄ Γ π΅ = π π΄ Γ π(π΅), where π π΄ gives the size of the set π΄. Also, π΄ Γ π΅ = π΅ Γ π΄.
Just for your interestβ¦
Have you ever wondered what happens if you βmultiplyβ two vectors or two sets?
ME-WOW!
Ermβ¦
In KS2/3 you probably only experienced variables holding numerical values. You since saw that variables can represent other mathematical types, such as sets or vectors:
π₯ = 3
π =1β2
π = {2,5,9,10}
But when we do we need to define explicitly what operators like β+β and βΓβ mean.
1 + 3 = 43β2
+10
=4β2
32β1
Γ101
=? π, π, π Γ π, π =?
Often these operators are defined to give it properties that are consistent with its usage elsewhere, e.g. βcommutativityβ: π + π = π + π for vector addition just as 2 + 4 = 4 + 2 for numbers.
In FM, you will see that βmultiplyingβ two 3D vectors (known as the cross product) gives you a vector perpendicular to the two.Vector multiplication is not commutative, so π Γ π β π Γ π (however it is βdistributiveβ, so π Γ π + π = π Γ π + π Γ π )The Casio Classwiz can calculate this in Vector mode!
32β1
Γ101
=2β42
π
π
π Γ π
Vector Basics
E Vector subtraction is defined using vector addition and negation:
π β π = π + βπ
π ππ
βπ
π β π
The zero vector π (a bold 0), represents no movement.
ππ + ππ = π
In 2D: π =00
F
G A scalar is a normal number, which can be used to βscaleβ a vector.β’ The direction will be the same.β’ But the magnitude will be different
(unless the scalar is 1).
π
2π
Any vector parallel to the vector π can be written as ππ, where π is a scalar.
H
The implication is that if we can write one vector as a multiple of another, then we can show they are parallel.
βShow 2π + 4π and 3π + 6π are parallelβ.
3π + 6π =3
2π + 2π β΄ parallel
β1
2π
If scalar is negative, opposite direction.
Example
Edexcel GCSE June 2013 1H Q27
ππ = βπ + π
For (b), thereβs two possible paths to get from π to π : via π or via π. But which is best?In (a) we found πΊ to πΈ rather than πΈ to πΊ, so it makes sense to go in this direction so that we can use our result in (a).
a
b ππ =2
5ππ + π
=2
5βπ + π + π
=2
5π +
3
5π
Fro Tip: This ratio wasnβt in the original diagram. I like to add the ratio as a visual aid.
Fro Workings Tip: While youβre welcome to start your working with the second line, I recommend the first line so that your chosen route is clearer.
ππ is also πbecause it is exactly the same
movement as ππ.
Test Your Understanding
π΄π΅ = π΄π + ππ΅= βπ + π
ππ = π +3
4π΄π΅
= π +3
4βπ + π
=1
4π +
3
4π
Edexcel GCSE June 2012
Exercise 11A
Pearson Pure Mathematics Year 1/ASPages 234-235
Representing Vectors
You should already be familiar that the value of a vector is the displacement in the π₯ and π¦ direction (if in 2D).
3β2 π =
3β2
, π =0β1
π + π =3β3
2π =6β4
! A unit vector is a vector of magnitude 1. π and π are unit vectors in the π₯-axis and π¦-axis respectively.
π =10
π =01
e.g. 43
= 410
+ 301
= 4π + 3π
π
π
Fro Side Notes: This allows us to write any vector algebraically without using vector notation. Any point in 2D space, as a vector from the
origin, can be obtained using a linear combination of π and π, e.g. if π 5,β1 , ππ = 5π β π. For this reason, π and π are known as basis
vectors of 2D coordinate space. In fact, any two non-parallel/non-zero vectors can be used as basis vectors, e.g. if π =52
and π =30
, itβs
possible to get any vector π₯π¦ using a linear combination of these, i.e. we can always find scalars π and π such that
π₯π¦ = π
52
+ π30
.
Examples
If π = 3π, π = π + π, π = π β 2π then:1) Write π in vector form.2) Find π + 2π in π, π form.
π =30
π + 2π = π + π + 2 π β 2π= π + π + 2π β 4π= 3π β 3π
1
2
Exercise 11B
Pearson Pure Mathematics Year 1/ASPages 237-238
Magnitude of a Vector
! The magnitude |π| of a vector π is its length.
34
π΄
π΅
π΄π΅ = 32 + 42 = 5
3
4
1β1
= 2 β5β12
= 13
π =4β1
π = 42 + 12 = 17
π =20
π = 22 + 02 = 2
! If π =π₯π¦ π = π₯2 + π¦2
Unit Vectors
! A unit vector is a vector whose magnitude is 1
Thereβs certain operations on vectors that require the vectors to be βunitβ vectors. We just scale the vector so that its magnitude is now 1.
π =34
π = 32 + 42 = 5
ΰ·π =3/54/5
Test Your Understanding: Convert the following vectors to unit vectors.
π =12β5
π = πππ + ππ = ππ
β΄ ΰ·π =
ππ
ππ
βπ
ππ
If π is a vector, then the unit vector ΰ·π in the same direction is
ΰ·π =π
π
π =11
π = ππ + ππ = π
β΄ π =
π
ππ
π
Exercise 11C
Pearson Pure Mathematics Year 1/ASPages 240-242
Position Vectors
Suppose we started at a point 3,2
and translated by the vector 40
:
π΄ 3,2
40
You might think we can do something like:
3,2 +40
= 7,2
But only vectors can be added to other vectors.If we treated the point π, π as a vector, then this solves the problem:
32
+40
=72
?
π΄ 3,2π =
32
π
! The position vector of a
point π΄ is the vector ππ΄,
where π is the origin. ππ΄is usually written as π.
π₯
π¦
A vector used to represent a position is unsurprisingly known as a position vector.A position can be thought of as a translation from the origin, as per above. It enables us to use positions in all sorts of vector (and matrix!) calculations.
Example
The points π΄ and π΅ have coordinates 3,4 and (11,2) respectively.Find, in terms of π and π:a) The position vector of π΄b) The position vector of π΅
c) The vector π΄π΅
ππ΄ = 3π + 4π
ππ΅ = 11π + 2π
π΄π΅ = 8π β 2π
a
b
c
You can see this by inspection of the change in π₯ and the change in π¦:
π΄ 3,4 β π΅(11,2)
More formally:
π΄π΅ = ππ΅ β ππ΄
=112
β34
=8β2
+8 β2
Further Example
ππ΄ = 5π β 2π and π΄π΅ = 3π + 4π. Find:a) The position vector of π΅.
b) The exact value of ππ΅ in simplified surd form.
ππ΅ = ππ΄ + π΄π΅ =5β2
+34
=82
ππ΅ = 82 + 22 = 2 17
a
b
Either a quick sketch will help you see this, or
thinking of ππ΄ as the
original position and π΄π΅as the translation.
Exercise 11D
Pearson Pure Mathematics Year 1/ASPages 243-244
Solving Geometric Problems
ππ΄
πΆ π©
πΏ
π
π¨ πͺ
π is a point on π΄π΅ such that π΄π: ππ΅ = 3: 1. π is the midpoint of π΅πΆ.
Show that ππ is parallel to ππΆ.
ππΆ = π + π
ππ =1
4βπ + π +
1
2π =
1
4π +
1
4π
=1
4π + π
ππ is a multiple of ππΆ β΄ parallel.
The key is to factor out a scalar such that we see the same vector.
For any proof question always find the vectors
involved first, in this case ππ and ππΆ.
The magic words here are βis a multiple ofβ.
Introducing Scalars and Comparing Coefficients
π
π
π
π΅
π΄
πΆ
π
Remember when we had identitieslike: ππ₯2 + 3π₯ β‘ 2π₯2 + ππ₯we could compare coefficients, so that π = 2 and 3 = π.We can do the same with (non-parallel) vectors!
ππ΄πΆπ΅ is a parallelogram, where ππ΄ = π and ππ΅ = π. The diagonals ππΆ and π΄π΅ intersect at a point π. Prove that the diagonals bisect each other.
(Hint: Perhaps find ππ in two different ways?)
By considering the route π β π΅ β π:
ππ = π + π π΅π΄= π + π βπ + π = π β ππ + ππ= ππ + 1 β π π
Similarly, considering the line ππΆ:
ππ = π ππΆ = π π + π = ππ + ππβ΄ ππ + ππ = ππ + 1 β π π
Comparing coefficients of π: π = πComparing coefficients of π: π = 1 β π
β΄ π = π =1
2β΄ π is the midpoint of each of the diagonals.
We donβt know what fraction of the way across π΅π΄the point π is, so let π be the fraction of the way
across. Weβre hoping that π =1
2, so that π is exactly
halfway across and therefore bisects π΅π΄.
We need to use a different scalar constant, this time π.It is common to use the letters π and π for scalars.
Test Your Understanding
π
π
π
π΅
π΄
πΆ
π
π
In the above diagram, ππ΄ = π,ππ΅ = π and ππ =1
3π. We wish to find the ratio ππ: ππΆ.
a) If ππ = π ππΆ, find an expression for ππ in terms of π, π and π.
b) If π΅π = π π΅π, find an expression for ππ in terms of π, π and π.c) By comparing coefficients or otherwise, determine the value of π, and hence the ratio ππ: ππΆ.
ππ = π ππΆ = π π + π = ππ + ππ
ππ = π + π π΅π = π + π βπ +1
3π =
1
3ππ + 1 β π π
Comparing coefficients: π =1
3π and π = 1 β π, β΄ π =
1
4
If ππ =1
4ππΆ, then ππ: ππΆ = 1: 3.
a
b
c
Expand and collect π terms and collect π terms, so that we can compare coefficients later.
Area of a Triangle
π΄π΅ = 3π β 2π and π΄πΆ = π β 5π. Determine β π΅π΄πΆ.
Strategy: Find 3 lengths of triangle then use cosine rule to find angle.
π΄
πΆ
π΅
3β2
1β5
23
π΄π΅ = 13
π΄πΆ = 26πΆπ΅ = 13
A clever student might at this point realise that we can divide all the lengths by 13 without changing β π΅π΄πΆ,
giving a 1: 1: 2 triangle (one of our βspecialβ triangles!), and thus instantly getting β π΅π΄πΆ = 45Β°.But letβs use a more general method of using the cosine rule:
π = 13, π = 13, π = 26
13 = 13 + 26 β 2 Γ 13 Γ 26 Γ cos π΄
cos π΄ =1
2β π΄ = 45Β°
Exercise 11E
Pearson Pure Mathematics Year 1/ASPages 246-247
[STEP 2010 Q7]Relative to a fixed origin π, the points π΄ and π΅ have position vectors π and π, respectively. (The points π, π΄ and π΅ are not collinear.) The point πΆ has position vector π given by
π = πΌπ + π½π,where πΌ and π½ are positive constants with πΌ + π½ < 1. The lines ππ΄and π΅πΆ meet at the point π with position vector π and the lines ππ΅and π΄πΆ meet at the point π with position vector π. Show that
π =πΌπ
1 β π½and write down π in terms of πΌ, π½ and π.Show further that the point π with position vector π given by
π =πΌπ + π½π
πΌ + π½,
lies on the lines ππΆ and π΄π΅.
The lines ππ΅ and ππ intersect at the point π. Prove that ππ
π΅π=
ππ
π΅π.
Extension
1
Click here for the solution:http://www.mathshelper.co.uk/STEP%202010%20Solutions.pdf(go to Q7)
Modelling
Velocitye.g.
34
ππ/β
In Mechanics, you will see certain things can be represented as a simple number (without direction), or as a vector (with direction):
4
3
Vector Quantity Equivalent Scalar Quantity
Speed= 5 ππ/β
5
Remember a βscalarβ just means a normal number (in the context of vectors). It can be obtained using the magnitude of the vector.
This means the position vector of the object changes
by 34
each hour.
β¦which is equivalent to moving 5km each hour.
Displacemente.g.
β512
ππ
Distance= 13 ππ
Example
[Textbook] A girl walks 2 km due east from a fixed point π to π΄, and then 3 km due south from π΄ to π΅. Finda) the total distance travelledb) the position vector of π΅ relative to π
c) ππ΅
d) The bearing of π΅ from π.
2
3
N
ππ΄
π΅
π
2 + 3 = 5ππ
ππ΅ =23
ππ
ππ΅ = 22 + 32 = 13 = 3.61ππ (3sf)
90Β° + tanβ13
2= 56.3Β°
a
b
c
d
Further Example
[Textbook] In an orienteering exercise, a cadet leaves the starting point π and walks 15 km on a bearing of 120Β° to reach π΄, the first checkpoint. From π΄ he walks 9 km on a bearing of 240Β° to the second checkpoint, at π΅. From π΅ he returns directly to π.Find:a) the position vector of π΄ relative to π
b) ππ΅
c) the bearing of π΅ from πd) the position vector of π΅ relative π.
9 ππ
15 ππ
N
π
π΄
π΅
π
N
240Β°
a 120Β°
30Β°
ππ΄ =15 cos 30Β°15 sin 30Β°
=13.07.5
ππ
b
9 ππ
15 ππ
N
π΄π
N
240Β°
30Β°60Β°
60Β°
π΅
π
Using cosine rule on Ξππ΄π΅,
ππ΅ = 13.1ππ (3sf)
c Using sine rule on Ξππ΄π΅:π = 36.6β¦ Β°
β΄ Bearing = 120 + 36.6Β° = 157Β° (3sf)
I have no specific advice to offer except:1. Draw a BIG diagram.2. Remember bearings are measured
clockwise from North.3. Donβt forget units (on vectors!)
Further Example
[Textbook] In an orienteering exercise, a cadet leaves the starting point π and walks 15 km on a bearing of 120Β° to reach π΄, the first checkpoint. From π΄ he walks 9 km on a bearing of 240Β° to the second checkpoint, at π΅. From π΅ he returns directly to π.Find:a) the position vector of π΄ relative to π
b) ππ΅
c) the bearing of π΅ from πd) the position vector of π΅ relative π.
d
9 ππ
N
π
π΄
π΅
ππ. ππΒ°
N
120Β°
30Β°
23.41Β°
ππ΅ =13.07 sin 23.41Β°β13.07 cos 23.41Β°
=5.19β12.0
ππ
(I think the textbook made a rounding error due to use of
rounded values. They put 5.1
β12.1)
Exercise 11F
Pearson Pure Mathematics Year 1/ASPages 250-251