vectors
DESCRIPTION
Vectors. Introduction. This chapter focuses on vectors Vectors are used to describe movement in a given direction They are also used to describe straight lines in 3D (in a similar way to y = mx + c being used for 2D straight line graphs). Teachings for Exercise 5A. Vectors. Q. - PowerPoint PPT PresentationTRANSCRIPT
Vectors
Introduction• This chapter focuses on vectors
• Vectors are used to describe movement in a given direction
• They are also used to describe straight lines in 3D (in a similar way to y = mx + c being used for 2D straight line graphs)
Teachings for Exercise 5A
VectorsYou need to know the difference between a scalar and a vector, and how to write down vectors
and draw vector diagrams
A scalar quantity has only a magnitude (size)
A vector quantity has both a magnitude and a direction
5A
Scalar VectorThe distance from P to Q is
100m P
Q
From P to Q you go 100m
north
N
60°
ScalarA ship is sailing
at 12km/h
VectorA ship is sailing at 12km/h on a bearing of 060°Direction and
Magnitude
VectorsYou need to know the difference between a scalar and a vector, and how to write down vectors
and draw vector diagrams
Equal vectors have the same magnitude and direction
5A
P
Q
R
S
PQ = RSA Common way of showing vectors is using the letters
with an arrow above
a
b
Alternatively, single letters can be used…
VectorsYou need to know the difference between a scalar and a vector, and how to write down vectors
and draw vector diagrams
Two vectors can be added using the ‘Triangle Law’
5A
a b
a + b
It is important to note that vector a + b is the single line from the start of a to
the end of b.
Vector a + b is NOT the two separate lines!
VectorsYou need to know the difference
between a scalar and a vector, and how to write down vectors and
draw vector diagrams
Draw a diagram to show the vector a + b + c
5A
a b
a + b + c
c
a b
c
VectorsYou need to know the difference between a scalar and a vector, and how to write down vectors
and draw vector diagrams
Adding the vectors PQ and QP gives a Vector result of 0.
Vectors of the same size but in opposite directions have opposite signs (eg) + or -
5A
P
Q
P
Q
a
-a
VectorsYou need to know the difference between a scalar and a vector, and how to write down vectors
and draw vector diagrams
The modulus value of a vector is another name for its magnitude
Eg) The modulus of the Vector a is |a|
The modulus of the vector PQ is |PQ|
Question: The vector a is directed due east and |a| = 12. Vector b is directed due south and |b| = 5. Find |a + b|
5A
a
ba + b
|𝑎+𝑏|2=122+52
|𝑎+𝑏|2=169
|𝑎+𝑏|=13
Use Pythagoras’ Theorem
Square the shorter sides…Square
Root
VectorsYou need to know the difference between a scalar and a vector, and how to write down vectors
and draw vector diagrams
In the diagram opposite, find the following vectors in terms of a, b, c and d.
a) PS
b) RP
c) PT
d) TS
5A
P
Q
R
T
S
abc
d
= -a+ c Or c - a
= -b+ a Or a - b
= -a+ b+ d Or b + d - a
= -d- b+ c Or c - b - d
Teachings for Exercise 5B
VectorsYou need to be able to perform simple vector
arithmetic, and know the definition of a unit vector
The diagram shows the vector a. Draw diagrams to show the
vectors 3a and -2a
Vector 3a will be in the same direction as a, but 3 times the size
Vector -2a will be twice as big as s, but also in the opposite
direction
5B
a
3a
-2a
VectorsYou need to be able to perform simple vector
arithmetic, and know the definition of a unit vector
Any vector parallel to a may be written as λa, where λ (lamda) is a non-zero scalar (ie - represents a
number…)
Show that the vectors 6a + 8b and 9a + 12b are parallel…
5B
6𝒂+8𝒃 9𝒂+12𝒃32
(6𝒂+8𝒃 )Factorise
The second Vector is a multiple of the first, so they are parallel.
In this case, λ is 3/2 or 1.5
VectorsYou need to be able to perform simple vector
arithmetic, and know the definition of a unit vector
A unit vector is a vector which has a magnitude of 1 unit
Vector a has a magnitude of 20 units. Write down a unit vector
that is parallel to a.
5B
The unit vector will be:120 𝒂
This will be in the same direction as a with a magnitude of 1 unit
As a general rule, divide any vector by its magnitude to obtain
a parallel unit vector
¿𝑎|𝑎|
VectorsYou need to be able to perform simple vector
arithmetic, and know the definition of a unit vector
If:
And the vectors a and b are not parallel and non-zero, then:
and
Effectively, if the two vectors are equal then the coefficients of a
and b must also be equal
5B
λ 𝒂+𝜇𝒃=𝛼 𝒂+𝛽𝒃
λ=𝛼 𝜇=𝛽
Given that:5𝒂−4𝒃=(2𝑠+𝑡 )𝒂+(𝑠−𝑡 )𝒃Find the values of the scalars s and t
Comparing coefficients:
2𝑠+𝑡=5𝑠−𝑡=−4
1)2)
3 𝑠=1𝑠=13
𝑡=4 13
Add the equations together
Divide by 3
Sub into either of 1) or 2) to find the value of t
VectorsYou need to be able to perform
simple vector arithmetic, and know the definition of a unit vector
In the diagram opposite, PQ = 3a, QR = b, SR = 4a and PX = kPR. Find in terms of a, b and k:
a) PS
b) PX
c) SQ
d) SX
5B
P Q
RS
X
3a
4a
bb-a
= 3a + b – 4a
= b – a
= kPR
= k(3a + b)
= 4a - b
= -b + a
k(3a+b)
+ k(3a + b)= -b + a + 3ka +
kb= (3k + 1)a + (k – 1)b
Multiply out the bracketGroup up and factorise the ‘a’ and ‘b’ terms
separately
VectorsYou need to be able to perform simple vector arithmetic, and know the definition of a unit
vector
5B
P Q
RS
X
3a
4a
b
e) Use the fact that X lies on SQ to find the value of kSQ = 4a - b SX = (3k + 1)a + (k – 1)b
Since X is on SQ, SX and SQ are parallel, ie) one is a multiple of another!
(3𝑘+1 )𝒂+(𝑘−1 )𝒃¿ λ (4𝒂−𝒃) Use the lamda symbol to represent
one being a multiple of the
other…(3𝑘+1 )𝒂+(𝑘−1 )𝒃¿ 4 λ𝒂− λ𝒃3𝑘+1=4 λ𝑘−1=− λ
3𝑘+1=4 λ4𝑘−4=−4 λ
1)2) x4
7𝑘−3=0𝑘=
37
Add together
Solve for k
Multiply out the bracket
Teachings for Exercise 5C
VectorsYou need to be able to use
vectors to describe the position of a point in 2 or 3 dimensions
The position vector of a point A is the vector OA, where O is the
origin. OA is often written as a.
AB = b – a, where a and b are the position vectors of A and B
respectively.
5C
O
A
a
aB
A
b
b - a
O
VectorsYou need to be able to use
vectors to describe the position of a point in 2 or 3 dimensions
In the diagram, points A and B have position vectors a and b
respectively. The point P divides AB in the ratio 1:2.
Find the position vector of P.
5C
A
B
P
O
a
b
1
2b - a
𝐴𝐵=𝒃−𝒂 Using the rule we just saw…
1/3(b – a)2/3(b – a)
If the line is split in the ratio 1:2, then one part is 1/3 and the other is 2/3
𝑂𝑃=𝒂+13 (𝒃−𝒂)
𝑂𝑃=23 𝒂+
13 𝒃 The position vector of P is how we get from O to P
Teachings for Exercise 5D
VectorsYou need to know how to write
down and use the Cartesian components of a vector in 2
dimensions
The vectors i and j are unit vectors parallel to the x and y axes, in the
increasing directions
The points A and B in the diagram have coordinates (3,4) and (11,2)
respectively. Find in terms of i and j:
a) OA
b) OB
c) AB
5D
5 10 1500
5
10
A
Ba
b¿3 𝒊+4 𝒋¿11𝒊+2 𝒋¿𝒃−𝒂¿ (11𝒊+2 𝒋 )−(3 𝒊+4 𝒋 )¿8 𝒊−2 𝒋
VectorsYou need to know how to write
down and use the Cartesian components of a vector in 2
dimensions
You can write a vector with Cartesian components as a column matrix:
Column matrix notation can be easier to read and avoids the need to write
out lots of i and j terms.
5D
𝑥 𝒊+𝑦 𝒋=(𝑥𝑦 )
Given that:a = 2i + 5jb = 12i – 10jc = -3i + 9j
Find a + b + c
𝒂+𝒃+𝒄=¿(25)+( 12−10)+(−39 )𝒂+𝒃+𝒄=¿(114 )
Be careful
with negatives
!
¿11𝒊+4 𝒋
VectorsYou need to know how to write
down and use the Cartesian components of a vector in 2
dimensions
The modulus (magnitude) of xi + yj is:
This comes from Pythagoras’ Theorem
5D
√𝑥2+ 𝑦2
xi
yjxi + yj
The vector a is equal to 5i - 12j. Find |a| and find a unit vector in the same direction as a.
5i – 12j
5i
12j
|𝒂|=√52+ (−12 )2
|𝒂|=√169|𝒂|=13
¿𝒂
|𝒂|𝑈𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 ¿5 𝒊−12 𝒋13
¿113 (5 𝒊−12 𝒋 )
¿ 113 ( 5−12)
Alternative notation…
VectorsYou need to know how to write
down and use the Cartesian components of a vector in 2
dimensions
The modulus (magnitude) of xi + yj is:
5D
√𝑥2+ 𝑦2
Given that a = 5i + j and b = -2i – 4j, find the exact value of |2a + b|
2𝒂+𝒃=¿2(51)+(−2−4)¿ (102 )+(−2−4)¿ ( 8−2)
¿2𝒂+𝒃∨¿√82+(−2)2
¿√6 8¿2√17
Use x = 8 and y = -
2
‘Exact’ means you can leave in surd form
Teachings for Exercise 5E
VectorsYou need to know how to use
Cartesian coordinates in 3 dimensions
Cartesian coordinates in three dimensions are usually referred to as
the x, y and z axes, each at right-angles to the other.
Coordinates in 3 dimensions are given in the form (x, y, z)
5Ex
y
z
Imagine the x and y-axes have fallen down
flat, and the z-axis sticks up vertically out
of the origin…
Find the distance from the origin to the point P(4, 2, 5)
y
z
x4
5
2
You can use the 3D version of Pythagoras’ Theorem
The distance from the origin to the point (x, y, z) is given by:√𝑥2+𝑦2+𝑧 2
¿√ 42+22+52¿6.71(2dp)
VectorsYou need to know how to use
Cartesian coordinates in 3 dimensions
Cartesian coordinates in three dimensions are usually referred to as
the x, y and z axes, each at right-angles to the other.
Coordinates in 3 dimensions are given in the form (x, y, z)
5Ex
y
z
Find the distance between the points A(1, 3, 4) and B(8, 6, -5)
𝐴𝐵=¿𝒃−𝒂¿ ( 86−5)−(134)¿ ( 73−9)
¿ 𝐴𝐵∨¿√72+32+(−9)2
First calculate the vector from A to B
Then use 3D Pythagoras
¿√139¿11.8(1dp)
VectorsYou need to know how to use
Cartesian coordinates in 3 dimensions
Cartesian coordinates in three dimensions are usually referred to as
the x, y and z axes, each at right-angles to the other.
Coordinates in 3 dimensions are given in the form (x, y, z)
5Ex
y
z
The coordinates of A and B are (5, 0, 3) and (4, 2, k) respectively. Given that |AB| is 3 units, find the possible
values of k
𝐴𝐵=(42𝑘)−(503)𝐴𝐵=( −12𝑘−3)|𝐴𝐵|=√(−1)2+22+(𝑘−3)2
|𝐴𝐵|=√𝑘2−6𝑘+143=√𝑘2−6𝑘+149=𝑘2−6𝑘+140=𝑘2−6𝑘+50=(𝑘−5)(𝑘−1)𝑘=5 𝑜𝑟 𝑘=1
Calculate AB using k
Use Pythagoras in 3D
Careful when squaring the bracket
|AB| = 3
Square both sides
Solve as a quadratic
Teachings for Exercise 5F
VectorsYou can extend the two
dimensional vector results to 3 dimensions, using k as the unit
vector parallel to the z-axis
The vectors i, j and k are unit vectors parallel to the x, y and z-axes in the increasing directions
The vector xi + yj + zk can be written as a column matrix:
The modulus (magnitude) of xi + yj + zk is given by:
5F
(𝑥𝑦𝑧 )
√𝑥2+𝑦2+𝑧 2
The points A and B have position vectors 4i + 2j + 7k and 3i + 4j – k respectively. Find |AB| and show
that triangle OAB is isosceles.
¿ 𝐴𝐵∨¿ √(−1)2+22+(−8)2
𝐴𝐵=( 34−1)−(427 )𝐴𝐵=(−12−8)
𝐴𝐵=𝒃−𝒂
¿ 𝐴𝐵∨¿ √69
¿𝑂𝐴∨¿√ 42+22+72¿𝑂𝐴∨¿√69¿𝑂𝐵∨¿ √32+42+(−1)2
¿𝑂𝐵∨¿ √26
Find the vector AB
Now find the magnitude of
AB
Find the magnitude of OA and OB using
their position vectors
Isosceles as 2 vectors are equal…
VectorsYou can extend the two
dimensional vector results to 3 dimensions, using k as the unit
vector parallel to the z-axis
The points A and B have the coordinates (t, 5, t-1) and (2t, t, 3)
respectively.
a) Find |AB|
b) By differentiating |AB|2, find the value of t for which |AB| is a minimum
c) Hence, find the minimum value of |AB|
5F
a) Find |AB|
𝐴𝐵=𝒃−𝒂𝐴𝐵=(2 𝑡𝑡3 )−( 𝑡
5𝑡−1)
𝐴𝐵=( 𝑡𝑡−54−𝑡)
|
|
|
Careful with the bracket
expansion!
Calculate the vector AB
Find the magnitude of AB in terms of t
VectorsYou can extend the two
dimensional vector results to 3 dimensions, using k as the unit
vector parallel to the z-axis
The points A and B have the coordinates (t, 5, t-1) and (2t, t, 3)
respectively.
a) Find |AB|
b) By differentiating |AB|2, find the value of t for which |AB| is a minimum
c) Hence, find the minimum value of |AB|
5F
b) By differentiating |AB|2, find the value of t for which |AB| is a minimum
|
|
|
𝑑𝑝𝑑𝑡 =6 𝑡−18
0=6 𝑡−183=𝑡
Square both sides
Differentiate (often p is used to represent the
vector)Set equal to 0 for a
minimumSolve
It is possible to do this by differentiating |AB| rather than |AB|2, but it can be more difficult!
The value of t = 3 is the value for which the distance between the points A and B is the
smallest..
VectorsYou can extend the two
dimensional vector results to 3 dimensions, using k as the unit
vector parallel to the z-axis
The points A and B have the coordinates (t, 5, t-1) and (2t, t, 3)
respectively.
a) Find |AB|
b) By differentiating |AB|2, find the value of t for which |AB| is a minimum
c) Hence, find the minimum value of |AB|
5F
c) Hence, find the minimum value of |AB|
|
𝑡=3
|
|
|
| (2dp)
Sub in the value of t
So for the given coordinates, the closest that points A and B could be is 3.74 units apart, when
t = 3.
Teachings for Exercise 5G
VectorsYou need to know the definition of
the scalar product of two vectors in 2 or 3 dimensions, and how it can be used to calculate the angle between
2 vectors
On the diagram to the right, the angle between a and b is θ. The two vectors must be directed away from point X
On the second diagram, vector b is directed towards X. Hence, the angle
between the two vectors is 160°.
This comes from re-drawing the diagram with vector b pointing away from point X.
5G
X
a
b
30°
X
a
b20°
X
a
b
20°160°
b
VectorsYou need to know the definition of
the scalar product of two vectors in 2 or 3 dimensions, and how it can
be used to calculate the angle between 2 vectors
The scalar product of two vectors a and b is written as a.b (‘a dot b’) and is
defined by:
The scalar product can be thought of as ‘the effect of one of the two vectors on
the other’
5G
𝒂 .𝒃=|𝒂|∨𝒃∨𝑐𝑜𝑠 𝜃
Vector multiplication
a.b = |a||b| By multiplication
a b
ab
In this case, the vector a can be split into a
horizontal and vertical component
Here we only consider the horizontal component
as this is in the direction of vector b
|a|cosθθ
By GCSE trigonometry
a.b = |a|cosθ|b|
a.b = |a||b|cosθ
This is the formula for the scalar ‘dot’
product of 2 vectors
VectorsYou need to know the definition of
the scalar product of two vectors in 2 or 3 dimensions, and how it can
be used to calculate the angle between 2 vectors
The scalar product of two vectors a and b is written as a.b (‘a dot b’) and is
defined by:
This formula can be rewritten in order to find the angle between 2 vectors:
5G
𝒂 .𝒃=|𝒂|∨𝒃∨𝑐𝑜𝑠 𝜃
𝑐𝑜𝑠𝜃=𝒂 .𝒃
|𝒂|∨𝒃∨¿¿
ab
If two vectors are perpendicular, then the angle between them is 90°. As cos90° = 0, this will cause the dot
product to be 0 as well
Hence, if vectors are perpendicular, the dot product is 0
If the dot product is 0, the vectors are perpendicular
VectorsYou need to know the definition of the scalar product of two vectors in 2 or 3 dimensions, and how it can be used to calculate the angle between 2 vectors
The scalar product of two vectors a and b is written as a.b (‘a dot b’) and is defined by:
This formula can be rewritten in order to find the angle between 2 vectors:
If we are to use this formula to work out the angle between 2 vectors, we therefore need
an alternative way to calculate the scalar product…
5G
𝒂 .𝒃=|𝒂|∨𝒃∨𝑐𝑜𝑠 𝜃
𝑐𝑜𝑠𝜃=𝒂 .𝒃
|𝒂|∨𝒃∨¿¿
If a = x1i + y1j + z1kand b = x2i + y2j + z2k
Then:
𝒂 .𝒃=(𝑥1𝑦1𝑧1) .(𝑥2𝑦2𝑧2)¿ 𝑥1𝑥2+ 𝑦1 𝑦2+𝑧1 𝑧2
This is a way to find the dot product from 2 vectors
VectorsYou need to know the definition of
the scalar product of two vectors in 2 or 3 dimensions, and how it can
be used to calculate the angle between 2 vectors
The scalar product of two vectors a and b is written as a.b (‘a dot b’) and is
defined by:
This formula can be rewritten in order to find the angle between 2 vectors:
5G
𝒂 .𝒃=|𝒂|∨𝒃∨𝑐𝑜𝑠 𝜃
𝑐𝑜𝑠𝜃=𝒂 .𝒃
|𝒂|∨𝒃∨¿¿
If a = x1i + y1j + z1kand b = x2i + y2j + z2kThen:
𝒂 .𝒃=(𝑥1𝑦1𝑧1) .(𝑥2𝑦2𝑧2)¿ 𝑥1𝑥2+ 𝑦1 𝑦2+𝑧1 𝑧2
Given that a = 8i – 5j – 4k and b = 5i + 4j – k:a) Find a.b
𝒂 .𝒃=(𝑥1𝑦1𝑧1) .(𝑥2𝑦2𝑧 2)
𝒂 .𝒃=( 8−5−4 ) .(54−1)
𝒂 .𝒃=(8×5 )+ (−5×4 )+(−4×−1)𝒂 .𝒃=24
Use the dot product formula
VectorsYou need to know the definition of
the scalar product of two vectors in 2 or 3 dimensions, and how it can
be used to calculate the angle between 2 vectors
The scalar product of two vectors a and b is written as a.b (‘a dot b’) and is
defined by:
This formula can be rewritten in order to find the angle between 2 vectors:
5G
𝒂 .𝒃=|𝒂|∨𝒃∨𝑐𝑜𝑠 𝜃
𝑐𝑜𝑠𝜃=𝒂 .𝒃
|𝒂|∨𝒃∨¿¿
If a = x1i + y1j + z1kand b = x2i + y2j + z2kThen:
𝒂 .𝒃=(𝑥1𝑦1𝑧1) .(𝑥2𝑦2𝑧2)¿ 𝑥1𝑥2+ 𝑦1 𝑦2+𝑧1 𝑧2
Given that a = 8i – 5j – 4k and b = 5i + 4j – k:a) Find a.b 𝒂 .𝒃=24
b) Calculate the angle between vectors a and b
𝑐𝑜𝑠𝜃=𝒂 .𝒃
|𝒂|∨𝒃∨¿¿
|a |b
|a |b
𝑐𝑜𝑠𝜃=𝒂 .𝒃
|𝒂|∨𝒃∨¿¿
𝑐𝑜𝑠𝜃=24
√105√42
𝜃=68.8 °
Use the angle formula – you will need to calculate the
magnitude of each vector as well…
Sub in the values
Solve, remembering to use inverse Cos
VectorsYou need to know the definition of
the scalar product of two vectors in 2 or 3 dimensions, and how it can
be used to calculate the angle between 2 vectors
The scalar product of two vectors a and b is written as a.b (‘a dot b’) and is
defined by:
This formula can be rewritten in order to find the angle between 2 vectors:
5G
𝒂 .𝒃=|𝒂|∨𝒃∨𝑐𝑜𝑠 𝜃
𝑐𝑜𝑠𝜃=𝒂 .𝒃
|𝒂|∨𝒃∨¿¿
If a = x1i + y1j + z1kand b = x2i + y2j + z2kThen:
𝒂 .𝒃=(𝑥1𝑦1𝑧1) .(𝑥2𝑦2𝑧2)¿ 𝑥1𝑥2+ 𝑦1 𝑦2+𝑧1 𝑧2
Given that the vectors a = 2i – 6j + k and b = 5i + 2j + λk are perpendicular, calculate the value of λ.
𝒂 .𝒃=(𝑥1𝑦1𝑧1) .(𝑥2𝑦2𝑧 2)
𝒂 .𝒃=( 2−61 ) .(52λ)𝒂 .𝒃=(2×5 )+(−6×2 )+(1×λ)𝒂 .𝒃=−2+ λ0=−2+λλ=2
Calculate the dot product in terms of λ
As the vectors are perpendicular, the dot product
must be 0Solve
Only this value of λ will cause these vectors to be perpendicular…
VectorsYou need to know the definition of
the scalar product of two vectors in 2 or 3 dimensions, and how it can
be used to calculate the angle between 2 vectors
5G
Given that a = -2i + 5j - 4k and b = 4i - 8j + 5k, find a vector which is
perpendicular to both a and b
Let the required vector be xi + yj + zk
𝒂 .(𝑥𝑦𝑧)=0 𝒃 . (𝑥𝑦𝑧 )=0andThe dot products of
both a and b with the required vector will be
0
(−25−4 ). (𝑥𝑦𝑧 )=0 ( 4−85 ) .(𝑥𝑦𝑧 )=0
−2 𝑥+5 𝑦−4 𝑧=0 4 𝑥−8 𝑦+5 𝑧=0−2 𝑥+5 𝑦=4 4 𝑥−8 𝑦=−5
Let z = 1
Let z = 1
−4 𝑥+10 𝑦=84 𝑥−8 𝑦=−5
2 𝑦=3𝑦=
32 𝑥=
74
Now solve as simultaneous
equations
So a possible answer would be:
x2
74 𝒊+
32 𝒋+𝒌 7 𝒊+6 𝒋+4𝒌
x4
Choosing a different value for z will lead to a vector that is a different size, but which is still pointing in the same direction
(ie – perpendicular)
However, this will not work if you choose z = 0
VectorsYou need to know the definition of
the scalar product of two vectors in 2 or 3 dimensions, and how it can
be used to calculate the angle between 2 vectors
5G
Given that a = -2i + 5j - 4k and b = 4i - 8j + 5k, find a vector which is
perpendicular to both a and b7 𝒊+6 𝒋+4𝒌
The 3D axes show the 3 vectors in question. The green vector is perpendicular to both the others,
but you can only see this clearly when it is rotated!
VectorsYou need to know the definition of
the scalar product of two vectors in 2 or 3 dimensions, and how it can
be used to calculate the angle between 2 vectors
5G
On this example, the second picture shows the diagram being viewed from the top of the
red vector
VectorsYou need to know the definition of
the scalar product of two vectors in 2 or 3 dimensions, and how it can
be used to calculate the angle between 2 vectors
5G
On this example, the second picture shows the diagram being viewed from the top of the
red vector The vectors do not need to be touching – it
is always possible to find a vector that is perpendicular to 2 others!
Teachings for Exercise 5H
VectorsYou need to be able to write
the equation of a straight line in vector form (effectively the
equation of a 3D line!)
Let us first consider how this is done in 2 dimensions
So any linear 2D graph needs a direction, and a point on the line
With just the direction, the line wouldn’t have a specific path and
could effectively be anywhere
With only a given point, the line would not have a specific direction
5H
y
x
𝑦=𝑚𝑥+𝑐
m is the gradient of the line This can also be thought of as
the DIRECTION the line goes
c is the y-intercept This is a given point
on the line
VectorsYou need to be able to write the equation of a straight
line in vector form (effectively the equation of a
3D line!)
In 3D, we effectively need the same bits of information
We need any point on the line (ie – a coordinate in the form
(x, y, z))
We also need to know the direction the line is travelling (a vector with terms i, j and
k)
5H
A vector equation of a straight line passing through the point A with position vector a (effectively the
coordinate), and parallel to the vector b, is:
𝒓=𝒂+𝑡𝒃where t is a scalar
parameter
VectorsYou need to be able to write the equation of a straight
line in vector form (effectively the equation of a
3D line!)
5H
A vector equation of a straight line passing through the point A with position vector a (effectively the coordinate), and parallel to the
vector b, is:
𝒓=𝒂+𝑡𝒃where t is a scalar
parameter
Find a vector equation of the straight line which passes through a, with position vector 3i – 5j + 4k,
and is parallel to the vector 7i – 3k
𝒂=( 3−54 ) 𝒃=( 70−3)This is the position vector we will use
This is the direction vector we will use
𝒓=𝒂+𝑡𝒃
𝒓=¿( 3−54 )+𝑡 ( 70−3) This is the vector equation of the
lineThe value t remains unspecified at this point, it can be used later to calculate points on the vector itself,
by substituting in different values for t
VectorsYou need to be able to write the equation of a straight
line in vector form (effectively the equation of a
3D line!)
5H
A vector equation of a straight line passing through the point A with position vector a (effectively the coordinate), and parallel to the
vector b, is:
𝒓=𝒂+𝑡𝒃where t is a scalar
parameter
𝒓=¿( 3−54 )+𝑡 ( 70−3)Some alternative
forms𝒓=3 𝒊−5 𝒋+4𝒌+𝑡 (7 𝒊−3𝒌)
𝒓=(3 𝒊+7 𝑡 ) 𝒊+(−5) 𝒋+(4−3 𝑡 )𝒌
𝒓=( 3+7 𝑡−54−3 𝑡)
(By writing in a different form)
(By multiplying out the brackets and then re-grouping i, j and k terms)
(By rewriting again in the original column vector form)
VectorsYou need to be able to write the equation of a straight
line in vector form (effectively the equation of a
3D line!)
5H
A vector equation of a straight line passing through the points A and B, with position vectors a and b respectively, is
given by:𝒓=𝒂+𝑡 (𝒃−𝒂)where t is a scalar parameter
As you aren’t given the direction vector in this type, you have to work it out by
calculating the vector AB (b – a)
Find a vector equation of the straight line passing through the points A and B, with coordinates (4, 5, -1)
and (6, 3, 2) respectively.
𝒂=( 45−1) 𝒃=(632)𝒃−𝒂=(632)−( 45−1)𝒃−𝒂=( 2−23 )
Calculating b – a will give you the vector AB, ie) the direction vector that passes through A
and B
𝒓=𝒂+𝑡 (𝒃−𝒂)
𝒓=( 45−1)+𝑡(2−23 )
Then use (b – a) along with either of the 2 coordinates/position vectors you’re given
Working in 2D – the equation of the line can be calculated by using either:a) The gradient (direction) and a
coordinate (like we just did)b) Two coordinates (since you can
calculate the gradient between them)
3D can also be done either way…
VectorsYou need to be able to write the equation of a straight
line in vector form (effectively the equation of a
3D line!)
5H
A vector equation of a straight line passing through the points A and B, with position vectors a and b respectively, is
given by:𝒓=𝒂+𝑡 (𝒃−𝒂)where t is a scalar parameter
As you aren’t given the direction vector in this type, you have to work it out by
calculating the vector AB (b – a)
The straight line l has a vector equation: r = (3i + 2j – 5k) + t(i – 6j – 2k)
Given that the point (a, b, 0) lies on l, calculate the values of a and b
𝑟=( 32−5)+𝑡 (1−6−2)
3+𝑡=𝑎2−6 𝑡=𝑏−5−2𝑡=0
The top numbers give the x coordinate, the
middles give the y, and the bottom gives the z,
all for an unknown value of t (at this point)
We can use the bottom equation to find the value of t
−5−2𝑡=0𝑡=−2.5
3+𝑡=𝑎 2−6 𝑡=𝑏3+(−2.5)=𝑎0.5=𝑎
2−6 (−2.5)=𝑏17=𝑏
The coordinate itself is (0.5, 17, 0)
VectorsYou need to be able to write the equation of a straight
line in vector form (effectively the equation of a
3D line!)
5H
A vector equation of a straight line passing through the points A and B, with position vectors a and b respectively, is
given by:𝒓=𝒂+𝑡 (𝒃−𝒂)where t is a scalar parameter
As you aren’t given the direction vector in this type, you have to work it out by
calculating the vector AB (b – a)
The straight line l has vector equation:r = (2i + 5j – 3k) + t(6i – 2j + 4k)
Show that an alternative vector equation of l is:r = (8i + 3j + k) + t(3i – j + 2k)
𝑟=( 25−3)+𝑡 (6−24 ) 𝑟=(831)+𝑡 (
3−12 )
If you look at the direction vectors, one is a multiple of the other
This means they are parallel and hence it does not matter which you use…
𝑟=( 25−3)+𝑡 (3−12 )
Original vector
updated with a different ‘b’
part
𝑟=( 25−3)+( 6−24 )If t = 2
𝑟=(831) So a coordinate on the line is (8, 3, 1)
1) Rewrite the original straight line equation with a different direction
vector2) Then try to find a value for t that will
give you the given coordinate as an answer
This shows that the given coordinate is on the line and hence, can be used in
the vector equation𝑟=(831)+𝑡 (
3−12 )
Teachings for Exercise 5I
VectorsYou need to be able to determine whether two given straight lines
intersect
Up until now we have used t as the scalar parameter
If we have more than one vector equation, then s is usually used to the other
Eg)
Sometimes the Greek letters λ and μ are used as well.
It is important to note that in 3 dimensions, 2 straight lines may pass each other without
intersecting!5I
𝒓=(5 𝒊+2 𝒋−3𝒌 )+𝑡 (2 𝒊−3 𝒋+𝒌)𝒓=(4 𝒊−5 𝒋+2𝒌 )+𝑠 (𝒊− 𝒋+6𝒌)
𝒓=(2 𝒊− 𝒋+2𝒌 )+λ (𝟒 𝒊−2 𝒋−2𝒌)
𝒓=(3 𝒊−5 𝒋+4𝒌 )+𝜇 (3 𝒊−3 𝒋+2𝒌)
VectorsYou need to be able to determine whether two given straight lines
intersect
It is important to note that in 3 dimensions, 2 straight lines may pass
each other without intersecting!
5I
1a) Show that the lines with vector equations:
r = (3i + 8j – 2k) + t(2i – j + 3k)
and r = (7i + 4j + 3k) + s(2i + j + 4k)
intersect.
𝑟=( 38−2)+𝑡 (2−13 )
𝑟=(743 )+𝑠(214)
3+2𝑡8− 𝑡−2+3 𝑡
7+2𝑠4+𝑠3+4 𝑠
Find the x, y and z coordinates in terms
of t and s
If there is a point of intersection, then at this point the equations for the x, y and z coordinates in terms of t and s
will be equal… Solve 2 of the equations simultaneously, and then check if
the answers also satisfy the third
3+2𝑡=7+2𝑠8− 𝑡=4+𝑠
2 𝑡−2 𝑠=4−𝑡−𝑠=−4𝑠=1 𝑡=3
Solve simultaneously by making either the t or s terms
‘equal’rearrange
−2+3 𝑡=3+4 𝑠−2+3(3)=3+4 (1)7=7
Sub s and t into the 3rd pair – if it ‘works’ then the lines intersect. If not, then they
don’t…
So the lines DO intersect
VectorsYou need to be able to determine whether two given straight lines
intersect
It is important to note that in 3 dimensions, 2 straight lines may pass
each other without intersecting!
5I
1a) Show that the lines with vector equations:
r = (3i + 8j – 2k) + t(2i – j + 3k)
and r = (7i + 4j + 3k) + s(2i + j + 4k)
intersect.
We have just calculated that the above lines intersect for the values of t = 3 and s = 1
b) Calculate the position vector of the point of intersection
𝑟=( 38−2)+𝑡 (2−13 )
𝑟=(743 )+𝑠(214)
𝑟=( 38−2)+3 (2−13 )
𝑟=(957)
𝑟=(743 )+1(214)
𝑟=(957)
Sub t = 3 into the first equation and
calculate the position vector
Sub s = 1 into the second equation and calculate the position
vector
You only need to choose one of the equations for the substitution, as you can see, it works for both!
Teachings for Exercise 5J
VectorsYou need to be able to calculate the angle between any 2 straight lines
The acute angle θ between two straight lines is given by:
Where a and b are the direction vectors of the two lines.
The lines do not have to be intersecting – the angle is the angle between them if
one was moved along so they do intersect
Eg) The lines to the right do not intersect, but the angle calculated is the angle
between them if one was translated such that they do intersect
5J
𝑐𝑜𝑠𝜃=¿
VectorsYou need to be able to calculate the angle between any 2 straight lines
The acute angle θ between two straight lines is given by:
Where a and b are the direction vectors of the two lines.
The lines do not have to be intersecting – the angle is the angle
between them if one was moved along so they do intersect
5J
𝑐𝑜𝑠𝜃=¿
Modulus is used so that you get the acute angle rather than the obtuse
one
90 180
270 360
y = Cosθ
10
-1
For example, calculating cos-1(-0.5) would give us the angle 120°
For example, calculating cos-1|(-0.5)| would give us the angle 60° since -0.5 would be
replaced with 0.5 Each pair will always add up to 180°
Acute
Obtuse
This is because when 2 lines cross, you will always get a straight line with an acute and an obtuse angle on it
VectorsYou need to be able to calculate
the angle between any 2 straight lines
The acute angle θ between two straight lines is given by:
Where a and b are the direction vectors of the two lines.
The lines do not have to be intersecting – the angle is the
angle between them if one was moved along so they do
intersect
5J
𝑐𝑜𝑠𝜃=¿
Find the acute angle between the lines with vector equations:
r = (2i + j + k) + t(3i – 8j – k)and r = (7i + 4j + k) + s(2i + 2j + 3k)
To do this, you only need the direction vectors
𝒂=( 3−8−1)𝒃=(223)𝒂 .𝒃=( 3−8−1) .(
223)
𝒂 .𝒃=(3×2 )+ (−8×2 )+(−1×3)𝒂 .𝒃=−13
¿𝒂∨¿√32+(−8)2+(−1)2
¿𝒂∨¿√74¿𝒃∨¿√22+22+32¿𝒃∨¿√17
Calculate the dot product, a.b
Calculate the magnitude of a and
b
VectorsYou need to be able to calculate
the angle between any 2 straight lines
The acute angle θ between two straight lines is given by:
Where a and b are the direction vectors of the two lines.
The lines do not have to be intersecting – the angle is the
angle between them if one was moved along so they do
intersect
5J
𝑐𝑜𝑠𝜃=¿
𝒂 .𝒃=−13¿𝒂∨¿√74¿𝒃∨¿√17
𝑐𝑜𝑠𝜃=¿𝑐𝑜𝑠𝜃=| −13
√74 √17|𝑐𝑜𝑠𝜃=|−0.3665…|
𝑐𝑜𝑠𝜃=0.3665
𝜃=68.5 °
Sub in the values we have just calculated
Work out the sum
Since the answer is negative, we need to ‘make it positive’ by
multiplying by -1
Summary• We have learnt a great deal about vectors this chapter
• We have seen that when vectors are perpendicular, their dot ‘scalar’ product is equal to 0
• We have looked at the vector equation of a straight line
• We have seen how to calculate the angle between 2 lines
• We have learnt how to calculate whether 2 vectors intersect