FC311E
Intermediate Mathematics
Vectors
LEARNING OBJECTIVES
Vectors
After studying this theme you should be able to:
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•
•
•
•
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Understand and use vectors
Add and subtract vectors and perform scalar multiplication
Find the dot (or scalar) product of two vectors
Find the angle between two vectors
Find the cross product of two vectors
Understand and use vector line equations
Determine if two lines intersect
Vectors
▪ In this session we will learn what is meant by a vector, how to add
and subtract vectors and how to multiply a vector by a scalar (i.e. a
number). We will also learn how to find the magnitude of a vector.
Vectors
A vector has magnitude and direction
Examples: weight, force, velocity, and momentum.
Drawing Vectors
Magnitude (size) is shown by the length of a line
A
Direction is shown by an arrow.
D
Vectors are written as
B
AB
CD
C
Vectors
Vectors are parallel to each other if one is a multiple
of the other
F
D
B
AB
A
GH
EF = 2CD
− AB = BA
C
= 3GH
Vectors can be multiplied by a scalar. E
Direction is the same, but scale has changed
Position Vectors
A position vector gives the position of a point relative to the
origin, O
A
OA
O
3
4
(3,4)
 3
OA =  
 4
Called a column vector /
component form
Position Vectors
A position vector gives the position of a point relative to the
origin, O
 3
A (3,4)
OA =   = a
 4
OA
O
3
4
A position vector is normally given
by a single letter
Arrow is not used but letter must
be underlined to identify it as a
position vector
Might be in bold in textbooks
Vectors
If A and B have position vectors a and b respectively,
AB can be written in terms of a and b
A
AB = −a + b
AB
O
AB = b − a
B
Vectors
If A and B are given by (3, -4) and (-1, 2),
express
AB as a column vector:
3
OA = a =  
 − 4
− 1
OB = b =  
2
− 1  3  − 4
AB = b − a =   −   =  
 2   − 4  6 
3
i is the unit vector in the
i
x-direction
2
j
1
Y
i=
j is the unit vector in the
y-direction
0
-1
[]
1
0
0
1
2
3
j=
-1
X
[]
0
1
All vectors can be expressed as a linear
combination of these 2 vectors
1
0
8
e.g.
=8
+5
5
[] [] []
0
1
Vectors: Another notation
Unit vectors are given by
1
i= 
0 
1 0
3
OA = a =   = 3  − 4 
0 1
 − 4
= 3i − 4 j
0 
j= 
1
Adding & Subtracting vectors
a=
()
5
3
3a + 6b
2a - b
( )
= 3( ) + 6( )
= ( )+ ( ) = ( )
b=
2
−4
5
3
2
−4
15
9
12
− 24
( )− ( )
= ( )− ( )
=2
5
3
2
−4
10
6
2
−4
27
−15
=
( )
8
10
Two vectors are given by 𝒖 =
3
5
and 𝒗 =
. Find the following:
2
−1
a) 𝒖 + 𝒗
b) 3𝒖 − 2𝒗
c) Write 𝒖 in terms of the unit vectors 𝒊 and 𝒋.
Two vectors are given by 𝒖 =
3
5
and 𝒗 =
. Find the following:
2
−1
a) 𝒖 + 𝒗
=
3
5
+
2
−1
b) 3𝒖 − 2𝒗
=3
=
3
5
−2
2
−1
=
8
1
9
10
−
6
−2
=
c) Write 𝒖 in terms of the unit vectors 𝒊 and 𝒋.
3
𝒖=
2
3
0
=
+
0
2
1
0
=3
+2
0
2
= 3𝒊 + 2𝒋
−1
8
Magnitude of a vector
Magnitude is given by the length of vector
A
OA
O
3
4
(3,4)
OA = 3 + 4
2
=5
2
Magnitude of a vector
If A and B are given by (3, -4) and (-1, 2),
find the magnitude of AB
− 1
3
OA = a =   OB = b =  
2
−
4


 
 − 4
AB = b − a =  
6
AB =
(− 4)
2
+6
2
Vectors in 3D
Vectors in 3D will have a third component
4


a = 2
3
a = 4i + 2 j + 3k
k is the unit vector in the
z-direction
Scaling, adding and subtracting still
works, but now there’s a third bit
Magnitude of 3d vector
To find the magnitude of a 3 dimensional vector,
we extend Pythagoras’ theorem.
B
If
C
A
2
4
4
→
AB =  2 
 
 3 
AC 2 = 4 2 + 2 2
Magnitude of 3d vector
To find the magnitude of a 3 dimensional vector,
we extend Pythagoras’ theorem.
B
If
3
C
A
4
→
AB =  2 
 
 3 
AC 2 = 4 2 + 2 2
AB 2 = AC 2 + CB 2
AB 2 = (4 2 + 2 2 ) + 3 2
 AB = 29
Magnitude of 3d vector
To find the magnitude of a 3 dimensional vector,
we extend Pythagoras’ theorem.
a 
AB =  
b 
a 


AB = b 
 c 
AB = a + b
2
2
AB = a + b + c
2
2
2
3
Q. Find the magnitude of the vector 𝒂 = 1
5
3
Q. Find the magnitude of the vector 𝒂 = 1
5
Magnitude 𝒂 = 𝒂 = 32 + 12 + 52
= 35
Scalar Product
▪ In this session we will learn how to find the scalar product (also
called the dot product) of two vectors and how to use this to find the
angle between two vectors.
So far we have learnt:
• How to add and subtract vectors
• How to find the magnitude of a vector
• How to multiply a vector by a scalar (i.e. multiplying by a number)
But what about multiplying vectors by each other?
There are two types of vector multiplication we will learn about:
The scalar product (or “dot” product)
The vector product (or “cross” product)
The Scalar Product or Dot product
𝑎 ⋅ 𝑏 = 𝑎 𝑏 cos 𝜃 = 𝑎1 𝑏1 + 𝑎2 𝑏2 + 𝑎3 𝑏3
Geometrically
Based on distance
(magnitude) and
angle between
vectors
Algebraically
Sum of the products
of the components
𝑎⋅𝑏
= 𝑎 𝑏 cos 𝜃
Magnitude 𝒂
x
Magnitude 𝒃 x cos(angle between them)
𝒂
𝜃
𝒃
= 𝑎1 𝑏1 + 𝑎2 𝑏2 + 𝑎3 𝑏3
Sum of product of components
Perpendicular Vectors
a b = 0 → a b cos  = 0
If a and b are non-zero vectors, then
𝜃 = 90°
cos  = 0
If the angle is 90°, then a and b
must be perpendicular
Find an angle
We can use the scalar product to find the angle between two vectors:
a  b = a b cos  = (a1b1 + a2b2 + a3b3 )
(
a1b1 + a2b2 + a3b3 )
cos  =
ab
3
5
a) Find the scalar product of the vectors 𝒂 = 2 and 𝒃 = 0 .
7
−4
b) Find the angle between the vectors.
3
5
a) Find the scalar product of the vectors 𝒂 = 2 and 𝒃 = 0 .
7
−4
𝒂 ∙ 𝒃 = 3 ∙ 5 + 2 ∙ 0 + 7 ∙ −4
= 15 + 0 − 28
= −13
b) Find the angle between the vectors.
𝒂 = 32 + 22 + 72 = 62
cos 𝜃 =
𝒂∙𝒃
𝒂 𝒃
=
−13
62 41
𝒃 =
52 + 02 + (−4)2 = 41
= −0.258
𝜃 = cos −1 (−0.258) = 104.9°
Vector Product
▪ In this session we will learn how to find the vector product (also
called the cross product) of two vectors.
Vector Product
Last time we learnt how to find the scalar when multiplying
two vectors using the scalar product.
Today we will be looking at the cross product (or vector
product).
Vector Product
Sometimes when you multiply two vectors, you want the
answer to be a vector as well.
In this case you use the vector product.
ഥ = 𝐹ത × 𝑟ҧ
𝑀
The resulting vector is perpendicular to both vectors.
Vector Product
To find the vector (cross) product, you need to use the
following formula.
𝑎2 𝑏3 − 𝑎3 𝑏2
𝒂 × 𝒃 = 𝑎3 𝑏1 − 𝑎1 𝑏3
𝑎1 𝑏2 − 𝑎2 𝑏1
3
5
Find the vector (cross) product of the vectors 𝒂 = 2 and 𝒃 = 0 .
7
−4
3
5
Find the vector (cross) product of the vectors 𝒂 = 2 and 𝒃 = 0 .
7
−4
𝑎2 𝑏3 − 𝑎3 𝑏2
𝒂 × 𝒃 = 𝑎3 𝑏1 − 𝑎1 𝑏3
𝑎1 𝑏2 − 𝑎2 𝑏1
2 ∙ −4 − 7 ∙ 0
= 7 ∙ 5 − 3 ∙ (−4)
3∙0−2∙5
−8
= 47
−10
Vector Line Equations
▪ In this session we will learn how to use vectors to write the equation
of a line in both 2 and 3 dimensions, and how to determine if two
lines ever intersect (i.e. cross each other).
y
Vector line equations
(any point
it passes
A through)
a
o
Any parallel vector (to line)
x
A line can be identified by a linear combination
of a position vector and a free vector
y
Vector line equations
Any parallel vector to line
a
o
A
(any point
it passes
through)
x
A line can be identified by a linear combination
of a position vector and a free vector
y
Vector line equations
A line can be identified by a linear combination
of a position vector and a free vector
A
b = pi + qj
parallel vector to line
a = xi + yj
o
E.g. a + b
= (xi + yj) + (pi + qj)
x
𝜆 is a scalar (i.e. any number)
The value of 𝜆 tells us how far along the line we are.
For any value of 𝜆 we will be on the line.
How to write the vector equation of a line – Example 1
1. Position vector to
any point on line
𝒚 = 𝒙 + 𝟐
[]
1
3
2. A free vector parallel
to the line
[]
[]
3. linear combination
of a position vector
and a free vector
[][] []
2
2
2
2
x
1
2
=
+
𝜆
y
3
2
Equation
Scalar (any number)
[]
1
3
Example 1 using a different combination of vectors
1. Position vector to
any point on line
𝒚 = 𝒙 + 𝟐
[]
4
6
2. A free vector parallel
to the line
-3
[]
-3
-3
[]
-3
3. linear combination
of a position vector
and a free vector
[][] []
x
4
-3
=
+𝜆
y
6
-3
[]
4
6
How to write the vector equation of a line – Example 2
1. Position vector to
any point on line
[]
2
4
2. A free vector parallel
to the line
4
2
3. linear combination
[]
of a position vector
and a free vector
[][] []
x = 2 +𝜆 4
y
2
4
𝒚 = 𝟏/𝟐 𝒙 + 𝟑
[]
4
2
[]
2
4
Finding the equation without a sketch
[][] []
x
1
1
=
+
𝜆
y
2
3
When 𝜆 = 0
[] []
x
y
=
x=1, y=2 is a
coordinate on
the line.
1
2
Tells you about the gradient
[]
1
3
Means go 1 right and 3 up.
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑦 3
𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡 =
= =3
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑥 1
Now use 𝑦 – 𝑦1 = 𝑚(𝑥 – 𝑥1)
𝑦 – 2 = 3(𝑥 – 1)
𝒚 = 𝟑𝒙 − 𝟏
a) Find a vector equation for the line 𝑦 = 2𝑥 + 3
𝑥
2
4
b) Write the equation for the line 𝑦 =
+𝜇
1
6
in the form 𝑦 = 𝑚𝑥 + 𝑐
a) Find a vector equation for the line 𝑦 = 2𝑥 + 3
1
satisfied the equation.
5
1
Any vector with 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = 2 satisfied the equation. Choose
.
2
𝑥
1
1
=
+
𝜆
𝑦
5
2
The position vector
𝑥
2
4
b) Write the equation for the line 𝑦 =
+𝜇
1
6
6
2
Point (4,1) is on line.
𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = = 3
𝑦 − 1 = 3(𝑥 − 4)
𝑦 = 3𝑥 − 11
in the form 𝑦 = 𝑚𝑥 + 𝑐
In 2 dimensions, there are 3 possible
relationships between two lines:
They are parallel
They are the same line
They intersect
Intersection of 2D lines in vector form
For example
x
1
2
=
+
s
y
3
2
[ ][ ] [ ]
and
[][] []
x
6
-1
=
+
t
y
-2
4
If the lines intersect, there must be values of s and t such that they give the same
value of 𝑥 and 𝑦.
x
1
2
x part: 1 + 2s = 6 - t
=
+
1.5
y
3
2
y part: 3 + 2s = -2 + 4t
x
1 + 3
Subtract x from y : 2 = -8 + 5t
=
y
3
3
5t = 10
t=2
4
x
= 6
y
Substitute: 1 + 2s = 6 - 2
2s = 3
position vector of the point of intersection
s = 1.5
[] [] []
[ ] [ ][ ]
[] []
Intersection of 2D lines in vector form
What is the point
of intersection?
a = (i + 2j) +  (4i - 2j)
b = (2i - 6j) +  (-2i + j)
i coefficients : 1 + 4 = 2 − 2
j coefficients : 2 − 2 = −6 + 
x2 : 4 − 4 = −12 + 2
add
5 = −10
… this doesn’t work!
The direction vectors:
(4i - 2j) and (-2i + j) are parallel
The lines are parallel so they never intersect!
z
Vectors for 3D lines
The position vector of point A
A line in 3D
b
a
y
o
3
 
a= 2 
10 
 
a = 3i + 2j + 10k
A parallel vector
 1 
  b = i + 2j - 4k
b= 2 
 − 4
 
A linear combination of a
position vector and a parallel vector
x
r
=
3
 
2
10 
 
+
 1 
 
 2 
 − 4
 
z
Example
A line in 3D
=
b
a
y
3
 
2
10 
 
+
 1 
 
 2 
 − 4
 
The position vector (r)
of any point on the line
 x
 
r =  y
z
 
r = (3i + 2j + 10k) + (i + 2j - 4k)
If  = 4
r = (3i + 2j + 10k) + 4 (i + 2j - 4k)
r = (3i + 2j + 10k) + (4i + 8j - 16k)
o
r = 7i + 10j -6k
x
In 3 dimensions, there are 4 possible
relationships between two lines:
They are
parallel
They are the
same line
They intersect
Or they
are skew
Skew lines
In 3D lines that are not parallel and do not intersect are called skew lines
z
Don’t meet
b
y
a
x
Checking if lines are parallel
Check whether the direction vectors are scalar multiples of each other:
 4 
 
c =  − 1  Is parallel to ….
 − 2
 
 12 
 
 − 3
 − 6
 
 4 
 
= 3 − 1 
 − 2
 
 2 
 4 


1 
 − 0 .5  =  − 1 
2 
 −1 


 − 2
 − 8
 4 
 
 
 2  = −2 − 1 
 4 
 − 2
 
 
=3c
= 1/2 c
= -2 c
Example
The lines r and s have the equations ...
 2  4 
   
r =  3  +   − 1
5  3 
   
and
 4
 2 
 
 
s =  7  +  − 2
 2
 3 
 
 
Show they intersect and
find the point of intersection
Example
The lines r and s have the equations ...
 2  4 
   
r =  3  +   − 1
5  3 
   
and
 4
 2 
 
 
s =  7  +  − 2
 2
 3 
 
 
Show they intersect and
find the point of intersection
If the lines intersect, there must be values of  and  that give the position
vector of the point of intersection.
 2  4   4
 2 
     
 
 3  +   − 1 =  7  +   − 2 
 5  3   2
 3 
     
 
x+y : 5 + 3 = 11
3 = 6
=2
Substitute
x : 2 + 4x2 = 4 +2
 =3
x : 2 + 4 = 4 +2
y : 3 -  = 7 - 2
z : 5 + 3 = 2 +3
Check the values in the 3rd equation
z : 5 + 3x2 = 2 + 3x3
11 = 11
Satisfied!
Hence a point exists
common to both lines
Example
The lines r and s have the equations ...
 2  4 
   
r =  3  +   − 1
5  3 
   
=2
and
 =3
 4
 2 
 
 
s =  7  +  − 2
 2
 3 
 
 
Show they intersect and
find the point of intersection
Values satisfy all equations
Hence a point exists
common to both lines
 2   4   2   8  10 
         
Substitute r =  3  + 2 − 1 =  3  +  − 2  =  1 
 5   3   5   6   11 
         
The 2 lines intersect at (10, 1, 11)
Skew Example
2 lines have the equations ...
r = (2i + 3j + 6k) + t(4i - j + 6k)
and r = (4i + 7j + 8k) + s(2i - 2j + k)
Show they are skew
Skew Example
2 lines have the equations ...
r = (2i + 3j + 6k) + t(4i - j + 6k)
and r = (4i + 7j + 8k) + s(2i - 2j + k)
Show they are skew
If the lines intersect, there must be values of s and t that give the position vector
of the point of intersection.
i : 2 + 4t = 4 +2s
j : 3 - t = 7 - 2s
Check the values in the 3rd equation
k : 6 + 6x2 = 8 + 3
18 = 11
k : 6 + 6t = 8 + s
i+j : 5 + 3t = 11
3t = 6
t=2
Substitute
i : 2 + 4x2 = 4 +2s
s =3
Not Satisfied! So do not intersect.
Direction vectors:
(4i - j + 6k) and (2i - 2j + k)
are not parallel
Therefore lines are skew
Intersecting Lines Summary
• Step 1 – Set up 3 simultaneous equations; one for x, y
and z.
• Step 2 – Solve a pair of your equations.
• Step 3 – Put your solution into the third equation and see
what happens.
– If it works then they intersect and you can find the point of
intersection.
– If it does not work the lines are either parallel or skew. Check
to see if they the direction vectors are parallel then decide.
Q. Determine if the following two lines intersect, are parallel or are skew:
3
2
8
5
𝒂= 1 +𝜆 2
𝒃 = −1 + 𝜇 1
5
−3
3
7
Q. Determine if the following two lines intersect, are parallel or are skew:
3
2
8
5
𝒂= 1 +𝜆 2
𝒃 = −1 + 𝜇 1
5
−3
3
7
Setup simultaneous equations
𝟏. −𝟐.
2 = 6 + 7𝜇
Substitute
3 + 2𝜆 = 5 + 8 × −
Test in 𝟑.
5−3×−
Test if parallel
9
7
𝜇=−
4
7
3 + 2𝜆 = 5 + 8𝜇
1 + 2𝜆 = −1 + 𝜇
5 − 3𝜆 = 7 + 3𝜇
4
7
= 7+3×−
𝜆=−
4
7
2
8
2 is not a multiple of 1
−3
3
𝟏.
𝟐.
𝟑.
9
7
62
7
=
37
7
not true!
Lines don’t intersect.
so lines are not parallel => SKEW