INTRODUCTION TO VECTORS AND VECTOR ARITHMETIC
A scalar is a quantity that has magnitude but no direction. You can have signed scalar
quantities as well.
A vector is a quantity that has both magnitude and direction.
Some examples of scalar quantities are length, temperature (signed), and mass. Some
examples of vector quantities are force, acceleration, displacement, and velocity.
To specify the velocity of a moving point in the coordinate plane, we must give both the rate
at which the point moves (speed) and the direction of that motion. The velocity vector
incorporates both pieces of information. The length of the vector represents its magnitude
and its direction is represented by the direction of the arrow that represents the vector.

A vector u is represented in boldface, but for our purposes, we will use u . There are a
number of different ways to express a vector.
When we describe vectors, it is usual to do so by expressing them in terms of a basis.
To describe vectors in two dimensions, we use two basis vectors, i, and j.
j
i
 
To describe a vector that is “2 steps to the right and 3 steps up” we write a  2i  3 j .


To describe a vector that is “1 step to the left and 4 steps up” we write b  i  4 j .


The vectors  i and 4 j are known as components of the vector b. It can also be written as a
  1
column vector   .
4
 0
 1
In column representation (in two dimensions), i    and j    .
1
 0
Other possible representations for vectors are i, j , <a,b> , and (a,b). These are for twodimensional vectors.
1
To describe vectors in three dimensions, we use three basis vectors, i, j, and k.
j
k
i
Note that the component vectors are all at right angles to each other. (Imagine k is coming
out of the page towards you).
To represent vectors in higher dimensional space, simply add more basis vectors.
a
  
In three dimensions, the notation for the vector u is u, u ,  b  , i, j , k , <a,b,c>, or (a,b,c).
c
 
1
0
0
  
     
In column representation, i   0  , j   1  , and k   0  .
0
0
1
 
 
 
Drawing Vectors in 2D

The vector A B is a position vector of B relative to A. It is a displacement.
B
A



  3
Ex. Draw the vector b  3i  4 j . Note: In column form, this is   .
 4 
2



Ex. Draw the vector a  2i  5 j .
Note: Vectors do not have to be drawn starting from the origin. If they are, it is a special
case that we will talk about later.
Two vectors are equal only if they have the same magnitude (length) and direction.
Therefore, equal vectors must be parallel.

A negative vector has the same magnitude but opposite direction. For example, A B and


B A have the same length but have opposite directions. Therefore, B A is the negative of

AB .
Note: Two vectors can have the same magnitude but different directions. Two vectors can
also have the same direction (parallel) but different magnitude, or two vectors can have the
same magnitude, be parallel, but be in opposite directions. (Show each case).
3
Vector Arithmetic - Sum and Difference
We can add vectors graphically or algebraically.






To add vectors graphically, we add them “nose to tail”. If a  2i  3 j and b  i  4 j
 
represent a  b in the plane.
0 
The zero vector is 0    and for any vector a,
0 
a + (-a) = (-a) + a = 0.
As mentioned last day, the negative of a vector is the same magnitude but in the opposite
direction. Subtraction of a vector therefore is the same as adding the opposite.


 
Ex. Represent a  b in the plane. (This is a  ( b ) )
 
 
Is b  a the same as a  b ? Check by graphing.
Vectors can be added algebraically by adding their respective components.
4
Ex. Add a + b algebraically.
Check to see if this looks like the graph.
Ex. Find a – b.
Check the graph.
Vectors can also be added in column representation (just like adding matrices).
   1
  4
 
Ex. If a    and b    , find a  b .
5
2
Vectors in three dimensions are added and subtracted the same way that vectors in two
dimensions are.
3
  2
 
  
 
 
Ex. If p    1 and q   0  , find p  q and q  p .
4
 3 
 
 
5
SCALAR MULTIPLICATION
In addition to being added and subtracted, vectors can be multiplied by a scalar (non-vector
quantity). Multiplication by a scalar affects the length or magnitude of a vector, but not the
direction (unless the scalar is negative).
 ka 
If k is a scalar, then ka =  1  . Each component is multiplied by the scalar.
ka2 
  2
  1
Ex. If a    and b    , find:
5 
3

a)  4b


b) a  2b
c)

1
b  2a
2
6
POSITION VECTORS
We can find a vector that represents the translation from a point E to a point F.
F
E
EF  6i  3 j
or
6
EF   
 3
This vector can be called a displacement vector – but it is important to note that it is not
unique.
For example, the translation from the point G to H is exactly the same, even though G and H
are different points. The magnitude and direction of GH are the same as EF .
H
F
G
E
GH  6i  3 j
or
6
GH   
 3
7

A position vector is the directed line segment O P from the origin O to the point P(a,b).
a
The component form of the position vector OP is ai  b j or in column form   - and
b
therefore is related to the coordinates of point P – but it is important to note the difference
between a point and the position vector of a point.
We can find the displacement vector between two points by using position vectors. In the
 2
example above, E  (2,3) and F  (8, 6) . The position vector of E is OE    and F is
 3
8
8  2  6
OF    . To find the vector EF  OF  OE          .
 6  3  3
6
Ex. Find the displacement vector AB between points A (3, -5) and B (-1, 2).
Ex. A surveyor is standing at the top of a hill. Call this point “the origin” (O). A lighthouse
(L) is visible 4 km to the west and 3 km north. A town (T) is visible 5 km to the south and 2


km to the east. Find the position vectors of the lighthouse, O L , and the town, O T . Hence,

find the vector L T and the position of the town relative to the lighthouse.
8
DISTANCE BETWEEN POINTS, MAGNITUDE OF A VECTOR, AND UNIT VECTORS
To find the distance between two points, use the distance formula:
d  ( x2  x1 ) 2 ( y2  y1 ) 2
- for two dimensions
d  ( x 2  x1 ) 2  ( y 2  y1 ) 2  ( z 2  z1 ) 2
- for three dimensions
Ex. Find the distance between the points (-1, 4) and (-6, 11).
Ex. Find the distance between the points (3, -1, 4) and (4, 7, -2).
The length of a vector is the magnitude of the vector or its absolute value. The length of a
a 
2
2
vector a is represented by a   1   a1  a2 .
 a2 



Ex. Find the length of the vector b  2i  7 j .
9
 3
  

Ex. Find p if p   4  .
  1
A unit vector is any vector which is one unit long. The length of a vector is used to calculate
its unit vector. If we are given a vector a, the unit vector â , is a vector of length one unit in
the same direction as a. To change the length of the vector to one, divide the original vector
by its length.

a
aˆ  
a



Ex. Given q  5i  2 j , find its unit vector.
You could also write it in column representation:
Ex. Find unit vectors in the same direction as each vector.
  3
 
a)  6 
 4 
 
4
 
2
b)  
1
 
2
 
10
SCALAR PRODUCT, ANGLES BETWEEN VECTORS AND DIRECTION ANGLES
The scalar product (dot product) of two vectors is:
   
a  b  a b cos 
where  is the angle between the two vectors if they are drawn from the same point.
The angle may be acute, right, or obtuse. The result of a scalar (dot) product is a scalar
quantity. The scalar product is most often used to find the angle between two vectors –
therefore the equation can be rewritten as:
 
a b
cos    
ab
  b1 
  a1 
 
To find the product a  b , assume that a    and b    . Then the scalar product is
a 2 
b2 
a1b1  a2b2 . (Note that this is like multiplying matrices). The scalar product for vectors in
three or more dimensions is similar.
Ex. Find the magnitudes and scalar products for each pair of vectors. Hence, find the angle
between each pair of vectors.




a)  i  3 j and  i  2 j
0
b)  5 and
 4 
  5
  1
 
  3
11
If two vectors are perpendicular, what should the scalar product be equal to?
If two vectors are parallel, what should the scalar product be equal to?




Property: If a and b are non-zero vectors, and  is the angle between a and b , then
 
 is acute iff a  b >0

 is obtuse iff a  b <0
 
 is  2 or 90  iff a  b = 0
Perpendicular vectors are also called orthogonal vectors.
Properties of Scalar Products:
1.
2.
3.
   
v  w  w v
  
   
u  (v  w)  u  v  u  w
 
 
(kv )  w  k (v  w)
  2
4. v  v  v
 


  
5. If v  w and v  0, w  0 then v  w  0 .


 
 
 
6. If v w and v  0, w  0 , then v  w   v w
Area of a Triangle:
For any triangle, the area is:
1
a b sin  , where  is the angle between sides a and b.
2
 
a b
Prove that cos     using the diagram below. (Hint: use cosine law)
ab
a
a-b

b
12
Direction Angles
Direction angles are the angles that a vector makes with each of the unit coordinate vectors.
The angles 𝛼, 𝛽 and 𝛾 are the direction angles of u and cos 𝛼, 𝑐𝑜𝑠𝛽, and 𝑐𝑜𝑠𝛾 are called the
direction cosines.
𝒖 = 𝑥𝒊 + 𝑦𝒋 + 𝑧𝒌
If we create right triangles (where |𝒖| is the
hypotenuse) we can come up with some
useful expressions.
𝑐𝑜𝑠𝜃 =
𝑐𝑜𝑠𝜃 =
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
|𝒖|
so |𝒖|𝑐𝑜𝑠𝜃 = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
Then:
𝑥
𝑦
𝑧
𝑥 = |𝒖|𝑐𝑜𝑠𝛼, 𝑦 = |𝒖|𝑐𝑜𝑠𝛽, and 𝑧 = |𝒖|𝑐𝑜𝑠𝛾 (and cos 𝛼 = |𝒖|, cos 𝛽 = |𝒖|, cos 𝛾 = |𝒖|)
We can write vector u as 𝒖 = (|𝒖|𝑐𝑜𝑠𝛼)𝒊 + (|𝒖|𝑐𝑜𝑠𝛽)𝒋 + (|𝒖|𝑐𝑜𝑠𝛾)𝒌, or
𝒖 = |𝒖|(𝑐𝑜𝑠𝛼𝒊 + 𝑐𝑜𝑠𝛽𝒋 + 𝑐𝑜𝑠𝛾𝒌)
If we take the magnitude of both sides,
13
|𝒖| = ||𝒖|(𝑐𝑜𝑠𝛼𝒊 + 𝑐𝑜𝑠𝛽𝒋 + 𝑐𝑜𝑠𝛾𝒌)|
|𝒖| = |𝒖|√cos 2 𝛼 + cos 2 𝛽 + cos 2 𝛾
This means that cos2 𝛼 + cos2 𝛽 + cos 2 𝛾 = 1, the sum of the squares of the direction
cosines is always 1.
For a unit vector, we the expression will be 𝒖 = cos𝛼𝒊 + cos𝛽𝒋 + cos𝛾𝒌 because (|𝒖| = 1).
So for a unit vector, the x, y, and z-coordinates are its direction cosines.
Ex. Find the direction cosines of the vector 𝐯 = 3𝐢 + 2𝐣 − 4𝐤 and approximate the direction
angles to the nearest degree.
14
REPRESENTATION OF A LINE IN THE PLANE
When describing a line in two dimensions we are used to Cartesian form. This form gives a
direct relation between x and y. However, there are other different ways of describing a line
in two or three dimensions.
Cartesian form (in 2D) – coordinates of a point on the line and the slope of the line
Vector form – position vector of a point on the line and a vector parallel to the line (note: this
is also sometimes called parametric form)
Consider a line which passes through point A and is parallel to vector b.
b
R
A
a
To find the vector equation of this line, r, consider a general point R on the line. Let R have

 

 
position vector r. A R is parallel to b, therefore A R  tb , but A R  a  r . Therefore,


 
 
 a  r  tb and r  a  tb . (Note: r represents any and every point on the line).
  
The vector form of a line is r  p  td , where p is any point on the line, d is any vector in the
 x   p1   d1 
direction of the line, and t is any value. The parametric form is       t   so,
 y   p2   d 2 
x  p1  td1 and y  p2  td2 .
For example, the line y  3 x  2 goes through the point (0,2) and has slope 3, therefore a
1
vector in the direction of the line would be   . In vector form, the equation of the line is
 3
 0  1
r     t   . Substituting values for t gives other points on the line. For example, if t =  2   3
  1
1, the point generated is   .
  1
1   2
Note: The parametric form is not unique; you could also use the equation r     t   to
 5   6 
describe the line.
15
 
Ex. Find the vector equation of a straight line that is parallel to the vector 2i  j and passes


through the point with position vector 3i  2 j .
Ex. Find the vector equation of the straight line that passes through points A and B given the
  1
 2
position vectors A and B are   and   respectively.
1
7
The easiest way to represent lines in three dimensions is using the parametric form. In fact, it
is very difficult to write the equation of a line in three dimensions in Cartesian form. We will
not even try in this course.
Ex. Find the vector equation of the line passing through the points A (2, 1, 1) and B (4, 0, 3).
16
  1   3
  2    1
Ex. Find the angle between the lines r     t   and r     t   . Hint: Find the
  1  2 
1  2 
angle between the vectors in the direction of each line.
17
APPLICATIONS OF LINES
If a body has initial position vector a, and moves with constant velocity b, its position at time
  
t is given by:
for
r  a  tb
t0
 1  3 
Ex. A particle is moving along the line r     t   where t is in seconds and distance
1  4 
units are in meters.
a) Find the initial position of the particle.
b) Sketch a graph using t = 0, 1, 2, 3
c) Find the speed of the particle.
 x  7  6 
Ex.       t   is the vector equation of the path of an object where t  0 , t in
 y   5    8
seconds. Distance units are metres.
a) Find the object’s initial position.
b) Find the velocity vector of the object.
c) Find the object’s speed.
18
  5
Ex. Find the velocity vector of a speed boat moving parallel to   with a speed of 65
 12 
km/h.

 
Ex. An object is initially at (5, 10, 2) and moves with velocity vector 3i  j  2k . Find:
a) The position of the object at any time t (t in minutes).
b) The position at t = 3.
c) The time when the object is at (35, 0, 22).
19
COINCIDENT AND PARALLEL LINES AND INTERSECTION OF LINES
Two lines in space are either parallel, intersecting, or skew.

Skew lines are lines that are neither parallel nor intersecting.



If the lines are parallel, the angle between them is 0º.
If the lines are intersecting, the angle between them is   .
If the lines are skew, there is still an angle that one line makes with the other and if we
translate one line to intersect the other, the angle between the original lines is defined
as the angle between the intersecting lines.

 x    1  2 
     
Ex. Line 1 has equation  y    1   s  2  . Line 2 has equation
z  1   4 
     
Show that lines 1 and 2 are parallel.
 x  1  1
     
 y    0   t 1  .
 z   3   2
     
20
 x    1  2 
     
Ex. Line 1 has equation  y    1   s  2  , line 2 has equation
z  1   4 
     
 x  1  1
     
 y    0   t  1  , and
 z   3   2
     
 x  1   2 
     
line 3 has equation  y     1  u   1 .
z  4   3 
     
a) Show that lines 2 and 3 intersect and find the angle between them.
b) Show that lines 1 and 3 are skew.
21
CARTESIAN EQUATIONS OF LINES IN 3-DIMENSIONS
a) PARAMETRIC FORM OF THE EQUATION OF A LINE - THREE DIMENSIONS
y  a2  v2
x  a1  v1
z  a3  v3
b) CARTESIAN FORM OF THE EQUATION OF A LINE - THREE DIMENSIONS
From above:

x  a1
v1

So in Cartesian form:
y  a2
v2

z  a3
v3
x  a1 y  a2 z  a3


v1
v2
v3
Ex. Given a line who passes through the point
 4,2,5
and having the direction vector
i  j  2k , find :
a) the vector equation of the line
b) the equation of the line in parametric form
c) the Cartesian form of the equation of the line
d) two other points on the line
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Ex. Convert
x 1 y  2 z  6
into vector form.


3
2
4
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VECTOR PRODUCT OR CROSS PRODUCT

The vector product of two vectors is a vector which is perpendicular to both of the original vectors.
a b  c
 The MAGNITUDE OF
c
c  a  b  a b sin
 DIRECTION OF
a b
The direction of
c
is the same as the direction of
n where n is perpendicular to Plane
AB
 What would you know about the magnitude of the vector product of :
1) PARALLEL VECTORS
2) PERPENDICULAR VECTORS
3) OPPOSITE VECTORS
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VECTOR PRODUCT FROM ANOTHER POINT OF VIEW
DETERMINANT FORM :
Given
a  a1 i  a2 j  a3 k
a b
=
and
b  b1 i  b2 j  b3 k
i
j
k
a1
a2
a3
b1
b2
b3
=



(a 2 b3  a3b2 )i  (a1b3  a3b1 ) j  (a1b2  a 2 b1 )k
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APPLICATIONS OF THE VECTOR PRODUCT
AREA OF A PARALLELOGRAM
Area =
ab
units2
AREA OF A TRIANGLE
Area =
1
2
ab
units2
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EQUATIONS OF PLANES
INTRODUCTION
 In simple terms a plane can be described as an infinite flat surface such that a straight line joining any two
points on it will lie entirely in it. The surface can be horizontal, vertical or sloping.
 To describe or define a plane, we need :
- 3 non-collinear points on the plane; or
- a point and two non-parallel lines that lie on the plane; or
- a vector perpendicular to the plane and a point that lies on the plane
 Remember, the equation of a plane is . . .
VECTOR FORM OF THE EQUATION OF A PLANE
r  a  b   c
Thus we need: 1) the position vector of a point A in the plane
2) two non-parallel vectors in the plane
PARAMETRIC FORM
 x   a1 
 b1 
 c1 
   
 
 
 y    a2     b2     c2 
 z a 
b 
c 
   3
 3
 3
x  a1  b1   c1
y  a2  b2   c2
z  a3  b3   c3
CARTESIAN FORM
ax  by  cz  d
We get this by solving and substituting in the parametric equations.
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 2
3
 
 
Ex. 1. a) Find the vector equation of the plane containing the vectors 1 and 0 which also contains the
 
 
1
 1
 
 
point
(1,2,0).
b) Find the Cartesian form of this equation.
1
 2 
1
 
 
 
2. Find the Cartesian equation of the plane defined by the vector equation r  3   1   1
 
 
 
 4
1
 2
 
 
 
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NORMAL FORM OF THE EQUATION OF A PLANE
If
n  bc
where n is the normal vector and b and c are two non-parallel vectors in the plane,
( x0 , y0 , z0 ) is a particular point in the plane, and r
is a general point
( x, y , z )
a
in the plane then the normal
equation of a plane is:


r n  an
Ex.3) A plane is perpendicular to the line i – j – 2k and passes through the point (-1, 2, 5). Find the vector
equation of
the plane in Normal Form.
CARTESIAN FORM OF THE EQUATION OF A PLANE - REVISITED
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 2
3
 
 
Ex. 1 – Revisited : Find the Cartesian equation of the plane containing the vectors 1 and 0 which also
 
 
1
 1
 
 
contains the point (1,2,0).
Ex. 2) Find the Cartesian form of the equation of the plane that passes through the points A(1,1,1), B(2,3,4) and
C(-1,0,5).
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ANGLE BETWEEN TWO PLANES
33
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INTERSECTING LINES & PLANES
INTERSECTION OF TWO LINES
TWO LINES CAN :
KEY POINTS TO REMEMBER :
a) intersect at a point
* Determine where the lines intersect by solving for  and
then determine the coordinates of the point of intersection.
b) be parallel
* The direction vectors can be written as scalar multiples of each other.
c) be coincident
* The direction vectors can be written as scalar multiples of each other.
 and
* Show that the lines share one point in common and therefore an
infinite number of points in common.
d) be skew
* Show that the direction vectors cannot be written as scalar multiples
of each other.
* Show that the lines do not share a point in common.
Ex. 1a) Find the point of intersection for the two lines
 1
 3
 
 
r   4     2 
0
1
 
 
and
4
 2
 
 
s   4     3  .
 1 
 2
 
 
b) Find the acute angle between the two lines.
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INTERSECTION OF A LINE AND A PLANE
A LINE AND A
PLANE CAN :
a) intersect at
a point
KEY POINTS TO REMEMBER :
* Substitute the parametric
forms of the line into the
equation of the plane and
solve for  . Find the
point.
*To find the angle of intersection
between a line and a plane, find the
angle between the line and the
normal to the plane and then
subtract this from 90.
v  n  v n cos to find  .
Then   90  
b) be parallel
*The normal to the plane is also perpendicular to the line.
c) be
coincident
* If coincident, the parametric equations of the line will satisfy the
Cartesian form of the equation of the plane.
( i.e the line
lies on the
plane)
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INTERSECTION OF TWO PLANES
TWO PLANES CAN :
KEY POINTS TO REMEMBER :
a) intersect at a line
* Isolate one variable twice and set the three
expressions equal to each other.
*REMINDER : When two planes intersect, the
angle between the planes is defined as the
acute angle between their normals.
b) be parallel
* The normals of the planes are parallel.
* The planes share no point in common
c) be coincident
* *The normals of the planes are parallel.
* The planes share a point in common and
therefore an infinite number of points in
common
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J.4. INTERSECTION OF THREE PLANES
THREE PLANES
CAN :
KEY POINTS TO REMEMBER :
a) intersect at a point
* The system of equations formed by the
equations of these planes has a unique
solution.
* Det
b) intersect at a line
or be coincident
0
* The system of equations formed by the
equations of these planes has an infinite
number of solutions.
* Det
0
* Augmented matrix will have the form :
c) have no common
points
* The system of equations formed by the
equations of these planes has no
solution.
*
Det  0
* Augmented matrix will have the form :
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