Vectors Super Slide Show

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A vector is a quantity that has both magnitude and
direction. It is represented by an arrow. The length of
the vector represents the magnitude and the arrow
indicates the direction of the vector.
Blue and orange
vectors have
same magnitude
but different
direction.
Blue and purple
vectors have
same magnitude
and direction so
they are equal.
Blue and green
vectors have
same direction
but different
magnitude.
Two vectors are equal if they have the same direction and
magnitude (length).
How can we find the magnitude if we
have the initial point and the terminal
point? The distance formula
Q
x2 , y2 
Terminal
Point
Initial
Point
x1, y1 
P
How can we find the direction? (Is this all looking familiar
for each application? You can make a right triangle and
use trig to get the angle!)
Although it is possible to do this for
any initial and terminal points, since
vectors are equal as long as the
direction and magnitude are the
same, it is easiest to find a vector
with initial point at the origin and
terminal point (x, y).
Initial
Point
0x1,, 0y1
P
Q
xx,
2 , yy
2
Terminal
Point
A vector whose
initial point is
the origin is
called a
position vector
If we subtract the initial point from the
terminal point, we will have an
equivalent vector with initial point at
the origin.
To
Toadd
addvectors,
vectors,we
weput
putthe
theinitial
initialpoint
pointof
ofthe
thesecond
second
vector
vector on
onthe
theterminal
terminalpoint
pointof
ofthe
thefirst
firstvector.
vector. The
The
resultant
resultantvector
vector has
hasan
aninitial
initialpoint
pointat
atthe
theinitial
initialpoint
point
of
ofthe
thefirst
firstvector
vector and
andaaterminal
terminalpoint
pointat
atthe
theterminal
terminal
point
pointof
ofthe
thesecond
secondvector
vector(see
(seebelow--better
below--bettershown
shown
than
thanput
putin
inwords).
words).
Terminal
point of w
vw
Initial point of v
v
w
w
Move w over keeping
the magnitude and
direction the same.
The negative of a vector is just a vector going the opposite
way.
v
v
A number multiplied in front of a vector is called a scalar. It
means to take the vector and add together that many times.
3v
v
v
v
v
u
w
Using the vectors shown,
find the following:
uv
 3w
w
w
w
uv
u
2u  3w  v v
u
u
u
v
w
w
w
v
Vectors Worksheet #1
Head-Minus Tail Rule
Prove that the two vectors RS and PQ are equivalent.
1.) R = (-4, 7)
S = (-1, 5)
P = (0, 0)
Q = (3, -2)
2.) R = (7, -3)
S = (4, -5)
P = (0, 0)
Q = (-3, -2)
3.) R = (2, 1)
S = (0, -1)
P = (1, 4)
Q = (-1, 2)
4.) R = (-2, -1)
S = (2, 4)
P = (-3, -1)
Q = (1, 4)
Vectors are denoted with bold
letters
This is the notation for a
position vector. This means
the point (a, b) is the
terminal point and the initial
point is the origin.
a
v   
a
b
v     ai  bj
We use vectors that are only 1
b
(a, b)
j
i
 3
v   
 2
unit long to build position
vectors. i is a vector 1 unit long
in the x direction and j is a vector
1 unit long in the y direction.
(3, 2)
j
j
i i i
v  3i  2 j
If we want to add vectors that are in the form ai + bj, we can
just add the i components and then the j components.
v  2 i  5 j
w  3i  4 j
v  w   2i  5 j  3i  4 j  i  j
Let's look at this geometrically:
Can you
see from
this picture
5j
how to find
the length
of v?
3i
w
v
 2i i
 4j
j
When we want to know
the magnitude of the
vector (remember this is
the length) we denote it
v

 2  5
2
 29
2
Vectors Worksheet #2
Using the following vectors:
P = (-2, 2)
Q = (3, 4)
R = (-2, 5)
S = (2, -8)
Find:
PQ
RS
QR
PS
2QS
(√2)PR
3QR + PS
PS – 3PQ
Vector Worksheet #3
Performing Vector Operations
Let u = <-1, 3> and v = <4, 7>
Find the component form of the following vectors.
u+v
3u
2u + (-1)v
u + v = <-1, 3> + <4, 7> = <-1 + 4, 3 + 7> = <3, 10>
3u = 3<-1, 3> = <-3, 9>
2u + (-1)v = 2<-1, 3> + (-1)<4, 7> = <-2, 6> + <-4, -7> = <-6, -1>
Performing Vector Operations
Now it’s your turn!
Let u = <-1, 3> , v = <2, 4> and w = <2, -5>
Find:
u+v
u + (-1)v
u–w
3v
2u + 3w
2u – 4v
– 2u – 3v
–u–v
Performing Vector Operations
Now it’s your turn!
Let u = <-1, 3> , v = <2, 4> and w = <2, -5>
Find:
u+v
u + (-1)v
u–w
3v
2u + 3w
2u – 4v
– 2u – 3v
–u–v
=
=
=
=
=
=
=
=
<1, 7>
<-3, -1>
<-3, 8>
<6, 12>
<4, -9>
<-10, -10>
<-4, -18>
<-1, -7>
Unit Vectors and Direction Angles
• Any vector can be broken down into its
components: a horizontal component and a
vertical component.
• In addition, any vector can be written as an
expression in terms of a standard unit vector.
• Unit vectors help us separate vectors into
components—a scalar and a unit vector.
Unit Vectors and Direction Angles
•
•
•
•
The standard unit vectors are i and j.
i = <1, 0>
j = <0, 1>
Using this style, we can now express
vectors as a linear combination.
Unit Vectors and Direction Angles
• Vector v = <a, b>
= <a, 0> + <0, b>
= a<1, 0> + b<0, 1>
• We now have linear combination.
v = a•i + b•j
Unit Vectors and Direction Angles
In vector v = a•i + b•j
a and b are now scalars and express the
horizontal and vertical components of
vector v.
We can use Trigonometry to calculate a
direction angle for our vector.
A unit vector is a vector with magnitude 1.
If we want to find the unit vector having the same
direction as a given vector, we find the magnitude of the
vector and divide the vector by that value.
w  3i  4 j
w 
 3   4
2
What is w ?
2
 25  5
If we want to find the unit vector having the same direction
as w we need to divide w by 5.
3 4
u i j
5 5
Let's check this to see if it really is
1 unit long.
2
2
25
3  4
u      
1
25
5  5
If we know the magnitude and direction of the vector, let's
see if we can express the vector in ai + bj form.
v  5,   150
5
As usual we can use the trig
we know to find the length
in the horizontal direction
and in the vertical direction.
150
v  v  cos  i  sin  j
5 3
5
v  5cos150i  sin 150 j  
i j
2
2
Acknowledgement
I wish to thank Shawna Haider from Salt Lake Community College, Utah
USA for her hard work in creating this PowerPoint.
www.slcc.edu
Shawna has kindly given permission for this resource to be downloaded
from www.mathxtc.com and for it to be modified to suit the Western
Australian Mathematics Curriculum.
Stephen Corcoran
Head of Mathematics
St Stephen’s School – Carramar
www.ststephens.wa.edu.au
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