Download Vectors-with

Vectors
Sections 6.6
Objectives
• Rewrite a vector in rectangular coordinates
(in terms of i and j) given the initial and
terminal points of the vector.
• Determine the magnitude and direction of a
vector given in terms if i and j.
• Add and subtract vectors given in terms of i
and j.
• Multiply a vector given in terms of i and j by a
real number (scalar multiplication).
• Find a unit vector for a vector given in terms
of i and j.
• Write a vector in terms of i and j the
magnitude and direction of the vector.
Vocabulary
•
•
•
•
•
straight line segment whose length is magnitude
vector aand
whose orientation in space is direction
scalars this is another name for a normal number
scalar multiplication multiplying a vector by a number
vector whose magnitude is 1 that has the same
unit vectors adirection
as the original vector
magnitude length of the vector (symbol: v )
Formulas
• Rectangular Coordinate
a vector v with initial point (x1 , y 1 ) and
terminal point (x 2 , y 2 ) can be
represented in rectangular coordinates
as v  (x 2  x 1 )i  (y 2  y 1 ) j
• Magnitude
a vector v with initial point (x1 , y 1 ) and
terminal point (x 2 , y 2 ) has a magnitude
of v  x2  x1 2  y2  y1 2
Write the vector v with initial point
P1 = (2, —9) and terminal point
P2 = (5, —6) in rectangular
coordinates.
To write a vector in rectangular coordinates, we need the following basic
set up:
v  (x 2  x 1 )i  (y 2  y 1 ) j
The coordinate x1 and y1 come from the initial point and x2 and y2 come
from the terminal point. When we plug these into the equation we get:
v  (5  2)i  ((6)  ( 9)) j
v  3i  ( 6  9) j
v  3i  3j
continued on next slide
Write the vector v with initial point
P1 = (2, —9) and terminal point
P2 = (5, —6) in rectangular
coordinates.
Although we are not asked, it is important to know in which quadrant our
vector v lies. All vectors in rectangular coordinates are in standard
position with their initial point at the origin (0, 0). The terminal point of
the vector is the point (a, b) where v = a i + b j.
Our vector is
v  3i  3j
This vector has terminal point (3, 3). Since this point is in quadrant I, our
vector lies in quadrant I.
Write the vector v with initial point
P1 = (3, 6) and terminal point
P2 = (3, —1) in rectangular
coordinates.
To write a vector in rectangular coordinates, we need the following basic
set up:
v  (x 2  x 1 )i  (y 2  y 1 ) j
The coordinate x1 and y1 come from the initial point and x2 and y2 come
from the terminal point. When we plug these into the equation we get:
v  (3  3)i  ((1)  6) j
v  0i  ( 7) j
v  7 j
continued on next slide
Write the vector v with initial point
P1 = (3, 6) and terminal point
P2 = (3, —1) in rectangular
coordinates.
Although we are not asked, it is important to know in which quadrant our
vector v lies. All vectors in rectangular coordinates are in standard
position with their initial point at the origin (0, 0). The terminal point of
the vector is the point (a, b) where v = a i + b j.
Our vector is
v  7 j
This vector has terminal point (0, -7). Since this point is on the y-axis
below the x-axis, our vector is point straight downward along the y-axis and
thus is not in any quadrant.
Write a Vector in Terms of Its
Magnitude and Direction
a vector v with magnitude ||v|| and
direction θ can be written
v  v cos  i  v sin  j
A vector in rectangular coordinates can be thought of a the terminal side
of an angle in standard position. The measure of the angle θ on the
interval [0, 2π) whose terminal side is the vector v is the direction of the
vector v.
Find the magnitude and direction
of the vector u = 8i — j.
Recall that magnitude was defined as:
In this problem, we have no initial point or terminal point given. Instead
our vector is in rectangular coordinates and thus has initial point (0, 0)
and terminal point (a, b) where in our case a is 8 and b is -1. When we
plug this initial and terminal point into the magnitude formula, we get
u 
8  0 
u 
82   12
2
  1  0 
2
You should notice that the numbers
in the parentheses to be squared
correspond exactly to the a and b
in our vector (form v = a i + b j).
continued on next slide
a vector v with initial point (x1 , y 1 ) and
terminal point (x 2 , y 2 ) has a magnitude
of v  x2  x1 2  y2  y1 2
Find the magnitude and direction
of the vector u = 8i — j.
This will give us an alternate formula for magnitude when the
vector is already in rectangular coordinates (form v = a i + b j).
v 
2


a   b
2
alternate magnitude formula.
Now back to our problem. If we continue to simplify the magnitude
calculation, we get
u 
82   12
u  64  1
u  65
Now that we have the magnitude, we can use the formula that defines
a vector in terms of its magnitude and direction to find the direction.
continued on next slide
Find the magnitude and direction
of the vector u = 8i — j.
a vector v with magnitude ||v|| and direction θ can be written
v  v cos  i  v sin  j
We know that the magnitude of our vector is u  65
We also know that a is 8 and b is -1. Thus we have the following two
equations.
8  65 cos 
 1  65 sin 
We can use either one to solve for θ. We will get the same answer.
The cosine equation turns out to be the easier equation to work
with when it comes to finding the direction of a vector.
If you want to see how to find the direction by solving the
sine equation, go to slide 30 of this slide show.
continued on next slide
Find the magnitude and direction
of the vector u = 8i — j.
We are going to use what we know about solving trigonometric equations
to solve the equation below.
8  65 cos 
8
 cos 
65
 8 
arccos 
 
 65 
We may be tempted to say that the direction is the angle found
above. But we run into one problem with this. The inverse cosine
function has a range of [0, π]. This means that vector would have to
lie in quadrants I or II for the angle θ to be the direction of the
vector.
This means that we have to check the quadrant in which our vector
lies to find the direction.
continued on next slide
Find the magnitude and direction
of the vector u = 8i — j.
Our vector has a terminal point at (8, -1). This point is in quadrant IV.
Thus the angle θ that we found is not the direction. Since θ is in
quadrant I, the reference angle for θ is θ itself. We need to find the
angle in quadrant IV that has the same reference angle. The formula for
finding a reference angle in quadrant IV is:
reference angle  2  angle
We now plug in the reference angle to the formula and solve for the
angle. The angle that we get will be the direction of the vector.
 8 
arccos 
  2  angle
65


 8 
arccos 
  2   angle
65


 8 
 arccos 
  2  angle
 65 
 8 
2  arccos 
  angle
 65 
Thus the magnitude is:
65
 8 
and the direction is: 2  arccos 

 65 
The process that we just went through in our problem to find the
direction of a vector that is in quadrant IV will also need to be done for
vectors in quadrant III.
When we solve the cosine equation for θ, we will get an angle in quadrant
II. We will need to reference angle for θ. In quadrant II, the reference
angle is found using
reference angle    angle
In our case that will be:
reference angle = π — θ
Our next step will be to find the angle in quadrant III that has that
same reference angle. The formula for finding a reference angle in
quadrant III is:
reference angle  angl e  
We now plug in the reference angle to the formula and solve for the
angle. The angle that we get will be the direction of the vector.
    angle  
2    angle
Notice that this is the same as when we had a vector in quadrant IV.
Thus if our vector is in quadrants III or IV, the direction can always
be found as direction  2  
Find the magnitude and direction
of the vector v = 1i + 2j.
We can use the alternate formula for magnitude when the
vector is already in rectangular coordinates (form v = a i + b j).
v 
2


a   b
2
alternate magnitude formula.
Plugging in for a and b we get
v 
12  22
v  14
v  5
Now that we have the magnitude, we can use the formula that defines
a vector in terms of its magnitude and direction to find the direction.
continued on next slide
Find the magnitude and direction
of the vector v = 1i + 2j.
a vector v with magnitude ||v|| and direction θ can be written
v  v cos  i  v sin  j
We know that the magnitude of our vector is
v  5
We also know that a is 1 and b is 2. Thus we have the following two
equations.
1  5 cos 
2  5 sin 
We can use either one to solve for θ. We will get the same answer.
Once again we will use the cosine equation just to be consistent.
continued on next slide
Find the magnitude and direction
of the vector v = 1i + 2j.
We are going to use what we know about solving trigonometric equations
to solve the equation below.
1  5 cos 
1
 cos 
5
 1 
arccos 
 
 5
Since the inverse cosine function has a range of [0, π]. Thus the angle
θ is in quadrant I. Our vector has a terminal point of (1, 2). This
 1 
point is also in quadrant I. Thus the direction of the vector is arccos 

 5
Given the vectors u = —3i — 7j
and v = 10i — 8j, find
• ||u||
This is asking us to find the magnitude of the vector u. We can
use the alternate magnitude formula for this
v  a 2  b 
2
u 
alternate magnitude formula.
 32   7 2
u  9  49
u  58
continued on next slide
Given the vectors u = —3i — 7j
and v = 10i — 8j, find
• u+v
This is asking us to add the vectors v and u. We add vectors by
adding like terms
u  v  ( 3i  7 j)  (10i  8 j)
u  v  3i  10i  7 j  8 j
u  v  ( 3  10)i  ( 7  8) j
u  v  7i  15 j
continued on next slide
Given the vectors u = —3i — 7j
and v = 10i — 8j, find
• u—v
This is asking us to subtract the vector v from the vector u. We
subtract vectors by combining like terms
u  v  ( 3i  7 j)  (10i  8 j)
u  v  3i  7 j  10i  8 j
u  v  3i  10i  7 j  8 j
u  v  ( 3  10)i  ( 7  8) j
u  v  13i  1 j
continued on next slide
Given the vectors u = —3i — 7j
and v = 10i — 8j, find
• 4v
This is asking us to multiply the vector v by the scalar 4. This is
done just as we do distribution of multiplication over addition
4 v  4(10i  8 j)
4 v  ( 4 * 10)i  ( 4 * ( 8)) j
4 v  40i  32j
continued on next slide
Given the vectors u = —3i — 7j
and v = 10i — 8j, find
• 10u + 7v
This is asking us to first do the scalar multiplication and then add
the results
10 u  7 v  10( 3i  7 j)  7(10i  8 j)
10 u  7 v  30i  70 j  70i  56j
10 u  7 v  30i  70i  70 j  56j
10 u  7 v  40i  14 j
Definition and Formulas
• Unit Vector
the unit vector of v is a vector of
magnitude 1 that has the same direction
as the vector v
v
v
Find the unit vector of the
vector v = — 1i — 2j.
The formula for find the unit vector is
v
v
This requires that we first find the magnitude of the vector and
then multiply the vector by the scalar that is 1 divided by the
magnitude. We can find the magnitude using the alternate
magnitude formula.
v  a   b 
2
2
u 
alternate magnitude formula.
 12   22
u  14
u  5
continued on next slide
Find the unit vector of the
vector v = — 1i — 2j.
Now we divide the vector by
5
This will give us
( 1i  2 j)
v

v
5
v
1

( 1i  2 j)
v
5
v
1
2

i
j
v
5
5
or multiply the vector by
1
5
Write the vector v in terms of
the i and j components if
||v|| = 3 and θ = 60 °.
a vector v with magnitude ||v|| and direction θ can be written
v  v cos  i  v sin  j
v  3 cos( 60) i  3 sin( 60) j
 3
1
j
v  3  i  3

2
2


3 3 3
v  i
j
2
2
Write the vector v in terms of
the i and j components if
||v|| = 5 and θ = 225 °.
a vector v with magnitude ||v|| and direction θ can be written
v  v cos  i  v sin  j
v  5 cos( 225) i  5 sin(225) j


2
2
 i  5 
j
v  5 



2
2




5 2
5 2
v
i
j
2
2
Write the vector v in terms of
the i and j components if
||v|| = 5 and θ = 180 °.
a vector v with magnitude ||v|| and direction θ can be written
v  v cos  i  v sin  j
v  5 cos( 180) i  5 sin(180) j
v  5 1i  50  j
v  5 i
Find the magnitude and direction
of the vector u = 8i — j.
From the example starting on slide 10, we got the following two
equations that we could solve.
8  65 cos 
 1  65 sin 
We can use either one to solve for θ. In the example starting on
slide 10, we solved the cosine function. Here we will go through
the process of solving the sine function.
 1  65 sin 
1
 sin 
65
 1 
arcsin 
 
 65 
The angle θ found by solving this equation
must be in the interval [-π/2, π/2] since this
is the range of the inverse sine function.
Since the argument of the arcsine function is
negative here, the angle θ must be in the
interval [-π/2, 0). Such an angle cannot be a
direction of a vector (remember direction
must be in the interval [0, 2π)).
continued on next slide
Find the magnitude and direction
of the vector u = 8i — j.
We need to take the angle that we have and move it to the interval
[0, 2π). One thing that we can do to find an angle in the interval [0, 2π),
it to find an angle coterminal to θ that is in the interval [0, 2π). We do
this by adding 2π to the angle we have.
 1 
angle coterminal to   arcsin
  2
 65 
This process will always work for a vector in quadrant IV when you use
the sine equations to solve for the direction.
continued on next slide
This process will not work for a vector which falls in any other quadrant.
If the vector is in quadrant I, solving the sine equation for the direction
will give you the answer without further work.
For all other quadrants, we need to remember that two angles which have
the same sine value will have the same reference angle.
If our vector fall in quadrant II, we know that the sine of the direction
will be positive. To find out what the direction is, we would first solve
the sine equation. This will give us the reference angle. We then plug the
reference angle into the equation for finding the reference angle in
quadrant II.
reference angle    angle
If we solve this equation for the angle, we will have the direction of a
vector in quadrant II.
reference angle    angle
  reference angle  angle
  reference angle  direction of vector in quadrant II
This basic equation will always produce the direction angle of a vector
in quadrant II.
continued on next slide
Once again we must remember that two angles which have the same sine
value will have the same reference angle.
If our vector fall in quadrant III, we know that the sine of the direction
will be negative. To find out what the direction is, we would first solve
the sine equation. This will give us angle, θ, in the interval [-π/2, 0] (the
part of the range for the inverse sine function comes from negative
inputs).
To find the reference angle for this angle, we first need for it to be in
the interval [0, 2π). To do we need to add 2π to the angle θ. Once there
we need to use the formula for finding the reference angle in quadrant
IV.
reference angle  2  (  2 )
reference angle  2    2
reference angle  
Now we need to put this reference angles into quadrant III by
plugging it into the equation for finding a reference angle in quadrant
III and solving for the angle that is the direction of the vector.
reference angle  angle  
  (  )  angle
    direction of vector in quadrant III
This basic equation will always produce the direction angle of a vector
in quadrant III.
continued on next slide
Having gone through all this for finding the direction of the vector by
solving the sine equation that is produced, you can see that this is more
complicated than using the cosine equation.
Each quadrant produces a different process for finding the direction of a
vector:
quadrant I: θ
quadrant II: π – reference angle
quadrant III: π – θ
quadrant IV: θ + 2π