Vectors

1
Vectors

• Sections 6.6

2
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.

3
Vocabulary
a straight line segment whose length is magnitude
and whose orientation in space is direction

• vector
• scalars
• scalar multiplication
• unit vectors
• magnitude

this is another name for a normal number
multiplying a vector by a number
a vector whose magnitude is 1 that has the same
direction as the original vector
length of the vector
(symbol )
4
Formulas

• Rectangular Coordinate
• Magnitude

a vector v with initial point and
terminal point can be represented in
rectangular coordinates as
a vector v with initial point and
terminal point has a magnitude of
5
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
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
continued on next slide
6
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
This vector has terminal point (3, 3). Since
this point is in quadrant I, our vector lies in
quadrant I.
7
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
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
continued on next slide
8
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
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.
9
Write a Vector in Terms of Its Magnitude and
Direction
a vector v with magnitude v and direction ?
can be written
A vector in rectangular coordinates can be
thought of a the terminal side of an angle in
standard position. The measure of the angle ? on
the interval 0, 2p) whose terminal side is the
vector v is the direction of the vector v.
10
Find the magnitude and direction of the vector u
8i j.
Recall that magnitude was defined as
a vector v with initial point and
terminal point has a magnitude of
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
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
11
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).
alternate magnitude formula.
Now back to our problem. If we continue to
simplify the magnitude calculation, we get
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
12
Find the magnitude and direction of the vector u
8i j.
a vector v with magnitude v and direction ?
can be written
We know that the magnitude of our vector is
We also know that a is 8 and b is -1. Thus we
have the following two equations.
We can use either one to solve for ?. 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
13
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.
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, p. This means that vector
would have to lie in quadrants I or II for the
angle ? 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
14
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 ? that
we found is not the direction. Since ? is in
quadrant I, the reference angle for ? is ?
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
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.
Thus the magnitude is and the direction is
15
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 ?, we will get an angle in
quadrant II. We will need to reference angle for
?. In quadrant II, the reference angle is found
using
In our case that will be reference angle
p ?
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
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.
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
16
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).
alternate magnitude formula.
Plugging in for a and b we get
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
17
Find the magnitude and direction of the vector v
1i 2j.
a vector v with magnitude v and direction ?
can be written
We know that the magnitude of our vector is
We also know that a is 1 and b is 2. Thus we
have the following two equations.
We can use either one to solve for ?. We will
get the same answer. Once again we will use the
cosine equation just to be consistent.
continued on next slide
18
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.
Since the inverse cosine function has a range of
0, p. Thus the angle ? is in quadrant I. Our
vector has a terminal point of (1, 2). This
point is also in quadrant I. Thus the direction
of the vector is
19
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
alternate magnitude formula.
continued on next slide
20
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
continued on next slide
21
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
continued on next slide
22
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
continued on next slide
23
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
24
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
25
Find the unit vector of the vector v 1i 2j.
The formula for find the unit vector is
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.
alternate magnitude formula.
continued on next slide
26
Find the unit vector of the vector v 1i 2j.
Now we divide the vector by or multiply
the vector by
This will give us
27
Write the vector v in terms of the i and j
components if v 3 and ? 60 .
a vector v with magnitude v and direction ?
can be written
28
Write the vector v in terms of the i and j
components if v 5 and ? 225 .
a vector v with magnitude v and direction ?
can be written
29
Write the vector v in terms of the i and j
components if v 5 and ? 180 .
a vector v with magnitude v and direction ?
can be written
30
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.
We can use either one to solve for ?. In the
example starting on slide 10, we solved the
cosine function. Here we will go through the
process of solving the sine function.
The angle ? found by solving this equation must
be in the interval -p/2, p/2 since this is the
range of the inverse sine function. Since the
argument of the arcsine function is negative
here, the angle ? must be in the interval -p/2,
0). Such an angle cannot be a direction of a
vector (remember direction must be in the
interval 0, 2p)).
continued on next slide
31
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, 2p). One thing that we
can do to find an angle in the interval 0, 2p),
it to find an angle coterminal to ? that is in
the interval 0, 2p). We do this by adding 2p to
the angle we have.
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
32
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.
If we solve this equation for the angle, we will
have the direction of a vector in quadrant II.
This basic equation will always produce the
direction angle of a vector in quadrant II.
continued on next slide
33
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, ?, in the interval -p/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, 2p). To do we
need to add 2p to the angle ?. Once there we
need to use the formula for finding the reference
angle in quadrant IV.
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.
This basic equation will always produce the
direction angle of a vector in quadrant III.
continued on next slide
34
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 ? quadrant II p reference
angle quadrant III p ? quadrant IV ? 2p