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

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

• vector • scalars • scalar

multiplication • unit vectors • magnitude

a straight line segment whose length is magnitude and whose orientation in space is direction 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 v(symbol: )

a vector v with initial point and terminal point has a magnitude of

a vector v with initial point and terminal point can be represented in rectangular coordinates as

Formulas• Rectangular Coordinate

• Magnitude

),( 11 yx

),( 22 yx

2122

12 yyxx v

jiv )()( 1212 yyxx

),( 11 yx

),( 22 yx

Write the vector v with initial point P1 = (2, —9) and terminal point P2 = (5, —6) in rectangular coordinates.

jiv

jiv

jiv

33

)96(3

))9()6(()25(

continued on next slide

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:

jiv )()( 1212 yyxx

Write the vector v with initial point P1 = (2, —9) and terminal point P2 = (5, —6) in rectangular coordinates.

jiv 33

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.

Write the vector v with initial point P1 = (3, 6) and terminal point P2 = (3, —1) in rectangular coordinates.

jv

jiv

jiv

7

)7(0

)6)1(()33(


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:

jiv )()( 1212 yyxx

Write the vector v with initial point P1 = (3, 6) and terminal point P2 = (3, —1) in rectangular coordinates.

jv 7

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.

a vector v with magnitude ||v|| and direction θ can be written

Write a Vector in Terms of Its Magnitude and Direction

jvivv sincos

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:

a vector v with initial point and terminal point has a magnitude of 212

212 yyxx v

),( 11 yx

),( 22 yx

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

22

22

18

0108

u

uYou 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).

con

tin

ued

on

next

slid

e

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).

22 ba v

Now back to our problem. If we continue to simplify the magnitude calculation, we get

65

164

18 22

u

u

u


alternate magnitude formula.

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.


65u

sin651

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.


jvivv sincos

cos658

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.

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, 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.

65

8arccos

cos65

8

cos658

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:


65

angle2anglereference

angle65

8arccos2

angle265

8arccos

angle265

8arccos

angle265

8arccos

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:

65

8arccos2

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

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

eanglanglereference

angleanglereference

angle2

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:

2direction

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.

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).

22 ba v

Plugging in for a and b we get

5

41

21 22

v

v

v


alternate magnitude formula.

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.

Find the magnitude and direction of the vector v = 1i + 2j.

5v

sin52

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.


jvivv sincos

cos51

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.

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, π]. Thus the angle θ 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

5

1arccos

cos5

1

cos51

5

1arccos

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

22 ba v alternate magnitude formula.

58

499

73 22

u

u

u



• u + v


This is asking us to add the vectors v and u. We add vectors by adding like terms

jivu

jivu

jjiivu

jijivu

157

)87()103(

87103

)810()73(


• u — v


This is asking us to subtract the vector v from the vector u. We subtract vectors by combining like terms

jivu

jivu

jjiivu

jijivu

jijivu

113

)87()103(

87103

81073

)810()73(


• 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

jiv

jiv

jiv

32404

))8(*4()10*4(4

)810(44


• 10u + 7vThis is asking us to first do the scalar multiplication and then add the results

jivu

jjiivu

jijivu

jijivu

1440710

56707030710

56707030710

)810(7)73(10710

the unit vector of v is a vector of magnitude 1 that has the same direction as the vector v

Definition and Formulas• Unit Vector

vv

Find the unit vector of the vector v = — 1i — 2j.The formula for find the unit vector is

vv

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.

22 ba v alternate magnitude formula.

5

41

21 22

u

u

u


Find the unit vector of the vector v = — 1i — 2j.

5Now we divide the vector by or multiply the vector by

This will give us

jivv

jivv

ji

vv

5

2

5

1

)21(5

1

5

)21(

5

1

Write the vector v in terms of the i and j components if ||v|| = 3 and θ = 60 °.


jvivv sincos

jiv

jiv

jiv

233

23

23

321

3

)60sin(3)60cos(3



jvivv sincos

jiv

jiv

jiv

225

225

22

522

5

)225sin(5)225cos(5



jvivv sincos

iv

jiv

jiv

5

0515

)180sin(5)180cos(5


sin651cos658


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.

From the example starting on slide 10, we got the following two equations that we could solve.

65

1arcsin

sin65

1

sin651The 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)).

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.



265

1arcsintocoterminalangle

This process will always work for a vector in quadrant IV when you use the sine equations to solve for the direction.

This basic equation will always produce the direction angle of a vector in quadrant II.

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. angleanglereference

IIquadrantinvectorofdirectionanglereference

angleanglereference

angleanglereference

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 III.

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.

anglereference

22anglereference

)2(2anglereference

IIIquadrantinvectorofdirection

angle)(

angleanglereference

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.


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

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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