For more information, please see full course syllabus of Trigonometry

For more information, please see full course syllabus of Trigonometry

## Discussion

## Study Guides

## Practice Questions

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### Vectors

**Main formulas**:

- All vectors have
*magnitude*and*direction*(except the zero vector, which has no direction). - Magnitude:
*r*= √{*x*^{2}+*y*^{2}} - Direction: θ = arctan(
*y*/*x*)*or*θ = arctan(*y*/*x*), depending on the signs of*x*and*y* - Components:
*x*= *r*cosθ*y*= *r*sinθ

**Example 1**:

**Example 2**:

^{°}ramp. Find the components of the force parallel to the ramp and perpendicular to the ramp.

**Example 3**:

^{°}W (this means 60

^{°}west of north) at an airspeed of 160 km/h. Meanwhile the wind is blowing

*from*the direction N60

^{°}E at 40 km/h. Assuming that Great Falls is 180km south of the US/Canadian border and the border runs due east/west, how long will it be until the plane enters Canadian airspace?

**Example 4**:

^{°}west of south.

**Example 5**:

^{°}east of south at 40km/h relative to the water. Find the true speed and course of the ship.

### Vectors

- Magnitude: r = √{x
^{2}+ y^{2}} - r = √{( − 7)
^{2}+ (12)^{2}} ⇒ r = √{193} - r = 13.9
- Direction Angle: θ = arctan[y/x] + 180
^{°}since the x value is negative and the vector is in quadrant II. - θ = arctan( − [12/7]) + 180
^{°}⇒ θ = − 59.73 + 180^{°} - θ ≈ 120.3
^{°}

^{°}

- Magnitude: r = √{x
^{2}+ y^{2}} - r = √{( − 5)
^{2}+ ( − 12)^{2}} ⇒ r = √{169} - r = 13
- Direction Angle: θ = arctan[y/x] + 180
^{°}since the x value is negative and the vector is in quadrant III. - θ = arctan([( − 12)/( − 5)]) + 180
^{°}⇒ θ = 67.38014 + 180^{°} - θ ≈ 247.4
^{°}

^{°}

^{°}East of South.

- 60
^{°}East of South means θ is 300^{°} - Components of a vector are: x = rcosθ ⇒ y = rsinθ
- x = 10cos300
^{°}, y = 10sin300^{°}

^{°}North of West.

- 30
^{°}North of West means θ is 150^{°} - Components of a vector are: x = rcosθ, y = rsinθ
- x = 12cos150
^{°}, y = 12sin150^{°}

^{°}ramp. Find the components of the force parallel to the ramp and perpendicular to the ramp.

- Use SOHCAHTOA to solve this problem
- cos60
^{°}= [x/20], sin60^{°}= [y/20] - x = 20cos60
^{°}, y = 20sin60^{°}

^{°}ramp. Find the components of the force parallel to the ramp and perpendicular to the ramp.

- Use SOHCAHTOA to solve this problem
- cos45
^{°}= [x/60], sin45^{°}= [y/60] - x = 60cos45
^{°}, y = 60sin45^{°}

^{°}with the horizontal. Find the vertical and horizontal components of the velocity.

- Horizontal component of velocity: x = rcosθ
- x = 70cos50
^{°} - Vertical component of velocity: y = rsinθ
- y = 70sin50
^{°}

^{°}with the horizontal. Find the vertical and horizontal components of the velocity.

- Horizontal component of velocity: x = rcosθ
- x = 50cos70
^{°} - Vertical component of velocity: y = rsinθ
- y = 50sin70
^{°}

^{°}above the horizontal. (Use the work formula W = FD where F is the component of the force in the direction of motion; D is the distance)

- F = 70cos45
^{°}, D = 35 - W = (70cos45
^{°})(35)

^{°}above the horizontal. (Use the work formula W = FD where F is the component of the force in the direction of motion; D is the distance)

- F = 65cos30
^{°}, D = 10 - W = (65cos30
^{°})(10)

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Vectors

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Vector Formulas and Concepts
- Vectors as Arrows
- Magnitude
- Direction
- Drawing Vectors
- Uses of Vectors: Velocity, Force
- Vector Magnitude Formula
- Vector Direction Formula
- Vector Components
- Example 1: Magnitude and Direction of Vector
- Example 2: Force to a Box on a Ramp
- Example 3: Plane with Wind
- Extra Example 1: Components of a Vector
- Extra Example 2: Ship with a Current

- Intro 0:00
- Vector Formulas and Concepts 0:12
- Vectors as Arrows
- Magnitude
- Direction
- Drawing Vectors
- Uses of Vectors: Velocity, Force
- Vector Magnitude Formula
- Vector Direction Formula
- Vector Components
- Example 1: Magnitude and Direction of Vector 8:00
- Example 2: Force to a Box on a Ramp 12:25
- Example 3: Plane with Wind 18:30
- Extra Example 1: Components of a Vector
- Extra Example 2: Ship with a Current

### Trigonometry Online Course

I. Trigonometric Functions | ||
---|---|---|

Angles | 39:05 | |

Sine and Cosine Functions | 43:16 | |

Sine and Cosine Values of Special Angles | 33:05 | |

Modified Sine Waves: Asin(Bx+C)+D and Acos(Bx+C)+D | 52:03 | |

Tangent and Cotangent Functions | 36:04 | |

Secant and Cosecant Functions | 27:18 | |

Inverse Trigonometric Functions | 32:58 | |

Computations of Inverse Trigonometric Functions | 31:08 | |

II. Trigonometric Identities | ||

Pythagorean Identity | 19:11 | |

Identity Tan(squared)x+1=Sec(squared)x | 23:16 | |

Addition and Subtraction Formulas | 52:52 | |

Double Angle Formulas | 29:05 | |

Half-Angle Formulas | 43:55 | |

III. Applications of Trigonometry | ||

Trigonometry in Right Angles | 25:43 | |

Law of Sines | 56:40 | |

Law of Cosines | 49:05 | |

Finding the Area of a Triangle | 27:37 | |

Word Problems and Applications of Trigonometry | 34:25 | |

Vectors | 46:42 | |

IV. Complex Numbers and Polar Coordinates | ||

Polar Coordinates | 1:07:35 | |

Complex Numbers | 35:59 | |

Polar Form of Complex Numbers | 40:43 | |

DeMoivre's Theorem | 57:37 |

### Transcription: Vectors

*We are learning about vectors and we are trying some more examples.*0000

*We are trying to find the components of a vector of magnitude 8 that points 30 degrees West of South.*0003

*First thing to do is to draw a picture here, let me put my North, East, South, West (Never Eat Shredded Wheat).*0009

*I want a vector that points 30 degrees West of South and that would be down there.*0023

*That is 30 degrees but I like to think of these things in terms of a reference angle from a positive x axis.*0029

*That would be a 60 degree angle there, the reference from the positive x axis would be a 180 + 60 that would 240 degrees.*0037

*Now I want to find the components of that vector, I know its magnitude is 8 so r=8.*0050

*Remember we got the components, these useful formulas x=arcos(theta), y=arcsin(theta).*0058

*We are going to use those, x=r(8)cos(theta) the reference angle is 240 degrees.*0069

*That is 8(cos) is the x coordinate of that angle and the x coordinate is -1/2.*0080

*Negative because its on the left hand side where the x value is negative so that is -4.*0089

*The y component is (8)sin(240degrees), again common value, I memorized the answers for multiples of 30 degrees so I do not need to check the calculator for this.*0097

*This is (8)(-square root of 3/2) that is -4 square root of 3.*0112

*The components there are xy=-4 square root of 3.*0122

*That one was one of the simpler ones, we are given a vector, magnitude, direction.*0136

*The magnitude of course is the r, the direction essentially is the theta but we want to draw a picture and make sure what theta we are talking about.*0140

*It says that this direction is given in terms of West of South*0150

*That is why I drew my vector down here, 30 degrees to the West of South and then I worked that out to be 240 degrees from the x axis.*0155

*Then I dropped those into arcos(theta) and arcsin(theta).*0165

*Simplify those down using the common values and I get the components of the vector.*0170

*Finally we have another one, it is a ship this time, sailing at a current that flows at 5km/hr NE.*0000

*The ship’s captain steers the course 30 degrees East of South at 40km/hr relative to water.*0009

*I want to find out the true speed and true course of the ship.*0017

*This is again one or more we are going to have two velocities.*0021

*There is the ship’s velocity and the current’s velocity, I want to figure out each one separately.*0025

*Find the components and add them up and then convert that back into a vector.*0030

*We are given the current first, I’m going to draw a graph of that and find its components.*0037

*The current flows 5km NE so this is the current, I will do this in black.*0046

*NE so that makes that this is N, this is E, that is a 45 degree angle.*0057

*The components we can get using arcos(theta) and arcsin(theta).*0064

*The current is arcos(theta), (phi)cos(45), (phi)sin(45) which is a common value that I recognized.*0072

*Remember that 45 is pi/4 and its cos is square root/2, so it its sin.*0092

*Now, I’m going to add up the vectors and the numbers are not going to be very nice.*0103

*I’m going to have to throw that in to my calculator.*0107

*Let me find the decimal approximation for (phi)(square root of 2)/2 it is about 3.54.*0111

*That is the components of the vector for the component.*0133

*Now, I will do the ship in blue, the ship is going 30 degrees East of South.*0137

*There is South down there, the ship is going 30 degrees East of South so that is a 30 degree angle.*0144

*It is steering a course at 40 km/hr relative to the water so the magnitude is 40 km.*0154

*I want to find the components of that vector.*0165

*I want to get a reference angle from the positive x axis, I know that it is 270 degrees down to the -y axis and then 30 more degrees, 270 + 30 would be 300 degrees.*0169

*To find the components of the ship’s vector we are going to use arcos(theta) and arcsin(theta).*0182

*The ship is (40)cos(300 degrees) and (40)sin(300 degrees).*0190

*300 degrees common value, I recognize that one that is 5pi/3 and its cos is +1/2 because its x coordinate is positive, its sin is (–square root of 3)/2 because its y coordinate is negative.*0204

*That is a common value, I do not need to check it on a calculator.*0228

*That simplifies down to 20 and -20 square root of 3.*0232

*I am going to be adding that together with the component for the current.*0237

*I am going to go ahead and throw that in to my calculator.*0243

*Of course 20 is still 20, 20 square root of 3 is 34.64, those are the components of the vectors for the ship.*0247

*Now, what I want to do is add these two up to find the net components of the ship’s travel.*0266

*Let me move the current vector down, I am going to draw a copy of the current vector, same magnitude and same direction.*0276

*I will draw the net in red here, I add these two vectors together and I will get the net.*0285

*Let me remind you that the blue one is the ship and the red one is the net.*0294

*For the net, I will just going to add those two together.*0302

*I take those two and I add them together, I get 23.54 for the (x).*0308

*For the (y) I get -34.64, so I get -31.10, that is the x and y components of the net travel of the ship.*0316

*We had its travel relative to the water but the water is also moving so the net travel, you add up those two velocities and you will get the component of the ship’s motion.*0343

*We are asked to find the true speed, speed is the magnitude of the velocity.*0353

*Remember velocity is a vector and speed is a number, and the true direction.*0361

*The speed is the (r), the magnitude of the velocity so that is the square root of (x ^{2} + y^{2}).*0369

*Where this is my (x), the (x) I am talking about is 23.54 and the y is -31.10.*0378

*I am going to work those out.*0386

*It works out to be just about 39.0 and it looks like our units here are km/hr.*0400

*That is the magnitude of that vector which is the true speed of the ship, that is half of our answer there.*0413

*Remember, we find the direction by finding (theta) is arctan(y/x) and then there is this question of whether you have to add 180.*0425

*I said you have to add 180 depending on which quadrant you are in.*0439

*You have to add 180 if you are in the second or third quadrant.*0444

*But we are in the fourth quadrant here, our (x) coordinate here so we are in the fourth quadrant, we do not have to add 180 here.*0452

*That is a small mercy there this is because we are in quadrant for or alternatively you can think of that as because the (x) is greater than 0, the (x) is positive there so you do not have to add 180.*0464

*We can just figure out the arctan(y/x), tan inverse, make sure you are in degree mode here.*0484

*The arctan(-31.1/23.54) I will work that out.*0492

*It works out to -52.9 degrees, of course I’m rounding there.*0514

*That really tells you the angle from the positive (x) axis, I do not really like giving a direction in terms of a negative angle.*0525

*If this is a real ship I would not want to say it is sailing in the negative angle.*0532

*Let me draw that out, that tells us that, that angle right there is 52.9 degrees.*0537

*A useful way to describe that would be as some angle East of South and so I can find the other complimentary angle by doing 90-52.9.*0553

*I worked that out and what it tells me is that its 37.1 degrees as reference in terms of East of South.*0581

*The direction that this ship is actually moving is 37.1 degrees East of South.*0607

*Just to remind you there, we calculated this angle right here that is where the 37.1 came from.*0621

*That was a pretty tricky problem, we had two different things going on here.*0630

*We had a ship sailing and a current, so we had a velocity vector for the current and then we had another velocity vector for the way the ship is steering relative to the water.*0634

*What we have to do there was to take the current’s vector 5km/hr NE and use arcos (theta) and arcsin(theta) to find the components of the current.*0646

*We took the ship’s velocity vector relative to the water and we use arcos(theta) arcsin(theta) here, the (r) is 40 because that is the magnitude, the speed of the ship relative to the water.*0663

*The 300 we got from the fact that it was 30 degrees East of South.*0677

*We graph that out and we found that it was 300 degrees from the positive (x) axis, that is where we get 300.*0682

*Plot that through arcos(theta) and arscin(theta) and we get a vector for the ship.*0690

*We add those two set of components together and we get the net components of the ship’s motion in terms of (x) and (y).*0697

*But we want to find the speed and direction, so we know the (x) and (y) components to find the speed that is the magnitude of the velocity.*0708

*We use square root (x ^{2} + y^{2}), that came out to be 39.10, and it is km/hr, because we solved it all our distances were in terms of km.*0716

*To find the direction, we used arctan(y/x) and there is always this issue of do you have to add 180? We are in quadrant 4 so we do not have to add 180.*0730

*We add 180 when if we were in quadrant 2 or 3, we do not have to add 180.*0740

*Another way to check that is to check whether the (x) is positive or negative, but our (x) is positive so ok no adding of 180.*0747

*We find the arctan (y/x) and it comes out to be -52.9 degrees.*0754

*That means that your 52.9 degrees below the positive (x) axis.*0761

*But if you want to give that in terms of the major compass directions North, South, East, West, we looked down there and discovered that is 37.1 degrees East of the South direction.*0766

*That is how we finally gave our answer as 37.1 degrees East of South.*0780

*That is the end of the vector lecture on trigonometry as part of the trigonometry series on www.educator.com.*0786

*Thanks for watching.*0792

*Hi, these are the trigonometry lectures on educator.com, and today we're going to talk about vectors which is a big topic in most trigonometry classes, and they also come up in a lot of physics applications as well.*0000

*You really want to think of vectors as being a way of measuring physical concepts that have both magnitude and direction, and kind of the graphical way you think about vectors is as arrows.*0012

*Think of vectors as being arrows.*0029

*Every vector has both magnitude and direction.*0033

*Those are the two things that are important about a vector.*0035

*The magnitude is the length of the vector.*0039

*Magnitude, that's the length of the arrow.*0044

*The direction is if you put it in the coordinate system, there's an angle θ that determines the direction of the vector.*0051

*Every vector has both magnitude and direction.*0061

*Of course, if the magnitude is 0, then you have an arrow that just reduces down to a dot, that doesn't really have a direction.*0063

*Most vectors, except for the zero vector, have magnitude and direction.*0071

*You can draw these vectors anywhere you want, you can move them around as long as you don't change the length or the magnitude, and as long as you don't change the direction.*0075

*You can move them around but you aren't allowed to move their direction, and you aren't allowed to stretch them or shrink them.*0087

*That's the idea of vectors.*0095

*They're used primarily to represent lots of physical concepts about things like velocity.*0097

*Why is velocity a vector?*0106

*We think of velocity as being synonymous with speed but that's really not quite the idea.*0108

*Velocity tells you which direction you're going and how fast you're going.*0115

*There's two ideas imbedded in velocity, one is how fast you're going, and one is the direction you're going.*0122

*That's something that's very natural to measure with a vector.*0130

*Another typical example of a physical concept that is measured with a vector is force.*0135

*If you push on something, you're pushing it in a particular direction and with a particular amount of force.*0141

*There's a magnitude that tells how hard you're pushing it, and there's a direction which tells you in which direction you're pushing it.*0148

*These are two examples of physical concepts.*0157

*There are lots and lots that when you try to describe them, you really need to talk about both magnitude and direction, so we keep track of them using vectors.*0161

*There's some equations to be associated with vectors.*0171

*If you think of vectors having components x and y, think of it as the terminal point of a vector where the arrow is going to as x and y in the initial point as the origin, then the magnitude ...*0178

*We often use r for the magnitude, you can figure that out using the Pythagorean theorem, it's just the square root of x ^{2}+y^{2}.*0197

*The direction is slightly more tricky.*0206

*To find the direction of a vector, you have to remember that tanθ=y/x, so it's tempting to say that θ=arctan(y/x), that's sometimes true but more subtle than that.*0209

*Let me show you why it's more subtle than that.*0227

*Remember that arctangent always gives you an angle between -π/2 and π/2.*0232

*Arctangent always gives you an angle in the fourth quadrant or the first quadrant, -π/2 to π/2, which means that if you take a vector in the second quadrant or the third quadrant ...*0240

*Let me label my quadrants here.*0262

*If you take a vector in the second quadrant or the third quadrant, and you take y/x, then you take arctan of that, it will give you an answer, an angle in the fourth or first quadrant and that will clearly be wrong.*0265

*The way you fix that is you add 180 to the arctan(y/x).*0282

*You want to do that whenever the vector is in the second or third quadrant, that in turn turns out to be whenever the x-coordinate is negative.*0288

*You use this formula when in the quadrants 2 or 3, which is the same as saying, when x is less than 0.*0306

*This standard formula, θ=arctan(y/x).*0331

*I'm running out of space here.*0338

*Let me make a little space over here.*0342

*Use when in quadrants 1 or 4, when x is bigger than 0.*0349

*That's a little tricky.*0367

*The formula for θ, it's usually arctan(y/x), but if you're in the second or third quadrant, or in other words, if your x-coordinate is less than 0, then it's arctan(y/x)+180.*0369

*That's probably the trickiest formula.*0381

*If you know r and θ, then it's easy to find x and y, this just comes from SOH CAH TOA.*0385

*There's the x-coordinate and the y-coordinate.*0402

*Remember sin(θ) is equal to the opposite, this is from SOH CAH TOA, over the hypotenuse.*0404

*Here, the opposite is y, the side opposite to y, the hypotenuse is r, you can solve that out to y=rsin(θ).*0413

*Similarly, cos(θ) is equal adjacent over hypotenuse, which is equal to x/r, you can solve that out to x=rcos(θ).*0431

*Those just come from old-fashioned right triangle trigonometry and SOH CAH TOA, x=rcos(θ), y=rsin(θ).*0448

*Those are always true no matter what is positive or negative.*0457

*You can be very safe with the x and y formulas.*0462

*The only tricky one is when you know x and y and you're solving for θ, it's a little tricky.*0465

*You have to check the sin(x), and you either have to use the arctan(y/x) or arctan(y/x)+180.*0470

*Let's practice that with some actual vectors and some examples.*0478

*The first problem is find the magnitude and direction of a vector whose horizontal and vertical components are -3 and 4.*0483

*Let me draw that vector.*0490

*X component is -3, y component is 4.*0495

*There's that vector (-3,4).*0505

*We want to find the magnitude and direction, that's r and θ, for that vector.*0508

*I'll use the formulas that we learned on the previous slide, r is equal to the square root of x ^{2}+y^{2}, which is the square root of -3^{2}+4^{2}.*0518

*That's 9+16, 25.*0533

*The magnitude of that vector is 5 units.*0540

*The direction of that vector, θ=arctan(y/x), 4/-3.*0547

*Remember that the formula for θ is tricky because it can be arctan(y/x) or, depending on where the vector is, it can be that plus 180.*0563

*This vector is in the second quadrant, we have to do plus 180 here.*0574

*That's because the x-coordinate is negative.*0579

*That's not a common value, let me work that out on my calculator.*0583

*I'm going to set degree mode here because I'm talking in terms of degrees.*0591

*When I said the 180, it's kind of determined that I have to be using degrees here.*0597

*Arctan(4/-3)=-53.1+180.*0603

*Of course, -53.1 degrees would be down there, that's -53.1 degrees.*0613

*That's certainly wrong because that's 180 degrees away from the vector that we're looking for, that's why we have to add the 180.*0623

*When we add the 180 to that, we get 126.9 degrees.*0629

*If you want to give that as an angle from the x-axis, there's your answer, 126.9 degrees.*0644

*In a lot of these applications, we're going to give North, South, East, West compass directions for our directions.*0650

*If you think about that, that's 90 degrees plus 36.9 degrees, that means that angle right there is 36.9 degrees.*0655

*If you want to give your answers in terms of a compass direction, this will be 36.9 degrees west of north.*0673

*It kind of depends on the kind of answer you're looking for.*0686

*If you're looking for an answer as an angle around from the x-axis, there's your answer right there, 126.9.*0691

*If you want a compass direction oriented from the North Pole, then I would call this 36.9 degrees west of north.*0698

*Let's recap there.*0706

*We were given the two components of a vector, the x and y, we find the r just by this Pythagorean formula, we throw those in and we get the magnitude of the vector.*0708

*We find the direction of the vector by using arctan(y/x), but because the vector is in the second quadrant, because its x is negative, we have to add 180.*0718

*We worked that through, we get an angle, and then if we want to describe it in terms of compass directions, we write it as 36.9 degrees west of north.*0732

*Let's look at another example here.*0746

*We're told that a horizontal force of 30 Newtons is applied to a box on a 30-degree ramp, and we want to find the components of the force parallel to the ramp and perpendicular to the ramp.*0749

*Certainly, a problem like this, you want to draw a picture.*0760

*You definitely want to draw a picture.*0763

*Let me draw a 30-degree ramp, there's the ground, there's my 30-degree ramp.*0768

*I've got a box on this ramp.*0779

*Apparently, I'm pulling on this box horizontally with a force of 30 Newtons.*0785

*There's my horizontal force and I know that that's 30 Newtons.*0795

*What I want to do is find the components of the force parallel to the ramp and perpendicular to the ramp.*0802

*I want to break this vector up into two pieces, one of which is parallel to the ramp, one of which is going straight up the ramp, and one of which is perpendicular to the ramp.*0808

*One of which goes straight in to the ramp.*0822

*If you see, I'm finding two vectors that are parallel and perpendicular to the ramp that add up to the force there.*0824

*Then I want to figure out what those vectors are.*0834

*I'm going to translate this angle 30 degrees.*0840

*This is probably a little easier if I get rid of some of the elements of the picture.*0844

*Let me just write this as a right triangle like that, with a 30-degree angle there.*0849

*That's the hypotenuse of the triangle, it's 30 Newtons there.*0862

*And I'm trying to find the lengths of the other sides.*0866

*This is a right triangle, it's okay to use SOH CAH TOA, I don't have to use law of sines or law of cosines because I have a right triangle.*0868

*SOH CAH TOA, that says that sin(θ) is equal to the opposite over the hypotenuse.*0881

*The opposite angle to the angle θ here or the opposite side is that.*0889

*That's the adjacent angle to angle, if we call that angle θ.*0897

*Of course, this is the hypotenuse.*0905

*Sin(30) is equal to the opposite over the hypotenuse, is 30, the opposite is equal to 30sin(30).*0908

*That's one of my common values, I remember the sin(30).*0924

*I don't have to go to the calculator for this one.*0927

*The sin(30), that's π/6, that's equal to the sin(π/6)=1/2, 30×1/2 is 15.*0929

*To find the adjacent side, I'm going to use the cosine part of SOH CAH TOA.*0951

*Cosine is equal to adjacent over hypotenuse.*0955

*Cos(30) is equal to ...*0964

*I don't know the adjacent yet but I know that the hypotenuse is 30.*0969

*If I solve that for the adjacent side, I get 30cos(30).*0972

*Again, 30 is one of my common values, it's π/6.*0980

*I know that the cos(30) is root 3/2, so I get 15 root 3.*0982

*If I fill those in, this is 15 root 3, and this is 15, the answer to the original question is that the parallel force, force parallel to the ramp is 15 root 3 Newtons, and the force perpendicular to the ramp is exactly 15 Newtons.*0994

*Let's recap what made that problem work.*1047

*Again, we started with the word problem, first thing you want to do is draw a picture.*1050

*I drew my picture of my box on a 30-degree ramp, I drew in my horizontal force as a vector, vector 30 units long, because it's a force of 30 Newtons.*1055

*I was asked to find the components both parallel and perpendicular to the ramp.*1065

*What I really had to do is break this arrow up into two arrows, one of which was parallel to the ramp, one of which is perpendicular to the ramp.*1071

*I broke it up into those two arrows, those two vectors then I abstracted into another triangle, then I could use SOH CAH TOA to find the lengths of those two sides of the triangles.*1081

*That's what I did here, I found the lengths of those two sides of the triangles, I plugged them back in to my original picture, then I interpreted them in the context of the original problem.*1096

*We have another word problem here.*1112

*It says a small plane leaves Great Falls heading north, 60 degrees west, that means 60 degrees west of north, an air speed of 160 kilometers per hour.*1116

*Meanwhile, the wind is blowing from north 60 degrees east at 40 kilometers per hour.*1127

*We're told the Great Falls is 180 kilometers south of the Canadian Border.*1134

*We want to know how long until the plane enters Canadian air space.*1140

*This is quite a complicated vector problem.*1142

*As usual with a problem, when you're overwhelmed by lots of words, you want to draw a picture, so let me try to draw a picture here.*1146

*We're told that the plane is heading 60 degrees west of north.*1158

*Let me just write in my compass directions here, North, West, South, East.*1164

*We're told that the plane is heading 60 degrees west of north.*1174

*There it is, that's a 60-degree angle.*1182

*It's heading at an air speed of 160 kilometers per hour, that means that the magnitude of that vector is 160.*1186

*Meanwhile, the wind is blowing from the direction north 60 degrees east at 40 kilometers per hour.*1196

*I think what we're going to be doing is we're going to be adding the contribution of the plane's engines and the contribution of the wind.*1203

*We want to figure those out separately.*1213

*I want to find the components of each one of those vectors.*1216

*Let's find the components of the plane.*1221

*If that's 60 degrees west of north, then that's really 60+90, that's 150 degrees.*1225

*The vector of the plane's motion is, remember x=arccos(θ) and y=arcsin(θ), we know our θ is 150 degrees, so x=160cos(150), and the y component is 160sin(150).*1239

*Let's figure out those first.*1280

*Cos(150), that's one of the common values, that's 5π/6.*1283

*I remember that it's cosine is negative square root of 3 over 2.*1289

*It's negative because the x-coordinate is negative, over there in the second quadrant.*1297

*Sin(150), that's sine of 5π/6, that's 1/2, that's a common value that I remember.*1300

*This is -80 square root of 3, and 80 for the plane's motion.*1312

*Those are the x and y components of the plane's motion.*1320

*Meantime, the wind, let me fork out the wind in blue.*1325

*The wind is blowing from north 60 degrees east.*1329

*The wind is coming from 60 degrees east of nort, it's coming from that direction.*1338

*What I'm going to do is move that vector down so that it starts at the origin.*1350

*There's the wind in blue.*1358

*We know that it's blowing at 40 kilometers per hour.*1361

*We want to figure out the components of the wind.*1365

*Let me figure out the wind.*1369

*Now, that is a 30-degree angle there, because that's 60.*1372

*The whole angle is 210, because it's 30 degrees past 180.*1381

*The angle is 210.*1388

*Using rcos(θ), rsin(θ), we have 40cos(210), and 40sin(210).*1390

*We got to work out what those are.*1406

*The cos(210), again that's a common value, that's 7π/6.*1408

*I can work that out, that's negative root 3/2, negative because we're over there in the third quadrant, the x-coordinate is negative.*1414

*Sin(210)=-1/2, it's negative because again third quadrant and the y-coordinate is negative there too.*1428

*This is -20 root 3, and -20.*1437

*That's the components of the wind's vector.*1447

*We found the plane's vector and the wind's vector, the net travel of the plane will be the sum of those two vectors.*1452

*We have the plane in black.*1468

*We have the wind blowing it in blue.*1474

*In red, I'll show the net travel is the sum of the two vectors.*1480

*I want to add up the x and y components of the plane and the wind there.*1486

*I'm going to do all these in red.*1492

*The net is equal to the plane plus the wind.*1495

*The net velocity of the plane, that's equal to ...*1501

*If I add up -80 square root of 3 and -20 square root of 3, I get -100 square root of 3, then 80-20=60.*1515

*That came from adding up those two vectors.*1527

*The net travel is -100 square root of 3 in the x direction, and 60 in the y direction.*1533

*Let's go back and read some more of the problem.*1547

*It says the Great Falls is 180 kilometers south of the Canadian border, and the border runs due east-west.*1550

*Let me draw a very rough map of the United States and Canada here.*1558

*Here's the US, and here's Canada.*1564

*We're told that this plane, now we know it's travelling according to that vector, we're really only interested when it crosses the Canadian border.*1568

*That is a matter of how long it takes, its y-component, to get us across the Canadian border, and we're told that the vertical distance is 180 kilometers.*1581

*Let me draw my net vector a little bit shorter here to make a little more to scale.*1604

*The y-component there is 60 kilometers per hour.*1611

*The question is how long will it take to go 180 kilometers north if its y-component is increasing at 60 kilometers per hour.*1616

*At 60 kilometers per hour, that's an easy division problem, it will take 3 hours to go 180 kilometers north.*1628

*Let's go back and look at that problem again and recap it.*1657

*Lots of words to start with, we're greeted by this problem that's very long and very wordy.*1660

*First thing to do is draw a picture and try to isolate the different quantities involved.*1667

*First we drew a picture of the plane's motion.*1671

*The plane's motion is this one in black here, this black vector we know it's 160, it has a 160 magnitude.*1674

*We're told that the angle is 60 degrees east of north, but in terms of a reference angle from the x-axis, that's really 150 degrees because it's 90+60.*1683

*That's where the 150 comes in.*1701

*To find its components we used x=rcos(θ), y=rsin(θ).*1702

*We plug those in and we get the components of the plane's motion.*1708

*Then we want to try to find the components of the wind's vector.*1712

*The wind is blowing from north 60 degrees east at 40 kilometers per hour.*1719

*That's where we get this vector is blowing from the northeast, I showed it blowing to the southwest.*1726

*From the northeast is the same as to the southwest.*1730

*It has magnitude 40, I put that in to rcos(θ) and rsin(θ), and I work out the components of the wind's motion.*1733

*The net motion of the plane is the way its flying, it's air speed, or it's velocity relative to the wind, plus the wind's velocity.*1744

*I add those two together, and in red, I get a vector representing the net velocity of the plane.*1757

*That's my vector representing the net velocity of the plane.*1765

*Finally, the question when you blow all the words out of it is asking how long does it take the plane to go 180 miles north.*1769

*Since the border runs east-west, it really doesn't matter how far the plane moves east or west during that time, we only care how long it takes to go 180 kilometers north.*1779

*I think I said miles before but our unit is kilometers.*1789

*We want to go 180 kilometers north, we don't care about the east-west motion which is why I never really did anything with the -100 square root of 3, instead we looked at their vertical motion, the 60 ...*1791

1 answer

Last reply by: Dr. William Murray

Wed Nov 11, 2015 9:22 AM

Post by Peter Ke on November 6, 2015

I am new to vectors, can you explain why sometimes when you do vector examples you have to add the degrees. For example, the last question I asked and the extra example 2 were you added 90 + 90 + 90 + 30 = 300 degrees. Why is that?

Because I did it this way:

Ship:

x = 40 cos(30)

x = 34.64

y = 40 sin(30)

y = 20

Answer = (34.64,20)

1 answer

Last reply by: Dr. William Murray

Wed Nov 11, 2015 9:22 AM

Post by Peter Ke on November 6, 2015

I REALLY REALLY don't get why the wind is 210 degrees for example 3. Please explain!

2 answers

Last reply by: Dr. William Murray

Mon Nov 26, 2012 6:51 PM

Post by William Davis on January 2, 2012

If the wind is blowing 60 degrees east of north, shouldn't it be blowing 30 North east?