WEBVTT mathematics/pre-calculus/selhorst-jones 00:00:00.000 --> 00:00:02.400 Hi--welcome back to Educator.com. 00:00:02.400 --> 00:00:06.900 Today, we are going to talk about vectors; this starts an entirely new section for us. 00:00:06.900 --> 00:00:11.100 We are getting into really new territory; we are going to talk about vectors, and then later on we will be talking about matrices. 00:00:11.100 --> 00:00:16.600 But first, let's talk about vectors: when we talk about the force on an object, we need to know two things: 00:00:16.600 --> 00:00:22.900 the magnitude of the force--how hard it is being pushed, and the direction--which way it is being pushed. 00:00:22.900 --> 00:00:26.700 If we are going to talk about something being pushed, we can't just talk about how hard it is being pushed. 00:00:26.700 --> 00:00:32.200 We also have to know which way it is being pushed; it is very different to push something this way than it is to push something this way. 00:00:32.200 --> 00:00:34.600 But even if we know the direction, we also have to know how hard it is. 00:00:34.600 --> 00:00:38.700 We are pushing really hard versus pushing very slightly--there are big differences here. 00:00:38.700 --> 00:00:43.800 We need to know both magnitude and direction; we need to know how hard and which way. 00:00:43.800 --> 00:00:48.100 Now, one number on its own won't be enough to get across both pieces of information. 00:00:48.100 --> 00:00:54.400 So, this leads us to the idea of a vector--a way to be able to talk about magnitude and direction at the same time-- 00:00:54.400 --> 00:00:58.200 be able to talk about both of these things in one piece of information. 00:00:58.200 --> 00:01:04.900 Vectors are massively useful: they are used throughout math; they are in the sciences everywhere, especially in physics; 00:01:04.900 --> 00:01:08.900 they are in engineering, computer programming, business, medicine, and more fields. 00:01:08.900 --> 00:01:15.700 Pretty much, if a field even vaguely uses math, it is going to use vectors; so this is a really useful thing that we are talking about in this lesson. 00:01:15.700 --> 00:01:20.600 Now, I want to point out that this lesson will use some basic trigonometry to figure out angles. 00:01:20.600 --> 00:01:27.200 So, make sure you have some familiarity with how trigonometry works, so that you can understand what is going on when we are figuring out angles. 00:01:27.200 --> 00:01:34.800 All right, let's go on: the idea of a vector: a vector is just trying to get across the idea of length and direction. 00:01:34.800 --> 00:01:42.100 Graphically, if we look at a picture of it, it is exactly that: it is a directed line segment, a length with direction. 00:01:42.100 --> 00:01:44.300 It is both of these things at once. 00:01:44.300 --> 00:01:52.200 So, in this case, we are starting at (0,0), and we have this length here; and the arrow at the end says that we are going in this direction. 00:01:52.200 --> 00:01:56.200 So, we have a chunk of length, and we see which way it is pointed. 00:01:56.200 --> 00:02:00.000 It is pointed in a very specific direction; it has some angle to it. 00:02:00.000 --> 00:02:02.900 How do we call out a vector, if we want to talk about a vector? 00:02:02.900 --> 00:02:11.200 Normally, we denote it with an overhead arrow like this: u with an arrow above it says that we are talking about the vector u. 00:02:11.200 --> 00:02:15.100 Or we can put it in bold face with a bold u. 00:02:15.100 --> 00:02:19.800 And once in a while, if we are talking about nothing but vectors, and there is nothing else showing up, 00:02:19.800 --> 00:02:24.700 sometimes it will just be assumed; but we are pretty much never going to see that--not in this course. 00:02:24.700 --> 00:02:30.200 Vectors are normally shown with lowercase letters; however, just like variables, you can use any symbol. 00:02:30.200 --> 00:02:37.000 But for the most part, we will stick with lowercase letters: u and v are very common letters for talking about them. 00:02:37.000 --> 00:02:45.700 One last thing: we use this u with the arrow on top, and that is how we will be denoting it in this course, although you might see bold face in textbooks. 00:02:45.700 --> 00:02:51.500 When you are actually writing it by hand, I write it like this, where I don't really have a full arrow. 00:02:51.500 --> 00:02:58.000 I have more of a harpoon, where it is just an arrow on one side--just this sort of arrow flange on one side. 00:02:58.000 --> 00:03:02.500 That is how I write it; you could write it with an actual arrow on top--there is nothing wrong with that. 00:03:02.500 --> 00:03:08.700 But I am lazy, like us all, and so I just tend to write it like that, because it is the fastest way that I know to write it. 00:03:08.700 --> 00:03:11.100 But it still gets across the idea of a vector. 00:03:11.100 --> 00:03:14.300 So, this is a perfectly fine way to write it by hand when you are working it out. 00:03:14.300 --> 00:03:20.200 But when we have it actually written out in this lesson, we will have it like that with an arrow on top. 00:03:20.200 --> 00:03:24.200 We can also give the vector algebraically by its location in the plane. 00:03:24.200 --> 00:03:28.700 We have these nice rectangular coordinates; we can break that into components. 00:03:28.700 --> 00:03:35.300 We call this the component form: we have a horizontal amount of 3 and a vertical amount of 4. 00:03:35.300 --> 00:03:39.800 So, we say that its horizontal component is 3, and its vertical component is 4. 00:03:39.800 --> 00:03:47.000 u, the vector u, equals (3,4); the first component is the horizontal one, and the second component is the vertical, 00:03:47.000 --> 00:03:50.600 just like when we are talking about points in the plane. 00:03:50.600 --> 00:03:58.200 We normally use angle brackets (these are angle brackets, because they are at an angle) to denote vectors. 00:03:58.200 --> 00:04:01.600 But we will often also see parentheses; parentheses are very common, as well. 00:04:01.600 --> 00:04:08.700 I personally actually tend to use parentheses more; but most precalculus/math analysis courses tend to use angle brackets. 00:04:08.700 --> 00:04:13.900 So, I am teaching with angle brackets; but personally, when I am just doing math on my own, I often tend to use parentheses. 00:04:13.900 --> 00:04:16.200 But either of them is just fine. 00:04:16.200 --> 00:04:20.800 After we learn about unit vectors, we will see one more way to talk about the component form, using i and j. 00:04:20.800 --> 00:04:25.600 But we will leave that until a little bit later, once we have actually talked about unit vectors. 00:04:25.600 --> 00:04:30.000 If we know the component form of a vector, we can figure out its length by the Pythagorean theorem. 00:04:30.000 --> 00:04:43.900 Remember: the Pythagorean theorem says that the hypotenuse to a triangle, squared, is equal to both of the legs, squared and then added together. 00:04:43.900 --> 00:04:47.000 So, that means that we can take the square root of both sides of the equation. 00:04:47.000 --> 00:04:54.900 We have that the hypotenuse is equal to the square root of each leg, squared and added together, the square root of (leg squared + leg squared). 00:04:54.900 --> 00:05:05.700 In this case, we know that our vector is (3,4); so we have that one leg is 3, and one leg is 4. 00:05:05.700 --> 00:05:12.300 So, we work this out; √(3² + 4²) becomes √(9 + 16), or √25, so we get 5. 00:05:12.300 --> 00:05:15.600 So, we see that the length of this vector is 5. 00:05:15.600 --> 00:05:20.200 When we want to denote the magnitude of a vector, when we want to talk about the length of a vector, 00:05:20.200 --> 00:05:24.300 the magnitude of a vector, we normally use the word "magnitude" to talk about length. 00:05:24.300 --> 00:05:29.600 But in either case, with "length" or "magnitude," we are just talking about how long the vector is. 00:05:29.600 --> 00:05:36.000 We normally use these vertical bars on either side of it, just like we do with absolute value. 00:05:36.000 --> 00:05:38.800 And we will talk about that in just a moment--why it is like absolute value. 00:05:38.800 --> 00:05:45.900 Also, sometimes you will see it written ||u||, vector u with double bars around it. 00:05:45.900 --> 00:05:54.200 In either case, whether it is single bar or single bar, what we are talking about is how long that vector is, if we measure it from the origin out to its tip. 00:05:54.200 --> 00:06:03.300 If u is equal to (a,b), then the length of u, the bars on either side of u (which would also be bars on either side of (a,b), 00:06:03.300 --> 00:06:08.000 because that is just a vector, as well), will give us the square root of a² + b², 00:06:08.000 --> 00:06:16.100 based on this exact same reasoning that the length of our hypotenuse is equal to the square root of each leg, squared and added together. 00:06:16.100 --> 00:06:20.800 So in this case, a and b are just there and there on our vector. 00:06:20.800 --> 00:06:25.600 Now, I would like to point out, really quickly, why in the world we are using vertical bars on either side, 00:06:25.600 --> 00:06:29.200 just like we did with talking about absolute value. 00:06:29.200 --> 00:06:38.600 Let's think about that for a bit: let's think about...when we work with absolute value, say we have 0 here, and here is +5, and here is -5. 00:06:38.600 --> 00:06:43.400 Well, if we talk about the absolute value of 5, and we talk about the absolute value of -5, 00:06:43.400 --> 00:06:47.800 in both cases, we are going to get the number 5 out of it. 00:06:47.800 --> 00:06:56.800 We get 5 in either case, because what the absolute value is telling us is: it is saying how far you are from 0--how far you are from the origin. 00:06:56.800 --> 00:07:03.900 So, the reason why the absolute value of 5 is 5 is because it takes 5 units of length to get from 0 to 5. 00:07:03.900 --> 00:07:11.500 And that is the exact same reason why the absolute value of -5 is 5: because it takes 5 units of length to get from 0 to -5. 00:07:11.500 --> 00:07:14.800 They go in different directions, but it is a question of how far away it is. 00:07:14.800 --> 00:07:17.600 They are both 5 units away from the origin. 00:07:17.600 --> 00:07:20.400 The same thing is going on when we are dealing with a vector. 00:07:20.400 --> 00:07:29.900 When we talk about the size of a vector, what the length of a vector is, we are asking how far the vector goes out from the origin. 00:07:29.900 --> 00:07:37.100 In both the case of the absolute value and the magnitude of a vector, what we are saying is, "How far are you from the origin?" 00:07:37.100 --> 00:07:42.700 We put bars around the u because it is basically doing the same thing as bars around a number with absolute value. 00:07:42.700 --> 00:07:48.400 Bars around a number are saying, "How far are you from 0?"; bars around a vector are saying, "How far are you from the origin?" 00:07:48.400 --> 00:07:52.400 So, they are both a question of length, effectively. 00:07:52.400 --> 00:07:57.500 All right, direction and angle: if we know the component form of a vector, we can figure out its angle for trigonometry. 00:07:57.500 --> 00:08:07.200 So, once again, we have u = (3,4); so, tanθ = side<font size="-6">opposite</font>/side<font size="-6">adjacent</font>, when we have a right triangle. 00:08:07.200 --> 00:08:12.100 And we have that here, because we know we are dealing with a nice rectangular coordinate system. 00:08:12.100 --> 00:08:22.600 We have a 3 here and a 4 here; so our side opposite to our angle θ is this side here, and our side adjacent is this side here. 00:08:22.600 --> 00:08:29.800 So, that gets us tanθ = 4/3; at this point, we can take the arctan of both sides 00:08:29.800 --> 00:08:34.300 (the inverse tangent--arctan and tan^-1 mean the same thing; I like arctan). 00:08:34.300 --> 00:08:39.900 So, θ = arctan (tan^-1) of 4/3; we plug that into a calculator or look it up, 00:08:39.900 --> 00:08:47.100 and we see that that is approximately equal to 53.13 degrees; so we see that that is the angle in there. 00:08:47.100 --> 00:08:52.300 Now, we normally talk about direction as the counterclockwise angle from the positive x-axis. 00:08:52.300 --> 00:08:55.700 It is a little bit confusing, but it just means that we start over here at the positive x-axis, 00:08:55.700 --> 00:09:02.100 and then we just keep turning counterclockwise until we get to whatever thing we are trying to measure out to, 00:09:02.100 --> 00:09:06.300 at which point we stop, and that is the measure of the angle that we are going with. 00:09:06.300 --> 00:09:11.900 It is just like we did with the unit circle: we started at the x-axis positive, and then we just kept spinning 00:09:11.900 --> 00:09:15.300 until we got to whatever angle we were trying to get to. 00:09:15.300 --> 00:09:18.300 But sometimes we won't be using the positive x-axis. 00:09:18.300 --> 00:09:23.900 That is what we normally end up using, but sometimes the reference location--sometimes--will change, 00:09:23.900 --> 00:09:27.600 and we won't be using this nice positive x-axis that we are used to using. 00:09:27.600 --> 00:09:34.500 But maybe we will be talking about how far we are off of the vertical y-axis to the right, 00:09:34.500 --> 00:09:38.900 or maybe some other thing, like we have this other angle here created, 00:09:38.900 --> 00:09:43.900 and we are talking about how far we are off of this other thing, some angle, going clockwise... 00:09:43.900 --> 00:09:45.400 So who knows how it is going to be done? 00:09:45.400 --> 00:09:48.300 We have to pay attention to what the specific problem is and how it is set up. 00:09:48.300 --> 00:09:53.600 It generally will be the positive x-axis, but it is not an absolute guarantee; you have to pay attention. 00:09:53.600 --> 00:09:59.000 This means, in any case, whatever you are dealing with, that I recommend always drawing a picture, 00:09:59.000 --> 00:10:01.900 because that is going to give you a way to be able to see what is going on. 00:10:01.900 --> 00:10:09.400 Draw a picture before you try to figure out or use angles; it will help you get a sense of what is going on, and really help clarify things. 00:10:09.400 --> 00:10:12.600 All right, now, what if we have the component from... 00:10:12.600 --> 00:10:19.400 We talked about (previously) taking the magnitude and angle and getting component forms. 00:10:19.400 --> 00:10:21.500 I'm sorry, we talked about the exact opposite of that. 00:10:21.500 --> 00:10:26.200 So now, we are going to say, "What if we know the magnitude and angle, and we want to get the component form?" 00:10:26.200 --> 00:10:31.200 So, if we are trying to find the component form from the magnitude and angle, we can figure that out. 00:10:31.200 --> 00:10:36.500 The first thing first: always draw a sketch--it will help keep things clear and help you understand what is going on. 00:10:36.500 --> 00:10:38.400 And then, from there, you just use trigonometry. 00:10:38.400 --> 00:10:46.800 So, in this case, we have a length of 6; the length of our vector u is 6; our angle θ is 120 degrees. 00:10:46.800 --> 00:10:52.500 We remember from trigonometry that the cosine of 120, the cosine of this angle here, is going to be equal to 00:10:52.500 --> 00:11:02.500 the horizontal amount of change, up until it drops down on its crossover, divided by the length of the segment we are dealing with. 00:11:02.500 --> 00:11:13.300 Cosine of 120 = x/6; multiply both sides by 6; 6cos120...cos120 is the same thing as -1/2, so 6 times -1/2...gets us -3. 00:11:13.300 --> 00:11:15.800 So, -3 is our value for x. 00:11:15.800 --> 00:11:24.200 A similar thing is going on for our y, the vertical component; sin(120), the sine of the angle we are dealing with, 00:11:24.200 --> 00:11:29.100 is equal to the vertical component, divided by the length of the entire segment (6 in this case). 00:11:29.100 --> 00:11:38.700 So, sin(120) = y/6; multiply both sides by 6; sin(120) is √3/2, so it simplifies to 3√3 = y. 00:11:38.700 --> 00:11:41.000 At that point, we just take both of these pieces of information. 00:11:41.000 --> 00:11:44.600 We slot them into our u, and we now have component form. 00:11:44.600 --> 00:11:48.600 Our u is equal to (-3,3√3). 00:11:48.600 --> 00:11:58.800 Now, I did tell us to do it with cosine 120; but most of us are probably used to dealing with trigonometry for degree angles under 90 degrees. 00:11:58.800 --> 00:12:03.500 It is a little bit confusing--maybe just a bit--to be able to work with things over 90 degrees. 00:12:03.500 --> 00:12:12.600 So, we also could have converted this to an angle of 120, so a total of 60 degrees over here, because it is 180 on the whole thing. 00:12:12.600 --> 00:12:15.500 So, we could have figured out that inside of the triangle is 60 degrees. 00:12:15.500 --> 00:12:20.100 Now, that is going to cause a little bit of difference here, because cos(120) brought that negative to the table, 00:12:20.100 --> 00:12:25.600 because indeed, our x is going to the left, and remember: this is the negative direction for x. 00:12:25.600 --> 00:12:32.700 Going this way gets us negative x values, when we go to the left, just like going down gets us negative y values. 00:12:32.700 --> 00:12:36.200 If that is the case, we have cos(60), what we can figure out... 00:12:36.200 --> 00:12:40.500 If you have things inside of a triangle, you are just going to figure out the length of each of those sides. 00:12:40.500 --> 00:12:51.700 So, cosine of 60 equals x/6; so we will get 6 times...cos(60) is just 1/2, so we get that 3 = x. 00:12:51.700 --> 00:13:00.500 But notice: what we are figuring out is...we are figuring out here to here as x, which is not the same thing as the x-value, as in the horizontal location. 00:13:00.500 --> 00:13:03.600 What we are figuring out is just the length of the side. 00:13:03.600 --> 00:13:12.500 We are figuring out how far it is from here to here; but we have to figure out the horizontal coordinates--not just the length, but the actual x-value. 00:13:12.500 --> 00:13:18.200 If that is the case, what we are really figuring out is the absolute value of x--how long our x is. 00:13:18.200 --> 00:13:23.900 And that means that, at the end, we have to look at this picture; and that is why we draw these sketches that are so handy. 00:13:23.900 --> 00:13:29.300 We look at the picture, and we see that our x has a length of 3. 00:13:29.300 --> 00:13:35.800 However, it is going to the left; so that means it has to be a negative thing. 00:13:35.800 --> 00:13:44.300 By paying attention, we see that it is -3...not just length 3, but -3; but it is because we drew a sketch that we are able to see this. 00:13:44.300 --> 00:13:48.200 You have this option: you can either just use the angle, and be able to be really good at trigonometry; 00:13:48.200 --> 00:13:50.800 or you can make it in a slightly simpler form, where it is easier to understand. 00:13:50.800 --> 00:13:54.700 But you have to be paying attention and realize that you have to set the sign at the end. 00:13:54.700 --> 00:13:58.800 I have to pay attention, based on this sketch: is this length going to come out to be positive? 00:13:58.800 --> 00:14:01.000 Is this length going to come out to be negative? 00:14:01.000 --> 00:14:03.300 You really have to pay attention to that. 00:14:03.300 --> 00:14:05.600 All right, we are ready to move on to a new idea. 00:14:05.600 --> 00:14:12.600 Scaling by scalars: if we have a vector u, we can scale it to a different length by multiplying by a scalar, a real number. 00:14:12.600 --> 00:14:17.700 A scalar is just some number; it is not a vector--it is just an actual, real number. 00:14:17.700 --> 00:14:21.800 Algebraically, the scalar just ends up multiplying each component. 00:14:21.800 --> 00:14:27.900 Let's say we start with this red vector on our picture, which is u = (3,-2). 00:14:27.900 --> 00:14:32.100 We can scale this by some other thing, by just multiplying it by some number. 00:14:32.100 --> 00:14:37.500 Like, for example: we could multiply u by 2, so we have 2; and now that is the blue vector that we see there. 00:14:37.500 --> 00:14:44.400 Now notice that the blue vector is double the length of the red vector, because it is 2 times u. 00:14:44.400 --> 00:14:50.600 So, it just takes that length, and it scales it by a factor of 2; it doubles that original length. 00:14:50.600 --> 00:14:59.300 Algebraically, we just end up having this 2 multiply on the 3, and multiply on the -2; so we get (6,-4) as the components. 00:14:59.300 --> 00:15:02.800 We can take this; we can try multiplying by something else--how about a negative number? 00:15:02.800 --> 00:15:06.500 What would a negative get us? A negative ends up going in the opposite direction. 00:15:06.500 --> 00:15:12.900 So, if we have a negative u, then the positive direction is the way it normally goes; negative will be the opposite way. 00:15:12.900 --> 00:15:19.300 So, negative goes in the opposite direction; so now, we are going opposite the direction that u went, as we can see pictorially here. 00:15:19.300 --> 00:15:24.000 And for how it is going to end up having it, it is just a negative now on each of the components. 00:15:24.000 --> 00:15:29.600 So, negative cancels out there, and we have (-3,+2). 00:15:29.600 --> 00:15:32.600 And that is what we see on our picture. 00:15:32.600 --> 00:15:39.600 We could also have something that is not just a whole integer number, like, say, -3/2. 00:15:39.600 --> 00:15:45.900 We end up having 3/2 times the length--1 and 1/2 the length of the original vector-- 00:15:45.900 --> 00:15:50.200 but then also, in the negative direction--opposite the direction of the first one. 00:15:50.200 --> 00:15:59.200 Once again, numerically, it just ends up being -3/2 times the first number times the second number; so we get (-9/2,3). 00:15:59.200 --> 00:16:00.300 That is what it ends up being. 00:16:00.300 --> 00:16:06.300 So, algebraically it just multiplies each component; graphically it is a question of stretching and maybe also flipping. 00:16:06.300 --> 00:16:16.100 In general, for any scalar k and some vector u that is (a,b), k times u...that is the same thing as k times (a,b), 00:16:16.100 --> 00:16:23.600 so that is the same thing as just that k getting distributed to both the a and the b, so it gives (ka,kb). 00:16:23.600 --> 00:16:30.600 Great; a unit vector is a vector with a length of 1; "unit vector" just means a length of 1. 00:16:30.600 --> 00:16:35.800 It still has a direction; it can have any direction, but it has to have a magnitude of exactly 1. 00:16:35.800 --> 00:16:40.600 Its magnitude--its length--how long it is...it is 1 in terms of length. 00:16:40.600 --> 00:16:44.500 We create a unit vector out of any vector u by dividing u by its length. 00:16:44.500 --> 00:16:51.200 Remember: if we divide by a number, that is the same thing as just using a scalar; we are multiplying by 1 over the number we are dividing by. 00:16:51.200 --> 00:16:59.900 So, previously, we could scale; if we know its length is 10, and then we divide it by 10, we are going to get something with length 1 now. 00:16:59.900 --> 00:17:09.600 We have some vector u that is some length; but then we divide it by that length, because the magnitude of u is just its length. 00:17:09.600 --> 00:17:15.300 So, we have that; we divide by that; we have scaled it back to what it would be if it was just at a length of 1. 00:17:15.300 --> 00:17:21.800 So, our unit vector is the original vector, divided by the length of the vector. 00:17:21.800 --> 00:17:30.200 This is the same direction as u; so it will be in the same direction, but it is going to have a length of 1--it will just be length 1. 00:17:30.200 --> 00:17:34.100 Now, the use of having a unit vector--the reason why it is so great to have a unit vector-- 00:17:34.100 --> 00:17:38.800 is that we can just take it later and multiply it by any scalar k that we want. 00:17:38.800 --> 00:17:43.100 And we will know that we will have created a vector that is length k, because we started at length 1. 00:17:43.100 --> 00:17:48.300 You scale that by k, and we are going to go with 1 times k; so we will just be at vector length k. 00:17:48.300 --> 00:17:53.900 And we are going to be scaling in the direction we already started with, so we know that we will be in the unit vector's direction. 00:17:53.900 --> 00:17:58.800 This gives us an ability to easily create vectors of any length we want in a known direction. 00:17:58.800 --> 00:18:04.500 And this ability comes in really, really handy in a lot of situations, and that is why unit vectors are important. 00:18:04.500 --> 00:18:09.700 All right, we can combine two or more vectors through addition or subtraction. 00:18:09.700 --> 00:18:12.900 It is actually not that difficult; we just add them component-wise. 00:18:12.900 --> 00:18:16.600 The horizontal components add together, and the vertical components add together. 00:18:16.600 --> 00:18:21.700 All of your first components go together; the second components go together, and so on and so forth, like that. 00:18:21.700 --> 00:18:30.400 In this case, if we add (3,4) and (2,-5), then what we end up doing is: we have the 3 and the 2 getting combined. 00:18:30.400 --> 00:18:35.800 And they become 3 + 2, because we are adding (3,4) + (2,-5). 00:18:35.800 --> 00:18:44.200 The same basic thing: the 4 and the -5 get combined, and so 4 + -5 is 4 - 5. 00:18:44.200 --> 00:18:49.900 We simplify that, and we get (5,-1); we are just taking the numbers at the beginning and adding them together, 00:18:49.900 --> 00:18:51.800 and the numbers at the end and adding them together. 00:18:51.800 --> 00:18:57.500 We are doing it component-wise: each component stays and only mixes with components that are the same type 00:18:57.500 --> 00:19:01.700 (first components with first components, second components with second components, and so on). 00:19:01.700 --> 00:19:07.100 So, in general, for some (u₁,u₂)...really, that is just saying the first component of u, 00:19:07.100 --> 00:19:13.100 and the second component of u...and then (v₁,v₂), the first component of v 00:19:13.100 --> 00:19:18.500 and the second component of v--if we add u and v, u + v, (u₁,u₂) + (v₁,v₂), 00:19:18.500 --> 00:19:23.400 it is going to be the exact same thing: the u₁ and the v₁, the first components, 00:19:23.400 --> 00:19:31.700 adding together, and then the second components, u₂ and v₂, adding together, as well. 00:19:31.700 --> 00:19:36.900 It is a very similar thing if we end up subtracting: u - v is just going to be u₁, 00:19:36.900 --> 00:19:40.100 the first component of u, minus the first component of v, v₁; 00:19:40.100 --> 00:19:43.900 u₂ - v₂ (the second component of u minus the second component of v). 00:19:43.900 --> 00:19:51.400 The components stay in their locations, but they do either addition or subtraction, depending on whether it is addition or subtraction; it makes sense. 00:19:51.400 --> 00:19:54.400 We can also do this geometrically and get an understanding of what is going on, 00:19:54.400 --> 00:19:58.300 because vectors are supposed to represent this thing--this directed line segment. 00:19:58.300 --> 00:20:05.000 So, we can see this idea geometrically: you add vectors by placing the tail of one at the head of the other. 00:20:05.000 --> 00:20:12.600 So, for example, in this one, you have u in red, (3,4); and then we have v in blue, (2,-5). 00:20:12.600 --> 00:20:18.200 So, we put (3,4) out to here; that goes out to (3,4). 00:20:18.200 --> 00:20:29.000 And then, this one goes over 2 and goes down 5; so (2,-5) is 2 to the right, down by 5. 00:20:29.000 --> 00:20:36.300 We put those together, and we get (5,-1), which is exactly what we see here in the purple vector that is the combination of them. 00:20:36.300 --> 00:20:39.200 So, u + v is the purple vector that we see there. 00:20:39.200 --> 00:20:44.700 We can see this algebraically and geometrically, and the two ideas match up completely. 00:20:44.700 --> 00:20:47.800 Not only that, but we get the same resultant vector. 00:20:47.800 --> 00:20:54.300 This resultant, what we get when we put the two things together, is the same whether it is u + v or v + u. 00:20:54.300 --> 00:20:59.700 It doesn't matter if we end up doing the blue one first, and then the red one. 00:20:59.700 --> 00:21:04.500 We can add them geometrically in any order we please, and it still comes out to be the same thing. 00:21:04.500 --> 00:21:11.500 We end up seeing this parallelogram; we can see this as another way of adding things--creating this parallelogram out of them, or just head-to-tail. 00:21:11.500 --> 00:21:15.100 But we see geometrically what is going on. 00:21:15.100 --> 00:21:20.000 All right, now that we have these two ideas, we can put together the idea of combining vectors 00:21:20.000 --> 00:21:24.600 with the idea of unit vectors to get a new way to express a vector's components. 00:21:24.600 --> 00:21:32.100 We start by creating two standard unit vectors, which is really just a fancy way of saying "things that make sense and are kind of fundamental." 00:21:32.100 --> 00:21:42.400 One horizontal, one vertical: i = (1,0); it is a unit vector that is purely horizontal; 00:21:42.400 --> 00:21:47.800 and j, which is (0,1), is a unit vector that is purely vertical. 00:21:47.800 --> 00:21:56.800 This one is just one unit long that way; and this one is just one unit long that way--purely vertical. 00:21:56.800 --> 00:21:58.600 And that is what we are seeing there. 00:21:58.600 --> 00:22:04.000 One other thing: if we want to write this bold thing, we can't really write bold on our paper; that is very difficult. 00:22:04.000 --> 00:22:11.000 You can end up writing it like an i, but instead of putting a dot on it, you put that little arrow on top (or in my case, the harpoon on top). 00:22:11.000 --> 00:22:14.500 The same thing with the j: you can make the j and then put a little arrow on top. 00:22:14.500 --> 00:22:19.300 And so, that is another way of talking about these units vectors, i and j, if you want to. 00:22:19.300 --> 00:22:23.600 All right, with this, we can express any vector in terms of i and j. 00:22:23.600 --> 00:22:30.200 If we have (3,4), well, that is the same thing as having 3 i's, three of these unit vectors that are horizontal, 00:22:30.200 --> 00:22:34.600 plus four of these unit vectors that are vertical; so that is what we have there. 00:22:34.600 --> 00:22:39.300 We can combine them; we can break them up into 3 horizontal motions plus four vertical motions. 00:22:39.300 --> 00:22:43.200 And we get to the same thing as if we had just done (3,4), all at once. 00:22:43.200 --> 00:22:51.300 3, our first component, matches up with the i's, because that is just our way of saying our horizontal, since our first component is our horizontal. 00:22:51.300 --> 00:22:58.700 And then, our 4 matches up with the 4j, because that is just the same thing as saying 4 vertical, or 4 units' worth of vertical. 00:22:58.700 --> 00:23:01.600 And this also works for numbers that aren't just whole integers, as well, 00:23:01.600 --> 00:23:09.900 because 4.7 times i just scales i by a factor of 4.7, so it will be in the same place horizontally. 00:23:09.900 --> 00:23:17.500 And π--we can also scale by a factor of π; as long as it is a real number, we can scale by it; so πj goes in there; great. 00:23:17.500 --> 00:23:24.400 All right, we can also talk about a zero vector; we denote the zero vector with that same arrow top on top of a 0. 00:23:24.400 --> 00:23:28.300 That has 0 in all of its components; so it will be nothing but 0's as the vector. 00:23:28.300 --> 00:23:31.700 It has no length; it is 0--it just lives at the origin. 00:23:31.700 --> 00:23:34.600 And so, since it has no length, its direction doesn't matter. 00:23:34.600 --> 00:23:40.900 So, here is an example: (0,0)...the zero vector is (0,0). 00:23:40.900 --> 00:23:45.700 It is just sitting at the origin; its distance from the origin is nothing, because it is currently at the origin. 00:23:45.700 --> 00:23:52.500 Notice: for any vector u whatsoever, if we add u and the zero vector together, we will just end up being there. 00:23:52.500 --> 00:23:56.000 We will not have moved anywhere, because head-to-tail we end up going someplace, 00:23:56.000 --> 00:23:58.700 and then we don't move, because the zero vector doesn't move at all. 00:23:58.700 --> 00:24:06.100 u - u: if we subtract u from itself, we will end up going out and then coming right back, so we will end up getting 0. 00:24:06.100 --> 00:24:09.800 And then finally, if we take any vector and multiply it by a scalar of 0, we will have some length, 00:24:09.800 --> 00:24:14.900 and then we bring that length to 0; so at a length of 0, we have the zero vector; great. 00:24:14.900 --> 00:24:17.500 All right, at this point, we have talked about a lot of different ideas with vectors. 00:24:17.500 --> 00:24:19.500 And we can turn this into a bunch of properties. 00:24:19.500 --> 00:24:22.700 Don't worry too much about understanding all of these properties right away. 00:24:22.700 --> 00:24:25.300 They will just make more sense, and you will be used to using them. 00:24:25.300 --> 00:24:29.900 The beauty of all of these properties is that they are very much what we are already used to using with the real numbers. 00:24:29.900 --> 00:24:35.000 So, vectors, as long as you remember to keep them in this form of components only interacting with other components, 00:24:35.000 --> 00:24:38.600 are very similar to working with numbers in many of the ways we are already used to. 00:24:38.600 --> 00:24:40.700 Let's talk through some of these properties. 00:24:40.700 --> 00:24:48.500 u + v is the same thing as v + u; this is the idea of commutativity, that 5 + 8 is the same thing as 8 + 5. 00:24:48.500 --> 00:24:50.100 We are used to that with the real numbers. 00:24:50.100 --> 00:24:56.600 We also have associativity, (u + v) + w is the same thing as u + (v + w). 00:24:56.600 --> 00:25:02.500 3 + 5 + 4 is the same thing, if you add the 3 and the 5 first, or if you add the 5 and the 4 first. 00:25:02.500 --> 00:25:08.900 3 + 5, then add 4, or 3 + (5 + 4) (where you add the 3 second)--it doesn't matter which way you do it; 00:25:08.900 --> 00:25:13.200 so once again, it is very similar to doing it with normal numbers. 00:25:13.200 --> 00:25:20.600 k times l...if we have two scalars, k and l, times u, well, that is the same thing as k times an already-scaled lu. 00:25:20.600 --> 00:25:23.500 So, we can either multiply our scalars together and then multiply the vector; 00:25:23.500 --> 00:25:30.000 or we can have them multiply the vector, each one after another--the same thing, either way, which is pretty much what we would expect. 00:25:30.000 --> 00:25:39.000 k times (u + v) is nice; it distributes: k times vector u plus vector v is k times vector u, plus k times vector v. 00:25:39.000 --> 00:25:48.200 It also distributes in the other way: k + l times vector u: the vector can distribute out onto them, so we have k times vector u, plus l times vector u; great. 00:25:48.200 --> 00:25:53.400 If we take u, and we add it to the zero vector, we end up just getting u; it has no effect. 00:25:53.400 --> 00:25:57.000 u - u is going to get us back to the zero vector. 00:25:57.000 --> 00:26:03.100 And then, a couple more: 0 times u is going to give us the zero vector; 1 times u has no effect-- 00:26:03.100 --> 00:26:08.200 we are scaling by just what we are already at; and then, -1 times u is going to flip us to the negative version; 00:26:08.200 --> 00:26:10.400 it will just cause everything in there to become negative. 00:26:10.400 --> 00:26:21.600 The only one that might be a little bit confusing is ku, the magnitude of a scaled u, k scaled on u, k scalar times vector u. 00:26:21.600 --> 00:26:27.800 The magnitude of that is equal to the absolute value of k, times the magnitude of u. 00:26:27.800 --> 00:26:30.100 Let's look at why that is the case. 00:26:30.100 --> 00:26:35.800 A really quick, simple example: let's consider if we had (0,1). 00:26:35.800 --> 00:26:43.400 We have this vector here that goes from here out to a length of 1. 00:26:43.400 --> 00:26:49.000 So, we could scale it by k; and let's say that k is equal to -2. 00:26:49.000 --> 00:27:05.100 So, we scale this; here is our u: u = (0,1), so we could scale it by k = -2; -2 times u will get us the same thing, 00:27:05.100 --> 00:27:08.700 but now it is going to be flipped, and it will be twice the length. 00:27:08.700 --> 00:27:11.200 We are going to go down two units now. 00:27:11.200 --> 00:27:20.200 So, whereas the first u went up by 1 unit (it had a length of 1), this has a length of 2. 00:27:20.200 --> 00:27:23.900 The fact that we are going down doesn't make it negative length; length always is positive. 00:27:23.900 --> 00:27:37.800 So, it has a length of 2, which is why we have the magnitude of (0,-2): well, that ends up being equal to +2, as we can see from this diagram right here. 00:27:37.800 --> 00:27:52.000 But this -2 times u--if we had separated this out into -2 times (0,1), what u started as, 00:27:52.000 --> 00:28:02.100 well, we could break this, by this rule, into the absolute value of our scalar, times the length of our initial thing, 00:28:02.100 --> 00:28:05.600 which would give us positive 2 times 1--the same thing. 00:28:05.600 --> 00:28:10.500 So, what it is doing is saying that the reason why we have absolute value on the scalar here 00:28:10.500 --> 00:28:14.400 is because it doesn't matter that we are flipping and pointing in a new direction; 00:28:14.400 --> 00:28:17.100 ultimately, length is always going to come out to be positive. 00:28:17.100 --> 00:28:23.500 So, we can't let a negative k cause our result to come out as a negative length, because that just doesn't make sense. 00:28:23.500 --> 00:28:28.100 So, we have to figure out a way for it to always stay positive, and that is why we have this absolute value here. 00:28:28.100 --> 00:28:32.300 All right, there is no multiplication between vectors. 00:28:32.300 --> 00:28:39.700 Even with everything we have seen so far about vectors, there has been no mention of vector multiplication, other than scalars multiplying on vectors. 00:28:39.700 --> 00:28:42.300 But other than that, we haven't talked about vector multiplication. 00:28:42.300 --> 00:28:46.800 That is because there is no good way to define vector multiplication. 00:28:46.800 --> 00:28:50.300 There is just no way to really do it that is going to make sense. 00:28:50.300 --> 00:28:56.600 So, we could make up some numerical way to multiply vectors: some vector times some vector makes some other vector. 00:28:56.600 --> 00:29:03.300 But it would probably be geometrically meaningless; we can't really come up with a good way that is going to have some deep geometric meaning. 00:29:03.300 --> 00:29:06.000 And that is the problem here: while we could come up with something numerically, 00:29:06.000 --> 00:29:09.700 we want all of this stuff to have a geometric connection, and all of this other stuff has. 00:29:09.700 --> 00:29:13.800 It makes sense to combine vectors; we are doing one vector, and then we are doing another vector. 00:29:13.800 --> 00:29:16.000 We are doing two pieces of motion. 00:29:16.000 --> 00:29:20.400 Or we stretch them: we have some piece of motion, and then we just elongate it or shrink it or flip it. 00:29:20.400 --> 00:29:23.400 Those things make sense geometrically; but what would it mean geometrically-- 00:29:23.400 --> 00:29:27.100 what would it mean as a picture--to multiply a line segment by another line segment? 00:29:27.100 --> 00:29:31.700 It just doesn't really make sense; and because of that, we do not define vector multiplication. 00:29:31.700 --> 00:29:35.600 There is just no vector multiplication, pretty much, to talk about. 00:29:35.600 --> 00:29:41.600 Now, all that said, there is an operation similar to multiplication that is called the dot product. 00:29:41.600 --> 00:29:46.400 That is different, though, because it will take two vectors and multiply them together (although it won't multiply them together)-- 00:29:46.400 --> 00:29:50.400 it will take two vectors, and this vector dot, this other vector, will give us a scalar. 00:29:50.400 --> 00:29:52.500 It will give us a single real number. 00:29:52.500 --> 00:29:55.000 Don't worry about that too much now; we are not going to talk about it in this lesson. 00:29:55.000 --> 00:29:59.300 But we will explore it in the next lesson; so we will see it soon. 00:29:59.300 --> 00:30:02.200 But for right now, there is just no way to multiply vectors. 00:30:02.200 --> 00:30:04.700 And even later on, once you see something that is kind of close to it, 00:30:04.700 --> 00:30:10.000 you will see that it is very different from actually multiplying vectors and getting a new vector out of it. 00:30:10.000 --> 00:30:15.700 All right, motion in a medium: a really common use of vectors is to analyze motion-- 00:30:15.700 --> 00:30:20.600 the location of an object, the velocity of an object, the acceleration of an object. 00:30:20.600 --> 00:30:26.000 However, what happens if something is moving relative to a medium, like water or air? 00:30:26.000 --> 00:30:29.900 Say we have a boat in a river, and so the boat is moving up the river. 00:30:29.900 --> 00:30:33.000 But at the same time, the water in the river is moving in another direction. 00:30:33.000 --> 00:30:35.500 We have to do something to take this into account. 00:30:35.500 --> 00:30:43.000 So, the object is moving relative to the medium, but the medium itself is also moving, like the boat in the water. 00:30:43.000 --> 00:30:46.900 So, to understand this, let's consider a fish swimming in an aquarium. 00:30:46.900 --> 00:30:49.600 I love this as an example. 00:30:49.600 --> 00:30:54.400 First, we have that the aquarium is completely still; we have some table like this; 00:30:54.400 --> 00:30:58.300 and imagine that my arm here is the aquarium, and so here is the table. 00:30:58.300 --> 00:31:03.600 The fish is here, and the fish is swimming forward; the fish swims forward, and it gets to the other end of the aquarium. 00:31:03.600 --> 00:31:07.200 That is how it starts in this picture here: the fish is the only thing moving. 00:31:07.200 --> 00:31:11.300 But what happens if we don't just let the fish be the only thing moving? 00:31:11.300 --> 00:31:17.100 If we grab the aquarium, and we actually slide it to the slide as the fish is swimming-- 00:31:17.100 --> 00:31:24.400 if we grab the aquarium, and we move it, we are going to see the aquarium move like this while the fish is moving like this, 00:31:24.400 --> 00:31:29.100 because the fish is moving inside of the aquarium, but now the entire aquarium is also moving. 00:31:29.100 --> 00:31:33.500 We see that the fish moves, but at the same time, the aquarium moves over. 00:31:33.500 --> 00:31:36.500 So, it is very different from the world where the fish ended over here. 00:31:36.500 --> 00:31:40.100 Now, the fish manages to move at the same time as the aquarium is moved. 00:31:40.100 --> 00:31:42.100 So, we have to take both of these things into account. 00:31:42.100 --> 00:31:51.000 We can see this pictorially here: our fish is moving to the side, but at the same time, the box is moving to the side, as well. 00:31:51.000 --> 00:31:55.900 The fish manages to get over to the left, but the box is now way over to the right. 00:31:55.900 --> 00:31:59.300 So, if we are going to talk about where the fish has gotten to, the velocity of the fish-- 00:31:59.300 --> 00:32:03.900 anything where we want to talk about the fish's motion and analyze the motion of the fish-- 00:32:03.900 --> 00:32:06.000 we have to take both of these things into account. 00:32:06.000 --> 00:32:10.800 We have to take into account the aquarium's motion, but also the fish moving inside of the aquarium. 00:32:10.800 --> 00:32:15.500 It is not enough to talk about just one of them; we have to combine these two ideas. 00:32:15.500 --> 00:32:20.600 Motion in a medium is the combination of the object's motion vector relative to the medium, 00:32:20.600 --> 00:32:27.800 the fish relative to being inside of an aquarium, and then the medium's motion vector--how the aquarium is moving. 00:32:27.800 --> 00:32:32.000 How does the fish move in the aquarium? How does the aquarium move in the larger world? 00:32:32.000 --> 00:32:36.100 That is what we mean by relative to the medium; it is just whatever the thing is inside of, 00:32:36.100 --> 00:32:40.000 and then how the thing that you are inside of is moving. 00:32:40.000 --> 00:32:47.300 So, we can break this down into the velocity of the object, plus the velocity of the medium, is equal to the velocity of the total motion. 00:32:47.300 --> 00:32:56.200 The fish relative to the table is the addition of the velocity of the fish in the water, plus the velocity of the aquarium, 00:32:56.200 --> 00:33:00.300 as vectors, because the velocity vector in one case is going to be positive; 00:33:00.300 --> 00:33:03.300 in the other...for one of them, it will be positive; for the other one, it will be negative, 00:33:03.300 --> 00:33:05.900 because they are going to be pointing in opposite directions. 00:33:05.900 --> 00:33:09.800 All right, we can also talk about more than two dimensions. 00:33:09.800 --> 00:33:15.200 At this point, we have only seen vectors in two dimensions; but we can expand this idea to any number of dimensions we want. 00:33:15.200 --> 00:33:19.400 For example, a three-dimensional vector could be (5,-2,3)--no problem. 00:33:19.400 --> 00:33:21.300 We will just keep putting in more components. 00:33:21.300 --> 00:33:27.000 By the way we define scalars and vector combinations, everything we have discussed so far about vectors still works just fine. 00:33:27.000 --> 00:33:29.900 It might get a little confusing to picture in our head in higher dimensions. 00:33:29.900 --> 00:33:33.400 But everything still makes sense; it is hard to picture higher than three dimensions, 00:33:33.400 --> 00:33:39.100 because we are used to living in a three-dimensional world, but it still makes sense in terms of the algebra of what is going on. 00:33:39.100 --> 00:33:44.800 That is great; also, one little thing: if we are talking about three dimensions...you remember that i and j-- 00:33:44.800 --> 00:33:48.200 we could talk about unit vectors as an alternate way of talking about component form. 00:33:48.200 --> 00:33:53.000 There is another standard unit vector; there is k, which is (0,0,1). 00:33:53.000 --> 00:33:57.600 So, i is the first component; j is the second component; k is the third component. 00:33:57.600 --> 00:34:06.100 With this, we can express three-dimensional vectors with (i,j,k), so (5,-2,3) would become 5i - 2j + 3k. 00:34:06.100 --> 00:34:11.500 5 becomes 5i; -2 becomes -2j; 3 becomes 3k; great. 00:34:11.500 --> 00:34:15.500 So, we can combine them in terms of standard unit vectors, as well. 00:34:15.500 --> 00:34:18.400 We can also talk about the magnitude of something that is higher than two dimensions. 00:34:18.400 --> 00:34:25.100 It might seem a little surprising at first, but it turns out that it is actually really easy to figure out the magnitude-- the length--of a vector in any dimension. 00:34:25.100 --> 00:34:32.200 Consider the n-dimensional vector x, where it is x₁, the first component of x, 00:34:32.200 --> 00:34:38.100 comma, x₂, the second component of x, all the way up until we get to x<font size="-6">n</font>, the nth component of x. 00:34:38.100 --> 00:34:44.800 Now, it turns out that the magnitude of our vector x is just the square root of the sums of each of its components, squared. 00:34:44.800 --> 00:34:46.700 Now, that seems a little bit confusing, but it makes sense. 00:34:46.700 --> 00:34:53.200 The length of our vector x is equal to the square root of the first component squared, plus the second component squared, 00:34:53.200 --> 00:34:57.800 plus...all the way up until the nth component, our last component, squared. 00:34:57.800 --> 00:35:00.600 So, that seems really surprising, the very first time we see this. 00:35:00.600 --> 00:35:04.500 We will actually explore why this is the case, and it will make sense why this has to always be the case, 00:35:04.500 --> 00:35:07.900 if we expand the idea that we will see in Example 3 of higher dimensions. 00:35:07.900 --> 00:35:14.600 Just think, "Oh, yes, that will just keep stair-stepping up, and that is why we see this square root of all of the components, squared and added together." 00:35:14.600 --> 00:35:17.800 But you can also just memorize this formula, if you want to. 00:35:17.800 --> 00:35:20.300 And it will be just fine, too, and work out. 00:35:20.300 --> 00:35:22.300 All right, let's see some examples. 00:35:22.300 --> 00:35:28.900 First, given that u = (1,3), v = (4,2), w = (-5,1), what are each of the following? 00:35:28.900 --> 00:35:37.700 (1,3)--if we are talking about u + v, then that is the same thing as talking about (1,3) + (4,2). 00:35:37.700 --> 00:35:43.200 So, remember: we add the components together, so it will be 1 + 4, because they are both the first components; 00:35:43.200 --> 00:35:46.200 and then 3 + 2, because they are both of the second components. 00:35:46.200 --> 00:35:53.800 1 + 4 gets us 5; 3 + 2 gets us 5 also, just by chance; so our answer here is (5,5). 00:35:53.800 --> 00:36:02.100 Next, we have 6u: well, 6 times...our vector is (1,3), so the 6 distributes effectively. 00:36:02.100 --> 00:36:06.900 It is not quite distribution--it is a little bit different--but it has the exact same effect and feel. 00:36:06.900 --> 00:36:15.600 So, the 6 multiplies each of the components; so we have 6, comma, 6 times 3 is 18; so (6,18) comes out of that. 00:36:15.600 --> 00:36:19.600 All right, we can probably start doing these scalars in our head; they are not too hard to do. 00:36:19.600 --> 00:36:28.800 1/2 times v...well, v was (4,2), so 1/2 times (4,2) is going to produce 2 (1/2 of 4 is 2), and 1/2 of 2 is 1. 00:36:28.800 --> 00:36:35.500 So, we have (2,1) for 1/2v; and then, minus...our w was (-5,1). 00:36:35.500 --> 00:36:40.300 We can distribute this negative here; so this becomes positive; this will become positive; this will become negative. 00:36:40.300 --> 00:36:47.900 We distribute that negative into there; and so, 2 + 5 is 7; 1 - 1 is 0. 00:36:47.900 --> 00:36:51.600 There we go; and the last one: if we have 2u - 3v + w... 00:36:51.600 --> 00:37:09.500 All right, 2 times u: 2 times (1,3) will become (2,6); minus...3 times (4,2) will become (12,6), plus w: w was (-5,1); great. 00:37:09.500 --> 00:37:23.300 (2,6) -...so that would make that -12 + -12, comma, -6, plus (-5,1). 00:37:23.300 --> 00:37:27.900 At this point, we could add them all together; we could add them one by one; it doesn't really matter how we approach this. 00:37:27.900 --> 00:37:39.200 Let's just add the first two; so 2 and -12 becomes -10; 6 and -6 becomes 0; plus...bring down the rest of it...(-5,1)... 00:37:39.200 --> 00:37:48.900 So, (-5,1) + (-10,0): -10 and -5 becomes -15; 0 and 1 becomes positive 1; and there it is. 00:37:48.900 --> 00:37:53.400 So, that is the basics of vector addition and multiplication by scalars. 00:37:53.400 --> 00:37:56.200 It is very similar to what we are used to doing with numbers normally. 00:37:56.200 --> 00:38:02.200 It is just that everything stays inside of its slot; they only interact with other things from the same slot as them. 00:38:02.200 --> 00:38:09.100 First slots interact; second slots interact; and if it is higher than two dimensions, third, fourth, fifth...whatever slots interact. 00:38:09.100 --> 00:38:17.200 All right, the next example: If u = 4, v = 6, and the two vectors make angles of θu = 30 degrees, θv = 120 degrees, 00:38:17.200 --> 00:38:21.700 to the positive x-axis, what is the component form of u + v? its length? its angle? 00:38:21.700 --> 00:38:28.800 All right, let's do u in red; over here, we first draw a sketch to be able to figure this out. 00:38:28.800 --> 00:38:33.100 So notice: if we are going to figure out u + v, if we are going to get the component form of u + v, 00:38:33.100 --> 00:38:36.400 well, it is hard to add angles and lengths together. 00:38:36.400 --> 00:38:41.700 We could draw...well, u is an angle of 30 degrees, so we will go out like this. 00:38:41.700 --> 00:38:49.200 And then, v is 6 at 120 degrees, so we will be a little bit longer...6...120 degrees...like that. 00:38:49.200 --> 00:38:53.500 And so, our final thing will end up being this; and we could measure what that is. 00:38:53.500 --> 00:38:58.100 But we would have to have really, really precise stuff; we would have to have a really accurate ruler, 00:38:58.100 --> 00:39:02.200 and be doing this with a protractor that was really good, and be really, really careful to get all this. 00:39:02.200 --> 00:39:05.000 So, it is not the sort of thing where we could draw it out and get a very good answer. 00:39:05.000 --> 00:39:10.000 So, our first step is to get a component format of each of them, because once we have a component form for u 00:39:10.000 --> 00:39:12.700 and a component form for v, it is easy to add them together. 00:39:12.700 --> 00:39:15.200 And we add them together, and then we can figure out the length and the angle. 00:39:15.200 --> 00:39:18.200 So, our first step is to get a component form. 00:39:18.200 --> 00:39:23.200 u = 4; our θu is 30 degrees; and remember, it was with the positive x-axis. 00:39:23.200 --> 00:39:29.500 So, our angle is 30 degrees like this. 00:39:29.500 --> 00:39:38.300 So, if we want to break down u into its x-component and its y-component, well, then, we know that cos(30)... 00:39:38.300 --> 00:39:47.900 remember, its hypotenuse was 4, so cos(30) is going to be equal to the x-component of u, 00:39:47.900 --> 00:39:56.200 the first component of u (here is u<font size="-6">x</font>); the side adjacent...cos(30) = side adjacent, u<font size="-6">x</font>, 00:39:56.200 --> 00:40:02.800 over the hypotenuse, 4; so 4 times cos(30) = u<font size="-6">x</font>. 00:40:02.800 --> 00:40:13.100 cos(30) is just the same thing as √3/2, so we have 4(√3/2); so we have 2√3 = u<font size="-6">x</font>. 00:40:13.100 --> 00:40:17.100 A very similar thing is going on if we want to figure out what u<font size="-6">y</font> is. 00:40:17.100 --> 00:40:28.300 It will be sin(30) = u<font size="-6">y</font>/4; multiply by 4 on both sides; 4sin(30) = u<font size="-6">y</font>. 00:40:28.300 --> 00:40:36.100 4sin(30)...sin(30) is just 1/2, so 4(1/2)...so we have 2 = u<font size="-6">y</font>. 00:40:36.100 --> 00:40:48.400 So, at this point, we have that u equals 2√3, its x-component, comma, and its y-component, 2. 00:40:48.400 --> 00:40:51.900 And that is what our values for u are. 00:40:51.900 --> 00:40:54.000 Now, I want to point something out before we keep moving. 00:40:54.000 --> 00:41:02.700 Notice right here: we were starting at sin(30) = u<font size="-6">y</font>/4, but we always get to this 4 times sin(30). 00:41:02.700 --> 00:41:10.700 If you know the length of your vector, and you know what angle it is at, you can actually just hop to length of the thing, 00:41:10.700 --> 00:41:17.800 times cosine (or sine, if it is side adjacent or side opposite, respectively) of the angle. 00:41:17.800 --> 00:41:22.100 And that will just give you what that side adjacent or side opposite is, respectively. 00:41:22.100 --> 00:41:24.100 So, we will end up doing that on the next one. 00:41:24.100 --> 00:41:31.700 If it was a little bit confusing, notice the parallel to how we just did it with the u vector while we are working on the v vector. 00:41:31.700 --> 00:41:37.000 But it is a really great way of being able to do this really, really quickly-- 00:41:37.000 --> 00:41:39.700 well, not really, really quickly, but it does help speed things up. 00:41:39.700 --> 00:41:42.900 And it is a good trick, because you end up seeing this quite a lot. 00:41:42.900 --> 00:41:47.300 So, we have 120 degrees here; we are at 120 degrees. 00:41:47.300 --> 00:41:52.700 Now, I think it is going to be a little bit easier to figure out in terms of this angle here, 00:41:52.700 --> 00:42:01.100 because it is easy to work under 90 degrees, so that is 60 degrees, because 120 + 60 = 180. 00:42:01.100 --> 00:42:09.300 So, now we want to figure out the v<font size="-6">x</font> component, the horizontal component, and the v<font size="-6">y</font>, the vertical component. 00:42:09.300 --> 00:42:16.600 So, v<font size="-6">x</font> is going to be equal to sine...it is 60...we have 60 in the angle, 00:42:16.600 --> 00:42:24.700 and it is going to be side adjacent, so it is going to be the length, 6, times cosine, side adjacent of the angle involved. 00:42:24.700 --> 00:42:27.000 However, there is one thing that we need to notice. 00:42:27.000 --> 00:42:32.800 What we are figuring out here is the length of that side of a triangle, because the angle is inside of a triangle. 00:42:32.800 --> 00:42:36.300 So, it is up to us to pay attention: is it going to be positive? is it going to be negative? 00:42:36.300 --> 00:42:40.900 Normally, we have the x direction going negative as we go to the left. 00:42:40.900 --> 00:42:47.800 That is negative x going this way, so that means we have to have a negative sign on our x-component for v. 00:42:47.800 --> 00:42:52.700 Otherwise, it won't make sense, because we can see from the picture that it is going negative on the horizontal. 00:42:52.700 --> 00:42:57.200 So, we have to pay attention to this; we have to notice this stuff happen, because otherwise it will be a mistake. 00:42:57.200 --> 00:43:08.900 OK, -6 times cos(60)...cos(60) is 1/2, so we get -3; v<font size="-6">x</font> = -3; v<font size="-6">y</font> = 6. 00:43:08.900 --> 00:43:19.500 This one is positive, because it is going up...sine of 60 is √3/2, so that gets us 3√3. 00:43:19.500 --> 00:43:25.800 So, we have that the x-component is -3; the y-component is 3√3. 00:43:25.800 --> 00:43:33.200 So, our v vector equals (-3,3√3). 00:43:33.200 --> 00:43:39.800 Now, if we want to add these two together, u + v, it is simply a matter of adding them together. 00:43:39.800 --> 00:43:55.700 Our u was (2√3,2); our v was (-3,3√3); we add them together; we have (2√3 - 3, 2 + 3√3). 00:43:55.700 --> 00:44:01.000 And if we want to, we could get what that is approximately, in terms of a decimal number, although that is exact. 00:44:01.000 --> 00:44:11.300 That is a perfect thing; this will just be approximate, because it is a decimal of square roots: (0.464,7.196). 00:44:11.300 --> 00:44:14.300 All right, so at this point, we want to figure out what its length is. 00:44:14.300 --> 00:44:25.500 The length of u + v is going to be equal to the square root of each of its components, squared and then added together. 00:44:25.500 --> 00:44:33.200 So, its two components were 0.464, squared, plus 7.196, squared. 00:44:33.200 --> 00:44:36.200 What does that end up coming out to be? 00:44:36.200 --> 00:44:48.100 We work that out, and that ends up coming out to be...sorry, I couldn't find it in my notes...7.211. 00:44:48.100 --> 00:44:50.400 It comes out to be approximately 7.211. 00:44:50.400 --> 00:44:56.300 If we want to figure out what the angle is that it is at, the first thing we probably want to do is draw a quick diagram, so we can see it. 00:44:56.300 --> 00:45:01.000 So remember: it was at this one right here, in terms of its components. 00:45:01.000 --> 00:45:07.100 0.464 is just a little bit over the x-component, and then 7.196 up...so it is like this. 00:45:07.100 --> 00:45:09.700 So, we can see that that should be what the angle is like. 00:45:09.700 --> 00:45:17.600 So remember: tan(θ) is equal to the side opposite, divided by the side adjacent. 00:45:17.600 --> 00:45:26.600 The side opposite in this case will be the vertical height, divided by the side adjacent of 0.464, the horizontal amount. 00:45:26.600 --> 00:45:36.300 Take the arctan of both sides, the inverse tangent of 7.196 over 0464. 00:45:36.300 --> 00:45:44.500 We plug that into our calculator or look it up in a table; and that comes out to be approximately 86.31 degrees. 00:45:44.500 --> 00:45:52.600 Great; all right, the next one: what is the magnitude of the vector u = (4,3,12)? 00:45:52.600 --> 00:45:57.200 And then, we want to figure it out by both the formula we were given and the Pythagorean theorem. 00:45:57.200 --> 00:46:02.000 First, the formula we have--this is the easy part, a nice, handy formula. 00:46:02.000 --> 00:46:12.000 It is going to be the square root of each of its components squared (4, 3, and 12), and all added together. 00:46:12.000 --> 00:46:26.300 We work this out; we get √(16 + 9 + 144) =...add those together; 16 + 144 gets us 160; 160 + 9 gets us 169, 00:46:26.300 --> 00:46:32.900 which simplifies to √169, is 13; so there is the length of our vector. 00:46:32.900 --> 00:46:35.600 Now, we also want to figure out the Pythagorean theorem. 00:46:35.600 --> 00:46:42.100 This is where we are going to understand why this formula--why this mystical formula actually works and makes sense all the time. 00:46:42.100 --> 00:46:45.900 First, let's see where this vector would get plotted out to. 00:46:45.900 --> 00:46:47.500 So, we are going to have to look at this three-dimensionally. 00:46:47.500 --> 00:46:50.700 And while we haven't talked about three-dimensional coordinates before in this course, 00:46:50.700 --> 00:46:53.300 you have probably seen them at some point previously in some course. 00:46:53.300 --> 00:46:57.600 Here are our x, our y, and our z-coordinates. 00:46:57.600 --> 00:47:06.100 We go out: (4,3,12); so 4 out on the x...a little way out on the x, a little less out on the y... 00:47:06.100 --> 00:47:17.600 We are out here: 4, 3, 4, 3...and then we go up by 12; so our vector is like that. 00:47:17.600 --> 00:47:23.500 Now, notice: we had to get out here to this place by the x and the y part first. 00:47:23.500 --> 00:47:28.700 We could figure out what the length is here, and then we have a square angle here, as well. 00:47:28.700 --> 00:47:35.600 So, let's figure out...we can break this down into two parts: what is happening in the (x,y) plane... 00:47:35.600 --> 00:47:43.100 in the (x,y) plane we have (4,3) as the cross-section here. 00:47:43.100 --> 00:47:51.300 I will color it: this part here is the same as this part here. 00:47:51.300 --> 00:48:00.300 We figure that out; that is 3 there, as well; we are going to end up getting (by the Pythagorean theorem) √(3² + 4²). 00:48:00.300 --> 00:48:09.400 That equals √(9 + 16), equals √25, equals 5; great. 00:48:09.400 --> 00:48:12.600 We have figured out what the lower part is on the bottom part. 00:48:12.600 --> 00:48:17.900 Now, we can do this cross-section with the z-axis included. 00:48:17.900 --> 00:48:28.100 So now, we look at a cross-section; using this cross-section, we can cut this, and we can see: 00:48:28.100 --> 00:48:32.200 here is the thing we are trying to figure out, the length of this. 00:48:32.200 --> 00:48:41.900 And we know that the z amount was 12; so a cross-section with the z-axis...this here maps to this part here. 00:48:41.900 --> 00:48:46.900 And then, here, our purple part shows up here. 00:48:46.900 --> 00:48:49.500 That was length 5, as we just figured out. 00:48:49.500 --> 00:48:58.300 We use the Pythagorean theorem here; so once again, the value of our hypotenuse is going to be the square root of 12² + 5². 00:48:58.300 --> 00:49:07.800 So, that is the square root of 144 + 25, or the square root of...add those together; we get 169, which equals 13. 00:49:07.800 --> 00:49:11.800 That is the exact same thing that we saw over here when we used that formula. 00:49:11.800 --> 00:49:13.100 Cool--it works out both ways. 00:49:13.100 --> 00:49:15.900 Now, let's understand why it works out both ways. 00:49:15.900 --> 00:49:21.900 Well, notice: the thing here for the purple line was the square root of 3² + 4². 00:49:21.900 --> 00:49:29.300 It was the square root of the x² plus the y², the first component and second component put together, squared. 00:49:29.300 --> 00:49:34.500 Now, notice: we end up just plugging in the purple part, because it makes up one of the parts of our triangle here. 00:49:34.500 --> 00:49:43.600 We could have alternatively looked at the square root of 12², plus the purple part, 00:49:43.600 --> 00:49:47.800 because that is what is going to go in there: the square root of 3² + 4², 00:49:47.800 --> 00:49:50.100 because that is what the purple part ends up being; 00:49:50.100 --> 00:49:54.800 and then, because we are going back to using the Pythagorean theorem, that whole thing is going to be squared, as well. 00:49:54.800 --> 00:50:00.600 Well, that means √(12² plus...if we have the square root then being squared, 00:50:00.600 --> 00:50:09.000 squared on top of square root, that cancels out, and we have + 3² + 4². 00:50:09.000 --> 00:50:12.400 And look: that is the exact same thing that we have up here. 00:50:12.400 --> 00:50:16.000 And so, that is where we are getting the ability to just put them all together--stack them all together. 00:50:16.000 --> 00:50:19.700 It is because we have to figure out one cross-section after another after another. 00:50:19.700 --> 00:50:23.600 But if we then plug in the way these cross-sections end up working out, each of the square roots 00:50:23.600 --> 00:50:27.500 that would go in from a cross-section would get canceled by the next cross-section it goes into. 00:50:27.500 --> 00:50:32.400 And so, ultimately, we end up getting this form of first component squared, plus second component squared, 00:50:32.400 --> 00:50:35.100 plus third component squared, until we get to our last component squared. 00:50:35.100 --> 00:50:40.400 And then we are adding them all and taking the square root, and that is why we have that formula for the magnitude. 00:50:40.400 --> 00:50:45.300 All right, next: A box weighing 300 Newtons is hung up by two cables, A and B. 00:50:45.300 --> 00:50:48.800 Using the diagram, figure out how much tension is in each cable. 00:50:48.800 --> 00:50:52.000 The first thing to do is to understand what this means. 00:50:52.000 --> 00:50:56.200 A lot of math problems get thrown at us like this, where they are actually pulling from physics. 00:50:56.200 --> 00:50:59.500 And they are sort of assuming that we know things about physics that we might have no idea about. 00:50:59.500 --> 00:51:01.500 This is a math course, not a physics course! 00:51:01.500 --> 00:51:05.100 Let's first get an understanding of what this means. 00:51:05.100 --> 00:51:12.400 If we have a box on a string--just imagine for a minute; now imagine that it weighs 100 Newtons. 00:51:12.400 --> 00:51:19.700 And if you didn't know, a Newton is the unit of force and weight in the metric system. 00:51:19.700 --> 00:51:24.800 They use kilograms for mass, but force, how hard something is being pushed or pulled--that is Newtons. 00:51:24.800 --> 00:51:32.600 In the English system, the British imperial system, it is pounds; so pounds is used for weight and force, 00:51:32.600 --> 00:51:35.600 although we actually have another unit for mass; but you almost never hear it. 00:51:35.600 --> 00:51:38.600 It is called slugs, if you are using the British imperial system. 00:51:38.600 --> 00:51:44.400 But Newtons are a way of measuring weight, which is just a question of how much gravity is pulling. 00:51:44.400 --> 00:51:49.700 OK, so imagine that this thing is being pulled down by 100 Newtons of gravity. 00:51:49.700 --> 00:51:52.100 We have 100 Newtons of force pulling down on this thing. 00:51:52.100 --> 00:51:56.300 Well, if we have some rope or some string that is holding this thing up, 00:51:56.300 --> 00:52:00.600 well, if it is being pulled down, it must be that the string is pulling back up. 00:52:00.600 --> 00:52:02.700 Otherwise, the thing would fall to the ground. 00:52:02.700 --> 00:52:04.300 So, how much is the string pulling up by? 00:52:04.300 --> 00:52:12.400 Well, the string must have a tension pulling up of 100 Newtons in the opposite direction; so there are 100 Newtons going up and 100 Newtons going down. 00:52:12.400 --> 00:52:18.200 Now, notice: as a vector, this would be a positive 100 Newtons, because it is in the up direction. 00:52:18.200 --> 00:52:21.800 As a vector, this would be a negative 100 Newtons, in the down direction. 00:52:21.800 --> 00:52:27.900 We take positive 100 Newtons, and we add that to -100 Newtons; we get 0. 00:52:27.900 --> 00:52:34.700 This makes sense, because no force means no acceleration. 00:52:34.700 --> 00:52:41.500 The thing is currently stopped; so as long as it doesn't have any acceleration to make it move somewhere, it is going to not pick up any motion. 00:52:41.500 --> 00:52:45.200 What we have to have: we have to have no force out of it for it to not move. 00:52:45.200 --> 00:52:50.800 Now, since it is hung up by two cables, it is perfectly reasonable to say, "Yes, if it is hung up, it is not currently moving anywhere." 00:52:50.800 --> 00:52:55.600 It is not falling to the ground; it is not swinging left and right; it is just hanging there in space--it is sitting there. 00:52:55.600 --> 00:53:00.300 So, that means that it must be a total of 0 for what is going down and what is going up. 00:53:00.300 --> 00:53:04.300 We know that it is 300 Newtons going down; so now it is a question of how much is going up. 00:53:04.300 --> 00:53:07.600 So now, we need to talk about these cables. 00:53:07.600 --> 00:53:14.400 We can think about A as being a vector pulling out and away, because it is pulling up on that box. 00:53:14.400 --> 00:53:16.200 Otherwise, it would be helping the box fall. 00:53:16.200 --> 00:53:26.200 So, this is some vector A, and it has some force tension; so we will say A is equal to the tension in A. 00:53:26.200 --> 00:53:31.400 And we will do the same thing over here with B; so now we have some vector B, 00:53:31.400 --> 00:53:36.700 and B will be the amount of the tension in B, how much it is being pulled up by. 00:53:36.700 --> 00:53:41.400 Now, we don't know what A and B are; that is what we are trying to figure out. 00:53:41.400 --> 00:53:47.000 But we want to figure out how to get to them; so we start looking at this, and we say, "Well, I don't know a lot." 00:53:47.000 --> 00:53:50.500 But they did tell us this information about the angles. 00:53:50.500 --> 00:53:56.400 So, maybe we can break these vectors up into component forms, based on these angles. 00:53:56.400 --> 00:54:04.000 So, if we have 40 degrees here, then we must have 90 - 40 here. 00:54:04.000 --> 00:54:08.100 And since this is another right angle, that means we have 40 here again. 00:54:08.100 --> 00:54:15.200 Similarly, with the same basic idea, since it was 70 up here, then we have 70 down here. 00:54:15.200 --> 00:54:21.600 With that in mind, we can figure out what the pieces are here. 00:54:21.600 --> 00:54:27.000 We know that A is the length of this hypotenuse; we don't know what the number is yet, but we are calling it A. 00:54:27.000 --> 00:54:33.400 So, the vector A is going to be broken into the components: the horizontal amount is the side adjacent to 40, 00:54:33.400 --> 00:54:39.400 so A, the length of the whole thing, times side adjacent of the angle; 00:54:39.400 --> 00:54:45.500 and then, A times opposite, if we want to talk about the vertical part right here. 00:54:45.500 --> 00:54:49.200 That will be Asin(40) over here. 00:54:49.200 --> 00:54:55.000 Next, we know that we can talk about vector B; we can break it down into very much the same way. 00:54:55.000 --> 00:55:03.800 That will be B times cos(70), because its side adjacent is 70; and B times its side opposite... 00:55:03.800 --> 00:55:08.600 I'm sorry, not its side adjacent; its side adjacent is not 70, but the angle connected to side adjacent is 70 degrees. 00:55:08.600 --> 00:55:13.800 And B times sin(70)...we can get this all just from basic trigonometry stuff. 00:55:13.800 --> 00:55:20.200 Here is B; we can multiply B, our hypotenuse, and figure out what the sides opposite are, based on this. 00:55:20.200 --> 00:55:22.800 We have two vectors here, A and B. 00:55:22.800 --> 00:55:29.200 One other thing that we know is the force vector: what is the force on this box? 00:55:29.200 --> 00:55:33.500 Well, it must be that the force on this box...is it moving up and down? Is it currently still? 00:55:33.500 --> 00:55:42.200 Well, it is currently still; it is not moving anywhere; and it is not moving anywhere horizontally; and it is not moving anywhere vertically. 00:55:42.200 --> 00:55:49.400 So, total force, in the end, once everything gets put together: it is not moving up and down; it is not moving left to right; so it must be (0,0). 00:55:49.400 --> 00:55:56.800 And they did tell us one other piece of information, 300 Newtons; so the weight is equal to 0 00:55:56.800 --> 00:56:02.600 (because it is perfectly down, so it doesn't have any horizontal) and -300, because it is moving down; great. 00:56:02.600 --> 00:56:07.000 At this point, we actually have enough information to solve it; let's work this thing out now. 00:56:07.000 --> 00:56:18.400 Remember: we have weight = (0,-300); and we know that when we combine the weight with each of the cables-- 00:56:18.400 --> 00:56:23.800 once we put all of these forces together, all of the forces put together must come out to be a force of nothing, 00:56:23.800 --> 00:56:26.900 because the box isn't moving anywhere--it is not being shoved around. 00:56:26.900 --> 00:56:47.900 We can plug all of these things together; we have A + B + the weight vector is going to equal our total force of (0,0). 00:56:47.900 --> 00:56:51.100 So, let's work out what that is: we have -A... 00:56:51.100 --> 00:56:58.000 Oh, that was one thing I didn't say on the previous slide; I put it as Acos(40), because Acos(40) is the length of this side. 00:56:58.000 --> 00:57:05.100 But it has to be negative, because remember: when we go to the left, it is negative in the x direction, so it is negative. 00:57:05.100 --> 00:57:13.100 Both of the verticals are positive, because they are both pointed up; but only the B is pointed to the right in this--pointed positively horizontally, as well. 00:57:13.100 --> 00:57:29.300 All right, we have (-Acos(40), Asin(40)), plus (Bcos(70),Bsin(70)). 00:57:29.300 --> 00:57:31.800 I'm sorry; I am going to have to continue onto the next line. 00:57:31.800 --> 00:57:37.400 Remember: this is just all one line put together...so + (0,-300). 00:57:37.400 --> 00:57:41.900 In the end, we will end up equaling (0,0). 00:57:41.900 --> 00:57:47.600 Let's combine this all together; we will switch to the color of green for everything together. 00:57:47.600 --> 00:58:00.000 We have -Acos(40) + Bcos(70) + the 0 from the weight; that comes out to be the total for our first components. 00:58:00.000 --> 00:58:12.400 And then, Asin(40) + Bsin(70) - 300 = (0,0). 00:58:12.400 --> 00:58:16.500 So, at this point, we know that the first component on the left side of an equation 00:58:16.500 --> 00:58:19.600 has to be the same as the first component on the right side of the equation. 00:58:19.600 --> 00:58:23.100 The same thing: the second component on the left side must be the same thing as the second component on the right side. 00:58:23.100 --> 00:58:25.100 So, we can break this down into two separate equations. 00:58:25.100 --> 00:58:38.700 We have -Acos(40) + Bcos(70) = 0; so we can get B on its own or A on its own, and then plug that into the other one. 00:58:38.700 --> 00:58:51.900 Let's solve for that first: Bcos(70) = Acos(40); at this point, we see that B is equal to A times cos(40), over cos(70). 00:58:51.900 --> 00:58:54.700 We could plug that into a calculator right now and get some number out of it. 00:58:54.700 --> 00:58:57.700 But then we would have to write all of these decimals; so we can just leave it like that for now. 00:58:57.700 --> 00:59:02.100 Next, we will swap to a new color for solving this part. 00:59:02.100 --> 00:59:14.500 We have Asin(40) + Bsin(70) - 300 = what was on the right-hand side, 0. 00:59:14.500 --> 00:59:24.600 So, we can move the 300 over; we see that we have Asin(40) + Bsin(70) = 300. 00:59:24.600 --> 00:59:32.500 Now, we see that B is the same thing as A times cos(40), divided by cos(70); so we have Asin(40) +... 00:59:32.500 --> 00:59:44.700 we swap out our B: (Acos(40))/cos(70), times sin(70), still equals 300. 00:59:44.700 --> 00:59:59.000 So, at this point, we pull out all of our A's; we have A times sin(40), plus cos(40)/cos(70), times sin(70), equals 300. 00:59:59.000 --> 01:00:04.000 Notice that there is nothing we can cancel out there, because we don't have any exactly matching things. 01:00:04.000 --> 01:00:22.100 But we can divide by it: 300 divided by sin(40) plus cos(40) over cos(70) times sin(70). 01:00:22.100 --> 01:00:25.800 We can plug that all into a calculator, and it will come up and give us an answer. 01:00:25.800 --> 01:00:40.400 And it will tell us that A is approximately equal to 36.40...oops, sorry, not 36.40, but times 3... 01:00:40.400 --> 01:00:48.500 give me just a second...the magic of video...it comes out to be approximately 109.2. 01:00:48.500 --> 01:00:53.900 And that is our value for A; to figure out our value for B, we have this handy thing right here. 01:00:53.900 --> 01:01:03.500 B equals A, which was 109.2, times cos(40), divided by cos(70). 01:01:03.500 --> 01:01:10.300 We work that out with our calculator, and we get approximately 244.6. 01:01:10.300 --> 01:01:29.400 So, the tension in B is 244.6 Newtons, and the tension in A is 109.2 Newtons; great--those are our solutions. 01:01:29.400 --> 01:01:37.500 All right, the last example: A plane has a compass heading of 75 degrees east of due north, and an airspeed of 140 miles per hour. 01:01:37.500 --> 01:01:44.000 If the wind is blowing at 20 miles per hour, and towards 10 degrees west of due north, what is the plane's direction and speed, relative to the ground? 01:01:44.000 --> 01:01:45.500 First, what does this first part mean? 01:01:45.500 --> 01:01:51.100 We have a compass heading of 75 degrees east, going at 140 miles per hour airspeed. 01:01:51.100 --> 01:01:56.500 Airspeed means your speed in the air--how fast you are moving, relative to the air right around you. 01:01:56.500 --> 01:02:05.700 So, if that is the case, then our plane is moving 75 degrees east of due north. 01:02:05.700 --> 01:02:15.600 Due north is this way, our vertical axis; so if that is the case, we need to curve 75 degrees down towards the east. 01:02:15.600 --> 01:02:19.600 So, east is this way, so that is 75 degrees here. 01:02:19.600 --> 01:02:23.600 If we want to figure out what is the thing in here, because we will probably want that for figuring out other things, 01:02:23.600 --> 01:02:29.700 that is going to be 15 degrees; so it is 15 degrees for our normal θ that we are used to. 01:02:29.700 --> 01:02:34.600 OK, so that is the direction that the plane is headed; it is going like that. 01:02:34.600 --> 01:02:38.000 It is going off in this way; but then, that is its airspeed. 01:02:38.000 --> 01:02:43.600 The air also is able to move; the air is going this way--the air is blowing the plane. 01:02:43.600 --> 01:02:48.400 So, the plane is going like this, but at the same time it is being blown off-course slightly by the wind. 01:02:48.400 --> 01:02:52.100 Or perhaps (hopefully) they have taken this into account, and it is not going off-course. 01:02:52.100 --> 01:02:57.900 We have the wind blowing at 20 miles per hour, and towards 10 degrees west of due north. 01:02:57.900 --> 01:03:04.800 So, once again, it is off of due north; and now it is just a little off, 10 degrees off, here. 01:03:04.800 --> 01:03:07.300 And it is total of 20 miles per hour. 01:03:07.300 --> 01:03:15.400 So, we have 140 miles per hour this way and 20 miles per hour, 10 degrees to the west of due north. 01:03:15.400 --> 01:03:22.100 So, notice that the total angle for that is going to end up being 100 degrees, if we wanted to figure that out. 01:03:22.100 --> 01:03:29.300 Or we could also look at it in terms of 80 degrees here, as well. 01:03:29.300 --> 01:03:32.900 It depends on which one you think is easier; I am going to go with the one inside of the triangle, 01:03:32.900 --> 01:03:37.700 and we will just have to remember to deal with the fact that our horizontal is going to be negative when we are working on the wind. 01:03:37.700 --> 01:03:43.400 So, the plane--what is the velocity of the plane in the air? 01:03:43.400 --> 01:03:48.200 To be able to figure out what the plane's direction and speed is relative to the ground, 01:03:48.200 --> 01:03:54.300 we have to combine its motion in the medium with the medium's motion relative to the ground. 01:03:54.300 --> 01:04:00.600 Its motion in the medium is the 75 degrees east of due north and an airspeed of 140 miles per hour, the blue part. 01:04:00.600 --> 01:04:06.600 And the red part, the wind blowing, is the air relative to the ground; so we have to combine those two things. 01:04:06.600 --> 01:04:14.700 So, the velocity of the plane in the air...its horizontal component is going to be the length of the vector, 140, times... 01:04:14.700 --> 01:04:18.600 the angle we have is 15, so if we are talking about the horizontal, that is going to be side adjacent. 01:04:18.600 --> 01:04:26.900 So, cos(15)...140 times sin(15) for the vertical, because that is side opposite...we work that out; 01:04:26.900 --> 01:04:35.600 and that ends up coming out to be approximately 135.2 and 36.23. 01:04:35.600 --> 01:04:44.000 And the units on both of those are miles per hour, because it is moving 135.2 miles per hour north and 36.23 miles east simultaneously. 01:04:44.000 --> 01:04:49.100 And then, we have the wind: what is the velocity of the air itself? 01:04:49.100 --> 01:04:56.500 The air is moving at a speed of 20; and if we want to figure out its horizontal component, it is going to be this part here, 01:04:56.500 --> 01:05:02.800 at which point we say, "Oh, right; it is going negative, so let's put in that negative sign, because it is moving to the left." 01:05:02.800 --> 01:05:05.700 We have to remember to catch those negatives. 01:05:05.700 --> 01:05:09.300 Pay attention to if it is going to be a positive or a negative direction for everything. 01:05:09.300 --> 01:05:16.200 So, that is going to be 20, the size of it, times cosine of 80 degrees, in this case; 01:05:16.200 --> 01:05:22.900 and then positive 20, because this one is going positively, times sine of 80. 01:05:22.900 --> 01:05:32.500 Work that one out with a calculator; and we get 131.7, 55.93. 01:05:32.500 --> 01:05:34.600 Oops, I'm sorry--I wrote the entirely wrong thing. 01:05:34.600 --> 01:05:44.000 I meant to write -3.47 (I read the wrong thing off of my notes), comma, 19.70; great. 01:05:44.000 --> 01:05:50.000 All right, so if we want to figure out the combination of the two--if we want to figure out what the total motion of the plane, 01:05:50.000 --> 01:06:00.400 relative to the ground, is, that is going to be the plane's motion relative to the air, plus the air's motion relative to the ground. 01:06:00.400 --> 01:06:08.000 We just figured out what each one of those is: (135.2,36.23) was our plane's motion, 01:06:08.000 --> 01:06:21.500 plus (-3.47,19.70); so in total, that gets us (131.5,55.93); great. 01:06:21.500 --> 01:06:28.600 So, that is what the velocity vector is going to be--what the component form is. 01:06:28.600 --> 01:06:34.500 However, it asked for speed and direction of the whole thing put together. 01:06:34.500 --> 01:06:37.200 If that is the case, we want to figure out speed. 01:06:37.200 --> 01:06:43.900 Well, speed is just the size of our velocity total vector. 01:06:43.900 --> 01:06:48.800 That is going to be the square root of the first component of our total vector, 131.7, squared, 01:06:48.800 --> 01:06:52.700 plus the second component, squared (55.93 squared). 01:06:52.700 --> 01:07:01.000 We take the square root of that and figure it all out, using our calculator; and we end up getting 143.1 miles per hour. 01:07:01.000 --> 01:07:07.700 The plane is actually going a little bit faster than its speed had been previously without the air connected to it. 01:07:07.700 --> 01:07:09.800 However, its direction will also change. 01:07:09.800 --> 01:07:14.700 To help us figure out direction, let's draw just a quick picture, so that we know what is going on. 01:07:14.700 --> 01:07:28.200 We have this right here as our motion: 131.7 is fairly horizontal, and a little bit...between 1/3 and 1/2 of our amount horizontally up. 01:07:28.200 --> 01:07:31.900 That is what our motion is like total. 01:07:31.900 --> 01:07:36.900 My picture is totally not relative; 140 up here should be much longer than the 20 here; 01:07:36.900 --> 01:07:40.300 and this 143.1 miles-per-hour long vector should be even longer. 01:07:40.300 --> 01:07:45.200 But that is OK; we are just trying to get a sense of what is going on; they are sketches, not perfect drawings. 01:07:45.200 --> 01:07:55.100 So, we are looking for this angle here: tan(θ) is going to be the side opposite, the vertical component, 55.93, 01:07:55.100 --> 01:08:00.600 divided by the side adjacent, the horizontal component, 131.7. 01:08:00.600 --> 01:08:05.500 We take the arctan of that; that gets us θ =...about 23 degrees. 01:08:05.500 --> 01:08:10.200 Now, notice: that is what θ is equal to: θ equals 23 degrees. 01:08:10.200 --> 01:08:15.200 All of this stuff was given in east of due north, so we have to put it in that same thing. 01:08:15.200 --> 01:08:22.600 If it is 23 degrees as our θ here, that means it is 23 degrees going up from our positive x-axis, which was east. 01:08:22.600 --> 01:08:32.100 Up would be towards the north; so we could phrase this as 23 degrees north of east. 01:08:32.100 --> 01:08:38.800 Alternatively, if we wanted to use the exact same thing that they had done, when they all talked about east of due north, 01:08:38.800 --> 01:08:46.900 90 - 23...if we want to figure out what this is here, 90 - 23 = 67. 01:08:46.900 --> 01:08:57.700 So, we could also talk about it as 67 degrees east of north. 01:08:57.700 --> 01:09:02.900 Either one would be fine; but we have to put it in the same format, because they didn't give us a θ previously. 01:09:02.900 --> 01:09:05.300 We don't know where θ is based off of, so we have to make sure 01:09:05.300 --> 01:09:10.000 that we are following this same pattern of east of due north, west of south, something like that. 01:09:10.000 --> 01:09:11.700 We have to go in that same pattern. 01:09:11.700 --> 01:09:15.800 All right, vectors are really, really useful; we will talk about them more when we talk about how matrices are connected to them. 01:09:15.800 --> 01:09:17.300 But this is really great stuff here. 01:09:17.300 --> 01:09:22.000 We have talked about a whole lot of things here, so if you had any difficulty understanding this lesson, 01:09:22.000 --> 01:09:25.400 just try watching piece-by-piece, and just work examples, one after another. 01:09:25.400 --> 01:09:29.600 There are a lot of things to digest here, but they all work together; and vectors are so useful. 01:09:29.600 --> 01:09:31.000 All right, we will see you at Educator.com later--goodbye!