WEBVTT mathematics/pre-calculus/selhorst-jones
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Hi--welcome back to Educator.com.
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Today, we are going to talk about vectors; this starts an entirely new section for us.
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We are getting into really new territory; we are going to talk about vectors, and then later on we will be talking about matrices.
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But first, let's talk about vectors: when we talk about the force on an object, we need to know two things:
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the magnitude of the force--how hard it is being pushed, and the direction--which way it is being pushed.
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If we are going to talk about something being pushed, we can't just talk about how hard it is being pushed.
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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.
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But even if we know the direction, we also have to know how hard it is.
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We are pushing really hard versus pushing very slightly--there are big differences here.
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We need to know both magnitude and direction; we need to know how hard and which way.
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Now, one number on its own won't be enough to get across both pieces of information.
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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--
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be able to talk about both of these things in one piece of information.
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Vectors are massively useful: they are used throughout math; they are in the sciences everywhere, especially in physics;
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they are in engineering, computer programming, business, medicine, and more fields.
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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.
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Now, I want to point out that this lesson will use some basic trigonometry to figure out angles.
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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.
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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.
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Graphically, if we look at a picture of it, it is exactly that: it is a directed line segment, a length with direction.
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It is both of these things at once.
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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.
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So, we have a chunk of length, and we see which way it is pointed.
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It is pointed in a very specific direction; it has some angle to it.
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How do we call out a vector, if we want to talk about a vector?
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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.
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Or we can put it in bold face with a bold u.
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And once in a while, if we are talking about nothing but vectors, and there is nothing else showing up,
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sometimes it will just be assumed; but we are pretty much never going to see that--not in this course.
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Vectors are normally shown with lowercase letters; however, just like variables, you can use any symbol.
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But for the most part, we will stick with lowercase letters: u and v are very common letters for talking about them.
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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.
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When you are actually writing it by hand, I write it like this, where I don't really have a full arrow.
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I have more of a harpoon, where it is just an arrow on one side--just this sort of arrow flange on one side.
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That is how I write it; you could write it with an actual arrow on top--there is nothing wrong with that.
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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.
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But it still gets across the idea of a vector.
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So, this is a perfectly fine way to write it by hand when you are working it out.
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But when we have it actually written out in this lesson, we will have it like that with an arrow on top.
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We can also give the vector algebraically by its location in the plane.
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We have these nice rectangular coordinates; we can break that into components.
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We call this the component form: we have a horizontal amount of 3 and a vertical amount of 4.
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So, we say that its horizontal component is 3, and its vertical component is 4.
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u, the vector u, equals (3,4); the first component is the horizontal one, and the second component is the vertical,
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just like when we are talking about points in the plane.
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We normally use angle brackets (these are angle brackets, because they are at an angle) to denote vectors.
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But we will often also see parentheses; parentheses are very common, as well.
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I personally actually tend to use parentheses more; but most precalculus/math analysis courses tend to use angle brackets.
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So, I am teaching with angle brackets; but personally, when I am just doing math on my own, I often tend to use parentheses.
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But either of them is just fine.
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After we learn about unit vectors, we will see one more way to talk about the component form, using i and j.
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But we will leave that until a little bit later, once we have actually talked about unit vectors.
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If we know the component form of a vector, we can figure out its length by the Pythagorean theorem.
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Remember: the Pythagorean theorem says that the hypotenuse to a triangle, squared, is equal to both of the legs, squared and then added together.
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So, that means that we can take the square root of both sides of the equation.
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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).
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In this case, we know that our vector is (3,4); so we have that one leg is 3, and one leg is 4.
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So, we work this out; √(3² + 4²) becomes √(9 + 16), or √25, so we get 5.
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So, we see that the length of this vector is 5.
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When we want to denote the magnitude of a vector, when we want to talk about the length of a vector,
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the magnitude of a vector, we normally use the word "magnitude" to talk about length.
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But in either case, with "length" or "magnitude," we are just talking about how long the vector is.
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We normally use these vertical bars on either side of it, just like we do with absolute value.
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And we will talk about that in just a moment--why it is like absolute value.
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Also, sometimes you will see it written ||u||, vector u with double bars around it.
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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.
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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),
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because that is just a vector, as well), will give us the square root of a² + b²,
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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.
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So in this case, a and b are just there and there on our vector.
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Now, I would like to point out, really quickly, why in the world we are using vertical bars on either side,
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just like we did with talking about absolute value.
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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.
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Well, if we talk about the absolute value of 5, and we talk about the absolute value of -5,
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in both cases, we are going to get the number 5 out of it.
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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.
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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.
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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.
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They go in different directions, but it is a question of how far away it is.
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They are both 5 units away from the origin.
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The same thing is going on when we are dealing with a vector.
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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.
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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?"
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We put bars around the u because it is basically doing the same thing as bars around a number with absolute value.
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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?"
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So, they are both a question of length, effectively.
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All right, direction and angle: if we know the component form of a vector, we can figure out its angle for trigonometry.
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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.
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And we have that here, because we know we are dealing with a nice rectangular coordinate system.
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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.
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So, that gets us tanθ = 4/3; at this point, we can take the arctan of both sides
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(the inverse tangent--arctan and tan^-1 mean the same thing; I like arctan).
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So, θ = arctan (tan^-1) of 4/3; we plug that into a calculator or look it up,
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and we see that that is approximately equal to 53.13 degrees; so we see that that is the angle in there.
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Now, we normally talk about direction as the counterclockwise angle from the positive x-axis.
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It is a little bit confusing, but it just means that we start over here at the positive x-axis,
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and then we just keep turning counterclockwise until we get to whatever thing we are trying to measure out to,
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at which point we stop, and that is the measure of the angle that we are going with.
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It is just like we did with the unit circle: we started at the x-axis positive, and then we just kept spinning
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until we got to whatever angle we were trying to get to.
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But sometimes we won't be using the positive x-axis.
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That is what we normally end up using, but sometimes the reference location--sometimes--will change,
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and we won't be using this nice positive x-axis that we are used to using.
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But maybe we will be talking about how far we are off of the vertical y-axis to the right,
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or maybe some other thing, like we have this other angle here created,
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and we are talking about how far we are off of this other thing, some angle, going clockwise...
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So who knows how it is going to be done?
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We have to pay attention to what the specific problem is and how it is set up.
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It generally will be the positive x-axis, but it is not an absolute guarantee; you have to pay attention.
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This means, in any case, whatever you are dealing with, that I recommend always drawing a picture,
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because that is going to give you a way to be able to see what is going on.
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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.
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All right, now, what if we have the component from...
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We talked about (previously) taking the magnitude and angle and getting component forms.
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I'm sorry, we talked about the exact opposite of that.
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So now, we are going to say, "What if we know the magnitude and angle, and we want to get the component form?"
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So, if we are trying to find the component form from the magnitude and angle, we can figure that out.
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The first thing first: always draw a sketch--it will help keep things clear and help you understand what is going on.
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And then, from there, you just use trigonometry.
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So, in this case, we have a length of 6; the length of our vector u is 6; our angle θ is 120 degrees.
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We remember from trigonometry that the cosine of 120, the cosine of this angle here, is going to be equal to
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the horizontal amount of change, up until it drops down on its crossover, divided by the length of the segment we are dealing with.
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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.
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So, -3 is our value for x.
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A similar thing is going on for our y, the vertical component; sin(120), the sine of the angle we are dealing with,
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is equal to the vertical component, divided by the length of the entire segment (6 in this case).
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So, sin(120) = y/6; multiply both sides by 6; sin(120) is √3/2, so it simplifies to 3√3 = y.
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At that point, we just take both of these pieces of information.
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We slot them into our u, and we now have component form.
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Our u is equal to (-3,3√3).
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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.
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It is a little bit confusing--maybe just a bit--to be able to work with things over 90 degrees.
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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.
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So, we could have figured out that inside of the triangle is 60 degrees.
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Now, that is going to cause a little bit of difference here, because cos(120) brought that negative to the table,
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because indeed, our x is going to the left, and remember: this is the negative direction for x.
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Going this way gets us negative x values, when we go to the left, just like going down gets us negative y values.
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If that is the case, we have cos(60), what we can figure out...
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If you have things inside of a triangle, you are just going to figure out the length of each of those sides.
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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.
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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.
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What we are figuring out is just the length of the side.
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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.
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If that is the case, what we are really figuring out is the absolute value of x--how long our x is.
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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.
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We look at the picture, and we see that our x has a length of 3.
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However, it is going to the left; so that means it has to be a negative thing.
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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.
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You have this option: you can either just use the angle, and be able to be really good at trigonometry;
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or you can make it in a slightly simpler form, where it is easier to understand.
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But you have to be paying attention and realize that you have to set the sign at the end.
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I have to pay attention, based on this sketch: is this length going to come out to be positive?
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Is this length going to come out to be negative?
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You really have to pay attention to that.
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All right, we are ready to move on to a new idea.
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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.
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A **scalar** is just some number; it is not a vector--it is just an actual, real number.
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Algebraically, the scalar just ends up multiplying each component.
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Let's say we start with this red vector on our picture, which is u = (3,-2).
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We can scale this by some other thing, by just multiplying it by some number.
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Like, for example: we could multiply u by 2, so we have 2; and now that is the blue vector that we see there.
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Now notice that the blue vector is double the length of the red vector, because it is 2 times u.
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So, it just takes that length, and it scales it by a factor of 2; it doubles that original length.
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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.
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We can take this; we can try multiplying by something else--how about a negative number?
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What would a negative get us? A negative ends up going in the opposite direction.
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So, if we have a negative u, then the positive direction is the way it normally goes; negative will be the opposite way.
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So, negative goes in the opposite direction; so now, we are going opposite the direction that u went, as we can see pictorially here.
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And for how it is going to end up having it, it is just a negative now on each of the components.
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So, negative cancels out there, and we have (-3,+2).
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And that is what we see on our picture.
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We could also have something that is not just a whole integer number, like, say, -3/2.
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We end up having 3/2 times the length--1 and 1/2 the length of the original vector--
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but then also, in the negative direction--opposite the direction of the first one.
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Once again, numerically, it just ends up being -3/2 times the first number times the second number; so we get (-9/2,3).
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That is what it ends up being.
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So, algebraically it just multiplies each component; graphically it is a question of stretching and maybe also flipping.
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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),
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so that is the same thing as just that k getting distributed to both the a and the b, so it gives (ka,kb).
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Great; a **unit vector** is a vector with a length of 1; "unit vector" just means a length of 1.
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It still has a direction; it can have any direction, but it has to have a magnitude of exactly 1.
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Its magnitude--its length--how long it is...it is 1 in terms of length.
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We create a unit vector out of any vector u by dividing u by its length.
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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.
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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.
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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.
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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.
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So, our unit vector is the original vector, divided by the length of the vector.
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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.
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Now, the use of having a unit vector--the reason why it is so great to have a unit vector--
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is that we can just take it later and multiply it by any scalar k that we want.
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And we will know that we will have created a vector that is length k, because we started at length 1.
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You scale that by k, and we are going to go with 1 times k; so we will just be at vector length k.
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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!