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 the dot product and the cross product.
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Now, we have some idea of vectors learned, so we can move on to looking at two new ways that vectors can interact: the dot product and the cross product.
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These ideas are very important and useful in advanced math, science (especially in physics), and engineering.
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If you have any interest in those fields, you definitely want to pay extra attention here.
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But you are also going to need it just for this course.
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All right, with each of these, we will start by looking at an algebraic definition,
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and then exploring what that means in a geometric interpretation, so we can get a sense
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of how this would look, as opposed to just a bunch of numbers.
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All right, let's go!
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First, the **dot product**: Given two n-dimensional vectors (that is, vectors that have n components,
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where n will just be some number), u, which will be our first component, u₁; our second component,
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u₂; up until the nth component, u<font size="-6">n</font>...so u₁, u₂, u₃,
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all the way up until we get to u<font size="-6">n</font>; and v, which is the same thing: v₁, v₂,
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up until v<font size="-6">n</font>--so the first component of v, the second component of v...up until the nth component,
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the last component, since it is just n-dimensional; then the dot product, which is just symbolized exactly as you would guess,
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a dot between the vectors, is the sum of the products of each pair of components.
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So, u dot v, which would be the same thing as u₁, u₂, up until u<font size="-6">n</font>,
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dotted with v₁, v₂, up until v<font size="-6">n</font>, would produce u₁v₁
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(the u₁, the first component here, and the first component here multiply together),
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and then we add that to the second component here and the second component here;
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and then we add that to all of the ones in between, and then we add that to u<font size="-6">n</font>,
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the last one, multiplied by the last one of v, as well.
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So, we have the first two components multiplied together, the second two components multiplied together...
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we do this for every one of them, until we get to the last two components, multiplied together.
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So, we add each of the possible component pairs, each of the component pairs between u and v, for everything that is in the same component location.
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We multiply each of those together, and then we sum it all up.
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Notice that the dot product will result in a scalar, not a vector.
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What we get out of this in the end isn't another vector; it is just a real number.
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And if we are talking about this, we say u, dot, v: so the symbols are vector u, dot product with, vector v.
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We just say it as "u dot v," as you have heard me repeatedly say.
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OK, so let's see an example in two dimensions.
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If we have (5,2) dotted with (-7,12), then what we do is multiply the first two components, 5 and -7.
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We have 5 times -7, and then we add that with 2 and 12, multiplied together.
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5 and -7 gets us -35, plus 24; and finally, we get -11 as our answer when we add it together.
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All right, it turns out that the dot product of two vectors is deeply related to the angle between the vectors and their magnitude.
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Now, for us to talk about the angle between the vectors, we can't have them in totally separate places.
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We have to put them together; so let's say we have two vectors like this and this.
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Before we can talk about the angle and dot product, we have to put them tail-to-tail.
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Now, once they are tail-to-tail, we can talk about the angle between them; so right from here to here, where my face is, would be the angle.
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So, u with θ in between, and then v, is the two angles.
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Now, it turns out that u dot v is equal to the length of u (remember, if we have vertical bars on either side,
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that says the length which we get by that magnitude formula, the square root of each component squared and added together),
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times the length of v, times cosine θ.
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If we want, we can shift this around to have the formula that makes the angle a little more accessible.
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Right now, we have cos(θ), but there is other stuff multiplied by it.
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So, we could solve for cos(θ); we would have cos(θ) = u ⋅v divided by the length of u, times the length of v.
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And then, at that point, if we want to find out the angle, we could just take the arc cosine of both sides, the cosine inverse of both sides.
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And we would figure out what our angle is.
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For the most part, though, I prefer this formula, because I think it is a little bit easier to remember
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that u ⋅ v is equal to length of u, length of v, times the cosine of the angle between them.
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All right, it turns out that we can actually prove this formula pretty easily.
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It is not going to be that hard for us to prove it.
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But it won't really help us directly understand vectors any better.
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As such, the proof is still going to be in this lesson, but it has been put at the end, after the examples.
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So, if you wait until we get through all of the examples, you will get to see the proof, if you are curious about it.
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If you have a few extra minutes, and you are interested, awesome--I would love to have you check it out.
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It is really cool to just get the sense of more things in math--all of the proofs.
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But if you are busy, and you don't have time for it, that is OK, too.
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It is really...if you had to miss one proof, this is probably the best one to just skip and take in faith.
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So, don't worry about it if you end up not being able to watch this proof.
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But if you have the time, and you are interested in proofs, it is really cool, and it is totally something we can actually manage pretty easily.
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OK, let's interpret this geometrically: we can break up our new formula, u ⋅ v = length u times length v times cos(θ),
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into length u times length v times cos(θ); so we think of it as two different pieces:
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the length of one vector, times the length of the other vector, multiplied by cosine θ.
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So, what that gives us: if we look at this, this length v times cosine θ is a way of thinking of the length of v
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if we projected it onto u--v as a projection onto the vector u.
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What does that mean? Let's look at these pictures here.
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We have v here; v is this vector here, so if v is shorter than u, we project it by dropping a perpendicular down to u;
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and then, what we have is: wherever u went, up until that perpendicular that we just dropped down,
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that is our length, length v times cosine θ.
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So, that gives us the length of the projection.
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We can think about this idea as how far we drop down; once we drop down this perpendicular onto u, we have created a projection of v onto u.
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It is like you take a flashlight and shine it directly down onto it.
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It is the shadow that v would cast on u.
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So, length of v, times cosine θ--well, remember: this is, after all...
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since it is a perpendicular that we dropped down, we have a right angle in there.
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So, since we have a right angle in there, and we have θ, it is just basic trigonometry.
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Cosine of θ times the length of the hypotenuse, length v, will give us the length that it is for that projection--the shadow that we just created.
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And we can also do this if our vector v ends up being longer than u.
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Instead of worrying about how long we are on u, it is if u had continued (this dotted line right here),
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and then we had dropped a perpendicular onto if-it-had-continued, where would we end up being on that continuation?
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So, we think of u, and then we think of u continuing off forever, and then v drops down onto that, and that gives us our projection of v.
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So, this gives us another way to think about where this is coming from.
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It is the length of the projection, multiplied by the length of the vector it is being projected onto.
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So, we see that u ⋅ v is the length of one vector (u in this case), multiplied by the length of the other vector's projection.
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Our other vector would be v, and its projection was this part right here; that is length v cosine θ.
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In general, we can interpret the dot product as a measure of how long and how parallel two vectors are.
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So, for example, if we have two vectors like this, we are going to get a larger dot product the smaller our θ is,
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because cosine of numbers close to 0 gets us numbers that will be close to 1.
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Cosine of 0 is 1; that is the absolute maximum.
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When they are perfectly parallel, that is going to be the largest possible projection that v can make onto u.
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As it gets less and less parallel, more and more perpendicular, though, cos(θ) is going to become smaller and smaller.
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This projection that is dropping down is going to become smaller and smaller, until finally, we eventually hit perpendicular,
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and it drops down, and there is no projection whatsoever.
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It is like shining a shadow on something pointing straight up; it doesn't cast a shadow at all--there is no projection that comes out of it.
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So, once they are perfectly perpendicular, we are going to get nothing out of it.
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But if they are parallel, the more parallel they are, the longer they are, the larger the value that we will get out of the dot product.
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This brings up a good point--this idea that, when they are perpendicular, when we have cosine of 90 degrees,
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or cosine of π/2 (degrees versus radians), we end up getting a cosine that comes out as a 0--it goes to 0.
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So, we will have nothing coming from the dot product, because there will be no projected shadow onto it.
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From our formula, u ⋅ v = length u(length v)(cos(θ)), we see that, if u ⋅ v = 0,
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then it must be the case that cos(θ) = 0, and if cos(θ) = 0, then θ = 90 degrees.
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So, if we have the dot product of two vectors as 0, we know we have perpendicularity; they have to perpendicular to each other.
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If u ⋅ v = 0, then u is perpendicular to v; and if u is perpendicular to v, then u ⋅ v = 0.
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I also want to point out that there are many equivalent words for "perpendicular."
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Perpendicular is...sometimes you will hear "normal," or maybe even "orthogonal,"
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although you normally don't hear that until later math classes, or "at right angles."
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But all of these might end up meaning the same thing: θ = 90 degrees, or alternatively, θ = π/2, if we are in radians.
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So, perpendicular, normal, orthogonal, right angles, θ = 90 degrees, θ = π/2--
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they all mean the same thing, this idea of "perpendicular," something that we are used to from geometry.
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Don't get confused if you end up hearing one of these synonyms; they just all mean the same thing.
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You are probably wondering why there are so many synonyms, and it is because it is a really important idea,
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and it shows up in a bunch of different places; so at different times, different people used different words.
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And so, we ended up having something like four different ways of talking about this; so, that is why we see so many.
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Also, I want to point out one other thing: technically, up here, if we have u ⋅ v = 0, well, u or v could be the zero vector, as well.
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If u or v was the zero vector, all 0's in all of the components, then u ⋅ v is going to come out as 0, also.
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We would get u ⋅ v = 0 from that, as well.
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But is our θ really going to be 90 degrees?
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Well, at that point, if we are just a dot, we are going to say that a dot, that 0 vector, is just going to be perpendicular to everything,
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because at that point, it is not really sticking out in any direction.
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So, it seems reasonable to say that it is perpendicular.
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And it works well with our new idea of "perpendicular" meaning that the dot product equals 0.
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So, we get around this by saying that the zero vector is going to be perpendicular to all vectors.
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So, if something is the zero vector, it automatically is going to be perpendicular.
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But mostly, when we think of this, it is going to work perfectly well.
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And because of this new thing about saying that the zero vector is always going to be perpendicular, it works out all the time.
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So, that is just always true; we can think of perpendicular as meaning that the dot product comes out to be 0.
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All right, let's talk about the cross product now.
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Unlike the dot product, the cross product will only work in three dimensions.
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When you are in space, like three-dimensional space, that is when you can use the cross product.
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It takes two vectors, and it produces a third vector that is perpendicular to both.
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What comes out of the cross product is something that will be perpendicular to our first vector and our second vector.
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If u is equal to u₁, u₂, u₃, and v is equal to v₁, v₂, v₃,
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it is this monster: u cross v equals u₂v₃ - u₃ v₂,
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u₃v₁ - u₁v₃, u₁v₂ - u₂v₁,
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where each one of those is multiplied and then subtracted from the other.
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So, those are our three components.
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We say u cross v as "u cross v," this vector u cross product with vector v; we just say it as "u cross v" to make it easy.
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We will calculate a product in Example 1; and basically, this formula is tough to remember, because it is just so many symbols at once.
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So, there is a pretty good mnemonic for this; but you might not have seen what we are about to use for it.
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So, we haven't learned about matrices or their determinants yet; but there is a great mnemonic, if you are familiar with it.
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If you learned about this in a previous math course, or if you go on and watch the lesson on determinants in a little bit--
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and matrices--there is a great mnemonic for remembering the cross product
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by using the determinant of a 3x3 matrix and the standard unit vectors.
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So, if we have u cross v, we can also write that as the determinant of the 3x3 matrix, ijk on the top,
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then the first vector, u₁u₂u₃, on the second one, and then v₁v₂v₃.
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At this point, we take ijk; we do a co-factor expansion on ijk, so when we look at i, that will knock out the u₁v₁ 0775.8 and the jk; so we have i along with the determinant, u₂u₃/v₂v₃,
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times the unit vector i, minus...next we do j; j will knock out u₂v₂; i knocked u₁v₁,
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and j and k; j will knock out i and k, and will also knock out u₂v₂,
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leaving us with u₁u₃v₁v₃.
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So, we take the determinant of that and multiply that by our j.
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And then finally, we get to k; k and its cofactor will knock out ij and u₃v₃.
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And so, we get u₁u₂v₁v₂k.
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And then, if you take the determinants of each of those 2x2 matrices, multiplying on the diagonal going down,
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that is the positive; and then subtract on the diagonal going up...so u₂v₃ - u₃v₂i,
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minus u₁v₃, minus u₃v₁j,
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plus u₁v₂ minus u₂v₁.
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That is another way of doing it; it is a pretty good mnemonic; if you are familiar with determinants, it works great.
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If you are not familiar with this, that all probably didn't make very much sense.
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So, you can go ahead and watch the later lesson on determinants, or you can just go back to that previous slide and just end up memorizing that formula.
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There is not really a very easy way to memorize it, other than this mnemonic, which works great.
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But if you don't know the mnemonic, it is a little bit difficult.
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Yes, it is really great if you learned determinants, though.
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And you will learn that later on, so it might be worth just learning it now,
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so you can have this stuck in your head if you end up having to do a lot of work with cross products.
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All right, what does this mean geometrically?
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If you are given two vectors, u and v, the cross product produces a third vector, u cross v, that is perpendicular to both.
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Say we have some u going off like this, and we have v going off like this.
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Now, what we have, u cross v, is this third vector that comes out of both that will be perpendicular to both.
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I think you can see this: so, u cross v would end up being perpendicular to both.
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It is perpendicular to u, but it is also perpendicular to v.
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And so, it comes out like that, and we have u cross v.
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However, how do we tell which way u cross v is going to point?
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The perpendicular vector can come out like this, but the perpendicular vector could also come out like this.
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There is nothing wrong with being perpendicular on the underside, as well.
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So, how do we tell which way you end up going?
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The trick to this is the right-hand rule: you point the fingers of your right hand in the direction of the first vector.
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And then, your palm goes in the direction of the second vector; you can also think of it as curling your fingers toward it.
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And the cross-product's direction is the direction your thumb points.
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In this one, if this is u and this is v, then you put u...my fingers are along u, and then my palm goes towards v.
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So, u, v...and u cross v comes out like this, which is exactly what you get from this picture right here.
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If you do u with your fingers, and then v with your palm, and then bring your thumb out, you will be able to see u cross v coming out in purple there.
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I really recommend trying this out right in front of you right now, because there is really no way to get this done geometrically,
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in your head, without actually seeing it visually in front of you.
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And then, if we wanted to see what v cross u would be, right-hand rule: this is our v; this is our u.
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So, v goes first; the fingers go in the v direction.
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And then, u: the palm goes in the u direction, so now our thumb is pointing down.
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v, u, down; v cross u will go down like that; that is the right-hand rule.
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Fingers go in the first vector's direction; palm (or fingers curling--either way you want to think about it) goes in the direction of the second vector.
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And whatever you have with your thumb, that is the direction that your cross-product is going to come out of it.
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It is definitely worth trying that; try it right now--make sure you end up seeing that you are getting the same thing,
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that you can do this with your hands, that you can see this, because it is really difficult to visualize
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purely with your mind; but if you use it with your hands, you will be able to see it very well.
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This stuff comes up all of the time in physics and engineering--it is really important stuff there.
00:16:59.100 --> 00:17:02.700
All right, cross-product...now, another way to interpret this geometrically...
00:17:02.700 --> 00:17:06.900
We talked about the direction, but we haven't talked about how long u cross v is going to be.
00:17:06.900 --> 00:17:15.000
The magnitude--how long u cross v will be--is equal to the area of the parallelogram that is enclosed by u and v.
00:17:15.000 --> 00:17:22.600
So, we have u and v; if we continue those out, u is down here, and so we also do a parallel one here.
00:17:22.600 --> 00:17:28.600
This is parallel to this, and then our v here is parallel to this.
00:17:28.600 --> 00:17:35.200
So, that makes a parallelogram; the area inside of that parallelogram is how long u cross v will be.
00:17:35.200 --> 00:17:39.100
So, notice that the more parallel u is to be, the more squished it becomes.
00:17:39.100 --> 00:17:44.500
If this is our u and v, the more parallel they are, the less area that there is going to be inside of them.
00:17:44.500 --> 00:17:50.400
They get squished more and more; as we open it up more and more, though, we have this larger area inside of it.
00:17:50.400 --> 00:17:55.800
So, as we squish it down, as u becomes more parallel to be, the less area it has.
00:17:55.800 --> 00:18:00.000
The more perpendicular, the more we open it up...we have this wide area.
00:18:00.000 --> 00:18:05.000
When we are perpendicular, we are going to have the maximum amount of area, because we will be a perfect square.
00:18:05.000 --> 00:18:10.700
So, the more it opens up, the more perpendicular it is; the more it opens up, the larger the area becomes.
00:18:10.700 --> 00:18:19.600
With this in mind, we can interpret the length of u cross v as a measure of how long and how perpendicular the two vectors are.
00:18:19.600 --> 00:18:26.600
If the vectors aren't very perpendicular at all, then we are not going to get much out of the cross product, in terms of its length.
00:18:26.600 --> 00:18:30.000
If they are really perpendicular, we are going to get a lot more out of it.
00:18:30.000 --> 00:18:34.600
And of course, we can just make the vectors longer in the first place to increase this area.
00:18:34.600 --> 00:18:36.400
All right, we are ready for some examples.
00:18:36.400 --> 00:18:45.300
The first one: A vector, a, equals (2,4,-5); vector b = (-3,1,2); the first thing to do is give the cross product of a and b, a cross b.
00:18:45.300 --> 00:18:54.500
Then, we want to show, by the dot product, that a is perpendicular to a cross b and that b is also perpendicular to a cross b.
00:18:54.500 --> 00:18:57.500
So, our first thing to do is just to figure out what a cross b is.
00:18:57.500 --> 00:19:09.900
We have (2,4,-5), (-3,1,2); so (2,4,-5) is crossed with (-3,1,2).
00:19:09.900 --> 00:19:18.300
We have this formula here: here is our formula; so for the first coordinate of our outcoming cross-product,
00:19:18.300 --> 00:19:24.800
it is going to be the second component of the first vector, u₂ (u₂ would be, in this case, 4),
00:19:24.800 --> 00:19:31.600
times the third component of the second vector, v₃ (so that would be 2 here), so 4 times 2;
00:19:31.600 --> 00:19:41.700
minus the third component of the first vector (that is a -5), times the second component of the second vector, v₂ (that is 1);
00:19:41.700 --> 00:19:44.500
the same thing is going on; see if you can follow along here...
00:19:44.500 --> 00:19:56.800
u₃ is -5, times v₁ is -3, minus u₁ (u₁, in this case, is 2); v₃ is 2,
00:19:56.800 --> 00:20:04.200
as well; comma, u₁v₂ (u₁ is 2; v₂ is 1),
00:20:04.200 --> 00:20:11.000
minus u₂ is 4...v₁ is -3; great.
00:20:11.000 --> 00:20:17.400
We start simplifying this out; we have 8 minus a negative--that cancels out, so we have 8 + 5.
00:20:17.400 --> 00:20:28.900
-5 times -3 becomes positive 15, minus 2 times 2...minus 4...2 times 1 is 2; minus 4 times -3...they cancel out, and we have addition there,
00:20:28.900 --> 00:20:42.900
as well; plus 12; simplify that, and we get (13,11,14), so that is a cross b.
00:20:42.900 --> 00:20:48.900
So, there it equals a cross b; there is our cross-product vector.
00:20:48.900 --> 00:20:53.700
Now, we want to verify this; we want to show that it is, indeed, going to be perpendicular to both a and b,
00:20:53.700 --> 00:20:57.700
because we know that the cross-product has to be perpendicular to both of them, so that had better come out.
00:20:57.700 --> 00:21:06.900
So, if a is perpendicular to a cross b, then that will be true if a dot a cross b comes out to be 0.
00:21:06.900 --> 00:21:11.600
So, if a dot a cross b comes out to be 0, we know that that is perpendicular by how the dot product works.
00:21:11.600 --> 00:21:16.200
Remember: that was one of our big realizations about the dot product--that if θ is equal to 90 degrees,
00:21:16.200 --> 00:21:21.000
if the two vectors are perpendicular to each other, then the dot product of the two vectors always comes out to be 0.
00:21:21.000 --> 00:21:35.200
a dot a cross b: our a is (2,4,-5); our b...sorry, not our b; we are dotting that with a cross b; a cross b is (13,11,14).
00:21:35.200 --> 00:21:47.000
2 times 13 is 26, plus 4 times 11 is 44, plus -5 times 14 is -70.
00:21:47.000 --> 00:21:53.300
26 + 44 becomes 70, so we have 70 - 70; that comes out to be 0, so that checks out.
00:21:53.300 --> 00:21:58.000
Our dot product came out to be 0, so we know that they must be perpendicular; great.
00:21:58.000 --> 00:22:12.100
The next one: Show that b is perpendicular with a cross b: so b, dotted with a cross b:
00:22:12.100 --> 00:22:23.700
b is (-3,1,2); our a cross b that we are dotting with is (13,11,14) (I am running a little bit out of room there);
00:22:23.700 --> 00:22:41.600
-3 times 13 becomes -39; plus 1 times 11; plus 2 times 14 is 28; -39 + 39...11 + 28...-39 + 39 becomes 0, so that checks out, as well.
00:22:41.600 --> 00:22:47.600
The dot product of b with a cross b comes out to be 0, so we know that those two vectors must be perpendicular.
00:22:47.600 --> 00:22:49.400
So, there we are; we have finished that one.
00:22:49.400 --> 00:22:56.900
All right, the second example: we have u = (5,2,8,k); v = (3,-4,1,3).
00:22:56.900 --> 00:23:09.400
What is k if u is perpendicular to v? If u is perpendicular to v, then that tells us that u dot v equals 0.
00:23:09.400 --> 00:23:26.600
Great; so if u dot v equals 0, then we have that (5,2,8,k), dotted with (3,-4,1,3), equals 0.
00:23:26.600 --> 00:23:40.400
So, we work this out: 5 times 3 is 15; plus 2 times -4 is -8; plus 8 times 1 is 8; plus k times 3 is 3k; equals 0.
00:23:40.400 --> 00:23:53.300
-8 + 8...they cancel each other out, so we have 15 + 3k = 0; 3k = -15, which gives us k must equal -5.
00:23:53.300 --> 00:23:57.600
So, there is nothing really difficult there; as soon as we realized that if they are perpendicular,
00:23:57.600 --> 00:24:03.400
then it must be that their dot product is 0, at that point, we can set something up that we can just solve through simple algebra.
00:24:03.400 --> 00:24:12.000
The third example: In physics, the work done by a force f over a distance d is defined as force dotted with distance.
00:24:12.000 --> 00:24:16.800
So, the work, W, is equal to the force vector, dotted with the distance vector.
00:24:16.800 --> 00:24:23.600
If you push a box with a mass of 20 kilograms with a force of 100 Newtons at an angle of 15 degrees above the horizontal,
00:24:23.600 --> 00:24:27.300
for 10 meters, how much work have you done on the box?
00:24:27.300 --> 00:24:33.100
So, the first thing we can do is figure out: All right, what is our force vector in terms of its components?
00:24:33.100 --> 00:24:41.200
So, we could get the force vector...force equals component stuff; and then we will figure out d = component stuff...
00:24:41.200 --> 00:24:45.700
well, that will be easy, because it is entirely horizontal; so we will have to use trigonometry to figure out what the force vector is,
00:24:45.700 --> 00:24:48.100
in its component form, and then we can dot the two together.
00:24:48.100 --> 00:24:50.400
But that is actually more work than we have to do.
00:24:50.400 --> 00:24:55.100
All we have to do is remember that we don't need component form at all, because we know that,
00:24:55.100 --> 00:25:05.100
if work equals the force vector dotted with the distance vector, then, well, u dotted with v is the same thing as length u times length v,
00:25:05.100 --> 00:25:09.000
times cosine θ, so this is the length of our force vector,
00:25:09.000 --> 00:25:14.400
times the length of our distance vector, times the cosine of the angle between them.
00:25:14.400 --> 00:25:19.400
We know what our force vector is--it came out to be 100 Newtons; the force was 100 Newtons.
00:25:19.400 --> 00:25:22.700
We know what our distance is: we go for a distance of 10 meters.
00:25:22.700 --> 00:25:26.400
And we know what our angle θ is; do we need the 20 kilograms of mass?
00:25:26.400 --> 00:25:33.000
20 kilograms of mass actually never shows up for figuring out the work; the mass of the object has no effect on the amount of work that goes in.
00:25:33.000 --> 00:25:40.600
It is all about the force, the distance that happens, and the angle between those two--how the two interrelate.
00:25:40.600 --> 00:25:45.600
So, we actually don't need to know the mass of the box at all to figure this one out; it is just a "red herring."
00:25:45.600 --> 00:25:55.800
So, force is 100 Newtons, times distance (is 10 meters), times cosine of the angle of 15 degrees.
00:25:55.800 --> 00:25:59.900
We work that out with a calculator: we have 1000 times the cosine of 15 degrees.
00:25:59.900 --> 00:26:06.400
And that comes out to be 965.93; now, what are the units of work?
00:26:06.400 --> 00:26:12.000
They told us in the problem that the unit is the Joule, or joules, which is signified with a J.
00:26:12.000 --> 00:26:15.800
So, we use the unit of J at the end of it.
00:26:15.800 --> 00:26:18.700
And there we are; there is our work; great.
00:26:18.700 --> 00:26:25.600
All right, the final example: Prove that u dot u equals the magnitude of u, squared.
00:26:25.600 --> 00:26:29.900
All right, the first thing to do: we have to have a way of talking about just some general vector u,
00:26:29.900 --> 00:26:31.800
because they didn't tell us much about u at all.
00:26:31.800 --> 00:26:37.800
They told us just "vector u"; so we need to be able to talk about what vector u is, in a way that we can actually work with it.
00:26:37.800 --> 00:26:44.000
Let's just give the components names; we will do it in the same way that has happened in all of the previous stuff,
00:26:44.000 --> 00:26:48.500
where we have just said that the first component is u₁ (so we will make this u₁);
00:26:48.500 --> 00:26:52.900
and then the second component will be u₂, and then the third component would be u₃,
00:26:52.900 --> 00:26:59.000
and all the way up until some u<font size="-6">n</font>, because every vector has to have some specific length.
00:26:59.000 --> 00:27:04.900
It is not allowed to go on forever; so we will stop at u<font size="-6">n</font>...n will be just the length of our vector.
00:27:04.900 --> 00:27:10.000
This is going to be the case for any vector at all; we could put it in this form of first component, second component,
00:27:10.000 --> 00:27:14.100
up until its last component, which we will say will be its nth location in the thing.
00:27:14.100 --> 00:27:20.000
All right, so now we have a way of doing this; let's just look at what u dot u is, and then what the magnitude of u².
00:27:20.000 --> 00:27:22.700
If they end up being equal, we have proved this thing.
00:27:22.700 --> 00:27:39.900
So, u dot u would be vector u₁, u₂, up until u<font size="-6">n</font>, dotted with u₁, u₂, up until u<font size="-6">n</font>.
00:27:39.900 --> 00:27:44.800
All right, so u₁ times u₁...well, u₁ is just some number, so that is (u₁)²,
00:27:44.800 --> 00:27:50.000
plus...u₂ times u₂...well, u₂ is just some number, so that is (u₂)²,
00:27:50.000 --> 00:27:53.600
plus...this is just going to keep happening, until we get to our final component.
00:27:53.600 --> 00:27:57.400
u<font size="-6">n</font> times u<font size="-6">n</font>...well, u<font size="-6">n</font> is just some number, as well, so that is (u<font size="-6">n</font>)².
00:27:57.400 --> 00:28:01.800
So, we have u₁ squared, plus u₂ squared, up until we get to u<font size="-6">n</font> squared.
00:28:01.800 --> 00:28:08.100
Now, there is not much we can do to simplify that there; so let's take a look at the magnitude of u².
00:28:08.100 --> 00:28:12.700
So, first, what is the magnitude of u? Well, the magnitude of any vector, remember,
00:28:12.700 --> 00:28:19.100
is the square root of each component squared added together underneath the square root.
00:28:19.100 --> 00:28:23.500
So, it would be the first component squared, plus the second component squared,
00:28:23.500 --> 00:28:28.200
up until the final component squared, and all of that underneath the square root.
00:28:28.200 --> 00:28:37.000
Well, if we square this, then we have the magnitude of u, squared, equals...well, if we square both sides of that equation above,
00:28:37.000 --> 00:28:44.300
the square root times the square root cancels out, and we just have u₁ squared plus u₂ squared,
00:28:44.300 --> 00:28:53.200
up until u<font size="-6">n</font> squared; well, look: this and this are the same thing.
00:28:53.200 --> 00:29:02.400
So, since they are the same thing, we have just shown that u dot u is equal to the magnitude of u, squared.
00:29:02.400 --> 00:29:03.900
Great; the proof is finished.
00:29:03.900 --> 00:29:07.700
All right, that finishes for the examples; thanks for coming to Educator.com.
00:29:07.700 --> 00:29:13.700
And we will see you in the next lesson, when we start talking about matrices--all right, goodbye!
00:29:13.700 --> 00:29:16.700
All right, I think they are gone--everybody who is not actually interested in the proof.
00:29:16.700 --> 00:29:23.800
So, are you ready for this: Bonus round--here we are, ready for the proof!
00:29:23.800 --> 00:29:30.500
Dot product formula: let's prove that u ⋅ v is equal to the length of u, times the length of v, times cosine θ.
00:29:30.500 --> 00:29:35.400
The very first thing that you want to do any time you are really trying to think about anything analytically is draw a picture.
00:29:35.400 --> 00:29:38.100
A picture is always a useful way to think about things.
00:29:38.100 --> 00:29:43.000
So, we start by drawing a picture: we have u and v, with θ in between.
00:29:43.000 --> 00:29:48.200
Now, we look at this, and we want to see: is there any way to connect the length of u, the length of v, and θ together?
00:29:48.200 --> 00:29:51.400
Is there some way to get these things to talk to each other?
00:29:51.400 --> 00:29:54.500
Do we know any way to say, "Yes, I know these are related"?
00:29:54.500 --> 00:29:58.100
Well, we look at this for a while, and we say, "Well, I don't see anything yet."
00:29:58.100 --> 00:30:03.300
But that looks kind of like a triangle; and I know a lot of things about triangles from trigonometry.
00:30:03.300 --> 00:30:07.900
So, let's say it looks like a triangle without a top; let's give that triangle a top!
00:30:07.900 --> 00:30:15.700
We draw in the top in this purple color; and we might realize that there are three sides to a triangle;
00:30:15.700 --> 00:30:23.500
I know an angle...sort of...oh, I can connect the length of u, the length of b, that angle θ, and the length of the top,
00:30:23.500 --> 00:30:27.300
together with something that we learned in trigonometry: the law of cosines.
00:30:27.300 --> 00:30:31.500
We might remember this; and if we remember this, we just go back and look up the law of cosines.
00:30:31.500 --> 00:30:33.400
There is no reason not to just go and look it up.
00:30:33.400 --> 00:30:39.600
We look up the law of cosines in a book; we refresh ourselves, and we get that the length a²,
00:30:39.600 --> 00:30:46.000
the side a, squared, is equal to the other two sides squared (b² + c²),
00:30:46.000 --> 00:30:57.000
minus 2 times b times c times cosine of capital A, where little a and capital A are the side and then the angle opposite;
00:30:57.000 --> 00:31:03.000
so little a is the length of one side, and then capital A is the angle opposite that side; so we have:
00:31:03.000 --> 00:31:07.900
a² = b² + c² - 2bc times cos(A).
00:31:07.900 --> 00:31:12.600
That is the law of cosines; now, we bring up our vectors picture, and we look at this.
00:31:12.600 --> 00:31:16.800
Now, one thing before we keep going: how are we going to have this top as u - v?
00:31:16.800 --> 00:31:23.200
We can see the top as the vector u - v, since it runs from the head of v to the head of u.
00:31:23.200 --> 00:31:31.500
And think about this: if we take v, and we add that to u - v, well, the -v and the +v would cancel out, and we would be left with just u.
00:31:31.500 --> 00:31:39.400
So, it must be the case: we can see graphically, through this algebra, that we can get from v to u by using u - v,
00:31:39.400 --> 00:31:44.800
because it will take away the v and give us the u, and we will manage to get from the head of v to the head of u.
00:31:44.800 --> 00:31:51.600
Cool; so that is how we have that vector u minus vector v is the way to be able to talk about the top as a vector.
00:31:51.600 --> 00:32:06.000
All right, using the law of cosines, then we have that the length of u - v, our a, squared, is equal to the length of u, our b,
00:32:06.000 --> 00:32:28.100
this part right here, squared, plus the length of v, this part right here, c, squared, minus 2 times the length of u,
00:32:28.100 --> 00:32:37.300
times the length of c, times cosine of the angle between them (cosine of θ).
00:32:37.300 --> 00:32:45.000
So, we have the length of vector u - v, squared, equals the length of u, squared, plus the length of v, squared,
00:32:45.000 --> 00:32:49.100
minus 2 times the length of u times the length of v, times cos(θ).
00:32:49.100 --> 00:32:52.600
All right, it looks like we are getting somewhere: we have some relationships going on.
00:32:52.600 --> 00:32:55.100
We even have that cos(θ), if we are trying to prove that thing.
00:32:55.100 --> 00:33:00.300
So, it looks like we are getting there; but we still have this problem, where we don't have any dot products there.
00:33:00.300 --> 00:33:04.900
So, how can we get u ⋅ v to show up in there? We want u ⋅ v in there.
00:33:04.900 --> 00:33:12.200
Well, remember: in Example 4, we just proved, for any vector a, a dotted with itself (a vector dotted with itself)
00:33:12.200 --> 00:33:17.900
is equal to the magnitude of that vector, squared; so a dot a equals the magnitude of a, squared.
00:33:17.900 --> 00:33:24.500
Thus, we can swap out each of these magnitude-squareds for a dot a.
00:33:24.500 --> 00:33:26.400
So, we have that relationship, whatever they are.
00:33:26.400 --> 00:33:34.000
So, u - v: the length of u - v squared will become the vector u - v, dotted with the vector u - v.
00:33:34.000 --> 00:33:38.400
u squared: the magnitude of u squared will become the vector u, dotted with the vector u.
00:33:38.400 --> 00:33:42.500
The magnitude of v squared will become v dot v.
00:33:42.500 --> 00:33:46.100
So, we have all of these swapping out right here.
00:33:46.100 --> 00:33:52.600
OK, at this point...we didn't show this technically, but you can prove it to yourself--it is not that difficult:
00:33:52.600 --> 00:33:58.200
the dot product is distributive, so we can actually distribute using this dot product.
00:33:58.200 --> 00:34:08.500
u - v dot u - v: well, then, we have u dot u minus u dot v, minus v dot u plus v dot v.
00:34:08.500 --> 00:34:20.100
u dot u minus u dot v, minus v dot u minus...times a negative again, so plus v dot v.
00:34:20.100 --> 00:34:22.300
Great; the stuff on the right just stays the same.
00:34:22.300 --> 00:34:24.800
At this point, we see that we have certain things on the right and the left.
00:34:24.800 --> 00:34:28.000
So, if we have v ⋅ v on both sides, let's just subtract it on both sides.
00:34:28.000 --> 00:34:30.900
We have u ⋅ u on both sides; let's just subtract it on both sides.
00:34:30.900 --> 00:34:33.600
So, we have - u ⋅ v - v ⋅ u.
00:34:33.600 --> 00:34:40.100
Well, notice: u ⋅ v is just the same thing as v ⋅ u, so we can combine them together.
00:34:40.100 --> 00:34:44.900
If we have u ⋅ v + v ⋅ u, then that is the same thing as 2 times u ⋅ v.
00:34:44.900 --> 00:34:50.100
So, if we have - u ⋅ v - v ⋅ u, then that is the same thing as -2(u ⋅ v).
00:34:50.100 --> 00:34:56.900
So, we have -2(u ⋅ v) there on the left; it equals -2 length of u, times length of v, times cosine θ.
00:34:56.900 --> 00:35:04.600
We have -2 on both sides; divide by -2 on both sides; those cancel out; we have u ⋅ v equals the length of u, times the length of v, times cosine θ.
00:35:04.600 --> 00:35:07.300
Cool; and our proof is finished--that was not too tough.
00:35:07.300 --> 00:35:10.200
All right, thanks for staying around; I think proofs are really cool.
00:35:10.200 --> 00:35:15.900
They are really, in my mind, the heart of mathematics--being able to show that this stuff is definitely always true.
00:35:15.900 --> 00:35:18.300
I think it is awesome; thanks for staying around--I was glad to share it with you.
00:35:18.300 --> 00:35:20.000
We will see you at Educator.com later--goodbye!