Geometric Algebra - Physics - 24 Aug 2025

Transcript

God damn. Okay. So, here's here's what I'm gonna do. So, I can I can actually type. So, we're gonna we're just doing this question and the question might just include the question somewhere here.

So, where was the 27 Titans? We should really change this class name to the One Piece or something. I feel like at this point it's kind of earned today. Um, so we'll just do imagine the one piece just turns out to be some dumb Probably will be. Okay, so here's what we're uh you're looking at this problem, right? So let me just These are always important to visualize. So I always like to I always like to visualize them for you.

So let's go ahead and plop this in here. Just text it to myself. Um text to Okay, let's do this. Where? Here we go. All right.

You all see this? So P is a point on A such that AP. So like just just drawing this picture, you have A, you have B. And so if you take a fraction of AB, so this is AB, right? If you take a fraction of AB, it will sort of be a scaled down version of it. Yeah. K AB.

Everyone agrees with that and so you understood the sort of construction of the question where did you struggle when you tried it? So you said part B, right? Find the value of K for which AP. So here say A is 32 of PB. Is that cool? This is PB. So you want three on two times PB. Fair enough.

Now you can be sort of you can you can sort of be cheeky here. So AP A equals 3 on 2 PB. Yeah. A is going to be final minus initial. Yeah.

So K AB. So you just want to rewrite it in terms of AB. Is that is that clear with everyone? Like you want to rewrite it as as AB. So that's going to equal 3 on 2 * B minus P. Yeah.

Everyone cool with that? So what I'm saying is, okay, AP is here and then you want PB. So three on two PB. Yeah. So if this is if you think about it, what is AB equal to? We can just be we could just be cheeky here and just say, okay, what's AB equal to? Do you all agree that AB is the same thing as A plus PB? Yes or not? Yeah. So PB is 2/3 of AP, isn't it? So I want to rewrite everything as see because I'm writing everything in terms of AP and AB.

Yeah. So I want I don't really like having an PB in there. PB is ugly, right? Okay, let's get get PB out of there. So I can say okay PB is going to be the same thing as 2/3 of AP. Everyone agrees with that? And then therefore AB is equal to AP plus 2/3 of AP.

Everyone agrees with that? And that's the same thing as AB equals 2/3 + 1, which is going to be 5/3 of AP. Yeah. So what we're saying is if you say here's AP, right? Or three fifths of AB. So moving that to the other side, three fifths of AB equals AP. You see that? It's just kind of nice to sort of write this in a different way.

So here's okay, this is AB. Three fifths of it is like right there. That's the idea. Find the value of K for which is the value of K threeths, right? See that? So K is three fifths. You see it? any problems with that? So, and that that kind of makes sense as well because if you're saying it's three three on two of PB is is AP, right? So, this this is PB and then three on two three on two of uh PB is this and this is just PB.

So, three on two of PB that means this is basically three times there's like three bits for every two bits of this. Okay. So that's like like it's like breaking it up into five intervals and then writing it like that. So three fifths of the total is going to be PB. Right? Now to just to give you a nice visual of this because I think it's nice to sort of see what the hell this question is even talking about.

So you have AB. We can draw the line from AB. So line array I think is segments. Okay. I'm going to go to Desmos geometry.

This is like a little bit better for this kind of stuff. So I can just put an A. You can you can still show the access and everything. So, grid access everything. So, A is like 32, 34, whatever you like.

Put it wherever you like. Then B is uh 81 A2. I'll just change that color to be red as well. Okay. So, I have I have these two here.

So, I can say say the line between A and B or I could do I think they call it ray or segments between A and B. This just draws just draws this over here, right? And I might as well make that Oops. Might as well make that one red as well. Cool. So, you can see this here.

Then I can pick P to be any anywhere along the line, right? So, if you think about what P is going to be, it's going to be let's go to A first, right? And then add on K time A, right? And then K * A minus B. Well, I'm going to write that as B minus A, right? That's A and then let K be here. So, let this be 0 to one. Let's let's see where K is going to be. You sort of see where it is.

It's like the slider. So again, just labeling these so you can see them. This one is A. This one's B. Yeah.

So if K is zero, what is that? That's kind of like starting at A, right? If K is half, then we're halfway, right? And then if K is K is one, then we're at B. See, it's like kind of kind of allows you to move from one to the other. Is that cool? See that? Now, this is all well and fun, but like since you guys did all this hard homework, I'm going to give you guys a bit of a bonus piece of math that's pretty cool, okay? that you don't really um so is that the only question you struggle with or is there something else you wanted? Yep. Everything else was okay. Cool.

So, let me show you a bonus p piece of math which is even cooler. So, bonus math. Ready? Bonus math. Pen tool. Has anybody ever used Photoshop? Has anybody used um like I'm sure you guys have seen before a pen tool.

But let's um let's hold on. Um we'll say uh I think it was a let me just show you this. So none of you guys have ever done any like im used image editing software or anything like that before. Nothing like that. Seriously, that's crazy.

Okay. Well, let me show you this. So here I have this um this is like a Photoshop clone. if you guys have ever sort of used Photoshop before, but let you just make a new let's say let's make a I don't know a new YouTube cover, whatever. Right? So, let's let's go ahead and do that.

So, I've got a YouTube color. I can go ahead and I can and on the layer just go ahead and I create a background or whatever the hell I want. One of the main things in Photoshop that you see is this pen tool over here, right? So, pen tool, I can sort of do this. Have you ever seen anyone do this before? Like a pen tool? You all seen this? And it draws nice curves like that. Have you all seen this before? Has anyone not seen this before? So you Charlotte Charlott saying yes.

Have you seen this before? Well, Chris, you haven't before. Right. So this is like one of the ways this is one of the ways that we draw nice smooth curves on a computer. This is the pretty much the most common way that people do it, right? So if you wanted to draw a nice smooth curve like this, even fonts like if you go into a font um font curves, right? This is a very very common thing where like even you probably never even realize this, but fonts on your on your computer are also specified with these like nice smooth curves, right? So every every font file is just a whole bunch of smooth curves. So one question you'd ask like in math class they taught you how to graph a bunch of stuff, right? Like you can graph like quadratics and and you can graph polomials and whatever the hell.

But then one fun thing you might ask is what if I wanted to draw the graph of like some random shape that I wanted to draw? Like what if I wanted to draw a random shape and I wanted to graph that? How could I do that? Right? So do you all kind of see the problem I'm trying to solve? So what if the what if it was some random shape that you've never seen before? Like I just want to graph some random shape. Like you do agree that this pen tool is kind of a good way to do it, right? So I can kind of just go step by step and just like if I wanted to draw any shape I wanted, like if I wanted to draw a love heart, like I could go ahead and and just go brand new. Like let's try a love heart, right? So put a point here. Okay. So I'm going to put a point here and a point there.

Right? And then I'm going to do that and I'm put there and I'm going to keep going. So like that. Okay. Awesome. Sort of see that? I made a bit of a wonky love heart because I tried to I tried to do it a bit too quickly, but I can make a new one here.

So, I can I can try again. So, I can make a I can make a pretty nice one as well. So, I can go here and then I could say this is like the piece of the heart. And then I can go up here also. See that? If I wanted to, I can even include that bit.

And then I can go up like that. You see how I'm doing that? I can just make any shape I like. This is the basic idea. Like if you ever became like a logo designer or something like that, this is how you would go about doing it. See, you can go ahead and you make that shape and then if once you've got the shape, it's really easy to do things like fill it in with a specific color and do all that kind of stuff, right? But you need to be able to define the shape before you do that, right? So that's difficult.

How are we going to do that? So this is a nice side application of vectors that people don't really realize is this idea of a pen tool or a bizier curve. In the 19 um I think it was the 1930s. Um, do you ever see like what cars used to look like in the 1930s versus like 1960s, etc.? So, like let's let's look at old cars. What do you notice about old cars versus new cars? Let's let's look at old cars versus new cars. Like really old cars kind of look a little bit like this, right? Compare them to like and then, you know, eventually they kind of started to look a little bit like this.

What's the difference between those cars and like modern cars like a Toyota or something? What's the main difference you notice? What's the big main difference? >> Yeah, like the see these ones are a lot more boxy, square, boxy, etc. So, the first car in the 19 I think it was the 1960s, I think, the Renault Renault was the first person to actually come up with this design, what we call a bezier curve, because if you're going to make a car, you want to be able to manufacture the car the same way all the time. You want to know the exact dimensions of the car. It's not good enough to sort of be like give someone a drawing because a drawing if you get off a drawing you're going to have like six copies and they'll look different. You have to have exact dimensions on there.

You know what I'm saying? So the first cars that ever came out with like these nice smooth curves were these were these Renault cars. Okay. So eventually like initially cars would look very very boring. Yeah. And then eventually like we just figured out how to do the maths of these like very nice smooth curves until these days we have like proper you know proper rounded edges on cars whereas before that was basically impossible.

Okay. So the question is then how are we going to do that? So the 1930s I think it was 19 so I think I'll look up the exact when when they were first done. So Pierre Bezier, he died in 1999, born in 1910. He became very famous um in 1960s. In the 1960s was the 19 first time we started to have those nice cars and that kind of kind of sounds about right, you know.

So you know, think of 60s cars, they like all these boxes. After that, we started to get like the Beatles, you know, the Volkswagen Beetle and all those kinds of things, right? So all those kinds of smooth cars, right? So 1960s is when we invented the Max, which isn't that long ago, right? invented the math for smooth curves, right? For smooth curves. So, the invention of this math directly led to basically smooth curves in architecture and in design and in cars and and even computers these days, right? Fonts still use these. So, the question is like how would we do that? So, the goal is pretty simple. We want to learn how to draw a nice smooth curve.

Fair enough. Or like a smooth curve that looks a little bit like this. Fair enough. So, the way we're going to do this is going to be pretty straightforward. What we're going to do is I'm going to slap on two points.

I'm going to say, okay, let's suppose that this point is here and this point is over here. It's very, very closely related to physics here, too. So, what I want to do is I want to imagine I'm moving in that direction there, right? And then at the end, I want to kind of imagine that I kind of make it in this way. And then I want to sort of figure out all the in between frames kind of like that and make it a little animation out of it. Can you all sort of see that? That's what I'm going to try to do.

This is going to be very, very clever. So, the way we do this is very interesting. So, let's let's think about this. So let's let's pick four points. Let's pick even just three points right now and we'll just keep it very very simple and then you'll see how we scale it up to four points because the the math is very cool.

So I want you to imagine just like before I have this segment let's call it uh P 0. I call this section P1 and I'll call the section P2. Everyone sees it. Simple enough. And I want you to imagine that I'm I've got a little car moving along these at a at a fixed interval.

So that you know one car goes from P 0 to P1 and then one car goes from P1 to P2. Do you all see that? So as this car is doing it, let's say that T is the interval. Yeah. This is going to be this point over here is going to be T times the distance from P 0 to P1. Right? P 0 to P1.

Everyone got that? And this over here is going to be T times the the interval from P1 to P2. Got it? Any any questions about that? So, okay, this I'm going to call I'm going to call this the point P 01 and this I'm going to call the point P12. Okay, that's the point between them. And then what I'm going to do is I'm going to be even more clever and I'm going to join these two together. And then again, just the same way, I'm going to take the same fraction.

So here you can see this is like 1/3. Does that roughly look about right to you? That's roughly 1/3. And the same thing here, this is about 1/3 of the distance from there to there. And then here also I've taken 1/3. Do you all see that? And what I'm going to do is I'm going to draw a nice curve.

So, I'm going to sort of imagine as this as this point moves from here to here, I'm going to sort of trace its path. You're going to kind of see that it's going to follow nicely there. This might remind you, have you guys ever done string art at school? You ever seen that thing where you like draw a string from there to there, then a string from there to there, then there, then there, then there? Did you ever do that in school? No. Who did do that in school? Did you? Yeah. Like in primary school, you would have been like put a bunch of nails from here to here, and then a bunch of nails from here to here, and then you connected the strings.

Yeah. You've never seen that before, Charlotte? Yeah. Yeah. So, it's like a cool thing that some primary schools will make you do the string. So, this is a pretty pretty common thing that people get you to do.

So, if you do string art, it's like that. You never seen that before? Ever? That's crazy. Yeah. So, you can make like you can make some pretty crazy I remember when I was in primary school, I made a they got us to literally make that sailboat. That exact sailboat right there.

So, it's pretty funny. It's a pretty common one. Do you all see how that that would work though? Right. So, the question would be then how would I how would I do that? So I want to be able to sort of make that nice smooth curve. So let's see it.

Let's see it in action. Can we find the point P 01 first? Right. So the point P 01, do you all agree that it's going to be the distance between P 0 and P1? Is that cool? So what we're going to do is we're going to start at P 0 and then we're going to move along this distance and take, you know, T times the distance between those two. Right? So if I wanted to go from here to here, what I'm going to what am I going to do? It's going to be Let me just write this as P 01. I'll just keep it simple.

I'll call this point P12. So let's let's see how well you've really understood the vector stuff. Let's how would I get to this point P 01 assuming that this was a fraction of T along the edge? What what would we do? So we're going to start by going So let's call it P 01. What's the equation of P 01? What do I do? I start at the point I start by let's say I start at this origin. I move to the point P 0.

Then what do I do? Any ideas? So, I'm going to move along the direction of P 01. Yeah. How do I find the the vector from here to here? Remind me. How do I find that vector? Any ideas? So, I got a vector. This is P 0.

This is P1. Yeah. How do I find the vector that points from here to here? What do we say we do? So, I've got a vector, let's say U, and I've got a vector V. And I want the vector to go from U to V. What do I do? Chris, you got this.

Sorry, I'm just picking up someone random. No, not sure. Emma, you got this. >> Sorry. >> Cosine rule.

Okay, cool. No, that's not what I had in mind. No, Charlotte, do you have any idea? >> Yeah, you minus it. You go final minus initial. Everyone remembers that? You go V minus U.

that that gives me the final minus the initial. Do you remember that? So that means this vector from here to here is going to be what? Final minus initial which is remind me >> P1 minus P 0. P1 minus P 0. Okay. And then what we're going to do is we're going to take some fraction of that.

So T, we're going to T is going to be like how far along it. T is going to be any variable from zero to one. You all kind of see that? So it's like first we're going to ask, okay, let's move along a line. Let's just move along the line. Just keep our life easy.

So I got this point P 0. I got this point P1. And I want to move along the line T. Is that cool? So the way I'm going to do that is I'm going to say, okay, this this point here, this point here, I'm going to call it P 01. P 01 should just be equal to um P 0, right? So just move to from your origin, move to P 0.

And then take a fraction of this where T is zero here, right? This is like kind of T equals Z right here. That's t=0 and that's t = 1 right there. Do you all see that? Anyone confused by that? So we're moving from one side to the other. T= half would therefore be in the middle. Everyone got it? Okay.

So we're moving from one side to the other. So what we're going to do is we're going to say P 0 plus T * P1 minus P 0. Can everyone see that? So P1 minus P 0 is the is the one that goes from here to here. Yeah. And then I'm going to multiply by t to lengthen shorten it.

Then I'm going to add on that vector and that should get me from there to there. Everyone see that? So that that equation should move me along the line. So let's let's see if that works. Let's start there and then we'll we'll make our way up to the harder stuff, right? So let's see Desmos. So I'm going to go back here and let's do this.

So let's say P 0 is going to equal say 34. I can put it anywhere I like. P1 is going to be 58. Okay. I want to make them all blue just to make my life easy, right? So, what I should find is that the point in between them, so let's call it P 01, right? That's going to be P 0 plus T * P1 minus P 0.

Okay? And then T could be any number from 0 to 1. And that should move between the p between them. You see it? So, this is this one over here. This is going to be P 0. Yep.

And this is going to be P1. all got it. Any questions? So then this this I can also if I wanted to I can draw a segment between P 0 and P1. Okay, this is this is what I'm dealing with here. Okay, so I've got this one here and then I've got this point over here which is my which is my motion point.

Okay, I can make that as as thick as I like. So I can go ahead and make that like 20. So you can see it. Okay, so this is at zero. It starts at P 0 and at one it goes to P1, right? If I wanted to repeat that again for another point P2, let's think about this.

P2 would be the next point, right? So P2 would be let's say 3 4. I'm going to draw that as blue again. So let's say P2 is like over here, right? You all see that? Now I'm going to do I'm going to copy this exact same equation to to find the point P 02. Do you all get that? So I'm going to copy the exact same equation to find the point P 02. So this is going to be.12.

Sorry, one, two. And here I'm going to start at 0.1 and then I'm going to go to 2 minus.1. You'll see it. And so now this one I want to also make it purple just to just keep my life simple. Also make it nice and thick.

Yep. So, let's see. At zero, they both start here, and now they're going to sort of go from there to there. Can you all see it? Everyone got it? So, they both move along a line. So, they're not doing anything particularly interesting.

They're just both moving along a line. So, I can go ahead and write the segment from P1 to P2. So, they're not doing anything interesting. So, these are these are still just kind of doing their thing. Cool.

So, everyone can see this so far. So, I'm moving from here to here and here to here. Now what I can do is I can draw the line between these two. So let's see this. So the line between these two watch.

So we're going to just keep going here. So draw the line between right. So the line between them would be a segment between P 01 and P12. Okay. So that's the segment between the pair of them.

Everyone gets this? So this is it's doing that. Cool. Yep. And then to find the line in between them to basically to find the middle point, I'm going to take this again. I'm going be really cheeky and we'll call this the final point, the point I'm after P here.

It's going to start at P12, right? Remember P12 is the point in the middle, right? That's that's this point over here. We're starting at this point over here, P12, the point, sorry, P 01. That's that point there. And then I'm going to do the exact same thing where I'm going to go up by a certain amount over there. Okay.

So, I'm going to say say the same thing here. So, I'm going to say, okay, P I'm going to start at P 01. Then we go to final point which is P12 and then minus the point P 01. Look at that. Can you all see it? Kind of goes like this.

And look what the green point's doing. See the green points kind of moving along on a curve. You all see that? Nice. It's moving along on some kind of nice curve like that. You'll see it.

If I could draw the curve, then that would be that would kind of be it. I'd be I'd be done. You all see it? So let's let's see if we can do that curve. Let's see if we can draw that curve out. Okay.

So let's see. So P 01 is the point P1 minus. So we said P 01 is going to be the point P 0 plus T * P uh P 1 minus P 0. Everyone agrees with that? If I want to expand and simplify that, watch this. I can write this as P 0 plus TP1 minus T P 0.

Everyone agrees? Take P 0 as a common factor. This is where it's going to be funny. P 0 out of 1 - T plus T * P1. Right? So, it's kind of an interesting this is a same equation, but it's just written a little bit differently. This idea, notice that this this 1 - T.

When T is equal to 0, what happens? 1 - 0 is just one. Do you all agree with that? And then when T is equal to 0, that just puts zero here and then just crosses this out. Now, when T is equal to one, what happens? If I put in a one minus one, that crosses this out and makes this one equal to one. So, notice how what we're doing is we're kind of mixing. We're sort of starting at P 0 and then we're mixing over to P1.

You sort of see that it's like over time we're sort of mixing over from P 0 to P1. This equation gives me a nice way to go from P 0 to P1. You sort of see that the way the way it works is you do one minus T of the thing you start with, right? So 1 minus P of the thing you start with and then 1 plus P of the thing you end with. Is that cool? So that's P 01. So let's do P12.

That's going to be the same thing. 1 minus T of the thing I start with. I'm going to start at P1. And then T of the thing I end with. P2.

Do you all see that? Now, I'm going to be clever and I'm going to mix these two again. So, I'm going to mix these two again. So, the final point P, I'm going to start with one minus T of the thing I started with, P 01. And then I'm going to say T of the thing I end with is P12. You follow? Now, I'm just going to sub and simplify.

Okay. So, let's see. P= 1 - T P 01. That's going to be 1 - T P 0 plus T P1. You'll see see how I did that.

So, I'm just plugging that in. Just be a bit careful. So, I just plugged in the equation there, right? We're going to need to be careful in just a second. And then I'm going to take this whole thing and multiply it by t. Okay.

So, plus t outside of all this. Okay. So, I'm going to expand and simplify here. Let's let's see this. P = 1 - t^ 2 * p 0 plus.

Let's see. 1 - t * t p1 + t * 1 - t p1 + t ^ 2 p2. Do you all see what I did there? I just expanded out the brackets. Now notice that there are two of these and that accounts for the fact that they could be in either order. 1 - t * t and t - 1 - t.

So then what I can do is in the last step I can just do this. Just duplicate them. Say bye-bye to these guys, right? This is P1 and put a two out the front of this. And this is cool. Do you notice this? This is kind of like the binomial theorem that you saw at school.

Does it look kind of similar? It's like something squared then 2 * something and then that's something squared. Do you all see that? It's kind of very similar to the binomial theorem. If I plug this in, theoretically, that should just draw the curve. Let's see it. Does that work? Let's see.

So I'm going to go to I'm going to go desmos over here again. So I'm going to just plug plug it right in. Right. So I'm going to go P = 1 - t ^ 2 right * p 0 right plus 2 * 1 - t * t of p1. Okay.

And then plus t ^ 2 outside of p2. Okay. Uh 2 * 1 - 2 * p. Oh, sorry. The t the has to go on the outside though.

Cool. Look at that. That's the curve. Isn't that cool? So, I could just go ahead and make all of these guys just a bit thinner. Just so just get them out of my way.

Just make them all thinner so they're all as out of my way as possible. Yeah. And then that's it. So, these are your control points. And then this is U.

So, I I don't actually even care about these points. I don't need to make these points so thick. I'll just make them thinner, too. Make it Five is enough. Five and five.

And then my last point is the is this green point over here which I do care about. So I'll keep it green. I'll make it 20. And I'll make that five. And I'll make it green again.

Okay. And this is this is our point. So I can control my P1, put it wherever I want. You will see that. And then I can go ahead and and get smooth curve like that.

Isn't that cool? That's nifty. So then what I can do is it allows me to draw any shape I want. And it doesn't even have to follow the vertical line test. You know the vertical line test that you got taught in in school? Notice that this definitely doesn't pass the vertical line test, but it's still, you know, it's a two-dimensional function. This is we're sort of like foreshadowing more things that are going to come up here.

But can you all see this? Does anyone have any questions about it? Yeah. So this is this over here is the is the you know Bezier curve. Bezier curve. We call this the third order Bezier curve. Right.

The reason we call it a third order bizier curve, by the way, and I I'll just drop this in for you. For some dumb reason, it's not like my copy paste has not been working all day today. So, I'm just going to like send things to myself. Um, yeah. So, third order bizier curve, I can sort of just drop this in here for you.

So, this is the third order bizier curve. So, you can just see that there. One way we can sort of also think about this is it's almost like you're expanding out the brackets of 1 minus t + t^ squ. Do you all see sort of see that? And that's how you how you get each term, right? That's that's kind of the idea. So if you wanted to the next order bezier curve, what you would then do is you would just do the same thing but just expand it to the power of three.

Does anyone everyone know how to expand a plus b cubed? Everyone knows it's a cubed plus 3 a 2 b + 3 a b 2. Anyone see seen this before? Plus b cub. Yeah. So therefore the next bezier curve would just be if you just think about it it would just be p 0 * t cubed right plus 3 and then * p1 * I should just write it I'll write it I'm going to be lazy so t cub * p 0 plus t ^ 2 * 1 - t I I also have to start by 1 - t sorry so just the same way so 1 - t cub would be the the first time, right? Just same just same as it was over here. It's going to be cubed and there's going to be three of these with one and so on, right? So, it's going to be 1 - t cub * p 0 plus three groups of 1 - t ^ 2 * t of p1 plus 3 1 - t * t ^ 2 of p2 plus t cub of p3.

You see this? And this is my this is going to be my final point with three control points. So, let me just sort of show you this. Um I don't want to I don't want to go into too much depth for this one, but you can derive it just by repeating that process one more time. Right? So you got this yellow point. You what you could do is then just make another point here, find its midpoint, find its midpoint again, and then connect these two points and then do it again.

Okay? You could do that again. If you had repeated it another time, you would see that you'd get the exact same thing, but this time it's even more flexible one. This is the one they use in Photoshop. This this exact equation is the equation that they use in Photoshop and pretty much every every 3D control program, which is pretty cool, right? All right. So, I'll just um I'm going to be a bit lazy here because it's going to get really ugly.

So, I'm going to just go ahead and add in a table. So, I'm going to call it um P. So, just because I want to I just want to be able to graph them all, right? So, XY. Okay. So, then I'm just going to do like two, three, two, four, one, five.

I'm just going to need three control points, right? Three, four. So, I'm going to need to be able to drag these around. So, I want to be able to drag these around. Yeah. So, I want to be able to drag them.

And you can sort of see that. So, the equation is just going to be let's see. So, to write the first point, I can just write x1 that gives me the two, right? x2 gives me the 0.16. You see that? So, I could just write them like this. So, I can just be lazy.

So, let's let's write them out. So, the x x equation is equal to 1 - t^ squ. Yeah. So, so let me call it a x equation. So I've called these X and Y.

So let me call them. No, I don't know. I wonder something. I've never tried this before. If I just type P= 3 4.

I've never actually tried this. So let me just try this. Can I drag these? No, I can't drag them. Stuffed. Okay, that's fine.

All right. So So I've got these ones over here. So, I'm just going to rewrite write the equation. So, I'm going I'm going to call this X of the point is going to equal 1 - t cubed. So, I'm just going to write this out.

X1, right, plus 1 - t^ 2, right? Time t and then there's going to be a three there. And then I have to do x2. You'll see that plus three. Then 1 - t * t ^ 2 then x3. Okay? And then I have to go t cubed x4.

Okay? That's going to be the first equation for X, right? Then I'm going to do the exact same thing for Y, right? I'm going to do the exact same thing for Y. You'll see that. I'm going to just make this a function of T just to keep my life simple. Yep. So, just swapping all the X's for Y's.

And then if nothing has gone wrong, everything should work. Okay, so this is the equation for x and this is the equation for y. And so then the point x p of t and the point uh y p of t is that curve. How cool is that? Right? I can even just draw the segments from segments from x1 y1 to x2 y2. Okay.

So, it's just a bit ugly there, but can you can all see what I did there? So, I just I just drew that. And then I just drew I'll do from X3 Y3 to X4 Y4. And that that should be the whole thing. Right. So, this is kind of how you do it.

This is literally the pen tool from Photoshop. See? Isn't that cool? So, you can just literally draw any shape you want. any shape you want just by connecting these together. Right, that's what Photoshop is doing. It just draws one and then it just draws another one right next to it and just draws another one right next to it as well.

Okay, that's what you do. Oh, sorry. I just realized my head is on the way. But yeah, that's that's I just drew the segment between the lines. Okay, so that's it.

That's the pen tool in Photoshop. So, Photoshop pen tool. Do you all see that? So, it's just using the same equation that you kind of had in your homework, but then just over and over and over again, just applying them over and over. So, you can see that vectors are pretty damn powerful, right? You can do some pretty crazy stuff with with vectors very very easily. Okay.

So, let me plug that in there. Cool. Did you find that? That's interesting. Not so interesting. How'd you how'd you find that? Right.

So, one of the things I I've I've given my students homework in the past where I'm like, "Okay, George, you know, it's the same stuff you get at schools like draw something on Desmos, but obviously like now that you have Bezos curves, you should be able to draw whatever the hell you want." Is that fair? So, you should you should be able to anything you can imagine. Do you guys all get to like draw something on Desmos homework? Did your school ever give you that? No, they will. They will. It'll come. But do you all do you all see this? Any problems with this? Okay, cool.

It's so funny. I'm just getting text from myself. Um, yeah. So, that's that's basically it. So, this is just like taking taking sort of the same ideas here and sort of taking them to the extreme.

You can do some really cool Is that cool? All right. So, that's it I have to say about that. And now we're up to the final thing, which is this idea of multiplying vectors. This is where all the magic happens, I will say. Right? This is where all the magic happens.

But it's magic that I don't think that you're quite primed to understand or appreciate just yet. Okay? But it is very, very cool. But it's going to be your sort of first time seeing it. Multiplying vectors. Okay, let's do that.

So we want to ask the question. So we know how to add two vectors together. We know so so far. You know the story so far. Just reminding ourselves.

Story so far. Okay. With vectors, we know how to do a few things. We know how to add two vectors. Let's say a and b, right? To make a sum, which looks like this, right? So, this is a plus b.

We know how to do that. Yes. The next thing we also know how to do is we know how to subtract. So, if we have a and b, we know how to subtract them. So, this is going to be b minus a.

Yeah. And the final thing we know is we know how to multiply it by a scaler. So, if you know that this is a, this is like 2 a. And then by that logic, this would be zero zero vector. Agreed? That's the story we we've said so far.

And you've seen some some good stuff. And to be honest, with just what you know so far, you can already solve so many freaking problems. One other thing we saw, by the way, was the length of a vector, right? We talked about the idea. One one last thing was that we talked about the length of a vector. So we have a vector A expressed in a basis E1 E2, right? We especially J was was the basis we're especially interested in right ig.

We then explained that you could always rewrite this as this, right? Ax a y and we could say that the length of a was equal to the square root of ax^2 + a y^2. Right? I think we've had a, you know, we've had a bit of a fanfare there and that's all been fun. Has anyone been confused by anything so far? Or do you feel like everything so far has been quite easy to understand? Right. So then the next obvious question is what we're going to ask. Okay.

How do we multiply two vectors together? And this is a question that um is very funny. Now first question you might ask is why would I want to in the first place? Why would I even want to multiply vectors in the first place? Right? This is a you know very valid question. So first first answer is this right? So the dude who first discovered this his name was Hamilton. Okay. William Hamilton.

Okay. Not the Hamilton from the from the um from the movie. >> Sorry. >> No no no. Not from Jurassic Park.

not the Hamilton from so if you've seen the there's a there's a play called the Hamilton Alexander Hamilton he's one of the forefathers of the US right but not that Hamilton but there's a different Hamilton right so this Hamilton was a he was an Irish guy and apparently he was so happy when he figured this out that he like walked up to a bridge and actually carved out his equation into a into the side of the bridge and it's like uh like because apparently he'd been working on this problem for like eight or eight or something years and till today it's been it's still considered one of the most important pieces of mathematics ever ever invented Right? Because it's um it's really crazy how much you can do when you just think about like how can I multiply two vectors like it's it you can't be properly primed to understand like how crazy this is. But just going to give you a few things that have been made possible because of it, right? Number one, the internet. Obviously, the internet um you're you're you're going to be communicating with someone on the other side of the world, right? And you want to be able to do that really really quickly. So, what you want to do is you want to transfer information really really quickly. And turns out a really good way to to store information is as a vector.

A file. You may or may not know this. the file on your computer is just a big multi-dimensional vector. If you think about what it what it is, right? Um if you think about a cell, right, I told you a two-dimensional vector is just a coordinate of two points, right? Like let's say 3 four. Yeah, that's a two dimensional vector.

But if you wanted to, you know, let's think about an image file for a second. What is an image file at the end of the day? Right? Isn't it just a grid? Isn't just a grid with a bunch of different brightness values on on different on different places? Yeah. If you just flattened the grid out into a big long line, wouldn't that just be a big vector? You sort of see that? So, so that's kind of the idea. Uh, another another one would be sound, right? If you ever wanted to transmit sound, if you think about what sound is for a moment, it's just a graph, isn't it? Just the sound intensity as a function of time. So, if you if you transmit all those numbers, so like imagine the intensity here is zero, the intensity here is like 50, and the intensity here is like minus 10.

Fair enough. And that makes a big vector. Yep. That's also, by the way, how you know how um you've ever used Spotify before? Oh, not Spotify, Shazam. You think like sometimes you're like, I don't know what the song is and you're like, you ask Siri like, hey, what's the song? And then it knows what the song is.

How do you think it does that? It's asking this really interesting question. It's asking, and this is the big question that Hamilton was interested in is how similar are these two vectors? Like, let's say I have these two vectors. I want to ask the question, how similar are they? Okay, it's a very good question. It's a very simple question but you know the the answer is going to be quite surprising quite quite interesting because vectors are fundamentally onedimensional things they're just line segments right and what I'm asking is like how close are they to each other do you kind of see that one problem though is that if you think of like an image file it's like how many dimensions is this one the coordinate has one number two numbers three numbers four five 6 7 8 9 10 11 you get that so if a one megapixel image is technically a 1 million dimensional vector if you think about it right That's technically what a what a image is, right? Another one that we use all the time is hearing aids, right? Hearing aids, you know, you obviously you hopefully should never need one, I hope. But if you need a hearing aid where they're completely relying on like vector math to basically figure out what which ears to which hairs in your ear to stimulate, you know, the way you hear is that you have these here uh hairs that vibrate in response to sound, right? And what they're doing is they record the sound and then they figure out, okay, which hairs should we stimulate? And that's again something you're going to need to multiply vectors for.

Another huge one is is rotation. And this is the one that Hamilton was trying to solve, rotation. And I just want to show you some funny things that come out of like multiplying vectors that you probably would not expect because that's that's always my favorite way to show these things. Where the hell did that go? Sorry, one second. Give me uh give me the tissue box over there for a second.

Yeah, I want to show you something funny. Actually, give me your calculator. Your calculator work better. Okay. So, let me let me sort of uh go over here and I want to show you an experiment.

Right. So, one of the funny things about about this one of the funny ideas about multiplication of vectors gives you it explains rotations of an object in 3D. Okay. Rotations of objects in 3D is is not very intuitive. I'm going to say this to you, right? You usually when you think of rotations, you think of like 20° plus 40°, yeah, is 60°.

And 40 degrees plus 20 degrees is still 60 degrees. So, the order of rotation does not matter. Correct. Okay. Well, that's not true for 3D.

Let me let me show you this. Suppose I did a I'm going to do a turn 90 degrees around this axis and then 90 degrees around this axis. Do you agree that? So, top axis, right axis. Easy peasy. So, here, let's start the calculator this way.

Top axis, right axis. The calculator is in this orientation. Everyone agrees with that? Put the calculator back. Now, if I do this axis first and then top axis, the calculator is in this orientation. The order of rotation in 3D is not commutative.

In other words, the order of rotation matters. Does that make some sense? One of the funny side effects of this is is actually it's this thing called the parallel axis theorem. And we figured this out very quickly when we were learning about um airplanes, right? So, if you want to keep an airplane stable in the air, it's pretty hard to do because let's think about this, right? The this calculator has three axis. It has this really short axis, has this medium axis, and then it has this really fat axis. Is that fair? So if I rotate the calculator along along the fat axis, let me show you this.

It's it's quite stable, right? So it's quite stable. It can rotate nicely. Fair enough. It just rotates nicely. If I rotate it across this short axis as well, it also rotates pretty nicely.

Fair enough. Like it rotates nicely. It sort of rotates in the spot. Just rotates nicely. But if you rotate it around the middle axis, so around this middle axis, and I try to throw it this way, it's going to tumble violently.

Okay? because what it's going to do is it's going to be wanting to go back to the short axis and the long axis and it's going to get confused like which one should I go on and that's because of the idea that the order of operation is not uh the order of rotation is not commutated so if I do this you can see it it just tumbles like crazy you see that see it's like it's like freaking out always does it you see that short axis uh short axis no problem see that long axis no problem right medium axis good luck good freaking luck Mark won't do it. I don't care how you try. Do do you sort of understand that? And that's one of the one of the side effects of like rotation. So, you know, I don't want to get into into weeds too much, but they are the point is why should I learn to rotate vectors? Because it's bloody useful. It's probably the most useful thing you're ever going to learn, right? Um later down the track, it'll lead you to if you want to study higher mathematics, it'll lead you to complex numbers.

Have you all heard of complex numbers before? So, if you ever study, if you study for unit, uh you're going to be talking about these things called complex numbers. It also leads you to this idea of of calculus as well. You know how I taught you guys about calculus. Calculus. I'm going to show you another way that actually gets you there using only vectors.

It's really cool because I taught you calculus and it seems pretty disconnected from vectors, doesn't it? They don't seem anything like each other. And what I will show you is that you can actually recover all of calculus by understanding vectors really well. It's really freaking weird. And that's kind of going to be the coupra here if we can get there today. But do you kind of understand that? So you you could explain calculus really really well.

Complex numbers are are going to be one and then a whole bunch of other stuff. I I don't really like you've probably heard before that the speed of light is constant. You've probably all heard that before and that comes straight from vector multiplication. If you understand how to multiply vectors, you will you'll say okay well I understand why the speed of light is constant or has to be constant. Right? So there's a whole bunch of reasons.

This is just motivation. Now we're just going to shut up and calculate so to speak. Okay? And we're going to base it off of this idea. Now, multiplying vectors is going to be based off of a simple idea. Here's the idea.

The length of a vector times itself. Yeah. Should be whatever the the length of that vector is squared. Are you happy with that? Or in other words, if I multiply a vector by itself, I want its length I want to get the the I want the answer to be the vector squared, right? Just the same way that I when I added a vector to itself, right? similar to remember when we did we defined scala multiplication to be v + v = 2 v remember that so analogously we're going to define v * v whatever the hell that's going to be to be v ^2 are you all happy with that any problems so this is going to be the big idea and we're going to base everything off of this big idea let's see how far we can take this okay can we do that any problems okay so let let's start here let's let's take a unit vector for a moment. Let's take let's take the unit vectors I, J, and K.

I, J, and K. Let's do it. Let's do most of our exploration in 3D, but then I'll sort of switch between 2D and 3D as we go. Actually, let's just do 2D first, and then we'll show you 3D. Okay, let's say I and J.

All right, let's ask this. What is I * J? What is I going to be? What should that be? >> Yeah, it should be equal to the magnitude of I Everyone agrees with that. I * I whatever the hell it should be. It should be I don't know what it should be but it should be equal to one. Do you agree with that? I * I should be equal to one.

So if I multiply I by itself I should expect one. That's pretty straightforward. Yes. Cool. What about I * J? Well, no.

Let's not go there. That's too hard. Let's go easier. What about J * J? the length of J squared. Yeah.

What's the length of J? It's a unit vector. So it should be one. So J J * J should be equal to one. I'm happy with that. Okay.

But then yeah, now we're going to I * J. How the hell do I multiply I * J? I don't know. I don't know. I actually have no idea. But what I do know is let's let's make a random vector V out of I and J.

You all agree that we can take the vector V and then write it as VXI plus V VYJ. Do you all agree with that? Yeah. So I can always rewrite let's say I have suppose I have a vector V. I don't know what the vector V is, but I can write it as VXI plus VYJ. Everyone agrees with that.

So I don't know what it should be, but I should know that V * V should be equal to the length of V ^2, shouldn't it? That's going to be my definition. That was my definition, right? And so okay, well I don't know what that's going to be on the left hand side, but I do know how to find the length of V square. That's easy. What's the length of V square going to be? I do know how to do the right hand side, but I don't know how to do the left hand side. Yeah, but we we can deal with that in just a minute.

Yeah. What do we do? Well, the length of V squared, that's just how do I find the length of a vector again? If it's Vx and VI, Vy, what do I have again? This is Vx. This is Vy. What do I do? How do I find the length of vector? What do I use again? Did I give you homework on this last week? How did you find the length of vectors? Use Pythagoras. Yeah.

So, V * V should just be Vx2 plus V Y^2, shouldn't it? I don't know what it should be, but I should know that V * V is equal to Vx^2 + V Y^2. That's what it should be. Am that makes sense to you as well? Yeah, Chris, you're same page as me. All right. So, let's let's take it a bit further.

Now, V is the same thing as VXI plus VYJ. Okay. And then we've got vxi plus vy j. And that's going to should be equal to vx^2 plus v y^2. Okay.

Let's expand the left hand side and let's and let's see what happens. Let's just let's just roll with it. vx * vx. Okay. Or vx 2 i * i.

Right? vx 2 i * i. Everyone everyone got that? And then I got what else I got? I got v y 2 j * j. That's all well and good. Yeah. And we know I * I is one.

We've already found that out. And J * J is just one. We've we already know that. Yes. All right.

What else have we got? The other term we have is we have VX Vyig J. Yeah. And another term we have is Vx Vy J. And that equals Vx^2 + V Y^2. All right.

Cool. What now? Right. So I I notice that I already have the vx squar here and the v y^2 here. But then I've got these two terms that are just kind of like hanging out there and I don't want them there. So what am I going to have? I'm going to have to say that whole thing has to be equal to what? If the yeah has to be zero because if the left hand side equals the right hand side this whole thing bloody will be equal to zero.

Do you agree with that? Otherwise it doesn't make sense. So therefore vx v y i j equals uh so let's say plus vx vy ji must equal zero everyone agrees that any questions about that okay so then if that's going to be the case I can lop off these guys I can lop off vx vy because they're just constants right I can divide both sides by vxy and then I'm forced to say that I j equals the negative of j Okie dokie. So, this is interesting. So, I J is the negative of Ji. I don't know what they are yet, but I know that whatever it is that they must be when they're backwards, they're not equal to each other.

Is that fair enough? So, when I flip the order, it must somehow swap the swap the sign. Do you get that? So, this over here is a very very important thing that we're going to talk about here. And this is what we're going to call a B vector or a units area. Okay, a B vector or a unit area. Is that cool? So in other words, what we're thinking about is when we multiply I* J.

Multiplying I* J, we're going to think of it almost like a an area, an area segment of size one, right? Oh, by the way, one of the other things that this does really well is if you if you read if you like it, is it does the perspective. If you've ever wondered, right, why all of a sudden in the 17th century, like you ever like looked at old art and realized that like it used to look like and then all of a sudden it just started looking good for some reason? Like what the hell happened? Did humans just like develop a new brain? No, they just developed the mathematics of perspective. Okay, that's the idea. Once we realized the mathematics of perspective, we knew how to like make things shorter as they got further away. Okay, that's basically the mathematics I used to um do that thing that I that I proposed my girlfriend with, right? I literally just use use the maths to make that work.

So, prior to the maths of perspective, it was um we had no idea how to draw things in in 3D on a 2D 2D plane. Is that cool? So, that's another huge thing that we get out of it, which is, you know, I like to say. All right. Are you cool? All cool with this. So, what I'm going to think of is I J.

So, now I'm going to have like I is the unit vector here. This is I. This is J. Yep. And then I J.

I'm actually going to think of it as the oriented segments. Basically, the unit square. You kind of see that it's like the unit square counterclockwise. This this answers very clearly to you, by the way, why counterclockwise is like the positive direction in in in maths, right? Because you're going from I to J. So that means you're going around that way.

You kind of see that? And if you go the opposite way, that would be considered negative, right? So positive turn and a negative turn will kind of cross each other out. Do you kind of see that? Any problems there? All right. So that's that's actually quite cool. Do you have any any questions about that so far? Okay. So then what we're going to do is we're going to come up with a with a nice simple thing.

So let's let's talk about talk about multiplying two vectors. Talk about now we're ready about multiplying two vectors in general. All right, let's do that. Can we do that? So let's say that I've got a vector u and I'm going to write that as ux i plus u y j and then I've got a vector v which is they're too similar to a and b shall we? Uh now if I get u is good just be careful don't get confused by this right u vxi plus v yj. Okay, if I multiply them together, what should that do? UV, when I multiply two vectors by each other, now I'm going to get uxi plus uyj multiplied by vxi plus v y.

Cool. So, if we expand these brackets, what are we going to get? >> Ux vx ii, which we know is just one. So I'm not going to include that, but let me just sort of like make a little table just so that we're we're very clear here. So I'm going to get the vector I and the vector J. Vector I, vector J.

And I'm multiplying them. This is the first one and this is the second one, right? The first is and the second, right? So I * I is one. J * J is one, right? I J is what we call the unit by vector. So it's like the like this. Some people call it capital I, but I'm just going to call it B for by vector.

Is that cool? Or A for area if you like. area segments. Is that cool? And then this is um this is the area segments as well. But usually the easiest way to write it is just J. Is that fair? I J and negative J.

Oh, sorry. I J and negative RJ, right? Because that's the first and that's that's the second. Do you kind of get that? All right. So, let's let's do it. UX Vx I * I plus UY VY J * J.

Cool. These guys just go away. So, I don't care about that, right? And then then what we have is another section. And then we're going to get ux vx sorry ux u v y. Yeah.

I j plus uy vx or vxux ji. Everyone got that? That's the whole thing. That's the whole expansion done. Yeah. So it's going to be uv Yeah.

happy. Okay. So, so then we have ux vx plus u y v y. Yeah. And then we're going to have plus Okay.

Let's let's write this ux vy. And then this one when we swap their order, they're backwards, right? They're backwards. So, I'm going to write this as minus uh uy vxig. Okay. So, here's the idea here.

We have here scalar part and then we have here a area. Do you all kind of see that? Right now this part is going to be super important. It's going to be called the dot product. Okay. And this part is going to be important as well.

This is going to be the well the oriented area or the cross productduct. We don't like to I don't like to use that word so much but this is the cross product. Okie dokie. This is this is going to be the idea. Are you guys all on the all on the same page as me here? This is just like completely mathematical.

Haven't really thought about this very much. Um, I'm just doing it. You all see it? Any problems there? Now, in in physics, this part is really related to energy. Um, because again, this isn't this is a um physics class at the end of the day, not a math class. So, this is part related to energy.

And this part is going to be really related to rotation. And that's something you're going to see over and over and over again. You're going to see these formulas pop up like a million times over the next two years. Okay? So this part is going every time you're going to see energy show up, you're pretty much going to see this formula pop up. And every time you're going to see rotation show up, you're going to see this this one show up.

Okay? So you're going to see them show up in one form or another. This is not going to be the form you usually see them in. But I'm going to like to give you a bit of a physical meaning to all of these. Like what the hell do these even mean? At the moment, we've just got symbols. So that that's not really useful.

So I'm going to try to give it a physical meaning so you understand well what the hell is going on here. Okay. So what I'm going to do is, you know, remember these and then we're going to try to understand this. Now, what did I say? say here, right? This if you like can be thought of as the area between U and V. Okay, I'm going to say this here.

I'm going to just tell you the answer and then we're going to prove it to you. This one can be thought of as the area between U and V. Is that cool? And this over here is the projection projected length of U onto V. Okie dokie. That's what we're going to say.

So projected length of U onto V. What I mean by that, so let me let me sort of tell you this. Suppose we got U and then we got V, right? This one, what we're going to do is we're going to project U onto V basically because what what the dot product is doing or what the scalar product is doing is it takes the part u in the direction of V. Fair enough. And then it multiplies basically the length of this.

So basically um we can we'll write this as U do V. This is going to be U u in the direction of V multiplied by the length of V. That's going to be what it is. Okay. You might Okay, why the hell would I want that? But you you'll see.

Okay, it's that's basically exactly the uh definition of energy in physics. You're going to see like when we define energy for you, because you probably heard the word energy before, when I define energy, it's going to be literally this formula. I'm going to just pop it up on the board and be like, you're good. Okay, that's that's basically it. But instead of U and V, I'm going to use force and distance.

Force times distance, that's work. Okay, so I'm going to I'm going to do that. Whereas this one over here is going to find the area between A and B. So say this time if I find u and v. Okay, I'm just basically going to find the area between them and I bet you it's going to equal this.

Do you kind of see that? So the area between u and v is going to literally equal that. Do you s that one's a nice vis visual easy one, right? Basically the area span between the two vectors or the area of that parallelogram will directly equal that. Do you all see it? So let's look at them both one by one so we can really understand it. Right? And this, by the way, easily translates into 3D. You don't need to do anything else to to make this work in 3D.

This works in 2D, works in 3D, works in 10D, works just the same way. Okay, so let's let's continue on. Let's let's try first we're going to try um the area one because I think that's So we're going to look at the cross productduct first. Okay, the crossroduct. Okay, sometimes, by the way, people like to write this part as um you know, UV.

I like to write it as UV equals U dot V. Okay, which is the scalar part. Okay, dot usually means scalar. And then and this is the main thing you need for high school. But then then for physics, we need this one.

U wedge V. This is the area part. Area part, scaler part. Are you all cool with that? So that that's just one thing to note and you're going to see both of them come up all the time. So let let's see let's see this in in action.

So the cross the cross productduct is going to be the the first one we're going to find. So how would I go ahead about how do how do I find the area of a parallelogram? This is the first question I like to ask you guys. So let's let's ask how do we find the area of parallelogram because I'm going to need that in order to find the area of a of a of a parallelogram here. Right? So I have a bit of a question for you guys before I start. Does that look like a parallelogram roughly? Yeah.

Okay. So before I do I don't actually care about parallelograms. I'm going to start with a rectangle. Okay, I'm going to start with a rectangle. I'm going to ask you a question.

Suppose for a moment that I put some hinges on the corner of each of the rectangles of the rectangle, right? I'm going to tilt the hinges. Okay. So, I'm going to tilt the hinges a little bit. So, you can kind of see that like I'm going to sort of twist this one and twist this one. Yeah.

So, that now all of a sudden it's like it's here and it's like here, right? You all kind of see that? I'm going to just twist the hinges, right? So, I want to ask you guys all a question. Has the area of the of the hinges increased? And I want you all to vote on this. Increased, stayed the same, or decreased? You get that? So, as I sort of tilt it a little bit, has the area increased, stayed the same, or decreased? I'm going to count down from 10, and I want you all to make a vote. Okay? Cuz I you learn more if you're wrong. So, just take a vote.

Do you all understand the problem? So, like, I start with a rectangle and then I tilt it a little bit. The question is, has the area increased, decreased, or stayed the same? Increased, decreased, stayed the same. Okie dokie. I want you to think about it and vote on it. I'm going to count down from 10 in a few seconds.

10 9 8 7 6 5 4 3 2 1 go. All right. So, you two say stay the same. Chris, what did you say? >> Stay the same. So, all of you think that the area stays the same, you know, and that that's um that's uh that's usually what people think, but it's not the right answer, right? So, let me let me prove to you that it's not the right answer, right? So, why did you get annoyed at yourself? I'm curious.

>> Yeah. Yeah. Right. Like, because I always find it funny like I actually kind of like being wrong. I don't I don't mind being wrong.

But I noticed that a lot of people when they start, they kind of, especially if they're like overachievers, they gen generally don't like being wrong. Like it feels bad almost, right? But um let me let me prove to you that it changed, right? If I was to take that triangle and chop it off, do you agree that it would sort of fit nicely there? Do you agree the width is the same, but the height has gone down by a little bit? No problems. Can you all see that or not? Yeah. So, if I wanted to find the area of a parallelogram, the general way I'm going to do it is I'm going to chop off this section. Yeah.

I'm going to chop off this section and then I'm going to paste over here. Can you see that? pasty pasty. Yes. Any problems? Okay. So then if I wanted to find the area of this parallelogram, I would find suppose I knew A and I knew B and I knew the angle between them.

That's that's what I know this time, right? Let's say that I wanted to find out because you know like I would know the ve the vector A, I would know the vector B and I would know the angle between them just like I did with the metric. Is that cool? How would I find the area of this parallelogram? Let's say area. What would it what would it be equal to? Or I could I could say suppose I knew the length U and that I knew the length V. Yeah. How would I find it? What would I do? Well, I'm going to need to find this vertical height, aren't I? Yeah.

And do you agree that that angle is actually just the same as that angle? Yeah. I know that length V. So, I can find the height H, couldn't I? How could I find the height H? So, area, do you agree that it's just H* V? How do I find h? Because I don't want to be dealing with h. I want to deal with u. So what do I do? So if I did sin theta, what's sin theta equal to? H u.

Therefore, h is equal to u sin theta. Everyone agrees? So therefore, this is equal to uh sin theta. Have you guys all seen this before? Yeah. This length time this length times the sign of theta. Do you all see where I got that? Do you have any problems with that? Right.

One of the funny coloraries of of the area of the parallelogram then is if you were to move along this line. Yeah. If you were to sort of move these corners along this line. Yeah. Does the area change? No.

Because the height stays the same, right? So the height stays the same and the base stays the same. This one changed because the height did not stay the same. See that? The hinge down. In fact, for this one, if you go all the way down, obviously, then it would collapse into a flat line and there's no area left. That's always a very good way to think about these problems.

By the way, when I give you a problem, always push it to the extreme. I'm always careful to not push it to the extreme to confuse you. Is that fair? But if you just push it to the extreme, you would immediately realize, oh yeah, it's just going to collapse into a straight line. So, it would have no area left. Can you all see that, Ch? That makes sense to you.

Are you are you um you're think you're deep in thought about something? Yes. Are we all on the same page here? Any problems? So, we're going to start with the cross productduct then. So, let's let's think about how we would find the cross productduct of these two vectors. Let's suppose that I have a vector u and then I have a vector v, right? Two vectors u and v. Yeah, I know the angle between them, but I don't really know the angle between them, but I could I could sort of figure it out, right? I know that the area between them is going to be uv sin theta, don't I? I'm going to say something like that.

Cool. So, I know the area between them is going to be equal to the length of u length of v time the sign of theta. Everyone's cool with that. How am I going to find those? That kind of seems a bit hard, right? But not really. Not really.

What could I do to figure figure this out? Now I have a few choices. There are a few choices for me. But I could just do this directly algebraically, right? If I wanted to find the area area there. One way I could do it is sort of like just by constructing this whole picture. So I could go UV.

Do you all see that? Cool. So I can constru construct this. Put that up there. That over there. Okay.

So let's see this. Now I can break this up into sections. So I'm going to break it here. This height over here is going to be vy. Everyone agrees that this width over here is vx.

This height over here is u y and this is ux. Yeah. Now since this is v, this again is just v y, isn't it? And this is just vx. Yeah. So how could I find that area? Can you all start to see ways I might be able to do that? How could I find the area? I could be a bit clever about it, couldn't I? I could be I could take that triangle and move it down.

Yeah. So, I'm just just being a bit cheeky and because I'm lazy. I don't want to do heaps of work. I want to make it minimal work for myself. Yes.

I want to make the work as easy as possible. So, what I'm going to do is I'm actually going to be clever and I'm going to take that top triangle and I'm going to put it because if you think about it, where where is that top triangle going to go? That top triangle is going to fit very nicely in there, isn't it? Yep. So, if I was to just delete that, it's going to sort of make a nice section like that. Okay. And then it's going to make a nice section just like that.

Yeah. By pulling that other triangle over there. But obviously, I ended up double counting some crap, which is a bit annoying, but it's not too bad. Right. So now, if you think about what what area is sort of filled in, I've kind of filled it in like this.

Right? So, initially, what I would have done is I would have taken the whole area and subtracted this, this, this, this, this and this. Do you all kind of see that? And then, but now what I've done is I've taken the whole area. Yeah. And then I'm just going to subtract these two and I'm going to subtract these two. Do you all get it? But then you have to just be careful because I've accidentally double counted.

So if I subtract twice, I have to add back on this section. Yeah. So I have to add back on that section. So just add add this back on. Can you all see it? So let's just work it out.

So that's going to be vx and it's going to be so you got vx and then ux. The height, the width is going to be ux plus vx. Do you all agree with that? Because you got the vx there and then you got the ux there. Yep. And you got uy plus vy.

So u y plus v y. That's going to be the whole area. Uy plus vy. Everyone agrees with that? And so then what I what I'm forced to say is I'm going to say okay the area is also equal to uy + v yep times um uy time ux plus vx right and then this one over here we have to be careful. So this this width over here is going to be ux.

Yeah. Times uh uy, isn't it? So minus ux v uh ux uy. Yeah. See that? And then I'm going to subtract off vx v y. That's going to be this bit.

So that's minus vx v y. Yeah. And then this height over here. So this height over here is going to be can you see this is vy and this is ux. Everyone sees it.

UX Vy. So I have to add that back on because I've subtracted it away twice. That's no good. So I have to add back on ux Vy, right? Plus ux vy. Everyone's cool with that or does anyone have any problems with it? Yeah.

Okay. So then let's go ahead and and solve for it. So if I expand this all out, let's let's see what happens. So area equals u y ux plus um V Y VX plus what? What else we got? 2 uh I got U X V Y + VX UY. Yeah.

Minus UX UY - Vx V Y plus UX VY. All right. So, let's see. U VY U. This one goes away.

No, no, it doesn't. >> So, any of these cross out >> UXUI? I don't I'm not saying it. It's This is why I didn't want to use >> Yeah, this is why I didn't want to use the same letter as two at the start. UX UY. Oh, yeah.

There you go. They go away. Cool. Nice. Anything else? >> Vy VX.

Okay. What are we left with? UX Vy. Two of those. plus Vx Vy. >> Something's gone very wrong here.

I think something's something's gone very wrong here. Hold on. Something's gone very wrong in this in this derivation of mine. So like something I'm actually like something I've done some mistake. I don't know what it is here.

Like I can't I can't see it to be honest. But let's let's do it an easy way because that way is too complicated anyway. All right. So let's try let's try coming up with a a different way. Do you do you kind of understand what we want to do though? Right.

We want to like chop pieces away. It's probably going to just be easier to just chop the pieces away directly to be honest. Like this is it's like it's like too complicated if we're going to do this. So, we've got these, you know, you got these like little cross pieces over here. And then you've got these pieces.

You got this piece and then this piece. There's two of them, right? So, you're going to chop away two of them. And you're going to chop away two of them, right? So, to chop two of these away, I can I can probably just do this very directly. Like, I can not even bother with this crap, right? And then I can just make it a lot simpler, right? Because this is all this is all too complicated. So I'm going to just make it simpler, right? So this is the whole area, right? This is the whole area.

Yeah, you'll agree with that. I've also got the expansion for it down here. But I'm going just I'll do it again anyway. We'll do it again. We're just going to be very direct here and we're not going to we're not going to be fancy at all.

Right. So then what I'm going to what I'm going to do, I'm going to chop away two groups of VXVY. Do you all see that? So I'm going to chop away two VXVYs. So I'm going to chop away two or just just really a VXVY, isn't it? Right. VXVY because there's there's the one down here and then there's the one over here.

Right. So join them together. That gives you a VXVY. Yeah. And then same thing, this one over here gives you this one plus this one gives you ux uy.

So minus ux uy. Okay. We're going to be quite careful with the way I write them. And then what I've got is I've got vy um this is v y and this is the same thing as ux isn't it? Right? So minus v y ux right and then I'm going to subtract off this part which is ux v y right and then if I didn't screw anything up here and I do the algebra it should work this time. Is that cool? So what what should happen here if I if I expand this out? So let's let's be careful this time.

U Y U X plus U Y VX. This is the last time I will try it. Um plus V Y UX plus Vy Vx. Okay. Minus all this stuff.

Okay. Any problems there? So you just have to be careful. But then let's let's look for all of them. Let's look for all the repeats. So, vy vx v y.

Do we see that? There it is. Cool. And then uxy. Bye-bye. Vxux.

I don't see that one. ux vy. Bye-bye. All right. What are we left with? Uy Vx minus Vy UX.

Exactly what we had last time. VX VX UY. You see that? Vy UX, right? That's exactly what we had on the previous page. See that? There he is. Do you all see that? So, bit annoying, but we we proved it to you either way.

So this is the very very important thing to say then is that u v u * v * the s of theta right whatever the hell that should be is going to be equal to vx u y minus v y ux okay this is this is going to be the very very important little formula that we're going to see and I'm going to prove this uh we're going to do this do another example with it just to show you like an actual example right let's do an actual example so we can see We'll do it the hard way first and then we'll do it the easy way. So the problem I'm going to give you is let's suppose that we have two vectors. So first vector is going to be like 3 I + 2 J and then let's do uh 2 I + 5 J. Yeah. And then what I'd like you to do is find the area area formed by the parallelogram between them.

Is that cool? You'll get the problem. So I want you to find the area of the parallelogram. Let's do it the hard way and then let's do it the easy way. The hard way. Okay, the hard way would be to to find use u v sin theta.

Okay, let's let's do it the hard way and then let's do it the easy way. The area would be uv sin theta. Let's see how are we going to do that. First we're going to say okay let's say that this one is u and this one is v. Okay, what's the length of u? How do we find the length of u? Remind me.

Yep. So let's do that. Square root of >> 22 + 5 2. What is it? >> Sure. Root 29.

Everyone gets that. Then the length of V. What's the length of V? Yep. So 3^ 2 + 2. Yep.

Equals. So try these ones in your book. Make sure you guys can sort of do them as well. So you get what is it? 13. Sounds good to me.

How am I going to find the angle between them, though? That sounds crazy. How the hell am I going to do that? That sounds annoying. How am I going to do that? Anybody know how to do the angle between them? >> Can't really can't really do that. Now, what are we going to do? How do we find the angle between two vectors? That sounds crazy. What are we going to do? >> Does anyone have any ideas at all? inverse sign.

>> You're close. How'd you get inverse sign? >> That's true. That's true. So, look what we're going to do is like let's let's let's first do you all agree that if we found the angle of U and then we subtracted off the angle of V, couldn't we find the angle between them? Do you agree with that? Like if I basically if I drew on my x-axis, if I found this angle and then I subtracted off this angle, wouldn't that give me the theta between them? Yeah. So, so let's do that.

How do I do that? How do I find the angle of a of a vector? We talked about it last week. That's that's pretty pretty easy to do. So, if I've got a vector and it's got two on this direction, five on this direction, and I want the theta, what am I going to do? We use trigonometry. Which one do I use? Tan theta equals what does it equal? Five on two. Right.

So the y component divided by the x component. Yeah. And that's going to give me the theta. So 10. So 10 inverse.

So basically the first one is going to be tan inverse of five on two. That's going to give me the angle of the bigger one. Yeah. Then what am I going to do? I'm going to subtract away. What am I going to do? Subtract away.

I want the angle of this one. Yeah, the angle of this one. What am I going to do that time? Any ideas? >> 10 inverse of two on three. Right. 10 inverse of two on three.

Everyone's on the same page as me. Okay. So then using my formula hard way is going to be roo 29 * roo3 times the tan oh sorry the sign disgusting sign of the tan inverse of 5 on 2 minus the tan inverse of the two on three. That should be the correct formula. Right? So if you plug all that in, what does it tell me that the area is going to be into your calculator? Roughly give me Give me roughly what the answer is.

You're going to be shocked at how nice the number is. All that all that crap we just did. What do we get? But 11. Oh, great. After all that, isn't that cool? Okay, but then let's do it the easy way and see why that happened.

What does the easy way tell us? Vx. What's Vx? What's vx? >> Three. And what's uy? Five minus v y. What's v y? >> Two. And then ux 2.

So 15 - 4. Ah, 11. I always love it when that happens, right? So you could do it the hard way or you could do it the easy way. I was like, it's up to you. But do you kind of all see that those two it's it's really not obvious that this should be equal to this.

That's freaking weird if you ask me. Right. It is that's not obvious at all. Do you do you all see that? Yeah. Everyone's happy with it though.

Right. So it is the case therefore that UV sin theta is equal to this area bit. Okay. We're going to talk about this a lot more in in some detail later down the track. Okay.

But this is really cool. It means that there's a nice cartisian formula as well as a nice geometrical formula. So there's a nice relationship between areas of parallelograms to the the direct coordinates themselves, which is not obvious. Like like I hope you can see that that's not obvious at all, but it is it is true, right? So I can find the area of a parallelogram just by taking the x coordinate subra times the y coordinate minus the ycoordinate time the x coordinate. That's that's not obvious at all.

I first time I saw it, I was like, "Ah, what the hell? That's weird." I mean, it's not it's not it's not something I expect at all. like if you understood it, it's like, you know, it's kind of hidden from you, but it's cool. Like we we discovered this a long time ago and it is a very very useful formula. Okay, sorry. It's just got a freeze here.

So, just just give me a second. It's my just freezes sometimes. Okay, there you go. All right, everyone everyone's on the same page there, right? So, that's that's where we are. That's what we got.

So, that's how you do the hard example and the easy example. So that at least tells us what's the what the cross part is doing. It's very easy. It's just finding the area of a parallelogram and you multiply that by by the unit area. This is a unit area and this is the actual area.

So this whole bit is just the area of the parallelogram. In other words, this is the just the parallelogram between U and V. Do you all see that? And one way to sort of think about this is this is like this is the scalar part, right? And this is just its J part, right? So this is like if you have UV and so because U starts here and V goes here. Here you're wrapping clockwise. So this would be a negative area.

You kind of see this. So this is a really good way to do all those like area problems that you had at school, right? If you wanted to solve if you wanted to find areas of any weird composite shape, this is kind of the way we do it in a computer, right? So because this can even account for the fact that areas are negative if you go the other way. So for example, we're not I had a job at CSRO and one of the things we had to do was have maps of all the cities in Australia because obviously we're the you know we're we're the dudes that need to do all the science on Australian things. So like one thing they did was they they wanted to find where floods would happen or even cooler one was like bush fires right we had a map in New South Wales as a vector tile. So vector tiles is basically a bunch of vectors connected to a thing just drawn as a bzier curve like I just showed you guys, right? And we wanted to ask the question, okay, where should we burn the trees, right? Where should we burn the trees to avoid bush fires in the next season? Fair enough.

If we get it wrong, well, you've seen what happens when we get it wrong. So the fact that we had don't have that every year is kind of kind of cool, right? That mean we don't get it wrong that often. Is is that fair? Um because Australia would just be on fire all the time given the current state of global warming right now, right? It would just like if we if we didn't have the scientific tools we had these days and we weren't back burning properly, Australia would just be on fire all the time. Like literally like it crazy. But anyway, funny to think about, but that's one of the one of the cool things.

So what what the reason we did that is what I'm trying to tell you is that when we if we found the area of the fire required, the easiest way to do that would be to just use these these methods here, right? And we could just write everything on an xy grid and then we could just use the these kilometers and coordinates and subtract them to find areas quickly, right? If we had to do this every time, we would be blowing up a bunch of computers. Can you all see that? Right. So, that's not that's not what we did in practice. Any any questions about areas or do you feel like you get the idea? All right. Because this part's a little bit more dense, but it's also a lot more it's a lot more um what's the word? It's a lot more pra practically applicable to you.

Is that fair? Like this part is a lot more applicable to what you're going to be learning. This part's just kind of like it's cool and cool to know. It's going to be very important year 12. So, it's going to come up again. I'm not saying it's not going to come up again, but this part is like you're going to even need it like in a few weeks, like you need this in year 11.

Okay, so this this is the work equation. So the next one we're going to talk about is the dot product. Okay, so that's that's the cross product, which is the same thing as u wedge v u wedge v, right? And we're going to talk about sometimes people call it the wedge product. Now we'll talk about the dot product. Okay, let's think about this one now.

So, what's the dot product on about? Okay, we're going to define the dot product as now I can do it a different way. There's there's a whole different um there's a whole different way to approach what we just did. I think it's worth seeing that there's an alternative approach just for the cross product because we're going to do the same thing with a with the dot product because it's a bit easier. So the alternative approach is just to say okay well what's the what's the area between the vector I and itself what's the area between the vector I and itself zero right there's no area between I and itself right so the area between I and I should just be zero yeah what's the area between I and J you have a one by one square what's the area of a one by one square yeah so I wedge J is just going to be one. You all see that? And then if you did J wedge J, that's also going to be zero, right? J wedge J, that's also going to be zero.

Everyone's on the same page there. So, one one thing you could do if you wanted to is you could also write uxi plus uyj. Okay, wedge vxi plus vxj. Are you all cool with that? If I wanted to. And then if I expand it out these brackets, I'm going to say, okay, ux i ux vx i wedge i plus ux Oh, sorry, this is vy.

My bad. Vyj. v y i wedge j, right? Plus u y vx j wedge i. Yeah. and then plus uy v y j wedge j.

Okay, now as you can kind of see this goes to zero and this goes to zero because they just don't count as anything. And since the area is assigned, this will just become ux v y minus u y vx i wedge j. That's kind of the idea. Do you all see that? And then if we use the formula like we could also just get straight to these from the formula. So if if it was just if you wanted the length of I wedge J, that would be the length of I time the length of J times the sign of the angle between them.

What's the angle between them? That's 90°. What's s 90? >> Yeah, s 90 is one. Okay, so that's just going to be one. These are one. One.

So I I J is one. But then if I asked you what's I wedge I, right? That's going to be the length of I times the length of I times the angle between them. Right? What's the angle? What's sign zero? >> Yeah. So this is zero and this whole thing go these are one but then zero. Do you see that? So I I should be zero.

That's that's going to be the idea. So straight from the formula you can see how the two are just like linked to each other. Okay. So now what I'm going to show you is the dot product. Dot product is even more interesting.

Um but not as geometrically obvious I think a lot of the time. Right. So what the dot product is asking you to do is technically it asks you to find the work done or finds you that finds the projection of one vector onto another. It's really useful if you wanted to for example um project something from 3D into 2D right if you wanted to project something from 3D into 2D dot product is pretty much going to be the only way you're going to be able to do that. Okay.

So, very very useful to do that and it's it's the main piece of piece of maths. For example, we we used to calculate lighting in video game scenes and and yeah, well, do product is used all over video games, but like you'd be surprised like do do you realize when you're playing a video game, you're technically playing a a graphing calculator. You're technically like giving arrow controls to a graphing calculator. It's just a very sophisticated graphing calculator. Like to to sort of prove that to you, like I've done some I've done some like Desmos is a pretty slow graphing calculator.

video game engines are very fast graphing calculators, right? So, just to sort of show you like funny examples is um like you can you can go ahead and make 3D shapes in there, you know, and you can even go ahead and um you can make I'm going to show you this 3D if I if I if I draw this is just using the 2D Desmos graphing calculator. This is using the perspectives, right? So you can kind of see um see how like works in perspective. So this is that this is the 2D calculator, not the 3D calculator, right? This is just the 2D calculator. I can like literally draw things in perspective. This is like something I do with my four unit students.

See that? You kind of see that that circle kind of looks like, you know, so I can do that. I can I can change P to increase that. And you can kind of see see how that works, right? I can change the radius of the sphere and sort of works in 3D. And if I wanted to, I can even change the size of the sphere. I can even bring it forward and then kind of see like you know you really get the perspective warping there.

You see that? So it's the if you're an artist they call it forcehortening. Have you guys ever done art before? Like have you heard the word forcehortening? Yeah. So you can kind of see it right there. The maths is all right there. Math is pretty cool.

Who doesn't like maths? >> Which bug? >> Oh, I'm sure I could. I can I can do some pretty pretty nutty stuff in desal be like I'll check it out but okay so dot products let's figure this one out so what what is the dot product going to do so if the cross productduct was uv cos theta dot product the easiest way I could say it is u the length of u * length of v * the cosine theta right the cross productd is uv sin theta right uv sin theta and the dot product is going to be uv cos theta is that fair so that's what that's going to be what the dot product is going to be so the definition of it is going to be u * v u dov is the same thing as u v cos theta. Okay, and it's going to be calculated just the same way that we did with these two. Okay, so I mean the hard way would be to do all of this crap just the same way. And the only difference is I would swap that sign for a tan for a cause.

You see that? That's the only difference. That's going to be the dot product. So swap that sign for a cause. That's going to be the dot product. But then I suppose there is probably an easy way and there is.

Let me show you the easy way. Okay. So if we're going to define u dov as u * v cos theta, what is i do i going to be? Right? It's going to be the length of i times the length of i times the cosine of the angle between them which is zero. What's the cosine of zero? One. This is why changing from a cosine to a sign actually changes the the story quite a lot.

You see that? So I do i actually becomes one. Okay. What's i j? length of I times the length of J times the cosine of the angle between them. 90 degrees. What's the cosine of 90? Yeah, that makes this go away.

You notice they're kind of like inverses of each other in many ways, right? So, okay, if that's going to be the case, let's let's do the same exercise. Let's suppose I got U do V this time. Okay, I've got U dot V. Got UV cos theta or U dot V. That's going to be uxi plus uyj dot this time vxi plus vy j.

Everyone got that? So u dot V therefore is going to be UX Vx I do I UX Vy I do J U Vx J do I plus UY V JJ Got it? Okay. By by by sound like a Backstreet Boy. ux Vx plus UYY VY. That's too easy, right? That's that's the formula. Basically, multiply all the X's together, multiply all the Y's together, add them all up.

That's it. That's how you do the dot product. Okay, that's what it that's how you do it. But then we'll try to figure out, okay, why you do it very soon. Okay, but right now we're just showing you how you do it.

Okay. So just similarly if I was to just do the exact same problem actual example. Okay. Click copy. All right.

The only thing that's going to change is we're not looking for the area of them anymore. We're looking for the projection. We're going to talk about what that means in in just a second. Okay. So uv cos theta.

So this only difference is that sin theta is going to change into a cos theta. Okay, the angle is still the same, but that's going to turn into a cos theta. Can you guys please put that into a calculator for me? Sorry, it's a bit of an annoying calculation. 60 60 sounds good to me. Okay, so projection here should be 60.

Let's see if that works, though. Multiply the x's. Oh, 16. Sounds good. Okay, so then here what it's telling me is multiply the x's.

So that's going to be 2 * 3 + 5 * 2. Does that work? Does 16. Great. Right. So just again we have a nice easier example.

So that's you know our simple actual example. But now let's actually assign some geometry. What does this mean? What does what does a U.V actually do? Okay. That's that's what I've you know alluded to. So let's try that.

So I've told you it's going to be that's u and that's going to be V, right? So let's see this for a second. I told you that u dov should be the same thing as it's a scalar first off and it's equal to the length of u time the length of v time the cosine of theta. Do you all agree with that? So let's think about this for a second. So this is the length of v, right? This is v right here. So if I dropped this triangle over here, do you agree that this one over here would be v? And let's say that I knew the angle between them.

Do you agree that this length over here would just be the V * the cosine of theta, right? So the length from here to here would just be V cosine of theta. Everyone's cool that Y and if I just multiply that. So that's V cosine of theta. Let's call this V parallel if you like. V parallel.

So U do V can just be thought of U do V is just thought of as the length of U parallel. Yeah. Times the length of V. That's it, right? That's what Sorry, times the length of U. So sorry.

called V parallel time length of Ven U. Cool. Or U* V parallel maybe is clearer. Okay. So what you're doing is you're just basically looking at the shadows.

Imagine like you're looking at the shadow of V. You look at the shadow of V first and then you just multiply like you multiply on a number normal number line. So the dot product is kind of like just normal multiplication. That's effectively what it is. Okay? So you're just dropping V onto onto U and then you're just multiplying it directly.

Can you all see that? And then by that logic then could I write V parallel as a vector? So let's write V parallel as a vector. Okay. Do you kind of understand? It's going to be in the direction of U. You'll see that. So it's going to be in the direction of U.

So what I'm going to need to do is I'm going to need to shrink it down so that it's just the right length. Do you all kind of see that? So what I'm going to do is I'm going to sort of want to think about this. U V parallel should be just equal to some constant times the vector U. Do you all agree with that? And the question is just which constant would that be? Yeah. Which constant should that be? Well, I know that if I found the length of V parallel, that should equal the K * the length of U.

Do you all agree with that? Whatever the hell that should be. Yeah. And so V parallel is the same thing as U V cos theta, isn't it? Do you agree with that? So V parallel should just be V cos theta. So therefore the k value I should use is just going to be v cos theta all on the length of u. Do you all see it? Cool.

And then this is then well I can be cheeky and then just type a length of u at the top and a length of u at the top. And then I can I can do something even nicer. So k this is just uv cos theta. So this is just the definition of my dot productduct that I started with. Right? That's just definition up there.

So I can just write this as u do.v and u dotu. Now this is this is another funny thing about the dot product right? What does the dot product kind of look like? Let's say I took the dot productduct of a vector with itself. What does that do? Right? So let's just take the dotproduct of u with itself. What does that do? So u dot u. What does that what does that do? Think about it.

Take the dot product of that vector with itself. What do you get? ux * ux which is ux 2 and plus >> u y^2 what does the dot product do okay pythagoras and in fact the other the idea is that the length of u right the length of u squared is just equal to u do that's actually back right back to the first definition that I gave you right so it just finds pythagoras it just finds the length of the vector so what dot product has always been doing is just finding lengths in 2D. It's like finding lengths and angles in 2D. That's the that's the cool thing about it, right? So then if that's the case, then u length of u that's u dotu. So hence v parallel is equal to u dov on u dot u * u.

Okay, that's how you would sort of write that one out. Okay, I'll just show you that on desmos. And I think that pretty much concludes us like I don't want to pump you with too much information in one day. So that give you some some practice problems so you can you can get a little bit better. Let's let's just see this, right? So I'm just going to go to a graphing calculator.

Okay. So let's just see this over here. So if I was to go the vector u so v parallel let's say u was equal to um let's say 34. Is that cool? So doesn't matter which one I use, but let's let's just let's say that the vector u is 34 and the vector v is um 28. 4.

Are you guys cool with that? So this is um again I can just draw them quickly. So just just things you don't need to see. So visuals uh UT and VT. Okay. So just just drawing them like this, right? Yeah.

So I I just want to project this blue vector onto the red vector. Meaning I want to take it shadow onto the red vector. Do you all kind of see what I'm trying to do? So according to my formula, so I I can say that it should be V parallel. Let's let's call it P for parallel. It should give me the the X component is U.

So it's going to be I'm going to multiply V by some scala. Do you all agree with that? And what was the scala on the top? It's going to have Ux u dox times what? Tell me u dox* what? So just looking at that looking at that thing we wrote down. Looking at that blue formula, u dov. How do we find U dov? Remind me u dov is >> uy vy. Exactly.

You all cool with that? And then what am I doing at the bottom? U right? Everyone gets that? So I'm projecting onto u here. So I'm projecting on to you. So what do I have to put down there? U 2. So there going to be U dox * u dox. So u ux^2 plus u doy squ.

Cool. And so now this is the projection of one vector onto the other. Okay. Um I don't know why it doesn't Something's going wrong. Oh, cuz I'm projecting on to you.

My bad. Okay, there you go. Cool. Can you see that? That's the shadow of the first vector on the second vector, right? And it can even go negative. That's fine.

See how it like projects on one. This is pretty satisfying. Come on. Right. That's pretty satisfying.

So this is the this is the shadow of one vector onto the other. Okay. You can kind of see that's that's the shadow of one vector onto the other. All right. So now you can just sort of do this here and it will work wherever.

Doesn't matter where the other vector is either. Right. Right. So you can even have the vector be longer. That's fine.

See this vector can be longer and it still works just fine. It's almost like the first vector makes a plane, right? And then you could you can do it like that. So that's that's going to be the projection, right? If you wanted the perpendicular version, by the way, like if you wanted the perpendicular vector, like you wanted the vector from here to here, obviously you could just subtract that vector from that vector and then you could get the perpendicular. Okay? So that's that's something else you can do. Just be aware of that.

So this is how you project one vector onto another. All right? So that's that's pretty much it. So let's just uh let's start saying a little bit of homework and I will um we'll do that there. Did you guys go okay today or was that a bit bit harder or how'd you find it? >> Yeah. Interesting.

More less interesting. Would you guys more interesting or less interesting? What do you guys think? >> Yeah, it definitely like gets very cool. Like the further I go down this rabbit hole, the cooler it's going to get, right? So if I if I will last week as sort of a coupra or next week as a coupra I want to talk to you a little bit about the jewel numbers right this is basically tying calculus to vectors right and that's going to be kind of the last thing I say once that's all done I've actually kind of already introduced them to you but you just haven't probably realized it yet but I'll I'll do that next time. All right but last thing I'm going to do is then just set you a bit of homework and and that's going to be it. So now um I gave you guys AB, didn't I? AB was pretty easy.

So now we're going to talk about the dot products component formula and then the length formula. Is that cool? So we we've just done that. That's going to be more or less all I need from you. Like the the geometric problems are cool, too. Like you can you can solve a few of those, but they're not super important.

So I I mainly want you to do projections. I I'm going to skip this one because this is more maths. they'll do it um they'll do it in maths class and I don't really care about them that much. So I just I just want to be able to do that and then applications to physical situations because that's going to be very relevant to physics. Okay, so this is just that's going to be all I'm going to really need.

Okay, so yeah, HC let's just um let's just find Teen Titans here. Where is Teen Titans? Okay. So, I'll just I'll just put this in here and if you have any any troubles. So, make sure you make sure you read it and it'll make sense to you. I love how like I haven't gotten any responses until I said something about One Piece.

You funny I'm still on the second one. I watch shows really slowly because uh I don't um I don't have a lot of time, but I I'll get I'll get around to it. It's more fun that way because then we can like talk about it slowly. So, he ate the he ate a fruit. Um did I spell that wrong? Exercise.

Okay. So, I'm just going to expect you to do um question let's do one B D question three AB. You just kind of have to get used to that notation, but it's pretty similar to what I did. Anyway, um question six B. Okay.

And then let's just try a few others. So, let's do question eight. Uh, question nine AC question 13 question 15 question 18. Okay, I don't want you to get too theoretical. I'm not going to even give you the enrichment questions even though like they're fun to do.

if you want to try some like you'll be preparing yourself quite well for for next year but I don't really care that much about it so I'm not going to do that exercise 8D I don't I don't care about at all right um it's not useful to us but what I do care about is the projections so 8e exercise 8 okay so we're going to do um going to do question uh three and do question five and then I'm going to do Question eight. Um, question 11 is fine if you want to give it a go. It's like 10 and 11. So that's forgot to b them. Okay.

So So that's going to be it's there. Okay. And then the last bit is 8F. So physical situations. Okay, so this one is where like you might struggle with this one because it's going to like jump throw you right into the physics deep end.

But you're going to start seeing how all this crap has been applicable to physics this whole time because so far it's probably felt like kind of theoretical, right? Um but hopefully this exercise will sort of make it clear to you. Um and why the hell I bothered to do this because in the past this has been a huge problem. So I just want you to be able to do a few of them. Um it's the maths book is pretty good because it kind of tells you how to you know do it to some extent and then 16 and then 17. So you might struggle with this last one a little bit but hopefully not too much.

Okay. So you guys could do all of that by next week we'd be golden. Okay. If you end up getting through all of this with no problem though, then we're like very ready to tackle mechanics. You guys will be bosses at that.

Okay? Okay, because then I can just focus on the physics rather than focusing on the maths, which is kind of the whole idea