Everyone welcome back. This is our lecture on the Dot Product in the Cross Product and with both of these attempt to do is chart to define some sort of multiplication on vectors. Will see why both sort of fail in their own regard and why he sorted need two of ' to try to get at since idea of multiplication of vectors. So first off let's define the Dot Product. We're going to start with some vector A. We're going to just give it some generic A1 components here A2 and A3. And then let's also start with some other vector B. And let's give it some generic components B1, B2 and B3 great. We then define the Dot Product between the two vectors appropriately written with a dot to be the sum of the product of a components. What does that mean? You take the first components multiplying together so A1 times B1 and he added to the product of the next components to an. You add that to the product of the 3rd component. So A3 and B3. This is called the Dot Product. It is important to note that this is sum number, When I multiply these 2 vectors, I'm not going to work multiply, but you can think of it this way with little Dot. It's important to read this when you do read this you say Dot to stress that it is not multiplication. We reserve multiplication for numbers and we say Dot Product for vectors, but this is a number and that's part of the reason why it's not a great way to multiply, you have this wonderful vector with the force magnitude direction. You have another vector with the magnitude direction. You do this operation. You get back a little number. You sort of lose the structure of vectors. Let's do a quick example just to see this with some real numbers here. So let's take 1 2 3 again. Bonus points for originality here, 4 5 6. Why not, Okay, I want a Dot Product, so students make mistake of handy back a vector. It's important. Realize this is going to be a number if you don't give me back a number here, something went amiss anyway. What do we do? We take 1 and multiply it by 4. We take 2 and multiply it by 5 and we take 3 and multiply it by 6, clean it up 4 plus 10 plus 18, 28 plus 4 32 is a couple of properties of Dot Products that I think it's easy to see if you take the zero vector zero vectors. The vector with all zeros in its components and you Dot it with any other vector. You will always get back to zero vector so anything Dot the zero vector. Is always zero and here's the relation to the last video. If I take a vector and dot it with itself. So I take the Dot product of vector in itself. You get back the components of itself squared. So like A 1 squared plus A2 squared plus A3 squared, that should look a little bit familiar. We saw this when we took the length, only the length had a square root. So to get rid of it we will square it so the vector Dot itself is the length squared this relationship. Between Dot Products and lengths. And one more property. That if I take a vector, A in vector B and I do a Dot Product, then the order does not matter. So you can take the data. You can see that from the formula it doesn't matter which one comes first, so this is commutative. We say Dot Product is commutative. Another special thing about the Dot Product is that it has an alternate formula. This is like version number 2, so another way to look at the Dot Product if I have A dot B, this is the magnitude of A times the magnitude of B. Times the cosine of Theta. Where Theta is the angle between the 2 vectors. So if I have A and B floating around where they live till always be a little angle Theta. Between the two vectors, notice how I read this. If I have A, I said magnitude of A times the magnitude B. If they realize that the magnitude is a number, this is the length of A times like to B. So I really want to say the A dot B is equal to the magnitude of a time dimension to be times cosine of the angle between them. Working this out as a unpleasant sort of. Journey through trig, which is I guess we'll skip some of that for now, but you can believe me that this is the formula if you want to investigate a quick search on Dot Products will show you how this formula is derived, but one thing to note is that it just gives an alternate form of it, and I still to this day whenever I see D for Dot Product and I always say, well, I notice that magnitudes because I knew numbers and then is it sign or cosine. Well so it's DC. I would say like Washington DC is my little mnemonic to help remember that the alternate formula for the Dot Product. Involves a cosine and what's nice about this formula is that you can use it to rearrange. So for example, find the angle between the Vector X which is 4 zero and two. And the vector Y which is 2- 1 zero. So you can use as you can rearrange things to find the angle between two vectors. This turns out to be a useful thing. So how do we rearrange some things here? Well, I can solve for cosine of Theta and get that that as a dot B divided by the product of the magnitudes. So to get all this information, which I'll do in a second. And then one last piece you can solve for Theta by using good old cosine inverse or arc cosine. So a dot B divided by the magnitude of a times the magnitude of being. So this is a nice little formula that comes from the alternate version. The alternate formula of the Doc product. So if I want to just do an example just so you can see some numbers behind this, let's go ahead and solve this thing. So I guess I call them X&Y, but it doesn't matter, so we take the dot product like. This is a think of this as b. So first let's take the dot product. So let's find a dot. If you want pause video, work it out. Just practice but she get 8 * 0 * 0. Plus 0 + 0, so of course that's just eight. So that was good ol' 8.b want the magnitude of a. I'm going to need that number from my denominator. That's the square root of 4 squared plus 0 squared +2 squared, so it's 16 + 4. That's for 20, and then I want the magnitude of the second vector. And that's the square root of 2 squared plus minus one squared plus 0 squared. That's 4 + 1 good old root 5. Alright, let's put it all together. So I have Theta will equal arc cosine of may not be. We said that was 8 divided by the magnitudes put together. So that's root 20 times. Root 5 cosine of eight over root 100, which of course is arc cosine of 8 over 10 or 4/5. Which running over the Calculator is about 6.64? Let's say 4 radians do most things in radiant, so let's leave it in radians. But you can find the angle between two vectors. You can imagine this pretty important one last definition if I could squeeze it here in the bottom of this slide. Is if the angle between two vectors is π over 2 or 90 degrees. Of course, then we say that the vectors. Are orthogonal, so think about this for a second. You know this is another word. If I say to you two things meet at a right angle π over 2 or 90 degrees prior. You would say that there perpendicular perpendicular is usually reserved for lines. Also, it's specific to like the plane. If I have vectors inside of our three were going to say that they are orthogonal orthogonal vectors. So any two vectors that meet at a right angle orthogonal? So for example, perhaps I&J is a nice little example, but there are lots of ' all right? So new words, new formulas for you, and let's give the second type of way to multiply vectors called across product. So the definition we're going to define the cross product, and again we read this as a cross be. We don't say times, times, is for numbers a Crosby, and here's what I'm going to write. It's going to put I. Jason makes sense in a second, so bear with me. I JK my vector eh EH1A2 and A3 my vector BB1B2 and B3. Whenever the numbers are whatever the components are with the formula down general. And then we'll do a product then will do an example. And it's going to be found by taking I times we do like a little X down here 8 two times B 3 minus a three times B2. So we do like a little X down here and then we move over to J. Now this is important it's minus J. So it goes plus for I minus for JI&J are your standard vectors. We do the same thing so we make a little X but you imagine that the row and column would J is removed so I'm only looking at. A1 B 3- 8 three B1 and last but not least. Is K same idea? We imagine that the column with K is removed, so we have this little box and the first 2 columns and that's we do our little X A1. B2 minus. A to B1 it's a nasty formula, but what ends up happening is you start following the pattern. More so then memorizing the formula, so let's just jump to an example with some actual numbers. So let's say that the vector a is 1, 2, and a 0, and then will say the vector b will be 0, 3, 1. Let's find their cross product, so a cross b, so how's it work? You do i, j, and k, give yourself some room. And we throw a across the middle so 1, 2, 0, and b goes last but not least along the bottom row, so 0, 3, 1. Okay, here we go, I take i the vector i, and now I imagine it taking my finger cover up the first column and I do my x pattern, with the remaining two columns so 2. So it's like 2 * 1- 0 * 3 like and you should draw a little x with your fingers you do this, so 2- 0, which of course is just 2. Be careful -j, again, imagine the middle row is completely gone, you don't even see it, so you're looking at 1, 0 and 0, 1. Do your ex pattern, so it's 1 * 1- 0 * 0. That's a fancy way to say 1 of course, and then + k. Same thing, so imagine that third row, third column is gone and we're just looking at the first four numbers. So I have to my ex pattern 1* 3- 2 * 0 is just 3, so put it all together and you have i * 2- j * 1 + k * 3. Remember i, j, and k there just your place holders for your components, so you have 2- 1 and 3. Supporting to notice that when you have a cross product, the answer is a vector. This is one of the things that distinguishes it from the dot product. If I take a dot product of two vectors, I get back a scalar. If I take the cross product of two vectors, I get back a vector. They say, well, this is great, right? Shouldn't this be like the best thing? Shouldn't this be the multiplication? Well, not so much, and you'll see why in a second. So let's look at a vector dot itself, so let's set this up with called components a1, a2, and a3. So I do i, j, k, a1, a2, a3 whatever the numbers are, and a1, a2, a3. So I repeat, the same rows, all right, let's go through the formula. So I take i little fast this time, and I look at a2 * a3- a2 * a3. So I'm doing my x pattern over here, and you notice that a2, a3- a2, a3 there exactly the same, so that's going to be 0. Then we'll do our- k and- k is a1 * a3, a2 sorry,- a2 * a1. Well wait a minute, that's the same as well. And then last but not least is j. I leave this, you can see it immediately was going to happen. You get a1 * a3- a1 * a3, and that's another 0. So all of a sudden you get back the vector, 0, 0, 0, better known as the zero vector. So if I take the cross product for any factor of itself, you always get back to zero vector. Isn't this is not great in terms of if, trying to cook up some multiplication property of vectors, and again note the difference versus the dot product. If I took a dot product of a vector with itself, I get this relation with the length turns out to be the length squared. If I take the cross product of a vector with itself, I get the zero vector back, when I take two vectors and I take their cross product, I get some new vector back, will call it c. So the question might ask is well, what's the relationship between the given two vectors and the new vector that I find. So let's draw a little picture, if I have some a and some b vectors, well it turns out that the cross product is going to be the one that sort of forms the third dimension, the third axis. This is going to be a cross b and I drew it downwards for reason, because if you have a sort of I guess, further north on the screen then b it points downward. However, if I drew the same picture, if I rotate the picture and put a, sort of lower than b, then the picture actually goes up the cross product, a cross b goes upward. They call this the right hand rule, if you take your hand and make it exactly straight, like you're going to high five someone. And you put your pinky along the direction of a, your and your fingers curl towards b then your thumb. Points in the direction of a cross b. So that property, it turns out that this vector, they form a right angle, not just with one of the vectors, but with both of them as well. Now, how can I confirm this? Well, I'd have to look at, how do you know the two vectors meet at a right angle, I have to look at the dot product. So you can check this, you can look at the dot product of c dot a. This would be, order doesn't matter inside a dot product, so you can look at a dot a cross b. It's a little bit of algebra, but take a little bit of your page, but you can check that this is 0. And for all the same reasons, if you took the cross product dot the vector itself, again, work this out, maybe pause the video and prove yourself this is true, you get back 0. Not 0 the vector, it's a dot product, so 0 the number, 0 the number. So when you think of a picture, if you ever need something that is orthogonal or perpendicular to two vectors, and this is going to be something we're going to want later, then we are going to start taking cross products. As the cross product is another vector, so let's just draw the picture here one more time, so I have a and b and it's a cross product. a cross b forms nice little right angles with everything. If I have this orthogonal vector, this cross product, I might want to know, well, how's the length related to the original lengths? So let's look at the absolute value here. Again, through another exercise in trig, which I don't really want to go through, but if you're dying to do it, you can try it. It turns out to be similar to the formula for the dot product, the alternate formula, but the length of the cross product is the length of a times the length of b, and again, those are both numbers, so I'm going to say times. And then times sine of theta, sine of theta. So this is a nice little formula to keep in mind as you go through things. The length of this just is a little, I'm going to do one more picture here for you. If I gave you two vectors, a and b, and I said, what is the sine theta part of it, it turns out it's the height of the parallelogram if you have theta as the angle between them. So you can think of it as a sine theta, you're forming this sort of base times height. The length of b is your base, and then your height is a, length of a sine theta. And so what this gives, this gives the area of the parallelogram with sides a and b. So here's a surprising little area formula that they don't teach you, of course, in good old geometry. But if you're ever looking for the side for the area of parallelograms, this is a nice little formula to know. It completely finds all the space that this shape takes up. In addition to just finding the area of a parallelogram, which might be nice to do, you can also use this to show when a is parallel to b, why? If I have two vectors that are parallel to each other, let's just say they go in the same direction, then the parallelogram that they will form has area 0. I can't draw a parallelogram with some area in here. So what you can do is you can show that they're parallel if and only if the area of the parallelogram they form, again, that's the magnitude of the cross product, would be 0. So you can compute the cross product, that's another vector, compute its length, we know how to do that. Then if you show that that's 0, you are now proving, you're showing that these two vectors are parallel. Before I move on to an example, and maybe I should do this, but at some point, there's just a lot of addition and multiplication, it's not super exciting. But one of the differences also about dot product, and I'll just put a star next to this, a little off topic here, but a dot b is b dot a, right? The dot product, the order does not matter, but if you do a cross product because of the right hand rule, you get that the orthogonal vector that results points in the different direction if you switch the order. So it's very important to realize that the cross product is not commutative. Order matters when you're doing cross products. So just be mindful of the order. Just be careful if you're trying to do some algebra, don't switch things up a little bit here. All right, with that being said, just wanted to mention that, that's an important property of the two, let's jump to our example. So let's find a non 0 unit vector orthogonal to the plane through the points P, which is 1, 0, 0, Q, which is 0, 2, 0, and R, which is 0, 0, 3. All right, one more time, find the non 0 unit vector. So we want a sum vector length 1, orthogonal to some plain through these three points. I like to do whenever I get a question with a lot of information, is trying to draw it out, we're in space. I'm not going to draw anything to scale, but I have some point P, have some point Q and I have some point R. And there's a plane that contains them, okay, fine and I want some vector that is perpendicular to this plane. I want a unit vector unit vectors. I can always get a unit vector by normalizing at the end, so I'm not really worried about finding unit vector right now. The question is how do I get, an orthogonal vector first? So let's break this up in a couple parts, if you want to pause the video and try to work this out, this would be a really good one to do it. So you can put all your formulas together. Okay ready, so step one. Let's create a couple of vectors on the plane. Obviously we wanted something orthogonal, that screams cross product. We're going to use cross product, but I don't have any vectors to begin with. I just have points. Well, how do you find a vector between two points? What is the vector from P to Q? We said this before, in last video we do head minus tail, so this is 0, 2, 0 minus 1,0,0. 0 minus 1 minus 1, 2 minus 0 is 2 and 0 minus 0 is 0. And then let's form the vector P to R, same thing. Head minus tail what matters here be careful, 0,0,3 minus 1,0,0. And that's the vector should wrangle brackets for all these vectors here, 0- 1 is -1, 0 minus 0 is 0 and 3 minus 0 is 3. Great. Now I can form the cross product, between these two vectors, so step one was get the vector. Step two is let's take a cross product. So PQ cross PR, it doesn't matter, and honesty which point you pick is like your vertex. We could have absolutely done Q and R to P Q. You will get the same answer or there's more than one answer here will get one of the answers. So as long as you following these steps, you're doing great. Okay, so let's do our cross product. Let's set it up, so we do I, we do J, we do K across the top row. We then grab our vectors we have -1,2,0, -1,0,3. We go a little quickly here since we've done these a few times so I. So you can follow with me, 6 minus 0, so let's just 6, -J, minus sign -3- 0- 3. So those two negatives become positive in a second, and then plus K, 0,--2. Right, you subtract off the negative, so that's just 2, and you get back the vector, 6, 3. It's probably fine vector, so let's maybe I'll call if so how to keep writing PQ cross PR. But it's a perfectly fine vector, so I have found a non zero vector that's orthogonal to the plane. I've done almost everything this question has asked. The problem is I haven't found a unit vector. So given a vector, how do we find a unit vector? We want to normalize. To normalize, we're going to use a formula that the unit vector, is 1 over the length of V times V. So do some scalar multiplication, to do this formula I need the length of V, so that's the square root of each component squared. And then add it up, so it's the square root of 6 squared, plus 3 squared, plus 2 squared, 36 plus 9 plus 4. Well, that's the square root of 49, better known as seven, and so therefore my final answer. My final vector is therefore going to be one seventh times the vector, I started with 6, 3, 2. You can leave your answer this way, most of the time though they do distribute and you write it as fractions six sevenths, three sevenths and two sevenths, that is a perfectly good answer. Remember, there's more than one, but this one is perfectly fine. If you got that and then pause the video, great job there. One more thing to talk about and it's called, the triple scalar product. Or sometimes called the scalar triple product, but it is a specific vector, will define this thing. And the specific vector is going to be a .b cross c. So you're given three vectors a,b and c, you take the cross product two of them and you take that vector across product is vector and you dot it with the other one. That's why this triple seeing three of them, and then this is key, this is why it's called the scalar product, so this gives back a number. This is a real number, why? because I take a vector another vector do a dot product I get back a number. This expression is just called the scalar triple product or triple scalar product or heard it both ways. So why is it useful? Well, it gives us a number, so what's the meaning of this number? Let's see if I could do a nice picture here for you. So, if I were to draw sort of like, it's not a cube, because you don't know that everything is the same, but it's a 3D parallelogram. It's called the parallelepiped. It turns out and you got to magine that the sides are a, there's b, and then we'll put c along the back here. So, if I take these three vectors and imagine that they create or span a parallelepiped, this 3D cube sort of thing, then the volume of this, the volume turns out to be related to this triple scalar product. Now, I'm going to write it down. There's one thing missing. Let's see if you can figure out what it is. So, the volume is this number, with this slide catch, remember dot product can be negative. You can certainly take two vectors and cook them up so the dot product is negative, volume is always positive. So, I want the absolute value to just make sure that they match, I don't want to get back in negative volume. This is the use of the triple scalar product, at least one of ' is to get volume. So just area form, those are important, so our volume formulas as well. And just as a side little note here, just like before, we industry wanted zero. So if the volume is 0, right, so if this triple scalar product or scalar triple product, whatever you want to call if this thing turns out to be 0, what does it tell you about a, b and c will it says, well, they have no volume. That means that one of ' has to sort of can't be pointing up in the opposite direction. It just must mean to say it a little more nicer is that a, b and c in the same plane or coplanar. So they ask you hear these three vectors in the same plane one way to do it, t there's a few ways to do it, but one way to do it is compute the scalar triple product. Show the to 0 and that gets you that they are in fact coplanar. All right you will do examples of these with some real numbers and put all these things together. Practice your formulas for dot products and cross products in the notes and quiz that follows. So just keep these formulas down another starting to add up. Just make yourself formula sheet as you go through these sections and then reference it as needed to wrap up this chapter. All right, great job on this one. Will see you next time.
