WEBVTT mathematics/multivariable-calculus/hovasapian
00:00:00.000 --> 00:00:04.000
Hello and welcome back to educator.com and multi variable calculus.
00:00:04.000 --> 00:00:08.000
Today's topic is going to be very, very important.
00:00:08.000 --> 00:00:12.000
We are going to be talking about parameterizing surfaces and then we are going to begin to discuss the cross product.
00:00:12.000 --> 00:00:18.000
Now we are moving away from 2-dimensions, and we are moving into 3-dimensions.
00:00:18.000 --> 00:00:29.000
We are going to revisit Green's theorem and the fundamental theorem of calculus in its 3-dimensional form, Stoke's theorem, divergence theorem, things like that.
00:00:29.000 --> 00:00:41.000
So, what we are going to do today is lay out the foundations, how to describe a surface, how to think about a surface, this thing called a cross product, what it means, so, let us just go ahead and get started.
00:00:41.000 --> 00:00:48.000
Okay. So, let us go quickly back to the parameterization of a line.
00:00:48.000 --> 00:01:15.000
A line or a curve is a 1-dimensional object, so the x axis, or the y axis, you just need one number... 5, 7, to tell you where you are along that line.
00:01:15.000 --> 00:01:41.000
Well, it is a one dimensional object, so to describe a curve parametrically, we require one parameter. That is it.
00:01:41.000 --> 00:01:53.000
We have been calling that parameter t. So, one dimensional object, a line or a curve, all you require is one parameter to describe that curve parametrically.
00:01:53.000 --> 00:02:06.000
An example would be... example... c(t) = let us say t, t², and t³.
00:02:06.000 --> 00:02:30.000
Okay. We have one parameter, that parameter is t. Well, we have 3-coordinate functions, 1, 2, 3.
00:02:30.000 --> 00:02:52.000
This means it is a curve in 3-space. If I had 7-coordinate functions, this would describe a curve in 7-space, you could not describe it geometrically because we cannot picture a 7-dimensional space, but mathematically it is a perfectly valid object. That is it.
00:02:52.000 --> 00:02:58.000
So, a curve, a 1-dimensional object, requires 1 parameter. I can define it in any dimensional space that I want to.
00:02:58.000 --> 00:03:40.000
Okay, now a surface, which is exactly what you think it is intuitively, a surface is -- actually, let me write one other thing, qualify it one other way -- a surface, whether flat or curved... so if I take a flat piece of paper and curve it, it is a surface but it is still a 2-dimensional object... flat or curved is a 2-dimensional object.
00:03:40.000 --> 00:04:06.000
So, let us see, to describe it parametrically requires -- you guessed it -- two parameters.
00:04:06.000 --> 00:04:18.000
I think I definitely need to write a little bit better here... two parameters. Let us call these parameters t and u.
00:04:18.000 --> 00:04:34.000
As it turns out, the parameters that you choose, it is going to depend on the problem itself, sometimes it is going to be φ θ, sometimes it is going to be xθ, sometimes it is actually going to be the xy that you happen to be dealing with in your function, but just for generic purposes, let us just call them t and u.
00:04:34.000 --> 00:04:45.000
So, a good way to think about this, you know... two parameters, surface... well, you are already familiar with the idea of one parameter and a curve... well, think of a surface this way.
00:04:45.000 --> 00:05:01.000
We have this, this, let us say that, and that. Well, that is 1 curve in the surface, let us say there is another curve in the surface, right?
00:05:01.000 --> 00:05:11.000
If this is a surface, if you hold one point you can follow a single curve, that is one parameter and you can go in another way along that surface, that is another parameter.
00:05:11.000 --> 00:05:22.000
You can think of it as once you -- you know -- sweep out one curve, and then you can go ahead and add another parameter and take that curve and sweep it this way.
00:05:22.000 --> 00:05:30.000
So, you sweep something this way, and then you sweep that curve along this way to form the surface, that is what is going on. That is it.
00:05:30.000 --> 00:05:47.000
Okay, so let us do an example of this one. An example might be s(t,u) = t and u and let us say t² + u².
00:05:47.000 --> 00:05:58.000
In this particular case this happens to be the parameterization of a paraboloid that is going up along the z axis, the standard paraboloid that you are used to.
00:05:58.000 --> 00:06:09.000
So, notice here we have two parameters, we have t and we have u. So, we have 2 parameters.
00:06:09.000 --> 00:06:21.000
Two parameters t and u, and again we have 3 coordinate functions, this one, this one, and this one. Each one a function of t and u.
00:06:21.000 --> 00:06:30.000
Now, granted the u does not show up here, the t does not show up here, the u and t show up here, that does not matter, it is still considered a function of the two variables t and u, t and u, t and u.
00:06:30.000 --> 00:06:40.000
So the three coordinate functions mean the same thing that they did before. This is a surface in 3-space.
00:06:40.000 --> 00:06:49.000
If I wanted to take some function of t and u, but if I had 8 coordinate functions, I would be defining a surface in 8-space.
00:06:49.000 --> 00:06:56.000
Obviously we do not know what the hell that looks like, but there it is. It is possible mathematically.
00:06:56.000 --> 00:07:08.000
So, 3-coordinate functions means a surface in 3-space, and of course that is what we are going to be concerning ourselves with. A 2-dimensional object in a 3-dimensional space.
00:07:08.000 --> 00:07:16.000
In other words, a surface in Euclidian 3-space. Okay.
00:07:16.000 --> 00:07:30.000
So, let us go ahead and just one thing I should say. Recall when we talk about the parameterization of a line, parameterizations are not unique. There is more than one parameterization for the same object.
00:07:30.000 --> 00:07:40.000
So, if I had a paraboloid, I might have one parameterization, somebody else might have another parameterization, in the long run it does not matter. You are going to end up getting the same answer.
00:07:40.000 --> 00:07:49.000
The final answer does not depend on the parameterization. The parameterization is just an artifact. It is something that we use to be able to deal with the object.
00:07:49.000 --> 00:08:00.000
When you integrate things, when you calculate certain mathematical quantities, those numbers are actually the same because they are deeper intrinsic properties of the objects themselves.
00:08:00.000 --> 00:08:10.000
That is what is amazing. So, a parameterization is just a representation. It is not just the object itself. You are just representing the object. It is not unique.
00:08:10.000 --> 00:08:22.000
So, parameterizations are not unique.
00:08:22.000 --> 00:08:33.000
Okay. So, let us just do an example and I think this is the best way to do it. You are going to see lots of parameterizations for lots of different objects. It just depends on what it is that you are dealing with.
00:08:33.000 --> 00:08:43.000
We are just going to throw out some basic ones, so let us just parameterize this here of radius r.
00:08:43.000 --> 00:09:03.000
So, we have a sphere of a given radius r, I am going to go ahead and use the letter p for parameterization. The same we used c for a curve, c(t), I am going to use p, p(t,u)... you can use any letter you want, it does not matter.
00:09:03.000 --> 00:09:17.000
Let us use p for parameterization. If you want you can use s for surface.
00:09:17.000 --> 00:09:48.000
Okay. So, the parameterization for a sphere, as it turns out, t and u is the following. It is rsin(t)cos(u), rsin(t)sin(u), and rcos(t), where t runs from 0 to π, and u runs from 0 to 2π.
00:09:48.000 --> 00:10:04.000
Does this look familiar? It should. It is basically spherical coordinates except now we fixed r, we set a specific radius for a given sphere, and then parameters are t and u, so it is t and u that vary. That is all that is going on here.
00:10:04.000 --> 00:10:17.000
Instead of using t and u here, let us just go ahead and use t and θ because that is what we are accustomed to and that is what... this spherical coordinate is.
00:10:17.000 --> 00:10:40.000
So the parameterization for a sphere, of given radius, is a function of the 2 parameters φ and θ, and it is equal to rsin(φ), cos(θ), rsin(φ)sin(θ), and rcos(φ). There you go.
00:10:40.000 --> 00:10:53.000
That is your parameterization for a sphere. At least it is one of the parameterizations, it is not a unique parameterization... it is probably the one that is used most often because it is very, very easy.
00:10:53.000 --> 00:11:15.000
Now we will do another example, so example 2. Now I am going to give you another parameterization for a sphere. So, another key for the sphere of radius r.
00:11:15.000 --> 00:11:29.000
Okay, well we know that the Cartesian equation for a sphere of radius r is x² + y² + z² = r².
00:11:29.000 --> 00:11:34.000
It is just a generalization in the 3-dimensions for the equation for a circle.
00:11:34.000 --> 00:11:41.000
So, here is what I am going to do, I am going to go ahead and actually solve for z, explicitly.
00:11:41.000 --> 00:11:55.000
When I move everything over, I am going to get the following: z = + or - sqrt(r² - x² -- y²).
00:11:55.000 --> 00:12:00.000
This is actually really, really, really convenient. Here is what I am going to do.
00:12:00.000 --> 00:12:14.000
I am going to say t = x, u = y, then my parameterization in terms of the generic variables, t and u, actually ends up being the following.
00:12:14.000 --> 00:12:24.000
t,u, and r² - t² - u², all under the radical.
00:12:24.000 --> 00:12:54.000
So, this is another parameterization for the sphere, except this time when you are given the sphere in terms of Cartesian coordinates, what you can do -- if you can half solve this implicit equation explicitly for the variable z, you can set the first parameter equal to something, the second parameter equal to the other variable, and you can just put the equation that you solve explicitly for z as your third coordinate function.
00:12:54.000 --> 00:13:00.000
I mean, clearly you do not need the t and the u, they are just generic variables, so why do we not just stick with the x and y?
00:13:00.000 --> 00:13:19.000
So, we can rewrite our parameterization as x and y, well, that is equal to, well x is the first parameter, y is the second parameter and the function itself, the function of x and y when you solve for the third variable, that is the third coordinate function.
00:13:19.000 --> 00:13:26.000
We get r² - x² - y² under the radical.
00:13:26.000 --> 00:13:38.000
This is our other parameterization for the sphere, that is it. 2 parameterizations, they describe the same object, it is not unique. Now, this is true in general.
00:13:38.000 --> 00:13:49.000
It is really very nice. So, it is actually very, very convenient, so that any time you are given an equation in x, y, and z, you automatically have a parameterization.
00:13:49.000 --> 00:14:00.000
So, let us say true in general.
00:14:00.000 --> 00:14:34.000
What is true is the following. If you are given z = to some function of x and y, well, then -- oops, here we go with these lines again -- then, your parameterization of x and y is equal to... well x is your first parameter, and y is your second parameter, and the actual function itself as a function of x and y is your third parameter.
00:14:34.000 --> 00:14:50.000
So, again, you are given a function, you could be given the function explicitly, you know... maybe you are given something like z = sqrt(xy), something like that, that is explicit, or you could be given the function implicitly like this.
00:14:50.000 --> 00:14:57.000
x² + y² = z², in which case if you can solve for z, then great. You can go ahead and use this parameterization.
00:14:57.000 --> 00:15:07.000
If you cannot solve for z, then you are going to have to search for another parameterization. Something with a t and a u, some other variables.
00:15:07.000 --> 00:15:16.000
So, let us do another example here... let me go... let me go to blue, how is that?
00:15:16.000 --> 00:15:28.000
So, example number 3. This time, I am going to parameterize a paraboloid.
00:15:28.000 --> 00:15:40.000
Again, that is just a 3-dimensional parabola, it goes up and around the z axis. Well the first parameterization for a paraboloid happens to be the following.
00:15:40.000 --> 00:15:46.000
I am going to write p₁, and it is going to be the parameters t and θ.
00:15:46.000 --> 00:16:16.000
Well, the parameterization is tcos(θ), tsin(θ), and t². This will give you, as you vary t and you vary θ... let us make sure we have the certain values... t > or = to 0, of course, and θ > or = to 0 and < or = to 2π, so as you vary t and θ, you are going to end up with this surface.
00:16:16.000 --> 00:16:23.000
Again, you are going to be given several different parameterizations.
00:16:23.000 --> 00:16:30.000
You are either going to be told what this is, or you are going to be asked to describe what this is.
00:16:30.000 --> 00:16:40.000
This is something that you just have to spend a little bit of time with. Pick some values of t, pick some values of θ, stop and think about where these points end up in 3-dimensional space.
00:16:40.000 --> 00:16:51.000
That is really the only way to wrap your mind around this geometrically. If you are somebody that absolutely needs pictures in order to make the math come together.
00:16:51.000 --> 00:17:00.000
If not, if the pictures do not matter, then do not worry about it, because ultimately we are just going to be operating on these functions the way we do -- you know -- everything else.
00:17:00.000 --> 00:17:12.000
We are going to be taking derivatives, and second derivatives, and partial derivatives, so if you just want to treat them as mathematical objects that have no picture associated with them, that is not a problem at all. Whatever works best for you.
00:17:12.000 --> 00:17:22.000
Okay. So, that is one parameterization right there. Well, let us go ahead and use our thing that we just discussed a minute ago. Something being true and general.
00:17:22.000 --> 00:17:36.000
The paraboloid, in terms of x and y is given explicitly by the following. x² + y², so if I want another parameterization, let us call this one p₂, and this we will call x and y.
00:17:36.000 --> 00:17:46.000
Well, it is equal to the following. x is the first parameter, y is the second parameter, very, very convenient and of course, x² + y².
00:17:46.000 --> 00:17:50.000
Now, you notice, this is the same thing as earlier in the lesson, when we had t and u.
00:17:50.000 --> 00:17:59.000
We said t, u, t² + u², we do not need t and u, we can just use the variables x and y that we are familiar with.
00:17:59.000 --> 00:18:06.000
So, this is a second parameterization of the paraboloid, and again, when you start doing the math with this, your final answer is going to be the same.
00:18:06.000 --> 00:18:12.000
These are just two representations of the same object. Okay.
00:18:12.000 --> 00:18:35.000
Now, let us do a final example. Example 4, a very, very important example. Something from single variable calculus. Remember when we did surfaces of revolution? We had some graph in the xy plane and we decided to rotate that graph either along the x axis or the y axis.
00:18:35.000 --> 00:18:42.000
Well, in this case, now we are going to talk about how to parameterize some surface described as a surface of revolution.
00:18:42.000 --> 00:19:03.000
So, a surface of revolution and our axis of revolution is going to always be the z axis. So, a surface of revolution around the z axis.
00:19:03.000 --> 00:19:34.000
Okay, now, we will let f be a function of 1 variable, let us go ahead and call that variable x for x > or = some value x₁, < or = some value x₂, like for example if x is between 3 and 7, something like that. Okay.
00:19:34.000 --> 00:20:13.000
Then, the surface of revolution generated by rotating the function z = f(x) around the z axis, is the following.
00:20:13.000 --> 00:20:35.000
Your parameterization is given by p, the two variables, the two parameters are going to be x and θ, always, and you are going to have the first coordinate function is xcos(θ), the second coordinate function is going to be xsin(θ), and the third coordinate function is just going to be your f(x).
00:20:35.000 --> 00:20:48.000
So, what you are doing is you are actually taking the z axis -- oops, let me get rid of this so it does not look like a triangle -- and you are taking x here.
00:20:48.000 --> 00:21:00.000
You can think of the y -- you know -- as going in and out of the page, and into the page, well if you are going to have some function, let us say, something that looks like that, function of one variable, z as a function of x, your normal single variable calculus.
00:21:00.000 --> 00:21:34.000
If you take this and you rotate it around the z axis, now you are going to end up with this flat surface that is above the xy plane. So, what this parameterization does is it gives you the x coordinate... I will do this in red... this arrangement right here, it gives you the x-coordinate of a point on the surface, it gives you the y coordinate of a point on the surface, and it gives you the z coordinate of a point on the surface.
00:21:34.000 --> 00:21:42.000
This is just a way of parameterizing a surface of revolution. That is it. Around the z axis.
00:21:42.000 --> 00:22:11.000
Okay. So, let us do an example. Our specific example will be the following. We will let f(x), or actually let us call it... yea, that is fine, we can call it f(x), let f(x) = sqrt(x)... and let us just say sqrt(x) is > or = 1, and < or = 3.
00:22:11.000 --> 00:22:27.000
Let us go ahead and draw this out. This is going to be something like that, okay? From 1 to 3. This is the x axis, this is the z axis.
00:22:27.000 --> 00:22:48.000
When I take this thing and I rotate it around the z axis, I am going to end up with something, some surface that looks like that... around the z axis, the parameterization for that is the following.
00:22:48.000 --> 00:23:03.000
The parameterization in terms of x and θ = xcos(θ), xsin(θ), sqrt(x). That is it.
00:23:03.000 --> 00:23:18.000
Very, very, very common. Many of the problems that you have will be surfaces of revolution. This will always give you a parameterization. Nothing more, nothing less. Okay.
00:23:18.000 --> 00:23:33.000
Now let us move on and develop a little bit more. I am going to introduce something called the cross product, and we are actually going to use this cross product in our next lesson when we talk about more properties of the surface.
00:23:33.000 --> 00:23:42.000
In particular the tangent plane to the surface, the normal vector to the surface, things like that, but let us go ahead and define it before we do anything else.
00:23:42.000 --> 00:23:59.000
So, let us see, given 2 vectors a and b, well we already define the dot product, the scalar product, right?
00:23:59.000 --> 00:24:35.000
We already defined a ⋅ b, we said that if a has... is a₁, a₂, a₃, and the vector b has coordinates b₁, b₂, and b₃, well we define their dot product as a₁b₁ + a₂b₂ + a₃b₃... we have been dealing with this forever now.
00:24:35.000 --> 00:24:58.000
Now, let us go ahead and define something called the cross product, or the vector product. Now, define -- I will do this in blue, I will go back to blue here -- so define the cross product, also called the vector product.
00:24:58.000 --> 00:25:16.000
The reason it is called the vector product is because you actually get a vector. Now when you multiply two vectors under the cross product operation, you are going to get a vector. Here you multiply this under the dot product operation, you got a scalar, a number. So, vector product.
00:25:16.000 --> 00:25:26.000
I am going to write out the definition, but then I am going to give you a symbolic way of actually solving it because the definition is actually a big unwieldy.
00:25:26.000 --> 00:26:00.000
So given these 2 vectors a and b, a cross b, again, that is why we call it cross... we use the × signal... is equal to a₂b₃ - a₃b₂, that is the first coordinate of the vector, a₃b₁ -a₁b₃, that is the second coordinate, and a₁b₂ - a₂b₁, that is the third coordinate.
00:26:00.000 --> 00:26:40.000
In terms of i,j,k, notation, you get the following... equals a₂b₃ - a₃b₂i + a₃b₁ - a₁b₃j, the unit vector in the y direction, i is the unit vector in the x direction, and that plus the final one, a₁b₂ - a₂b₁k. That is it.
00:26:40.000 --> 00:26:49.000
That is the definition of the cross product. Now, I am going to go ahead and give you a way of doing this symbolically given 2 vectors a and b.
00:26:49.000 --> 00:27:03.000
Okay. Now, what I am about to describe is definitely just symbolic. It is a way of remembering this so that you do not have to remember this, all of these indices 2, 3, 3, 1, you know, 2, 1.
00:27:03.000 --> 00:27:40.000
Okay. a cross b, so a × b is equal to the determinant -- let me go ahead and draw a matrix first before I... okay, we write i, j, and k on the top, we have a₁, a₂, a₃, b₁, b₂, b₃ -- we form this matrix and then we take the determinant of this matrix.
00:27:40.000 --> 00:27:57.000
Let me write it out in determinant notation, i, j, k, a₁, a₂, a₃, b₁, b₂, b₃. This is symbolic.
00:27:57.000 --> 00:28:05.000
When you take a 3 by 3 determinant, you are going to get a number, but again this is just a symbolic way of finding the cross product.
00:28:05.000 --> 00:28:29.000
Let us just do an example here. So, this is example 5. We will let a = the vector (3,1,-5), and we will let b, the vector, be (2,-2,6).
00:28:29.000 --> 00:28:45.000
So, we want to find a × b, well, let us make our little determinant here, so we have i, we have j, and we have k.
00:28:45.000 --> 00:28:53.000
We have (3,1,-5), (2,-2,6), and we are going to take this determinant.
00:28:53.000 --> 00:29:02.000
Again, with determinants and matrices, the biggest problem is going to be algebra, the positive and negative signs. What you are multiplying, there are a whole bunch of numbers floating around so just go slowly.
00:29:02.000 --> 00:29:14.000
Are you going to make mistakes? Yes, you are going to make mistakes. I make them all the time with this stuff, because again, it is just arithmetic. Arithmetic can be tedious and annoying.
00:29:14.000 --> 00:29:27.000
So, let us remember, when we do this we are actually expanding this determinant along the first row. That is why we have the i, j, k, so we do not just expand randomly, the way you would with any determinant.
00:29:27.000 --> 00:29:35.000
You can choose the row or column of your choice to expand along, but here we are definitely expanding along the first row.
00:29:35.000 --> 00:29:51.000
So, let us recall +, -, +, -, +, -, +, -, +, when you are expanding along the first row, your first term, you are going to have 3 terms.
00:29:51.000 --> 00:30:03.000
First term has a + sign, second term has a - sign, third term has a + sign. You have to account for all of this in addition to all of the pluses and minuses of the numbers.
00:30:03.000 --> 00:30:34.000
So, let us go ahead and do this one in red. So, for i, we are going to have the following. We are going to have 1 × 6, so let us actually write all of this out, so 1 × 6 - (-5) × -2, that is going to be i.
00:30:34.000 --> 00:30:50.000
Now, -, I will expand along j. We are going to have 3 × 6 - (-5) × 2, and that is going to be j.
00:30:50.000 --> 00:31:01.000
Then we have +, see this +, -, +, that comes from here, here, and here, and I keep everything separate. Do not do this in your head. Write it all out.
00:31:01.000 --> 00:31:17.000
3 × 6 -- no, this is that way so now I have 3 × -2 -- 3 × -2 - 1 × 2... and this is k.
00:31:17.000 --> 00:31:49.000
Okay. Hopefully I did all of this right. We have 6 - 10, so this is going to be -4i... 3 × 6 is 18, and this is -10, but this is - (-10), so it is + 10, so 18 + 10 is going to be 28... so we are going to have -28j.
00:31:49.000 --> 00:32:02.000
Then this is going to be -6 - 2, this is going to be -8, so it is going to be - 8 + -8, is a -8k. There you go. This is our vector.
00:32:02.000 --> 00:32:30.000
Given (3,1,-5), (2,-2,6), their vector product, their cross product is the vector (-4i,-28j,-8k), or if you prefer this... so the i,j,k stuff tends to be popular with the engineers and physicists in terms of coordinate functions, you get the vector (-4,-28,-8), which is more characteristic of mathematical work.
00:32:30.000 --> 00:33:01.000
Okay. Now, here is the interesting part. a cross b -- let me go back to black ink here -- let us do this, now the vector a cross b, it is a vector, it is perpendicular, it is orthogonal, perpendicular to both a and b.
00:33:01.000 --> 00:33:18.000
Its direction is given by the right hand rule.
00:33:18.000 --> 00:33:54.000
The right hand rule is as follows: so, we have got a cross b, so the order of a vector product is actually very, very important. Now, the first part, your fingers, your 4 fingers, they point in the direction of vector a, the first term in the multiplication.
00:33:54.000 --> 00:34:13.000
2. What you do is you swing your fingers, you curl your fingers, you swing them in or curl them in, swing your fingers in the direction of b. Towards b, the second.
00:34:13.000 --> 00:34:52.000
3. Whichever direction your thumb is pointing, when you actually do that, the direction of your thumb is the direction of a cross b.
00:34:52.000 --> 00:35:07.000
What you have is something like this. So, if you have a vector -- let us say this is that and this is that, let us call this vector a and this is vector b -- well, let us say these, both of these vectors are in the plane right here.
00:35:07.000 --> 00:35:17.000
Well, if I take a, my fingers are that way and I swing them in the direction of b, my thumb is point up.
00:35:17.000 --> 00:35:35.000
As it turns out, the vector a cross b is actually pointing out of the page. Well, there is another vector. If I do the other direction, if I change the order and do b cross a, swing it in the direction of a, now my thumb is pointing down. So that would be down this way.
00:35:35.000 --> 00:35:44.000
So, I am going to say this is a cross b, and this is perpendicular there, and it is perpendicular there.
00:35:44.000 --> 00:36:01.000
In the other direction, that would give me b -- excuse me -- cross a, so in normal multiplication, 2 × 3 is equal to 3 × 6 -- I am sorry, 2 × 3 is equal to 3 × 2, and the dot product is also commutative. Cross product is not commutative.
00:36:01.000 --> 00:36:20.000
You get the vector of the same magnitude, same length, but opposite direction. That is all that is going on here. Given 2 vectors, the first one, swing the fingers towards the direction of the second one, the direction that the thumb is pointing in, that is the direction that the actual a cross b vector is pointing in.
00:36:20.000 --> 00:36:27.000
It is perpendicular to both. That is what is important. Okay. Let us just do an example here.
00:36:27.000 --> 00:36:50.000
Oh, one last thing I should let you know. Remember we had the formula a ⋅ b = norm(a) × norm(b) × cos(angle between them), something like that if this is the angle θ.
00:36:50.000 --> 00:36:56.000
Well, there is a similar formula for the cross product, and it is good to know.
00:36:56.000 --> 00:37:10.000
a × b = norm(a) × norm(b), sure enough × sin(angle between them), so sin(θ).
00:37:10.000 --> 00:37:18.000
This formula is also good to know. Okay. Let us do an example here.
00:37:18.000 --> 00:37:28.000
Okay. Example 6.
00:37:28.000 --> 00:37:45.000
So, let us say we have a is the vector (1,2,1), and let us say b is the vector (1,4,1). Let us go ahead and draw this out just for the heck of it.
00:37:45.000 --> 00:37:59.000
So we have got this vector here, let us call that one b, let us call that one a, so this is vector a, this is vector b, we know that the cross product is going to be a vector that is going to be perpendicular to both.
00:37:59.000 --> 00:38:18.000
It is going to be coming out of the page if in fact -- well, in this particular case these are 2 vectors in 3-space, so we are just going to draw them out like this -- this particular vector which is a cross b, we do not know if that actually comes out of the page or not because we are dealing with 3-space, but it is perpendicular to both a and b.
00:38:18.000 --> 00:38:38.000
So, let us go ahead and do our symbolic calculation. So a cross b = i,j,k, and we have (1,2,1), (1,4,1).
00:38:38.000 --> 00:38:56.000
When we actually do this, expand it along the first row, we end up with following with -2i + 0j -- and I hope that you will confirm this for me because it is very, very possible that I made an arithmetic error -- and in terms of regular coordinates it is (2,0,2).
00:38:56.000 --> 00:39:04.000
That is our vector. The vector itself is (2,0,2), and now let us go ahead and find its norm.
00:39:04.000 --> 00:39:16.000
So, the norm of a cross b, well that is equal to 2² + 0² + 2² all under the radical.
00:39:16.000 --> 00:39:30.000
That equals sqrt(8), so the length of this vector is sqrt(8), its direction is perpendicular to both vectors. It is a vector. It has a magnitude and it has a direction.
00:39:30.000 --> 00:39:35.000
The dot product was a scalar product, it is only a magnitude, it is only a number.
00:39:35.000 --> 00:40:02.000
Okay. So, let us go ahead and find the angle between those 2 vectors. So norm(a) = sqrt(6), because it is 1² + 2² + 1², and norm(b) = sqrt(18), right? 1² + 4² + 1².
00:40:02.000 --> 00:40:21.000
So, now, we have sin(θ) = -- I am going to rearrange that formula -- = a cross b... actually, you know what I am going to go ahead and stop there. I am not going to do that.
00:40:21.000 --> 00:40:41.000
So, that is what is important here... being able to find the cross product given the vectors a and b, realizing that the cross product itself is a vector and the direction of that vector is going to be given by the right hand rule, where you take the first vector a cross b, the first, swing it into the second.
00:40:41.000 --> 00:41:02.000
The direction that the thumb is pointing, that is the direction. The first direction is b cross a, and you can find the norm of the vector like you would the norm of any other vector, and it also turns out if I draw a little parallelogram... complete the parallelogram for the vectors a and b.
00:41:02.000 --> 00:41:16.000
This particular parallelogram that is spanned by the vectors a and b, that actually happened... this cross product, the magnitude of the cross product happens to be the area of that parallelogram.
00:41:16.000 --> 00:41:22.000
Of course we will talk more about that next time when we discuss surface area and things like that.
00:41:22.000 --> 00:41:27.000
Okay, so that is it for today's lesson, parameterizing surfaces and cross products.
00:41:27.000 --> 00:41:29.000
Thank you for joining us here at educator.com, and we will see you next time.