For more information, please see full course syllabus of Multivariable Calculus

For more information, please see full course syllabus of Multivariable Calculus

### Parameterizing Surfaces & Cross Product

^{2}− 3x − 1 in one parameter t.

- A curve in two dimensions is described through two components, C(t) = (C
_{1},C_{2}). - To describe the curve we let x = t.
- Substituting yields f(t) = 4(t)
^{2}− 3(t) − 1 = 4t^{2}− 3t − 1.

^{2}− 3t − 1).

^{2})/4] + [(y

^{2})/16] for x > 0 and y > 0 in one parameter t.

- A curve in two dimensions is described through two components, C(t) = (C
_{1},C_{2}). - To describe the curve we let x = t.
- Substituting yields 1 = [(t
^{2})/4] + [(y^{2})/16]. We solve for y in order to have our second component for the curve. - Now, 16 = 4t
^{2}+ y^{2}and so y = ±√{16 − 4t^{2}} = ±2√{4 − t^{2}} . Since x > 0 and y > 0 we take the positive square root.

^{2}} ).

- The parametrization of a sphere with radius R is P(f,θ) = (Rsinfcosθ,Rsinfsinθ,Rcosf) for 0 ≤ f ≤ π and 0 ≤ q ≤ 2π.

^{2}= 4 using two parameters t and u.

- A surface in two dimensions is described through three components, C(t) = (C
_{1},C_{2},C_{3}). - To describe the surface we let x = t and y = u.
- Substituting yields 2z + tu − t
^{2}= 4. We solve for z in order to have our third component for the surface. - Now, 2z = 4 + t
^{2}− tu and so z = 2 + [(t^{2}− tu)/2]

^{2}+ y

^{2}− y using one parameter t.

- A curve in three dimensions is described through three components, C(t) = (C
_{1},C_{2},C_{3}). - To describe the curve we let x = sin(t) and y = cos(t).
- Substituting yields z = sin
^{2}(t) + cos^{2}(t) − sin(t) = 1 − sin(t).

^{2}, 0 ≤ x ≤ 1 around the z - axis.

- The parametrization for a surface of revolution around the z - axis is P(x,θ) = (xcosθ,xsinθ,f(x)) for x ∈ [a,b], θ ∈ [0,2π]

^{2}) for x ∈ [0,1], θ ∈ [0,2π].

i) →a = (0,0,1), →b = (0,0,1)

- To find the cross product of vectors →a and →b, →a ×→b, we compute det(

), that is, the determinant of the matrix formed by the vectors →i, →j, →k and the components of →a and →b.→i →j →k a _{1}a _{2}a _{3}b _{1}b _{2}b _{3} - Then det(

) = det(→i →j →k a _{1}a _{2}a _{3}b _{1}b _{2}b _{3}

) = (0 − 1)→i − (0 − 1)→j + (0 − 0)→k = − 1→i + 1→j + 0→k. Recall that →i = (1,0,0), →j = (0,1,0) and →k = (0,0,1).→i →j →k 0 0 1 0 0 1

ii) →a = (1,0,0), →b = (0,0,1)

- To find the cross product of vectors →a and →b, →a ×→b, we compute det(

), that is, the determinant of the matrix formed by the vectors →i, →j, →k and the components of →a and →b.→i →j →k a _{1}a _{2}a _{3}b _{1}b _{2}b _{3} - Then det(

) = det(→i →j →k a _{1}a _{2}a _{3}b _{1}b _{2}b _{3}

) = (0 − 0)→i − (1 − 0)→j + (0 − 0)→k = 0→i − 1→j + 0→k. Recall that →i = (1,0,0), →j = (0,1,0) and →k = (0,0,1).→i →j →k 1 0 0 0 0 1

- To find the cross product of vectors →a and →b, →a ×→b, we compute det(

), that is, the determinant of the matrix formed by the vectors →i, →j, →k and the components of →a and →b.→i →j →k a _{1}a _{2}a _{3}b _{1}b _{2}b _{3} - Then det(

) = det(→i →j →k a _{1}a _{2}a _{3}b _{1}b _{2}b _{3}

) = (6 − 30)→i − (4 − ( − 5))→j + ( − 12 − ( − 3))→k = − 24→i − 9→j − 8→k. Recall that →i = (1,0,0), →j = (0,1,0) and →k = (0,0,1).→i →j →k − 2 3 5 − 1 6 2

- To find the cross product of vectors →a and →b, →a ×→b, we compute det(

), that is, the determinant of the matrix formed by the vectors →i, →j, →k and the components of →a and →b.→i →j →k a _{1}a _{2}a _{3}b _{1}b _{2}b _{3} - Then det(

) = det(→i →j →k a _{1}a _{2}a _{3}b _{1}b _{2}b _{3}

) = ( [2/(√{15} )] − 0 )→i − ( [1/(√{15} )] − 0 )→j + ( − [1/(√{15} )] − [2/(√{15} )] )→k = [2/(√{15} )]→i − [1/(√{15} )]→j − [3/(√{15} )]→k. Recall that →i = (1,0,0), →j = (0,1,0) and →k = (0,0,1).→i →j →k 1 \mathord / phantom 1 √5 √5 2 \mathord / phantom 2 √5 √5 0 1 \mathord / phantom 1 √3 √3 − 1 \mathord / phantom 1 √3 √3 1 \mathord / phantom 1 √3 √3

- To find the cross product of vectors →a and →b, →a ×→b, we compute det(

), that is, the determinant of the matrix formed by the vectors →i, →j, →k and the components of →a and →b.→i →j →k a _{1}a _{2}a _{3}b _{1}b _{2}b _{3} - So det(

) = det(→i →j →k a _{1}a _{2}a _{3}b _{1}b _{2}b _{3}

) = ( − sinθ− cosθ)→i − ( − cosθ− sinθ)→j + (cos→i →j →k cosθ sinθ 1 sinθ cosθ − 1 ^{2}θ− sin^{2}θ)→k = − (sinθ+ cosθ)→i + (sinθ+ cosθ)→j + (cos^{2}θ− sin^{2}θ)→k.

^{2}θ− sin

^{2}θ)).

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Parameterizing Surfaces & Cross Product

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Parameterizing Surfaces
- Describing a Line or a Curve Parametrically
- Describing a Line or a Curve Parametrically: Example
- Describing a Surface Parametrically
- Describing a Surface Parametrically: Example
- Recall: Parameterizations are not Unique
- Example 1: Sphere of Radius R
- Example 2: Another P for the Sphere of Radius R
- This is True in General
- Example 3: Paraboloid
- Example 4: A Surface of Revolution around z-axis
- Cross Product

- Intro 0:00
- Parameterizing Surfaces 0:40
- Describing a Line or a Curve Parametrically
- Describing a Line or a Curve Parametrically: Example
- Describing a Surface Parametrically
- Describing a Surface Parametrically: Example
- Recall: Parameterizations are not Unique
- Example 1: Sphere of Radius R
- Example 2: Another P for the Sphere of Radius R
- This is True in General
- Example 3: Paraboloid
- Example 4: A Surface of Revolution around z-axis
- Cross Product 23:15
- Defining Cross Product
- Example 5: Part 1
- Example 5: Part 2 - Right Hand Rule
- Example 6

### Multivariable Calculus

### Transcription: Parameterizing Surfaces & Cross Product

*Hello and welcome back to educator.com and multi variable calculus.*0000

*Today's topic is going to be very, very important.*0004

*We are going to be talking about parameterizing surfaces and then we are going to begin to discuss the cross product.*0008

*Now we are moving away from 2-dimensions, and we are moving into 3-dimensions.*0012

*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.*0018

*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.*0029

*Okay. So, let us go quickly back to the parameterization of a line.*0041

*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.*0048

*Well, it is a one dimensional object, so to describe a curve parametrically, we require one parameter. That is it.*0075

*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.*0101

*An example would be... example... c(t) = let us say t, t ^{2}, and t^{3}.*0113

*Okay. We have one parameter, that parameter is t. Well, we have 3-coordinate functions, 1, 2, 3.*0126

*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.*0150

*So, a curve, a 1-dimensional object, requires 1 parameter. I can define it in any dimensional space that I want to.*0172

*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.*0178

*So, let us see, to describe it parametrically requires -- you guessed it -- two parameters.*0220

*I think I definitely need to write a little bit better here... two parameters. Let us call these parameters t and u.*0246

*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.*0258

*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.*0274

*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?*0285

*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.*0301

*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.*0311

*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.*0322

*Okay, so let us do an example of this one. An example might be s(t,u) = t and u and let us say t ^{2} + u^{2}.*0330

*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.*0347

*So, notice here we have two parameters, we have t and we have u. So, we have 2 parameters.*0358

*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.*0369

*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.*0381

*So the three coordinate functions mean the same thing that they did before. This is a surface in 3-space.*0390

*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.*0400

*Obviously we do not know what the hell that looks like, but there it is. It is possible mathematically.*0409

*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.*0416

*In other words, a surface in Euclidian 3-space. Okay.*0428

*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.*0436

*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.*0450

*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.*0460

*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.*0469

*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.*0480

*So, parameterizations are not unique.*0490

*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.*0502

*We are just going to throw out some basic ones, so let us just parameterize this here of radius r.*0513

*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.*0523

*Let us use p for parameterization. If you want you can use s for surface.*0543

*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 pi, and u runs from 0 to 2pi.*0557

*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.*0588

*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.*0604

*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.*0617

*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.*0640

*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.*0653

*Okay, well we know that the Cartesian equation for a sphere of radius r is x ^{2} + y^{2} + z^{2} = r^{2}.*0675

*It is just a generalization in the 3-dimensions for the equation for a circle.*0689

*So, here is what I am going to do, I am going to go ahead and actually solve for z, explicitly.*0694

*When I move everything over, I am going to get the following: z = + or - sqrt(r ^{2} - x^{2} -- y^{2}).*0701

*This is actually really, really, really convenient. Here is what I am going to do.*0715

*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.*0720

*t,u, and r ^{2} - t^{2} - u^{2}, all under the radical.*0734

*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.*0744

*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?*0774

*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.*0780

*We get r ^{2} - x^{2} - y^{2} under the radical.*0799

*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.*0806

*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.*0818

*So, let us say true in general.*0829

*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.*0840

*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.*0874

*x ^{2} + y^{2} = z^{2}, in which case if you can solve for z, then great. You can go ahead and use this parameterization.*0890

*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.*0897

*So, let us do another example here... let me go... let me go to blue, how is that?*0907

*So, example number 3. This time, I am going to parameterize a paraboloid.*0916

*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.*0928

*I am going to write p _{1}, and it is going to be the parameters t and θ.*0940

*Well, the parameterization is tcos(θ), tsin(θ), and t ^{2}. 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 2pi, so as you vary t and θ, you are going to end up with this surface.*0946

*Again, you are going to be given several different parameterizations.*0976

*You are either going to be told what this is, or you are going to be asked to describe what this is.*0983

*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.*0990

*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.*1000

*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.*1011

*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.*1020

*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.*1032

*The paraboloid, in terms of x and y is given explicitly by the following. x ^{2} + y^{2}, so if I want another parameterization, let us call this one p_{2}, and this we will call x and y.*1042

*Well, it is equal to the following. x is the first parameter, y is the second parameter, very, very convenient and of course, x ^{2} + y^{2}.*1056

*Now, you notice, this is the same thing as earlier in the lesson, when we had t and u.*1066

*We said t, u, t ^{2} + u^{2}, we do not need t and u, we can just use the variables x and y that we are familiar with.*1070

*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.*1079

*These are just two representations of the same object. Okay.*1086

*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.*1092

*Well, in this case, now we are going to talk about how to parameterize some surface described as a surface of revolution.*1115

*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.*1122

*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 x _{1}, < or = some value x_{2}, like for example if x is between 3 and 7, something like that. Okay.*1143

*Then, the surface of revolution generated by rotating the function z = f(x) around the z axis, is the following.*1174

*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).*1213

*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.*1235

*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.*1248

*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.*1260

*This is just a way of parameterizing a surface of revolution. That is it. Around the z axis.*1294

*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.*1302

*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.*1331

*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.*1347

*The parameterization in terms of x and θ = xcos(θ), xsin(θ), sqrt(x). That is it.*1368

*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.*1383

*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.*1398

*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.*1413

*So, let us see, given 2 vectors a and b, well we already define the dot product, the scalar product, right?*1422

*We already defined a · b, we said that if a has... is a _{1}, a_{2}, a_{3}, and the vector b has coordinates b_{1}, b_{2}, and b_{3}, well we define their dot product as a_{1}b_{1} + a_{2}b_{2} + a_{3}b_{3}... we have been dealing with this forever now.*1439

*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.*1475

*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.*1498

*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.*1516

*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 a _{2}b_{3} - a_{3}b_{2}, that is the first coordinate of the vector, a_{3}b_{1} -a_{1}b_{3}, that is the second coordinate, and a_{1}b_{2} - a_{2}b_{1}, that is the third coordinate.*1526

*In terms of i,j,k, notation, you get the following... equals a _{2}b_{3} - a_{3}b_{2}i + a_{3}b_{1} - a_{1}b_{3}j, the unit vector in the y direction, i is the unit vector in the x direction, and that plus the final one, a_{1}b_{2} - a_{2}b_{1}k. That is it.*1560

*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.*1600

*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.*1609

*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 a _{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3} -- we form this matrix and then we take the determinant of this matrix.*1623

*Let me write it out in determinant notation, i, j, k, a _{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}. This is symbolic.*1660

*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.*1677

*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).*1685

*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.*1709

*We have (3,1,-5), (2,-2,6), and we are going to take this determinant.*1725

*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.*1733

*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.*1742

*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.*1754

*You can choose the row or column of your choice to expand along, but here we are definitely expanding along the first row.*1767

*So, let us recall +, -, +, -, +, -, +, -, +, when you are expanding along the first row, your first term, you are going to have 3 terms.*1775

*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.*1791

*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.*1803

*Now, -, I will expand along j. We are going to have 3 × 6 - (-5) × 2, and that is going to be j.*1834

*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.*1850

*3 × 6 -- no, this is that way so now I have 3 × -2 -- 3 × -2 - 1 × 2... and this is k.*1861

*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.*1877

*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.*1909

*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.*1922

*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.*1950

*Its direction is given by the right hand rule.*1981

*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.*1998

*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.*2034

*3. Whichever direction your thumb is pointing, when you actually do that, the direction of your thumb is the direction of a cross b.*2053

*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.*2092

*Well, if I take a, my fingers are that way and I swing them in the direction of b, my thumb is point up.*2107

*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.*2117

*So, I am going to say this is a cross b, and this is perpendicular there, and it is perpendicular there.*2135

*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.*2144

*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.*2161

*It is perpendicular to both. That is what is important. Okay. Let us just do an example here.*2180

*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.*2187

*Well, there is a similar formula for the cross product, and it is good to know.*2210

*a × b = norm(a) × norm(b), sure enough × sin(angle between them), so sin(θ).*2216

*This formula is also good to know. Okay. Let us do an example here.*2230

*Okay. Example 6.*2238

*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.*2248

*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.*2265

*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.*2279

*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).*2298

*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).*2318

*That is our vector. The vector itself is (2,0,2), and now let us go ahead and find its norm.*2336

*So, the norm of a cross b, well that is equal to 2 ^{2} + 0^{2} + 2^{2} all under the radical.*2344

*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.*2356

*The dot product was a scalar product, it is only a magnitude, it is only a number.*2370

*Okay. So, let us go ahead and find the angle between those 2 vectors. So norm(a) = sqrt(6), because it is 1 ^{2} + 2^{2} + 1^{2}, and norm(b) = sqrt(18), right? 1^{2} + 4^{2} + 1^{2}.*2375

*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.*2402

*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.*2421

*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.*2441

*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.*2462

*Of course we will talk more about that next time when we discuss surface area and things like that.*2476

*Okay, so that is it for today's lesson, parameterizing surfaces and cross products.*2482

*Thank you for joining us here at educator.com, and we will see you next time.*2487

1 answer

Last reply by: Professor Hovasapian

Fri Feb 26, 2016 5:05 AM

Post by David LÃ¶fqvist on February 13, 2016

Hi! This might be a little bit off topic, but i'm wondering if since a X a = 0 does that mean that a X ä = 0 in general? a is a vector

0 answers

Post by Heinz Krug on October 25, 2013

Hi Raffi,

that makes sense now. In the lecture you did not mention the magnitude and you said "... there is a similar formula for the cross product" and you wrote:

a x b = ||a|| ||b|| sin Î¸

(a and b are vectors with arrows on top)

I was a bit worried then and looked it up in Wikipedia, which clarified the point by multiplying this with a normal vector n.

2 answers

Last reply by: Heinz Krug

Thu Oct 24, 2013 4:42 AM

Post by Heinz Krug on October 23, 2013

Hi Raffi,

why is the cross product also defined as norm of vector a times norm of vector b times sin of angle? Wouldn't that be a scalar? Isn't the norm always a scalar, so this product would be the product of three scalars, therefore a scalar? What is missing here?

0 answers

Post by Aaron Harper on November 23, 2012

Dr. Hovasapian,

I've triple checked, you are most certainly correct. The cross product is [-4, -28, -8]. I thought it was very strange that there was an error. Being I've watched you're entire series in this section and have not encountered an error yet.

You do a great job at teaching this subject and I appreciate your feedback. I apologize that I was the one that made the error. I rechecked it many times, but I was the one with the wrong setup.

So far, you and Educator have helped me pass Calculus 3 with a good understanding, and a decent grade thus far.

Thank you,

Aaron

0 answers

Post by Professor Hovasapian on October 29, 2012

Hi Aaron,

I hope all is well with you.

Aaron, I've double-checked the arithmetic -- both by hand and with Mathematical Software (Maple) -- and A X B is definitely (-4, -28,-8).

The values you calculated came from missing the negative sign on the 5 in vector "a" -- it is definitely -5.

This brings up a very important issue that I address several times in many lessons throughout the entire course, and I hope you'll forgive me if I philosophize a little.... Arithmetic is not important. It matters for a specific problem, but it is not Mathematics.

What I hope for regarding my students is that they transcend calculational mathematics and begin to see the important things that take place underneath.

I talk all the time about the thousands of Arithmetic errors I make all the time -- you'll find many of them in these lessons -- because arithmetic is tedious and troubling and meaningless in the big picture of real mathematics -- this is why we have Math Software to do the dirty work for us.

But math software can never think for us -- can never set up an Integral for us -- can never choose the vectors making up a cross product for us. Only WE can do these things.

There will always be someone there to check your arithmetic -- there will not always be someone there to check your mathematical structures. It's the structures that matter, because structures DO NOT CHANGE. Numbers do....

Now, I understand that for the purposes of a test for a class, YES, calculation matters. It is my hope that someday more and more professors will stop burdening their students with silly calculations and start helping them to become big picture thinkers.

Sorry for the philosophy lecture.

Take good care, Aaron, and Best wishes, always.

Raffi

1 answer

Last reply by: Aaron Harper

Fri Nov 23, 2012 11:01 AM

Post by Aaron Harper on October 28, 2012

The cross product is [16, -8, -8], it is not [-4, -28, -8]

Now I'm skeptical about the entire video.