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Lecture Comments (10)

1 answer

Last reply by: Professor Hovasapian
Fri Feb 26, 2016 5:05 AM

Post by David Löfqvist on February 13 at 05:08:14 AM

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 θ
(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.

Parameterizing Surfaces & Cross Product

Describe the curve f(x) = 4x2 − 3x − 1 in one parameter t.
  • A curve in two dimensions is described through two components, C(t) = (C1,C2).
  • To describe the curve we let x = t.
  • Substituting yields f(t) = 4(t)2 − 3(t) − 1 = 4t2 − 3t − 1.
Our curve is now C(t) = (t,4t2 − 3t − 1).
Describe the curve 1 = [(x2)/4] + [(y2)/16] for x > 0 and y > 0 in one parameter t.
  • A curve in two dimensions is described through two components, C(t) = (C1,C2).
  • To describe the curve we let x = t.
  • Substituting yields 1 = [(t2)/4] + [(y2)/16]. We solve for y in order to have our second component for the curve.
  • Now, 16 = 4t2 + y2 and so y = ±√{16 − 4t2} = ±2√{4 − t2} . Since x > 0 and y > 0 we take the positive square root.
Our curve is now C(t) = ( t,2√{4 − t2} ).
What is the parametrization of a sphere with radius 5 (centered at the origin) using two parameters f and θ ?
  • The parametrization of a sphere with radius R is P(f,θ) = (Rsinfcosθ,Rsinfsinθ,Rcosf) for 0 ≤ f ≤ π and 0 ≤ q ≤ 2π.
Hence the parametrization of a sphere with radius 5 is P(f,θ) = (5sinfcosθ,5sinfsinθ,5cosf) for 0 ≤ f ≤ π and 0 ≤ θ ≤ 2π.
Describe the surface 2z + xy − x2 = 4 using two parameters t and u.
  • A surface in two dimensions is described through three components, C(t) = (C1,C2,C3).
  • To describe the surface we let x = t and y = u.
  • Substituting yields 2z + tu − t2 = 4. We solve for z in order to have our third component for the surface.
  • Now, 2z = 4 + t2 − tu and so z = 2 + [(t2 − tu)/2]
Our curve is now C(t) = ( t,u,2 + [(t(t − u))/2] ).
Describe the curve z = x2 + y2 − y using one parameter t.
  • A curve in three dimensions is described through three components, C(t) = (C1,C2,C3).
  • To describe the curve we let x = sin(t) and y = cos(t).
  • Substituting yields z = sin2(t) + cos2(t) − sin(t) = 1 − sin(t).
Our curve is now C(t) = ( sin(t),cos(t),1 − sin(t) ).
Find a parametrization for the surface of revolution for f(x) = (x − 1)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]
Substituting yields P(x,θ) = (xcosθ,xsinθ,(x − 1)2) for x ∈ [0,1], θ ∈ [0,2π].
Find the cross product of a and b given:
i) a = (0,0,1), b = (0,0,1)
  • To find the cross product of vectors a and b, a ×b, we compute det(
    i
    j
    k
    a1
    a2
    a3
    b1
    b2
    b3
    ), that is, the determinant of the matrix formed by the vectors i, j, k and the components of a and b.
  • Then det(
    i
    j
    k
    a1
    a2
    a3
    b1
    b2
    b3
    ) = det(
    i
    j
    k
    0
    0
    1
    0
    0
    1
    ) = (0 − 1)i − (0 − 1)j + (0 − 0)k = − 1i + 1j + 0k. Recall that i = (1,0,0), j = (0,1,0) and k = (0,0,1).
In our usual vector notation a ×b = ( − 1,1,0).
Find the cross product of a and b given:
ii) a = (1,0,0), b = (0,0,1)
  • To find the cross product of vectors a and b, a ×b, we compute det(
    i
    j
    k
    a1
    a2
    a3
    b1
    b2
    b3
    ), that is, the determinant of the matrix formed by the vectors i, j, k and the components of a and b.
  • Then det(
    i
    j
    k
    a1
    a2
    a3
    b1
    b2
    b3
    ) = det(
    i
    j
    k
    1
    0
    0
    0
    0
    1
    ) = (0 − 0)i − (1 − 0)j + (0 − 0)k = 0i − 1j + 0k. Recall that i = (1,0,0), j = (0,1,0) and k = (0,0,1).
In our usual vector notation a ×b = (0, − 1,0).
Find the cross product of a = ( − 2,3,5) and b = ( − 1,6,2)
  • To find the cross product of vectors a and b, a ×b, we compute det(
    i
    j
    k
    a1
    a2
    a3
    b1
    b2
    b3
    ), that is, the determinant of the matrix formed by the vectors i, j, k and the components of a and b.
  • Then det(
    i
    j
    k
    a1
    a2
    a3
    b1
    b2
    b3
    ) = det(
    i
    j
    k
    − 2
    3
    5
    − 1
    6
    2
    ) = (6 − 30)i − (4 − ( − 5))j + ( − 12 − ( − 3))k = − 24i − 9j − 8k. Recall that i = (1,0,0), j = (0,1,0) and k = (0,0,1).
In our usual vector notation a ×b = ( − 24, − 9, − 8).
Find the cross product of a = ( [1/(√5 )],[2/(√5 )],0 ) and b = ( [1/(√3 )], − [1/(√3 )],[1/(√3 )] )
  • To find the cross product of vectors a and b, a ×b, we compute det(
    i
    j
    k
    a1
    a2
    a3
    b1
    b2
    b3
    ), that is, the determinant of the matrix formed by the vectors i, j, k and the components of a and b.
  • Then det(
    i
    j
    k
    a1
    a2
    a3
    b1
    b2
    b3
    ) = det(
    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
    ) = ( [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).
In our usual vector notation a ×b = ( [2/(√{15} )], − [1/(√{15} )], − [3/(√{15} )] ).
Find a ×b given that a = (cosθ,sinθ,1) and b = (sinθ,cosθ, − 1) for 0 ≤ θ ≤ 2π
  • To find the cross product of vectors a and b, a ×b, we compute det(
    i
    j
    k
    a1
    a2
    a3
    b1
    b2
    b3
    ), that is, the determinant of the matrix formed by the vectors i, j, k and the components of a and b.
  • So det(
    i
    j
    k
    a1
    a2
    a3
    b1
    b2
    b3
    ) = det(
    i
    j
    k
    cosθ
    sinθ
    1
    sinθ
    cosθ
    − 1
    ) = ( − sinθ− cosθ)i − ( − cosθ− sinθ)j + (cos2θ− sin2θ)k = − (sinθ+ cosθ)i + (sinθ+ cosθ)j + (cos2θ− sin2θ)k.
In our usual vector notation a ×b = ( − (sinθ+ cosθ),(sinθ+ cosθ),(cos2θ− sin2θ)).

*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 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

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, t2, and t3.0113

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 t2 + u2.0330

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 x2 + y2 + z2 = r2.0675

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(r2 - x2 -- y2).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 r2 - t2 - u2, 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 r2 - x2 - y2 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

x2 + y2 = z2, 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 p1, and it is going to be the parameters t and θ.0940

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 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. x2 + y2, so if I want another parameterization, let us call this one p2, 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, x2 + y2.1056

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

We said t, u, t2 + u2, 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 x1, < or = some value x2, 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 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.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 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.1526

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.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 a1, a2, a3, b1, b2, b3 -- 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, a1, a2, a3, b1, b2, b3. 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 22 + 02 + 22 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 12 + 22 + 12, and norm(b) = sqrt(18), right? 12 + 42 + 12.2375

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