We’ve learned a good bit

about the dot product. But when I first introduced it,

I mentioned that this was only one type of vector

multiplication, and the other type is the cross product, which

you’re probably familiar with from your vector calculus

course or from your physics course. But the cross product is

actually much more limited than the dot product. It’s useful, but it’s

much more limited. The dot product is defined

in any dimension. So this is defined for any two

vectors that are in Rn. You could take the dot product

of vectors that have two components. You could take the dot product

of vectors that have a million components. The cross product is

only defined in R3. And the other, I guess, major

difference is the dot produc, and we’re going to see this in a

second when I define the dot product for you, I haven’t

defined it yet. The dot product results

in a scalar. You take the dot product

of two vectors, you just get a number. But in the cross product you’re

going to see that we’re going to get another vector. And the vector we’re going to

get is actually going to be a vector that’s orthogonal to

the two vectors that we’re taking the cross product of. So now that I have you excited

with anticipation, let me define it for you. And you probably already have

seen this once or twice in your mathematical careers. Let’s say I have the vector a. It has to be in R3, so

it only has three components: a1, a2 and a3. And I’m going to cross that with

the vector b and it has three components:

b1, b2 and b3. a cross b is defined

as a third vector. And now this is going to seem a

little bit bizarre and hard to essentially memorize because

this is a definition. But I’ll show you how I think

about it when I have my vectors written in

this column form. If you watch the physics

playlist, I have a bunch of videos on the cross product

and I show you how I think about the cross product when I

have it in the i, j, k form. But when I have it like this,

the way you think about this first term up here, this is

going to be another three vector or another vector in R3,

so it’s going to have 1, 2, 3 terms. For the first term, what you do

is you ignore these top two terms of this vector and then

you look at the bottom two and you say, a2 times b3

minus a3 times b2. And I’ve made a few videos on

determinants, although I haven’t formally done them in

kind of this linear algebra playlist yet. But if you remember kind of

co-factor– finding out the co-factor terms for when

you’re determining the determinant or if you’re just

taking the determinant for a 2×2 matrix, this might

seem very familiar. So this first term right here is

essentially the determinant of– if you get rid of this

first row out of both of these guys right here, you take a2

times b3 minus a3 times b2. So it’s a2 times b3

minus a3 times b2. That was hopefully pretty

straightforward. Now, not to make your life any

more complicated, when you do the second, when you do the

middle row, when you do this one right here, so you

cross that out. And you might want to do a1

times b3 minus a3 times b1. And that would be natural

because that’s what we did up there. But the middle row you

do the opposite. You do a3 times b1 minus

a1 times b3. Or you can kind of view it as

the negative of what you would have done naturally. So you would have done

a1 b3 minus a3 b1. Now we’re going to do

a3 b1 minus a1 b3. And that was only for

that middle row. And then, for the bottom

row, we cross that out again or ignore it. And we do a1 times b2, just like

we do with the first row. Times a2 b1. Or minus a2 b1. This seems all hard to– and

it is hard to remember. That’s why I kind of have to get

that system in place like I just talked to you about. But this might seem pretty

bizarre and hairy. So let me do a couple of

examples with you, just so you get the hang of our definition

of the dot product in R3. So let’s say that I have the

vector– let’s say I’m crossing the vector. I have the vector 1,

minus 7, and 1. And I’m going to cross that

with the vector 5, 2, 4. So this is going to be equal

to a third vector. Let me get some space to

do my mathematics. So for the first element in

this vector, the first component, we just ignore the

first components of these vectors and we say minus 7

times 4 minus 1 times 2. And these are just regular

multiplication. I’m not taking the

dot product. These are just regular

numbers. Then for the middle term, we

ignore the middle terms here and then we do the opposite. We do 1 times 5 minus

1 times 4. Remember, you might have been

tempted to do 1 times 4 minus 1 times 5 because that’s how we

essentially did it in the first term. But the middle term

is the opposite. And then finally, the third

term you ignore the third terms here and then you do it

just like the first term. You start in the top left. 1 times 2 minus 7. Put that in parentheses. Minus minus 7 times 5. And so that is equal

to– let me see. What do we get? Minus 7 times 4. I don’t want to make a careless

mistake here. That’s minus 28. Minus 2. So this is minus 30 for

that first term. This one is 5 minus 4. 5 minus 4 is just 1. And then 2 minus minus 35. So 2 minus minus 35. That’s 2 plus 35. That’s 37. So there you go. Hopefully you understand at

least the mechanics of the cross product. So the next thing you’re

saying well, OK. I can find the cross product

of two things. But what is this good for? What does this do for me? And the answer is, is that this

third vector right here, and depending on whether I stay

in the abstract case or whether this case with numbers,

this is orthogonal to the two vectors that we took

the cross product of. So this vector right here is

orthogonal to a and b. Which is pretty neat. If you just go think about the

last video when we were talking about normal vectors

to a plane, we can define a vector by– we can define

a plane by two vectors. If we define a plane– let’s say

that I have vector a right there, and then I

have vector b. Let me do vector b like this. Those define a plane in R3. Let me define your plane. So all the linear combinations

of those two guys, that’s a plane in R3. You can kind of view it

as they might form a subspace in R3. That forms a plane. If you take a cross b, you

get a third vector that’s orthogonal to those two. And so a cross b will

pop out like this. It’ll be orthogonal to both of

them and look like that. And so this vector right

there is a cross b. And you might say, Sal, how did

you know– I mean, there’s multiple vectors that

are orthogonal. Obviously, the length of the

vector, and I didn’t specify that there, but it could pop

straight up like that or why didn’t it– you know, you just

as easily could have popped straight down like that. That also would be orthogonal

to a and b. And the way that a cross b is

defined, you can essentially figure out the direction

visually by using what’s called the right hand rule. And the way I think about it

is you take your right hand and let me see if I can draw

a suitable right hand. Point your index finger

in the direction of a. So if your index finger is in

the direction of a and then I point my middle finger in

the direction of b. So my middle finger, in this

case, is going to go something like that. My middle finger is going to

do something like that. And then my other fingers

do nothing. Then my thumb will go in the

direction of a cross b. You could see that there. My thumb is in the direction

of a cross b. And assuming that you are

anatomically similar to me, then you still get

the same result. Let me draw it all. So this is vector a. Vector b goes in

that direction. Hopefully you don’t have a

thumb hanging down here. You know that a cross b in this

example will point up and it’s orthogonal to both. To kind of satisfy you a little

bit, that the vector’s definitely orthogonal or that

this thing is definitely orthogonal to both of these,

let’s just play with it and see that that definitely

is the case. And what is orthogonal? What is in our context, the

definition of orthogonal? Orthogonal vectors. If a and b are orthogonal, that

means that a dot b is equal to 0. Remember, the difference

between orthogonal and perpendicular is that

orthogonal also applies to 0 vectors. So these could also

be 0 vectors. Notice that I didn’t say that

any of these guys up here had to be nonzero. Well, in a little bit, we’ll

talk about the angle between vectors and then you have

to assume nonzero. But if you’re just taking a

cross product, nothing to stop you from taking– no reason

why any of these numbers can’t be 0. But let me show you that

a cross b is definitely orthogonal to both a and b. I think that might be somewhat

satisfying to you. So let me copy a cross b here. I don’t feel like

rewriting it. OK. Let me paste it. OK, bundle up little other

stuff with it. Let me take the dot product of

that with just my vector a, which was just a1, a2, a3. So what does the dot

product look like? That term times that, so it’s

a1– let me get some space. It’s a1 times a2 b3 minus

a1 times that. Minus a1 times a3 b2. And then you have plus

this times this. So plus a2 times a3. So plus a2 times a3 times b1. And then minus a2 a1 b3. And then finally, plus– I’ll

just continue it down here. Plus a3 times a1 b2 minus

a3 times a2 b1. All I did I just took the

cross– the dot product of these two things. I just took each of this. This guy times that was equal

to those two terms. That guy times that was equal to the next

two terms, equal to those two terms. And then this guy

times that was equal to those two terms. And if these guys

are really orthogonal, then this should be equal to 0. So let’s see if that’s

the case. So I have an a1 a2 b3, a

positive here, and then I’m subtracting the same

thing here. This is the same thing as a1 a2

b3, but it’s just a minus. So that will cancel

out with that. Let’s see, what else

do we have? We have a minus a1 a3 b2. We have a plus a1 a3 b2 there,

so these two are going to cancel out. And I think you see where

this is going. You have a positive a2 a3 b1 and

then you have a negative a2 a3 b1 there. So these will also cancel out. Now, I just showed you that

it’s orthogonal to a. Let me show you that it’s

orthogonal to b. Let me get another version of

my– the cross product of the two vectors. Probably scroll down

a little bit. And let me go back. And let me multiply that

times the vector b. b1, b2 and b3. I’ll do it here just so

I have some space. So b1 times this whole thing

right here is b1 a2 b3 minus b1 times this. Minus b1 a3 b2. Let me switch colors. And then b2 times this thing

here is going to be b2– so it’s going to be plus. This is all really one

expression, I’m just writing it on multiple lines. This isn’t a vector. Remember, when you take the dot

product of two things, you get a scalar quantity. So plus b2 times this thing. So b2 a3 b1 minus b2 a1 b3. And then finally,

b3 times this. So plus b3 a1 b2

minus b3 a2 b1. So if these guys are definitely

orthogonal, then this thing needs to equal 0. And let’s see if that’s

the case. We have a b1, a2, b3. So b1 and a b3. b1 a2 b3, that’s a positive

one and this is a negative one. You have a b3, an a2 and a b1

so that and that cancel out. Here you have a minus

b1 a3 b2. So you have a b1 and a b2. It’s a minus b1 a3 b2. This is a plus the same thing. Just switching the order

of the multiplication. But these two are

the same term. They’re just opposites of each

other, so they cancel out. And then finally, you have

a b2, an a1 and a b3. It’s a negative. And then you have a positive

version of the same thing. So these two guys cancel out. So you see that this

is also equal to 0. So hopefully you’re satisfied

that this vector right here is definitely orthogonal

to both a and b. And that’s because that’s

how it was designed. This is a definition. You could do a little bit of

algebra and you could have without me explaining this

definition to you, you could have actually come up with this

definition on your own. But obviously this was kind

of designed to have other interesting properties to it. And I’ll cover those in the next

few videos, so hopefully you found that helpful.

This is all great stuff! I'm going to have a nice head start on vectors!

Thanks for making this!

Thank you alot Sal

thank you, your videos are great

Where would i be without Sal! Great hand btw.

that was a sick hand you drew

Yeah seriously, that was a good hand.

trivia: the cross-product also applies to vectors in R7

I must say, that hand is quite awesome.

I'm curious, what do you do your math on? What is the software? Great video btw! đ

i jumped over here from khan academy to say.. NICE HAND

awesome thank you

I like all khan tutorials…. but as almost 99% of the written and video tutorials over the internet is the disassociation between pure math, physics or any science AND the real life.

Almost none of the tutorials talks about the application of whatever topic they discuss with the real life.

I think any such tutorials should start with what could be the practical application i.e. where and how it can be used in the real life…

Nothing wrong with learning new concepts, but w/o real app.its devoid

Minor mistake: At 4:38 you say "dot product" when I think you mean "cross product".

I owe you my decent grade in linear algebra.

@eileenBrain Anyway on a computer the above does not work the way things cancel out on the board.

ur picture final helped me understand what the heck that right hand rule meant

@karevkarev Anatomical is position, not size.

There's something wrong with your calculations. The sole reason the cross product was created was to create a vector orthogonal to the two other vectors that were crossed in R3, and any vectors that are orthogonal, by definition, have a dot product of zero. So it just makes sense that (a Ã b) Âˇ a = 0, and (a Ã b) Âˇ b = 0, because (a Ã b) is, by definition, orthogonal to a and b.

Wow dude, you totally didn't understand. Watch that part again. He was introducing the definition of orthogonal.

my brain wasnt fully functioning at the time i watched this.

I think the crossproduct is easier to remember if you draw the a vectors underneath again. for the second element, you do a3 times the b1 you drew underneath the original b3, – a1 that you drew under a3, times b3. that makes it easier to remember without 'doing opposites and such', you just expand your box. for the third, basically you do the cross of the elements underneath again, which is again a1b2 – a2b1.

8:47 for hand drawing

The cross product is not limited to R^3 only. There is a generalization for any sample of (n-1) vectors in R^n…

Your an artist as well. That is awesome.

are you a professional hand artist or something

this guys is a fucking boss. not only can he do math, but he can draw

I just came here from your site to say that the hand you drew is AMAZING!

'now that i have you excited with anticipation' lol

i was writing my own notes while viewing this and the picture of the hand i drew looks like my ass

There was NOTHING overly complicated in this.

How old are people in 10th grade in america? it worries me that im in university doing this and theyre doing the same thing haha

grade 8-10 math is simple, and you can solve the equations doing just that

but he's explaining everything to the fundamental level, which is quite important for college or university since

we dont get easy questions where we only have to put variables on both sides.

After learning absolutely nothing from this 15+(47/60) minute video, I was able to master linear algebra by simply reading your comment. You, little 8th-grader, are clearly more knowledgeable than this dottering old MIT graduate. Look at how many more math videos you've made than him! Look at your accomplishments! 10th grade math! 10th grade social studies! Clearly, your innate aptitude for doing exactly what your teachers tell you cannot be the manifestation of anything short of pure genius.

Yeah, a lot of kids think that the only thing you can do with this is solve school math equations. Its getting sad.

People, stop whining about this being complex. This is colledge/uni level maths. (and dont go "i did this in tenth grade!" because you only did this, not the 125 vids after this) where you dont learn "you solve it like this because I, your teacher, tell you too" like in 10th grade, but you learn why it works. Also, I never use matrix notation to solve systems of 3 or 4 equations, and if there are more, I let a computer do it. No, I use this for calculating torque in mechanics, and so on.

hi sir, u said when proving that a.b are orthogonal ,then a.b=0 but in the example u did the dot product of the result of de axb.a.may you please explain that

As a college engineering freshman, this stuff is gold!!!! Thanks!

reading these hand comments, had to skip ahead to see this. and it was worth it.. haha, btw thanks heaps for the vid !

This is good stuff. Thanks.

Anime drawing style right there. I know you are a fan of anime. >.> Khan. Let us exchange some words and drink tea.Â

i like the hand you draw

I wanted you to complete the proof by deriving the original method that's used to calculate the cross product but you didn't .. I did it on my own , but i left with 3 equations one of them is non-linear so the equation gets complicated. i guess, thisÂ definition basically comes from solving 3 linear equations making the orthogonal vector the determinant of these 3 equations.

wow thats a really good hand drawing

HAND

Holy shit that thumb thing was a game-changer

There's an easier way to do it: instead of taking the terms all confusing together x=(x1, x2, x3) y=(y1, y2, y3) you can make a matrix with z=(i, j, k) of the form

| i , j , k |

det |x1, x2, x3| = i(x2*y3 – x3*y2)+j(x3*y1 –

|y1, y2, y3| x1*y3)+k(x1*y2 – x2*y1)

and you have your termsīģŋ

"hopefully you dont have a thumb sticking out here"

hahaha hillarious!!

And for all this time I was thinking math teachers couldn't draw for shit. This guy comes along and draws a near perfect hand.

that is one beautiful hand

hhh everyone talked about the hand

Just here to show some love for that sweet hand.

That hand drawing really woke me up. I was paying attention and nothing things down but I wasn't really into it until that beautiful drawing. Thanks, Sal! Who says math teachers can't draw!

i just wanted to ask , if i watch all these videos in this playlist of "linear algebra", will i be ready to take on Quantum mechanics and understand matrix mechanics and Dirac notations and relativistic equations?

you are always super in explaining things

Sal, you are HILARIOUS and it's interesting to see how you teach as you literally explain EVERY step of the way not only to the audience, but to yourself too as you do it to confirm your own knowledge. Just brilliant dude.

That hand, tho… smh… so good

where does the 'n' in the formula a x b = absinx n ….. come from ?

Just here for the hand drawing.

Almost broke my hand trying to make that gesture

I liked this video just for the hand drawing.

almost everyone here is commenting just for the hand. i'm just over here really glad you proved the cross product dot a or b cancel out back to 0. i never thought to prove that in the definition. it makes much more sense knowing that.

but btw, nice bloody hand drawing.

is this for calc 3?

that is one good drawing of a right hand. I had to pause the video and type this in.

what is r3?

4:38 You mean cross product, right?

Your drawing is AWESOME =))))

I Love your way of teaching Sal……….

Thank you so much………

"assuming that you're anatomically similar to me" that line cracked me up XD

so if the cross produce =0, then it is orthogonal?

looks like someone has a minor in art/graphic design

Why cross product is written in matrix form….

good video although did not quiet get it

10/10 would watch that hand drawn again

Hand drawing pretty good

And surprised to find comments r all about the hand haha

what kind of platform this video was made? using some kind of table with pen? or on microsoft surface?

jesus, what a draw, you could be the next picasso

You can do cross product in higher dimensions, you just need more than two vectors. In R4, you take three vectors and compute a vector orthogonal to all three. So in n-space a cross product requires n-1 vectors.

Can you cross two matrices that are not vector matrices (e.g. 2×2 matrix by 2×2 matrix)?

âHopefully you donât have a thumb hanging down hereâ, lol.

9:00 that drawing lol what a pro

so instead of memorizing a formula of an easy determinant whose columns are ijk, a and b, you would have me remember some sort of half determinant sign switched in the middle?

Hey sal isn't determinant only defined for square matrix

wow this is very understandably

I can visualise the dot product in the meaning of having a component acting in the direction of the other vector but I never understood the cross product or can never visualise it whatsoever, It's a definition but isn't real?. The cross product has to be the most difficult concept I have ever come accross and Ikeep asking people "what is it " and they all say the same thing more or less "it's the product of two vectors and acts orthogonal to these two vectors ", but I keep asking the back "Why does it become orthogonal? " and they look at me like I am a moron from the planet moronay.

The day someone actually explains the cross product so I can visualise it then I will kiss their feet.