# Cross product introduction | Vectors and spaces | Linear Algebra | Khan Academy 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,
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.

### 85 thoughts on “Cross product introduction | Vectors and spaces | Linear Algebra | Khan Academy”

• October 13, 2009 at 5:24 pm

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

• October 13, 2009 at 5:36 pm

Thanks for making this!

• October 13, 2009 at 8:10 pm

Thank you alot Sal

• October 14, 2009 at 3:38 am

thank you, your videos are great

• November 1, 2009 at 5:59 am

Where would i be without Sal! Great hand btw.

• November 9, 2009 at 9:04 am

that was a sick hand you drew

• November 10, 2009 at 9:52 pm

Yeah seriously, that was a good hand.

• January 26, 2010 at 3:18 am

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

• June 1, 2010 at 9:25 pm

I must say, that hand is quite awesome.

• June 9, 2010 at 2:38 am

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

• July 8, 2010 at 8:41 am

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

• August 13, 2010 at 10:30 pm

awesome thank you

• November 2, 2010 at 7:59 pm

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

• December 8, 2010 at 8:35 pm

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

• December 15, 2010 at 2:48 am

I owe you my decent grade in linear algebra.

• June 27, 2011 at 7:57 pm

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

• October 10, 2011 at 8:08 am

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

• October 21, 2011 at 3:04 am

@karevkarev Anatomical is position, not size.

• March 22, 2012 at 2:21 am

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 × b) · a = 0, and (a × b) · b = 0, because (a × b) is, by definition, orthogonal to a and b.

• April 10, 2012 at 6:25 pm

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

• April 13, 2012 at 1:43 pm

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

• April 15, 2012 at 3:00 pm

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.

• April 26, 2012 at 7:41 pm

8:47 for hand drawing

• April 28, 2012 at 9:49 pm

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

• May 7, 2012 at 9:54 pm

Your an artist as well. That is awesome.

• August 14, 2012 at 2:35 am

are you a professional hand artist or something

• September 9, 2012 at 2:43 am

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

• October 7, 2012 at 2:16 pm

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

• December 3, 2012 at 12:34 pm

'now that i have you excited with anticipation' lol

• December 3, 2012 at 1:09 pm

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

• December 9, 2012 at 1:45 pm

There was NOTHING overly complicated in this.

• December 10, 2012 at 12:53 pm

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

• December 19, 2012 at 12:22 am

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.

• January 4, 2013 at 7:10 am

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.

• January 10, 2013 at 8:08 am

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

• January 10, 2013 at 8:14 am

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.

• April 8, 2013 at 3:05 pm

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

• April 12, 2013 at 2:51 am

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

• June 2, 2013 at 1:20 pm

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

• May 23, 2014 at 7:48 pm

This is good stuff. Thanks.

• November 10, 2014 at 7:04 pm

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

• January 31, 2015 at 3:23 pm

i like the hand you draw

• March 25, 2015 at 8:31 pm

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, this definition basically comes from solving 3 linear equations making the orthogonal vector the determinant of these 3 equations.

• June 17, 2015 at 1:22 pm

wow thats a really good hand drawing

• August 10, 2015 at 10:19 pm

HAND

• November 5, 2015 at 11:48 pm

Holy shit that thumb thing was a game-changer

• December 6, 2015 at 10:20 pm

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)

• December 29, 2015 at 11:54 am

"hopefully you dont have a thumb sticking out here"
hahaha hillarious!!

• February 9, 2016 at 9:07 pm

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.

• March 21, 2016 at 4:30 am

that is one beautiful hand

• June 13, 2016 at 5:14 am

hhh everyone talked about the hand

• July 31, 2016 at 7:57 am

Just here to show some love for that sweet hand.

• November 15, 2016 at 6:40 pm

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!

• November 16, 2016 at 5:49 am

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?

• December 6, 2016 at 6:54 pm

you are always super in explaining things

• January 11, 2017 at 11:25 am

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.

• January 22, 2017 at 6:06 am

That hand, tho… smh… so good

• February 2, 2017 at 1:07 am

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

• February 8, 2017 at 5:25 am

Just here for the hand drawing.

• April 15, 2017 at 4:38 pm

Almost broke my hand trying to make that gesture

• May 11, 2017 at 5:06 am

I liked this video just for the hand drawing.

• October 3, 2017 at 4:40 pm

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.

• January 31, 2018 at 10:45 pm

is this for calc 3?

• March 12, 2018 at 1:31 am

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

• May 22, 2018 at 11:20 am

what is r3?

• May 31, 2018 at 1:44 am

4:38 You mean cross product, right?

• June 2, 2018 at 8:57 am

• June 16, 2018 at 10:48 am

I Love your way of teaching Sal……….
Thank you so much………

• July 3, 2018 at 4:10 am

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

• July 19, 2018 at 5:26 pm

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

• August 22, 2018 at 2:08 pm

looks like someone has a minor in art/graphic design

• September 12, 2018 at 10:54 am

Why cross product is written in matrix form….

• September 19, 2018 at 8:21 pm

good video although did not quiet get it

• October 8, 2018 at 5:50 pm

10/10 would watch that hand drawn again

• January 22, 2019 at 12:58 am

Hand drawing pretty good

• January 28, 2019 at 2:02 am

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

• February 18, 2019 at 10:09 pm

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

• February 21, 2019 at 4:50 am

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.

• April 2, 2019 at 4:51 pm

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

• June 26, 2019 at 6:21 pm

“Hopefully you don’t have a thumb hanging down here”, lol.

• June 29, 2019 at 9:36 am

9:00 that drawing lol what a pro

• July 13, 2019 at 5:32 pm

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?

• September 29, 2019 at 10:29 am

Hey sal isn't determinant only defined for square matrix

• November 4, 2019 at 2:05 pm
• 