Cross products in the light of linear transformations | Chapter 11, Essence of linear algebra

Why the formula for cross products matches the geometric intuition.
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Home page: www.3blue1brown.com/
For anyone who wants to understand the cross-product more deeply, this video shows how it relates to a certain linear transformation via duality. This perspective gives a very elegant explanation of why the traditional computation of a dot product corresponds to its geometric interpretation.
Minor error at 1:44, the third line of the matrix should read "v1 * w2 - w1 * v2"
*Note, in all the computations here, I list the coordinates of the vectors as columns of a matrix, but many textbooks put them in the rows of a matrix instead. It makes no difference for the result since the determinant is unchanged after a transpose, but given how I've framed most of this series I think it is more intuitive to go with a column-centric approach.
Full series: 3b1b.co/eola
Future series like this are funded by the community, through Patreon, where supporters get early access as the series is being produced.
3b1b.co/support
Thanks to these viewers for their contributions to translations
Hebrew: Omer Tuchfeld
Vietnamese: @ngvutuan2811
------------------
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[@mohandeshpande3201]

The next step after watching this video is to watch this video again.

[@troubletaker6614]

How's your brain now?

[@laurv8370]

@mohan, yeah, my first impulse too

[@TheLeontheking]

but with a pen and paper this time ;) (math of course, not role-games)

[@BentHestad]

and then once again, in my case at least:-)

[@Manu-xj4xy]

@@TheLeontheking lmao

[@ahmedkamal428]

I beg you, Please make "Essence of Probability" Series

[@shubhamroy3131]

try edx

[@shubhamroy3131]

mit course

[@benedictus5657]

You should check out statquest. It’s similar in giving you intuition instead of seemingly arbitrary formulas for everything related to statistics and probability

[@touisbetterthanpi]

Omg please!!

[@starinsky2873]

@@shubhamroy3131 Doesn't explain everything but explains better than any other university

[@SkunZielonyJakMech]

I'll try to explain it further (the way I understand the clue of this video). We define vector p to be such a vector, that for any given vector (x,y,z) dot product of (p) and (x,y,z) is equal to determinant of matrix [x v1 w1 / y v2 w2 / z v3 w3] (let's call this matrix M). As we have seen in previous videos determinant of a matrix is equal to volume of parallelepiped formed by its column vectors. From above facts and definitions we know that: Volume of parallelepiped = det(M) = vector (p) (dot) vector (x,y,z) Now let's look at the volume of parallelepiped more geometrically. It's defined by formula: area of base * height. We know that area of the base is equal to area of parallelogram of vectors v and w. Now, what is its height? It's the portion of vector (x,y,z) which is perpendicular to the parallelogram! How do we find it? From previous videos we know that we can find it simply by taking dot product of vector (x,y,z) with unit vector which is perpendicular to parallelogram. Let's call this vector u. Now our new formula for volume of parallelepiped is: Area of parallelogram * {vector (u) (dot) vector (x,y,z)} Compare both results: Area of parallelogram * {vector (u) (dot) vector (x,y,z)} = vector (p) (dot) vector (x,y,z) And you clearly see that: Area of parallelogram * vector (u) = vector (p). So vector (p) is vector (u) (which is, by definition, UNIT vector perpendicular to parallelogram of v and w) multiplied by *Area of parallelogram*. So vector p is: 1) perpendicular to both v and w 2) has a magnitude = area of parallelogram of v and w. Amazing result.

[@Miserlis]

Really helpful - I think what I failed to get was that we could define the volume of a parallelepiped by base * height.

[@Lasterlastful]

Thanks a lot!... Very good and clear summary

[@webgpu]

volume = A.(BxC) ("Triple Product") :-)) en.wikipedia.org/wiki/Scalar_triple_product

[@udaysinha8215]

That shows that the dot products of both the vectors with the xyz vector are the same. It doesn't mean both vectors are same.

[@relaxradio5795]

@@Miserlis ahhh that's why it didn't make sense. I forget

[@gibsonman507]

My brain short-circuited somewhere between 0:00 and 13:09. I just woke up. I don't know where I am and I'm scared

[@MrChangCJ]

glad im not the only one feeling like a dumbass, not having some epiphany like all the others in the comments.

[@jrpie1215]

People who respond well to the video are probably a lot more likely to comment

[@darklordvadermort]

try it again on .5 or .25 speed. Think it could be better with an example with numbers as well.

[@khaledkord8021]

nope, you are not alone i'm with u guys :"D

[@shayanmoosavi9139]

You were supposed to watch the previous episodes in this playlist.

[@taraspokalchuk7256]

Please tell me where did you learn this from? They do not teach this in uni nd the textbooks are awful. It's scary that my knowledge of linear algebra depends on some good man's will to share this videos.

[@ElchiKing]

Wisdom comes with time ;) On a more serious note: I guess, this understanding comes from using those objects really often.

[@ZardoDhieldor]

Very smart people find this out on their own (like one in a thousand, not only geniuses). I believe that 3b1b is one of them. Everyone else is dependent on them to spread their insights. Which in a lot of courses in school and uni is not happening. I guess this is because a lot of teachers and lecturers don't like their job. Those who do like their job - and are didactically competent - usually tell you things like this.

[@noamtashma2859]

but you have to remember that when these concepts were invented, the people who invented them thought of them basically in these ways. 3b1b isn't inventing it all (or maybe he found these independently?), but for some reason this knowledge isn't very widespread.

[@MrPiks0u]

That is the way I was taught about the cross product (in France, btw). I don't think I ever saw the computational trick of part 1. Basically, we proved that *p* in the formula at 7:12 existed (and was unique) if *v* and *w* were linearly independent, then we defined the cross product of *v* and *w* as that unique *p*.

[@taraspokalchuk7256]

Please answer, my love

[@ChumX100]

Tbh this one is a bit tough to grasp on the first try, but something tells me is gonna feel amazing when I finally get it, thank you so much for sharing! This stuff is gold.

[@3blue1brown]

It is notably heavier than other concepts, which is part of why I pulled it into a separate video. For my part, I can say that I probably was exposed to it 3 or 4 times in various contexts before it actually sank in, and it was well worth it once it did.

[@jamesgoodman5102]

+3Blue1Brown well that makes me feel a bit better for not really grasping it tbh lol

[@ZardoDhieldor]

It really does feel amazing eventually. This feeling is the reason people actually go and study math.

[@NomadUrpagi]

@@ZardoDhieldor it is the reason people go study science in general. And i mean STEM because i never got joy from understanding political studies compared to math or physics. All of the social sciences are pretty subjective and open to interpretation that is why they have little value to me. Whereas natural sciences are so universal and objective that it brings me joy understanding it.

[@ZardoDhieldor]

@@NomadUrpagi I have to disagree on the social sciences. Psychology and economics are two examples of really intriguing social sciences. They are much more difficult because they deal with human behaviour which is very complicated and not entirely predictable but I find the m very interesting nontheless!

[@prakhar242]

It feels so good to know that there's someone kind enough to put in so much effort and time to make these quality videos(and videos like these are very hard to find) and make them available free to all students around the world.

[@shoumikacharya8060]

please make a series of "Essence of Complex Analysis"

[@Joshiraxd]

Noice

[@mattraymond1497]

yes

[@phenomenalphysics3548]

Yes please!!

[@blureyes106]

Very nice if so

[@MadhurDevkota]

Yes please

[@GTGTRIK]

Okay, so the problem with linear algebra is that people write things in weird ways and then wink at you knowingly.

[@VestinVestin]

I think the process goes like this: * People use esoteric, incomprehensible vocabulary and notation. You have no idea what they mean. The expressions scare you. * Through elaborate explanations and intellectual effort you gain an intuitive understanding of underlying concepts. * You tie the concepts to commonly used shorthands, the terms of art everyone in the field uses and understands. * You grow increasingly enamored with the terms, since you know what they mean, they roll of the tongue easily, and everyone else understands you immediately. * You set out to introduce utterly green and ignorant newbies to what you know. At this stage it either occurs to you that your new, thoroughly transformed understanding and vocabulary utterly mismatch those of the uninitiated, or you treat them like you would your in-the-know colleagues and consistently terrify them with impossible to grasp concepts expressed in esoteric terms (and, possibly, explained using other, equally unfamiliar words). The cycle continues. ...or, more succinctly: To competently transform the ideas of people who don't understand into ideas of people who do, you need to understand what it's like to not understand AND what it is like to do, and then bridge the two. If you only understand what it's like to understand, you will have no clue why those who don't don't.

[@tadaiyoradima]

*unknowingly

[@tylern6060]

@@VestinVestin wow, I've had some inkling of this idea haunting me every time I try and fail to teach other people math. You just articulated it perfectly.

[@stephenday4834]

Exactly. I believe humanity has never found reasonable ways to express math concepts in letters and symbols so got stuck with archaic notation that you simply have to "get used to". Many of the concepts are easily understandable in themselves, but the mash-up of context sensitive symbols and terminology and the inconsistent use of these presents a high barrier to understanding. Humanity needs better languages that allow expressing everyday things and abstract concepts clearly and unambiguously.

[@ishworshrestha3559]

Ok

[@MikeAuerNixego]

Your series has let me glimpse perhaps a morsel of how theoretical physicists like Einstein see something in the "real world" in terms of its geometric essence, and then use mathematics to describe it. Thank you for that!

[@dayliss413]

Holy shit. I almost broke down crying about how much sense this makes. Thank you so much!

[@ZardoDhieldor]

This, my friend, is the beauty of math! :) Probably the sole reason why there are mathematicians.

[@romanski5811]

+Zardo Dhieldor No, I think that's also because it's useful.

[@Zekian]

I think that's why there are engineers.

[@cineblazer]

@@ENXJ I'm concerned, were you spying on OP while they watched the video? Can't imagine any other way you'd be able to confirm or deny the veracity of their claim.

[@hamorahime937]

I broke down crying about it's not making any sense (at least yet) to me ;-;

[@jamesjin1668]

If you don't understand it, all I can say is to not give up, because when you finally figure it out, the clarity of understanding and the delayed gratification are simply astounding. I had to watch the video 2 times and think for at least 30 minutes to finally get it, and I am really glad I persevered.

[@juustgowithit]

You will probably rewind several times, so: "what vector p has the special property, that when you take a dot product between the p and some vector [x,y,z] it gives the same result as plugging [x, y,z] to the first column of the matrix whose other two columns have the coordinates of v and w, then computing the determinant?"

[@paulbarton4395]

ha I paused and rewound at exactly that part to think about this sentence

[@qqcat3678]

if I find one p that satisfies this, how will I be sure that this is the only possible one?

[@qqcat3678]

ah, it's a dual of a linear transformation, it must be unique!

[@stealthemoon8899]

Hero

[@Someone-uw7je]

@@qqcat3678 I know I'm very late, but why is the various determinants created by f([x y x]) linear? How are we sure about it?

[@sathirasilva4958]

You just covered my 6 months course by only 15 videos. HUGE RESPECT! We need more series like these. Thank you sir!

[@pictzone]

This is why I think the current educational system is obsolete. I've been 100% committed to this "unconventional" learning approach (online self-learning) in the past few years and my skills have expanded exponentially. Anyone can achieve this if done right. Never returning to the legacy way.

[@rowangoebel-bain3447]

You tube is Revolutionizing education. They give us postdocs at Dartmouth who have barely taught before and all I need are these videos. Love the visual clarity

[@shreyjoshi4891]

This geometric interpretation of the dot product is brilliant, and when applied to define the computation for the cross product is very insightful. This is literally what I was missing in my HS precal class: reasons for why these methods for computing this stuff works. On an unrelated note, I would appreciate it if you made an "Essence of Classical Mechanics" series!

[@tequilajoe518]

So true. Classical mechanics never made sense to me bcoz of d way it is taught in colleges.

[@_ish]

oh

[@ianc226]

This is how I understand it: VOLUME(determinant of 3 3D vectors,i.e. volume of parallelepiped) = BASE (parallelogram of vectors v and w) x HEIGHT(vector [x,y,z] mapped from 3 dimensions onto 1 dimension that is perpendicular to the base) Hence, the magnitude of the vector P gives the BASE while mapping the vector [x,y,z] to the 1-dimensional direction of P gives the HEIGHT. Thus, the dot product between P and [x,y,z] == VOLUME Essentially, this means that P serves as a "special 3D vector such that taking the dot product between p and any other vector [x,y,z] gives the same result as plugging x,y,z into the first column of a 3x3 matrix, then computing the determinant" of that 3x3 matrix. Thank you, Grant, for this beautiful piece of knowledge.

[@jeffbronson3696]

+

[@KreationKills]

thank you for your comment, i felt like i got what he said from the video but even rewatching it a few times it was still a little bit nebulous. reading your comment clarified that. :)

[@cineblazer]

Thank you for this! I still have to re-watch the video another 2 or three times to hopefully make it "click" but this helped me feel like I'm getting closer!

[@jeffbronson3696]

@@cineblazer I hope u got it in the end! Have a great week

[@chibuzordesmond3937]

But P is not one dimensional? I’m trying hard to understand this.

[@kostathomas8732]

I'm a first year aerospace engineering student who's taking linear algebra and statics. It took me 5 attempts at watching before it finally started to make sense. I have to think about these things intuitively to know if what I'm doing is correct and I can't imagine learning the same things without having watched this series.

[@ProfFeinman]

As in my previous comment, if this were done in the context of some physical or engineering problem it would be easier to visualize. For your sixth try, can you apply this to something. (I could have done it when I was an undergraduate (I think) but I no longer study much physics).

[@JethroYSCao]

Absolutely brilliant. When doing the computations, I've always felt that finding the cross product and taking the determinant of a 3x3 matrix were somehow related, but couldn't quite put my finger on it. And now you've shown me the light. Much appreciated!

[@BharCode09]

This makes more sense in physics than in Math(geometry). Analogy for this cross product is winding/unwinding a screw, where the forces (screwing) applied is in 2D plane, the screw itself gets deeper inside or comes outside into or out of 2d plane, i.e 3D, which is ofcourse perpendicular. It's called Torque. This is also applicable in figuring out magnetic field on a current carrying conductor. Dot product is for work done (or displacement) given the direction and magnitude of forces applied. Now go through this video again and as he says "Pause and ponder" that might help. Thanks for these videos!

[@NierAutomata2B]

I think your comment is highly underrated. This video misses the part that "why we care about such a definition called cross product". Though it's neat that there is this duality of linear transformation and a vector, but I don't see how this definition is valuable in different use cases. From what I know, same as your comment, the cross product is used heavily in classical mechanics and electrodynamics where forces, work, energy and fields are cleanly expressed using cross product between physical quantities.

[@sulymanhossain4144]

@@NierAutomata2B , You are talking from a Physicist's perspective, But the instructor is a Mathematician and probably he is trying to explain those Mathematical operations from a Mathematician's perspective.

[@Riddhanand]

@@NierAutomata2B Why do physicists think they are at the center of the world xD. This video does an excellent job at taking such abstract topics such as vector spaces and making them highly intuitive. Surely, physics is a great way to visualize and apply math, but it is also nice to actually understand this topic in its most basic form.

[@NierAutomata2B]

@@Riddhanand It has nothing to do with what physicists think whatsoever. I also would like to know how it's used in other applications. Those mentioned are just examples I know. Not sure what "understanding" something means to you, to me it means knowing the motivation why it was created and what purpose it serves.

[@Riddhanand]

@@NierAutomata2B The video series is called essence of linear algebra. Not applications of linear algebra. I'm a mathematical physics student and to me understanding something is learning something from its most basic truth. Might be different for you but again, this video is to do with essence of linear algebra. Not it's applications.

[@nickiexu7259]

Amazing...I feel that all the previous knowledge from school are 2D and now I am able to get the understanding in another dimension! For people who could not understand by watching it once, please don't just let the understanding go! I literally watched more than 10 times just on this chapter today and I finally got it! So happy!

[@charlesd774]

this is really amazing, the height of the cross product embeds the information of the determinant of V and W (area) and meanwhile the dot product with a variable vector gives us the "height" of the parallelipiped so together they make the volume. It's also interesting that if we view the cross product as a linear transform, its null space is spanned by V and W. This corresponds to the fact that: 1) for any variable vector that we dot with it, if it has components from the null space, they add 0 to the volume 2) geometrically this means tilting the parallelipied, which doesnt change the volume (like a shear) today I learned about how the determinant function was derived from 3 constraints of being multilinear, alternating, and normalized. It's really hard to tell where there should be intuition, or "the numbers just work out", or it's just magic (PS part 1 is reflected in the determinant function, if we nudge the variable vector in the direction of the null space, by linearity of det, we can calculate that volume slice separately, and it is equal to zero because the nudge direction is linearly dependent on V and W, aka a flat additional volume)

[@sooryanarayana3929]

Appreciate your patience and your will to make these videos so that everyone is on the same page, this is one of the best mathematical proof made in a video, you have great respect from everyone, as you help us all learn. Thank you for making us learn and understand the real meaning. 👏👏👏👏👏

[@kristianemilkjrgaard531]

I have a written linear algebra exam tomorrow and this series is an awesome supplement to books, exercises, and previous exam sets. Love it

[@kaustubhxdd]

9:30 you have no idea the immense joy I felt when I was able to deduce mentally myself that its gonna have a length same as the volume of the parallelogram and perpendicular. Steps must be taken on a global level to normalize this type of teaching and making it easily accessible.

[@atklobas]

my physics classes said the mechanics of the cross product is outside the scope of the class, my vector calculus teacher mentioned that a dot (b cross c) is the volume of a parallelepiped but that was just as a 'refresher' at the course start. My linear algebra teacher never even mentioned the cross product (then again I learned more linear algebra in a week of diffEQs than the first half of his class). If I had ever seen it expressed as a determinate everything would have made more sense.

[@moglibora]

same story bro :)..

[@iwtwb8]

Translation: "I don't understand this well enough to teach it effectively so I'm just going to skip it."

[@breakprismatshell6270]

Took me years to figure exactly this out.

[@sebcodestheweb9385]

I think this is about the third time I watched this video, after writing everything out, explaining it for myself and doing some exercises. I now have a much deeper understanding of the cross product compared to my peers, whom I will gladly enlighten with this beautiful connection. Thanks so much!

[@MaxwellsWitch]

omg, thank you so much. I remember taking multivariate calculus and always wanting to know the mechanics of it. you always give me new insight to these things. keep going btw, your videos are the ones I always wait for and watch right as they come out.

[@duydangdroid]

6:25 Since vectors v and w are fixed, only one side of parallelepiped can change. Thus the function is linear

[@Masterofxioen]

My hero!! (Although why does he say to figure out from the properties of the determinant?...)

[@Masterofxioen]

This function is linear in the sense that it’s not quadratic: there is no X Square Y Square Z Square. It’s a bit confusing because up until now we had stuff like linear transformations and linear algebra, but here he is only talking about the ‘function’ being linear. The power of any variables is just one. Compute the determinant using the standard way to do so, and it will only have constants (v3w2 - v2w3 etc) Or variables with constant coefficients.

[@ryanwu8503]

The transformation(function) is linear if it has two properties after transformation: 1. all line must remain line 2. origin doesn't move In this case, it's a 3d to 1d transformation and the line will still be a line and the origin will still be the origin which makes it linear. Please someone correct me if I'm wrong 😅

[@luphiax4239]

yes but it is a little bit confusing since he pictured the determinant function in chapter 6 at 5:17 as non linear

[@JivanPal]

This is a sweet demonstration, and really intuitively shows why the signed volume of the parallelepiped defined by given vectors *a*, *b*, and *c* is equal to *a*•*b*×*c*.

[@newkid9807]

You pally

[@ashtravelerr.3895]

After watching this, I will immediately understand what a paper indicates when they use some weird vector dot product just by duality. This is the whole context behind the linear algebra calculation and can easily help relate what the author wants to achieve along their lines. This is amazing. I truly thank you.

[@vanshjhunjhunwala974]

Thanks for all the work you put into this Grant. I recently watched your essence of calculus series and I am now hooked into this. It took me two replays and several pauses to understand the concept but now that I somewhat do (I guess...) It feels like a completely different way of looking at vectors. Again, really appreciate you hard work.

[@shawaizhaider3978]

Finally, finally, finally, somehow I understood it after looping it for 3 hours.

[@blalabhel5848]

please do a video on tensors

[@jawadtahmeed9854]

I've started to watch this series to grab a better understanding about martices so that it helps to understand neural networks and those kind of stuffs. Eagerly waiting for a series on tensors.

[@Riddhanand]

YES PLEASE

[@sebastiandelgado6263]

It took me a long time to understand this, had to rewatch this multiple times, solve problems / normal linear algebra class, and I think it finally stuck with me today. Thanks Grant, you are a saint

[@kevinhyun5354]

After watching the video 3 times and reviewing duality and transformation into number line repeatedly, I finally get what's going on! Glad that I didn't give up halfway. I wanted to understand cross-product after learning about dot products in my year 11 class!

[@alvarol.martinez5230]

That was beautiful. Also, there's a nice coherence between the videos so far

[@lironkarpati9605]

It took me a while to piece together the arguments but eventually it made sense and it really is a cool way to look at the cross product -- for anyone working it out: keep going its worth it! I got caught up on why the unit vectors were still in the determinant because those cannot be values of x,y,z. It has nothing to do with plugging in but rather once you understand what p should be, the determinant computation places the unit vector components to the correct places to represent p. The possible motivation illuminating how one could arrive at this connection on their own could be: I want to find a vector that is perpendicular to two other vectors. What do I know to find such a thing. Well if I multiply the parallelogram formed by some height I get a parallelepiped. Ah well that means I have some arbitrary x,y,z vector that forms this parallelepiped. Ok so in relation to this new vector, , the perpendicular height is some projection of , onto some other perpendicular vector, we will call it P. What do I know that could help me find such a P. Well, the determinant of the vectors forming the parallelepiped is a constant. And I know that the dot product of two vectors is a constant. So lets define P as the vector that taken as the dot product with (because I was already thinking about it projecting x,y,z onto the height) to equal the determinant. Well, when you do that you see that there is a nice grouping of terms when you work out the dot product and determinant! You can now define P very nicely in terms of the two original vectors and your goal of finding a perpendicular vector was resolved.

[@joao_ssouza]

For me that took a linear algebra course last year, seeing all of this is just satisfying, now I'm finally getting why things are the way they are, not just accepting that "it works".

[@Darakkis]

This is insane dude, I don't think my teachers are ever gonna teach me linear algebra like this... Unbelievably good work!

[@kjekelle96]

0:00 intro 0:15 recap and geometric properties of a 3D cross-product 3:13 goal of the video 3:45 the cross-products in 2D and 3D 5:56 bringing in duality 7:28 computationally 9:21 geometrically 11:52 in sum

[@PiercingSight]

Holy crud. That blew my mind so hard I had to watch it again to make sure that the dots you connected were correct. And by george, they were! Fantastic series of videos. You've inspired me to produce my own series of mathematics videos that attempt to build foundational intuitions, rather than rote manipulations, regarding simple algebra, geometry and other things like imaginary numbers. That area is my specialty. Good on you sir. been subscribed for a while, and that's not changing anytime soon.

[@denelson83]

You mean he connected the dot _products,_ right?

[@peronianguy]

i really enjoy watching these videos over an over again, pausing and ponder as you have taught us ---and suffering a bit as well. then coming the next day and understanding everything better, life seeming bright overnight

[@TheAndreotte]

I would like to thank you for your videos. They help me to understand the math in a way that become natural and lot more easy. Thanks again! Regards.

[@rnash01]

It took me AGES to understand this, but it was certainly well worth it. Your explanation got me to be satisfied with the property that the vector p had to have a length equal to the area of the parallelogram formed by the span of the vectors v and w; perhaps I missed something but I couldn't get the same satisfaction to the property of p being perpendicular to those two vectors. But I did some writing on paper (and some looking around online), and it finally occurred to me that the reason it is perpendicular, is that plugging in the coordinates of v into x, y, and z will return a zero determinant for that function you defined, and so will plugging in the coordinates of w. This would imply that, if the determinant is 0, then the dot product of p and v and that of p and w will also be 0. Since p is non-zero, and v is non zero, then we can take the equation given by the dot product, (p dot v) = (abs(p)) * (abs(v)) * cos(theta)). cos(theta) MUST be 0 to satisfy this, therefore theta must be 90 degrees or pi/2 radians, hence it is perpendicular to v, and therefore perpendicular to w. Took me 10 watches of this video, but I'm glad it got me to think a little bit about the relevance of duality with all of this. Thanks Grant!

[@adsurbum]

I didn't get what you explained. But I think that it should be perpendicular just geometrically because the perpendicular will give you the height of the parallelepiped. So by multiplying it by the area of the parallelogram will give you the total value.

[@Craznar]

Where were you 35 years ago when I did this at university :)

[@ZardoDhieldor]

Not born then, I assume... :D

[@vedant6633]

Lepus Lunaris yeah! He's quite young!!

[@kushaljain9312]

To sum up such thing in just 13 min is really something else. Thank you.

[@learnbycoding]

It is amazing how well he explains the concepts. He makes things so clear.

[@khautukt4700]

01:47 I think the last row of "Numerical Formula" should be: v1w2 - v2w1

[@aminarahman2429]

I wish all the teachers were like you. :( Thanks A LOT for spreading your knowledge. These videos are priceless.

[@teodoranasz3531]

Sir I am forever grateful. I reeeally hope that your channel is still around when my future kids are ready for it.

[@andrecalado4652]

I liked the analogy given at 7:00 where we equate the cross product to a dot product between our new unknown vector and (x,y,z). However I think it becomes fairly easier and clear that when we just plug in the basis unit vector (x,y,z) = (i,j,k) that by solving the linear set of equations we can compute the new vector that is the cross-product. Great video!

[@rakittna90]

This series is so good! What about explanation on Clifford algebra?

[@parameshnat]

The bit that took me a couple of rewinds to understand is why the length of P is the same as the area of the base parallelogram. Here's how I understood it. The dot product between P and X results in a volume. The dot product between two vectors X and Y: 1. projects X onto Y 2. scales it by length Y (as discussed in 3b1b's video on the dot product). Therefore P.X is projecting X onto P and then scaling it by the magnitude of the length of P. I. E. Volume = P.X = length(projection of X on P) x mag(length(P)) The volume is also: Volume(parallelopiped) = area of parallelogram x height Comparing the two Area of parallelogram x height = length(projection of X on P) x mag(length(P)) Where height is by definition, length(projection of X on P). The area must therefore be equivalent to the magnitude of P. Hope that helps someone out there.

[@eduardojreis]

One of the most insightful lectures I ever watched. Thank you so much! Mind blowing.

[@wombat6]

I've worked my butt through 10 semesters of college math and I must say you're a better teacher than every prof I had in whose class we used the word "duality"

[@toadkatze5503]

For anyone struggling to understand - this might help you: a very important part is at 7:36 From there until 8:15, everything should be clear so far since both terms actually are the same. Note however (IMPORTANT) that the left side of the equation at the top, which is a vector dot product, gives us a scalar (number). Also does the right term, the derterminant. 👍🏻 8:33 - now, on the left, there is a vector cross product, which itself gives us A VECTOR. Now comes the critical part - if i, j and k would not be there, what would the result of the determinant be? A scalar! So it would be: a vector = a scalar That cannot be the case, so we need a workaround to make the result of the determinant become a vector also. That is where i, j and k come into play, since they sort of “collect” each of the components below, packing each of them into a vector: 8:41 (no comment) Now, you should be able to comprehend how both equations I was referring to correlate to each other :) Spoiler: determinant (with i, j, k) = p, leads to: determinant(i,j,k) * (x,y,z) = determinant(x,y,z)

[@arpanpiano]

This is what I understood after watching a couple times: Computational approach: We defined a function that took 3D input to a 1D output. Also this function is a linear transformation. These two facts help us find dual vector p such that p · = det(, v, w). Geometric approach: Consider the volume of the parallelepiped. Since volume = Base x Height = det(, v, w), we were able to find dual vector p' such that p' · = det(, v, w). Now we leverage two facts: p and p' are dual vectors that correspond to the same linear transformation. Also, a linear transformation has ONE corresponding dual vector. Thus p' = p. That is, the geometric approach is equivalent to the computation approach.

[@arpanpiano]

I'm taking a theoretical linear algebra course right now and wow have these videos kept on giving. Duality, and linear functionals (space to line linear transformations) is what my class if focusing on right now and these videos give good intuition!

[@pythagorasaurusrex9853]

It's 2020 and I just stumbled over this video. Excellent explanation of the geometric use of the determinant and the special products of vectors! EDIT: This is the first time, someone explains the meaning of a dual vector geometrically and not just in an abstract way what you see in so many (bad) text books.

[@RoselineJerryA]

love your name!!

[@JTCF]

I wish we really did learn about the cross product better in school. Because if I remember correctly, we interpreted a cross product of two 2d vectors as a 2d vector! And that didn't make any geometrical sense! Then, after the school was over, I learned about properties of the dot product, and recently had a problem where I needed "a dot product but sideways". What I meant by that thought is the projection of one vector to a perpendicular vector of another. Turns out, cross product is exactly what I needed! And then I also got to use it in 3d space for finding a perpendicular vector in a situation when I have "forward direction" vector and "down direction" vector. Thank you for putting those videos out, they really help solidifying the knowledge I get from various sources.

[@theflaggeddragon9472]

Where did you get this explanation from? Or did you invent this chain of logic? Also could you do a video on exterior calculus and clifford algebra?

[@TheKostalomac]

im just writing a comment so youtube algorithm thinks this is some popular scandal topics and recommends it to more people

[@siddharthviswanathan1671]

I wish I had a Math teacher like you in my school. Nevertheless, it is never too late. Its an eye-opening of how math is so beautiful and I can't stop smiling at the concepts explained by you. Its total bliss. Thank You

[@y2k898]

Changing the perspective from computational to geometrical really made the whole thing intuitive! Wish I had those lecture in uni

[@dark3l192]

watched 3 times and still my peanut size brain didnt understood anything, gotta rewatch i guess

[@nicholashassan8711]

At 6:18, how do we know the function is linear based on properties of the determinant?

[@sanghi286]

I had to watch this multiple times to get exactly what was being said but it was definitely worth it. And I have a college degree and considered myself above-average at mathematics. Perhaps I have to reconsider. And for those, who directly came to this video, please watch the earlier ones as well for a clear understanding of duality and the idea of treating a vector as a linear transformation.

[@angelsandemons25]

I had almost failed in math in my high school. Here I am, learning data science at my own pace (self taught). I had tears of joy when I finally understood this video completely. Sure, it did take several attempts! I just understood that you can never be bad at anything as long as you learn it from the right source!!! YOU ARE AMAZING!!!!!!!!!!!!

[@Heroni66a]

I had to watch this 6 times. THANKS to the INTEL CELERON IN MY BRAIN.

[@user-fs1iy8ux6t]

Holy shit, how can you explain this concept in such a brilliant and convincing way！

[@newkid9807]

How do I say your name?

[@ethernet764]

Rewatching works. I had to rewatch the video about Determinant, Dot product, and Cross product MULTIPLE TIMES across MULTIPLE DAYS before finally understanding what this video wants to show. To reinforce what I learned, I'll try to explain using my own words. If I understood correctly, what this video is trying to say is that there is a relationship between the Dot product and the Cross product. Since both concepts can be defined using the Determinant, we can establish a definition where the projected vector in the Dot product and the vector which the Cross product aims to find out are one and the same.

[@meinkea]

My mind has been blown, and than remade better than it was before. Thanks!

[@gaatutube]

I think I will have to watch this at least 10 times before it might begin to make sense.

Well done. Also, in 1:46 I think the correct is v1w3 instead of v2w3 in the last row of the rhs vector.

[@3blue1brown]

Good catch! Thanks.

Thank you for this nice series!

[@gorgolyt]

video ruined series ruined everything ruined

[@fossilfighters101]

**runs around screaming**

[@andreapaulsen3006]

Marcos Lourenço . No the right answer is v1w2 - v2w1 instead of v2w3 - v3w2 in the last "row" of computed cross production.

[@altolegato2493]

This is amazing! I have never peeked under the hood before, it is so satisfying!

[@vigneshkrishnaraj605]

OMG.. Who can teach linear algebra better than this.. Thanks for such an awesome video.

[@ElchiKing]

Hm, when I first learned about cross products some years ago, I realised that for any vectors u,w,v u.(v x w)=det(u v w), but I didn't realise (until now) that v x w is the _unique_ vector, such that this property holds. But I managed to finish lectures on linear algebra without ever having heard about cross products (I learned about them short time later).

[@jaoreir]

Woah, there's going to be a pink pi person in the next episode? I thought there were only blue and brown guys! haha

[@Ucedo95]

I got why this channel is named 3blue1brown because of this comment.

[@jaoreir]

Darth Armot Correct me if I'm wrong but I think the name came first

[@zairaner1489]

Yeah I believe he confirmed it somewhere else that the name has to do with his eye colours

[@jaoreir]

Don Sanderson I knew that it is related to eye color, but I have no idea about the details. I mean, he doesn't have four eyes right?

[@zairaner1489]

Maybe one of his eye is brown and one is half brown, half blue? Maybe it is just intensity of the colours.Jaoreir Music 焦耳音樂

[@asterisqueetperil2149]

Nice one ! I am personnaly used to think of it the other way, from its utility to its formula. So the question is : "how can I construct a vector which represent an oriented element of surface (to calculate the flux for exemple) ?" 1) The first way is to build it from the basic vector (i,j,k) by setting the following rules : i x j = - j x i = k , j x k = - k x j = i , k x i = - i x k = j (so i x i = j x j = k x k = 0) which gives you the good orientation and the good surface for the basic vectors, and we take the cross product ...x... to be linear on each side to allow you to extend this "good properties for basic vectors" to any linear combinaison of the basic vectors (thus extending it to any vectors). This set of rules has all the good properties and allow me to find the vector I want from any pair of vectors like this : (a.i+b.j+c.k) x (d.i+e.j+f.k) = ad(i x i) + ae(i x j) + af(i x k) + bd(j x i) + be(j x j) + bf(j x k) + cd(k x i) + ce(k x j) + cf(k x k) = ae(k) + af(-j) + bd(-k) + bf(i) + cd(j) + ce(-i) = (bf - ce)i + (cd - af)j + (ae - bd)k which is the classical formula 2) The second way is to decompose it with respect to each direction of space. If I have two vectors perpendicular to the vector i, I want their cross product to be collinear to i and to have a length equal to the oriented surface formed by the two vectors. I know a number which can represent an oriented surface, its just the determinant of the two vectors ! So in this case, their cross product is just the determinant of the two vectors times i. For two vectors (0, b, c) and (0, e, f), this gives us : (0, b, c) x (0, e, f) = det([(b,c),(e,f)]).i = ((bf - ce), 0, 0) Now I can apply the same reasoning to vectors perpendicular to j or k, and thanks to the linearity of the cross product I can sum up every contribution, which gives us : (a, b, c) x (d, e, f) = ((bf - ce), (cd - af), (ae - bd)) again the same formula This last approach can be used for higher dimensions, and is used for example in electrodynamics (electromagnetism + special relativity) to define a 4D vector which characterize a "3D surface". Forword : the important things are the properties of the cross product, and its the only thing that matters for any mathematical tool. Here we have : - anti-symetry : U x V = - V x U - bilinearity (linearity to the right and to the left) - transform two vectors to one vector Bilinearity gives "area like" properties ( area of a square of length 2a : (a + a)(a + a) = a² + a² + a² + a²= 4a² ). Anti-symetry gives "algebraic numbers like" properties (the order matters). Bilinearity+anti-symetry gives "algebraic area like" properties, just like the determinant. The third property is about orientation. Determinant associate two vectors to give a number, whereas cross product gives a vector. To form a vector we need a length, an orientation and a sign. The determinant gives us the length and the sign, so we need to add an orientation to have a vector. You can see how these things come into play in the two proofs I gave. These properties are also useful to define vectors that represents a rotation (something to do with how to transform such a vector when we reflect everything with respect to the plane of rotation, see pseudo-vectors) and to work with it. To name a few : L = r x p, uj = Nabla x B, B = Nabla x A ...

[@JavArButt]

I really want to thank you, it is amazing how much effort you put into your videos. This work really helps students like me to understand the real idea behind various mystical mathematical operations like Determinand, Crossproducts and so on. Furthermore, I think it is great that you do not go for stump combining two numbers, what I mean with that is, as an engineer most people can handle numbers, but only a few people know about the ideas behind them.

[@jbz3]

we have to go deeper.

[@conradschmitt2536]

9:54 - 11:01 best... analogy... ever... THAT IS A STROKE OF GENIUS!

[@yacinelaghmari]

The positive thing here is that this video was as diffiuclt as the previous ones. That gives some sort of feeling good feeling here where I understand that I know so little that I don't really even know how deep we are going, but I'm all here for it. Lets go

[@elijahperez2389]

this may be a dumb question but usually when I see a cross product the i j and k are on the top the the next 2 rows are the vectors. does this change the value? I haven't taken linear algebra but I'm in calc III and we use cross products a lot

[@ruffaa1995]

the determiant of a transposed matrix is equal to the determinant of that matrix (sorry for the bad english)

[@3blue1brown]

Good question, I meant to make a note of that in the description, so this comment was a good reminder. Most textbooks do seem to put things in the rows instead of columns, it's true, but it doesn't actually make any difference since the determinant doesn't change with transposes. However, I find it notably more intuitive to think in terms of columns.

[@JM-us3fr]

Yes I learned it this way too, but I think 3Blue's version is better for this particular explanation.

[@AuroraNora3]

3Blue1Brown It's standard notation in Norway to list vector coordinates horizontally, and transformations matrices vertically. It's kind of confusing for me to switch back and forth but the result of a determinant will be the same regardless. Excellent video. This really gave me a deeper understanding behind the computation of a cross product and why it's linked with the area

[@7chanconn7]

1:04 Wouldn't it be -J hat times some constant, or am I over thinking it?

[@3blue1brown]

It depends on whether you write that constant term as (v3w1 - v1w3) or as (v1w3 - v3w1). Different ways to think about the determinant will make a different one of these seem more natural.

[@doyouknowdawae2625]

Wonderful teacher, somehow understood this in one go through the relationship between the cross and dot products with both geometric and numerical understanding

[@syuhaichen3273]

I didn't realize the cross product is a great example for the duality, or the Riesz representation theorem. Thank you for your wonderful explanation.

[@theultimatereductionist7592]

I thank you for trying to correct a student's potential error of thinking a cross product would take three 3D vectors & spit out a number using a determinant as an analogy to using the determinant on two 2D vectors & spitting out a number. But, you went the wrong way with your analogy. The proper "generalization" of the cross product to two dimensions, or, rather, to an N-dim vectors sitting inside an N-dim world would be this: a cross product takes (N-1) N-dim vectors and produces an N-dim vector. So, for N=2, that the means the "cross product" acts on only N-1=1 vector, not on 2 or more vectors. The "cross product" would be a unary function in the N=2 case, taking the vector [a,b] to det[ [a,b],[i,j] ] = a*j-b*i = [-b,a], which is (one of) the (two) vector(s) perpendicular to [a,b] in 2-space with the same length as [a,b].

[@sripathykiran]

I hope you can answer my doubt. Assume a m x n transformation matrix from domain: A to domain: B where n is the number of dimensions present in domain A and m is the number of dimensions in the new domain B with m > n+1. Out of the m transformed basis vectors in the new domain B, the first n transformed vectors in domain B can be obtained from the product of the transformation matrix and the corresponding basis vectors in domain A. Am I right in assuming that the (m-n) basis vectors in the domain B can be obtained by the cross product of the n existing basis vectors in domain B? If it is true, how can it be accomplished, given that there are only n basis vectors in domain B of m dimensions, whereas the requirement for cross product as mentioned in the above comment is to have m-1 basis vector of m dimensions? Thanking you in advance.

[@david21686]

This is a pretty darn good explanation, but unfortunately, my mind is refusing to accept the idea of a 1D vector representing a 2D area. "Sorry, David21686, a unit of length and a unit of area doesn't match up. Time to eject all of this from your brain".

[@3blue1brown]

That's a really good point, and in a sense, it stems from the fact that the cross product doesn't really want to be a vector. It wants to be a transformation. The vector is just sort of a stand-in for that function. For example, multivariable calculus student will be familiar with the fact that in surface integrals, the cross product of differentials is naturally dotted with a different quantity (in a manner that makes computations much easier).

[@debasishraychawdhuri]

you drop the idea of unit to a vector - just think that force is a vector and length is a different kind of vector. they have different units, but they still are vectors. similarly, an area is a vector with a different unit. not all vectors have to be length.

[@vivekhebbs6340]

The units of a vector are not necessarily length. For example, a vector could be a force, a velocity, a momentum, or possibly even something with units of area. In physics, the unit of the product is always the unit of v times the unit of w, so there is no problem. And in math the vectors need not have units at all! Units are only necessary in the physical world.

[@Saltofreak123]

imagine a rectangle with the height of the vectors length and a width of 1. the value of the rectangle's area and the vector's length is the same, but now it's easier to mentally compare them.

[@totaltotalmonkey]

Numbers are 1d vectors

[@legr4ndk891]

"I finally grasped this after reviewing it a second time. To truly understand this chapter, a solid understanding of vector duality (the relationship between vectors and linear functionals on their dual space through the inner product) and the geometric interpretations of both the dot product (one vector projected onto another) and the cross product (the vector perpendicular to the area of both input vectors) is necessary". Once you get it, you'll see that the derivation is purely elegant!

[@ParthPaTeL-wm3kt]

I doesn't grasped after 1st watch then i watched previous video again and reminded all previous video's intuition, after that i watched this video again with pauses and rewatching some content 2 or 3 times, and now i got it. So if you don't understand, please remind previous videos and watch again carefully. Great work 3b1b, Keep it up. Thanks

[@ASLUHLUHCE]

Watching this a few more times should do the trick

[@ASLUHLUHCE]

(one day later) Mindblown

[@luffyorama]

I don't understand a thing. Need to rewatch the entire series back i guess :(

[@MIDOFCIS]

I kept replaying this like 7 times or more to understand the last part!!

[@maggielin8664]

Again, and again and...

[@srgkzy1294]

Just let it sit for a while :) digest it

[@alfcnz]

Rotflmao! 🤣

[@NomadUrpagi]

@@MIDOFCIS damn u have patience of a priest

[@Taricus]

i heart the quote at the beginning ^~^ in astrophysics and theoretical physics, I keep having to do proofs. It's really daunting sometimes, until you realize there is a very simple way of doing it. You feel lost and confused until you find that way. You really don't fully understand until that method clicks. :P

[@princeistalri7944]

I read a book that interpreted the cross product uxv as the dot product of an antisymmetric tensor (ux) and v, which looks like a linear transformation acting on a vector...

[@ZardoDhieldor]

Yes, you can do that. The transformation that maps _v_ onto _uxv_ is antisymmetric, in the sense that it's transposed is it's negative. Maybe 3b1b will talk about transpose matrices some time in the series. Wow, he sure doesn't lack topics to do videos about! :D

[@Magnasium038]

Are you guys referring to the double dot product between the permutation tensor (epsilon) and dyadic product of v and u?

[@ZardoDhieldor]

Tony Mathew That looks like a similar thing! _ux_ is just the (single) dot product of _u_ and _epsilon_.

[@Magnasium038]

That's an interesting 2nd order tensor... ux... what significance was given to it?

[@ZardoDhieldor]

Tony Mathew It is an endomorphism, so you can take the endomorphism exponential (=matrix exponential) _A(t)=exp(t ux)_ and get a uniform rotation around the axis spanned by _u_ with speed _||u||_. This is why the cross product is used in rotational physics.

[@AlexBitmap]

I think it would have been better to show that the vector p is really perpendicular to vectors v and w, because I think it is assumed that v and w are perpendicular to p. Otherwise, great insight in understanding the geographical meaning of the cross product! Thank you immensely for the video :)

[@AlexBitmap]

Also, at about 11:12 you mention the direction of the vector p, but just based on the video it doesn't seem evident to me (disclaimer: I'm a math idiot) that the vector p follows the right hand rule at the moment. I personally think that some more explanations about the direction of p can make things more clear.

[@johnfykhikc]

i had the same feeling

[@amineaboutalib]

because it's a volume

[@LAnonHubbard]

Phew. I got it. Eventually. Probably because I've already been working with dot and cross for a while. It was a relief to finally hear the music coming in at the end!

[@NomadUrpagi]

Excellent video to fall asleep to. I use it everytime i'm struggling!

[@sp4yke]

thanks very much for this series. one question, at 6:21 you say that this function is linear and I'm trying to find the intuition based on the determinant properties. I watched the determinant video and I don't understand how to put it in words.

[@mintyboi5543]

+1. I have this exact question. The brute force algebraic proof is straightforward, but not as insightful if there is a way to reason through that with the intuition from the former videos.

[@brianwahome5789]

I'm struggling with that too but Im assuming based on his definition of linear transformation retaining the area of the grids and keeping the grid lines parallel when space is morphed, Considering that the vectors U and V are held constant and what we have eliminated is Vector W, we replace it with a variable vector X so that any vector can be plugged in here and for some parallelopiped with the area held constant by U and V. The area of the parallelopiped, of course, can be dictated by the determinant of the matrice when actual valued for vector X are plugged in. The big question, as to how the function is linear, remember that the function basically gives us the area of the parallelopiped. Let's plug in some values into the function, be it a vector X, a vector Y or a vector Z, the parallelogram constant remains the same. The V, W grids always stay constant. The area being defined only by X means the grid spacing along the variable dimension X/Y or whatever will always be equal as they are dictated by just one Vector. Their grid lines will always stay parallel that is confirmed and depending on the input vector, The grid areas will always be equal post evaluation. This I guess qualifies the function as linear as linear. In fact, even think about it, if you add some magnitude m to the vector input, say mX, and assuming V and W are along i and j, you can basically come up with the volume mX * det(VW) as m(X * det(VW)). The effect of the added magnitude factor is transposed linearly in that .

[@DChuletaman]

A certain transformation f is linear if it verifies two conditions: f(u+v)=f(u)+f(p), u and p being vectors f(c*u)=c*f(u), with c being a scalar So we have to prove that the transformation f that for any given vector u=(x,y,z) returns the determinant of the matrix containing this vector as its first column and the other two columns containing the other two fixed vectors v and w, is linear. I'm going to do it in two parts, first let's prove that f(c*u)=c*f(u): c*u=c*(x,y,z)=(c*x,c*y,c*z)==>f(c*u)=det[c*x,c*y,c*z;v1,v2,v3;w1,w2,w3]=c*det[x,y,z;v1,v2,v3;w1,w2,w3]=c*f(u) This means that when you scale u by some constant c, the volume of the parallelepiped is scaled by that same constant, as you get c times the volume of the parallelepiped. Now let's prove that f(u+p)=f(u)+f(p), with p=(x',y',z'): u+p=(x,y,z)+(x',y',z')=(x+x',y+y',z+z')==>f(u+p)=det[x+x',y+y',z+z';v1,v2,v3;w1,w2,w3]=det[x,y,z;v1,v2,v3;w1,w2,w3]+det[x',y',z';v1,v2,v3;w1,w2,w3]=f(u)+f(p) This means that the volume of the parallelepiped spanned out by u+p, v and w is the same as the volume of the parallelepiped spanned out by u, v and w added with the one spanned out by p, v and w. So we have prooved that f is linear, at least this is what i got, i hope it's correct

[@adamjones7891]

Thanks Patrick Diego and Brian. Those explanations really helped! My roadblock is I was considering [x;y;z] as three different inputs and not 1 that is scaled altogether.

[@adarshkumar4336]

Not nearly enough people have seen this.