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.
You should check out statquest. It’s similar in giving you intuition instead of seemingly arbitrary formulas for everything related to statistics and probability
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.
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.
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.
@@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.
@@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!
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.
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.
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.
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*.
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.
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.
@@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.
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.
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?"
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!
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
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.
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).
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.
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!
@@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.
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.
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. :)
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!
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!
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.
@@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.
@@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.
@@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.
@@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.
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.
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.
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 😅
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)
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.
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!
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*.
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.
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.
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
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
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. 👏👏👏👏👏
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.
Some points which when emphasized made it easier for me to understand 1. The volume of our parallelepiped = det([xyz, v, w]) = area of the base * height (measured perpendicular from the base) 2. The area of the base of the parallelepiped = the length of p 3. The height of the parallelepiped = xyz projected onto the unit vector (length = 1) perpendicular to the base 4. The projection of xyz onto any vector v perpendicular to the base is equal to xyz projected onto the unit vector perpendicular to the base, multiplied by the length of v 5. vector p is equivalent to the unit vector perpendicular to the base scaled by the area of the base. 6. p dot xyz is equivalent to xyz projected onto a unit vector perpendicular to the base (height) then multiplied by the length of p (area of the base) The volume of our parallelepiped = det([xyz, v, w]) = height * base area = p dot xyz p dot xyz essentially bundles finding the height of xyz (its perpendicular component relative to the base) and multiplying that height by the base area (the length of p)
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.
quick tip: v2w3-v3w2 is the view of a parallelogram while looking down the x axis v3w1-v1w3 is the view looking down the y axis v1w2-v2w1 is the view looking down the z axis imagine bouncing something off of the parallelogram perpendicular to the surface, the force down each axis is proportional to the area that can be seen while looking down that axis. an easy way to think about this is take a square and face it flat against the X Y plane, you can see 100% area while looking down z axis. now rotate theta degrees towards z axis, now view down z axis is cos theta times area, but view from x axis is sin theta times area. since sin^2 plus cos^2=1 (100% of area) and the absolute value of a vector is the square root of its component vectors squared, the area seen from each axis is equal to the components of P (the cross product). i hope this helps someone and if you have any ideas about how to interpret this please comment thanks!
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!
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.
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.
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!
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.
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.
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)
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!
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.
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.
Writing it down, really helped me to understand it. It might help you too. First let’s define all necessary vectors / matrices in correspondence with the video. I encourage you to draw them. h = [x, y, z]^T: The free-to-choose vector, named h for convenience. v = [v1, v2, v3]^T: The vector representing one base of the parallelepiped. w = [w1, w2, w3]^T: The vector representing the other base of the parallelepiped. p = [p1, p2, p3]^T: An unknown vector that is perpendicular to the base of the parallelepiped formed by v and w. p’ = [a1, a2, a3]^T: A vector from the components of h that is perpendicular to the base of the parallelepiped formed by v and w. Its length equals (h dot p) / |p|, because the dot product adds an unknown scaling factor for the vector we are projecting onto. Introduced at 10:27 M = [h, v, w]: A 3 by 3 Matrix, representing the parallelepiped, named M for convenience. The first thing to understand is, that we can get Det(M), the signed volume of the parallelepiped, with the formula below. Because the volume of a parallelepiped is just width * depth * height and p’ is the component of h perpendicular to v and w. Det(M) = |p’| * |v| * |w| But if the length of p would be equal to |v| * |w|, the area of the parallelogram, then we could also just write it as below because the dot product is the length of the projection of h onto p, times |p| = |v| * |w|. Det(M) = (p dot h) Now, how can we find such a vector h and p such that this holds? We calculate the determinant of the matrix, leading us to: p1 = v2 * w3 - v3 * w2 p2 = v3 * w1 - v1 * w3 p3 = v1 * w2 - v2 * w1 Which we can just calculate. Hope this helps someone, it was worth figuring out.
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.
"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!
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.
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].
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.
I watched this video somehow confused and started to systematically learn linear algebra until I reached the chapter of duality. Damn amazing. The elegance is proportional to the confusion
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".
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).
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.
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.
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.
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!
When I studied maths in university, all concepts are separate and only exist by themselves, but I focused on calculation and still achieved great scores in tests. After watching these videos, I realized I just learned nothing...
Another helpful way to interpret the takeaway from this video is that the cross-product of two vectors can be used to calculate the volume of any parallelepiped that can be formed from the parallelogram of those two vectors as a base. Essentially, any two vectors form a parallelogram on a plane. Any third vector added can form a parallelepiped with the parallelogram as its base. The component of this third vector that is orthogonal to the plane of parallelogram will have a magnitude that is equal to the height of the 3-D shape. Taking the dot product of the third vector with the cross-product vector essentially then produces a simple base x height volume calculation, where the magnitude of the cross-product vector is the area of the base and the magnitude of the parallel component of the vector is equal to the height.
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
8 лет назад
Well done. Also, in 1:46 I think the correct is v1w3 instead of v2w3 in the last row of the rhs vector.
After having watched the video 4 times, I sat down determined(no pun intended) and finally got it on my 5th watch, and the main thing to do is unlearning all you know and following his lead. Thanks Grant.
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"
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".
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).
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!!!!!!!!!!!!
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
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.
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
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
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 :)
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.
You really helped me a lot with linear algebra. When I use these intuitional ideas to learn crystallography, I found someting interesting. Say, if we have an full rank 3×3 matrix {a1 | a2 | a3}（a,b,c are vectors）and its inverse transpose matrix {a1' | a2' | a3'} we can found that : a1' is perpendicular to the plane of plane span {a2,a3} while the dot product of a1 and a1' is 1 a2' is perpendicular to the plane of plane span {a1,a3} while the dot product of a2 and a2' is 1 a3' is perpendicular to the plane of plane span {a2,a3} while the dot product of a3 and a3' is 1 (Such propositions also appear in square matrixes of all size.) In fact, in crystallography, if a1, a2, a3 are the basis of some crystal's lattice, then a1'， a2'， a3' happen to be the basis of its reciprocal lattice. However in crystallography textbooks, the relationship of inverse transpose hardly mentioned, rather, a1' , a2', a3' are defined as: a1' = (a2 × a3) / det {a1 | a2 | a3} a2' = (a1 × a3) / det {a1 | a2 | a3} a3' = (a1 × a2) / det {a1 | a2 | a3} In fact, this reciprocal lattice idea can be an good example to show the inner relationship between cross product and inverse transpose matrix. And during the process I am proving this relationship, I found that adjoint matrix of {a1 | a2 | a3} can be written as the transpose of { (a2 × a3) | (a1 × a3) | (a1 × a2) } \ . Since the product of a matrix A and its adjoint matrix A* generates some determinant while a1·a2 × a3 also generates some determinant, it seems that adjoint matrix can be understood rather intuitionally. Could you make an video to visually all these ideas? I think it can help the world a lot. Thanks!!!
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
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 ...
i feel like this man has so much power over us like he could easily convince me of a very wrong idea being the true one since im kinda dumb uni student whos only taught to compute things the whole life, and i get too excited by beautiful maths concepts explained by an adorable pi fella but he wakes up and chooses to bring knowledge and peace to the world, not chaos, and its absolutely brilliant
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
