Hey folks! Where we left off, I was talking about how to compute a three-dimensional

cross product between two vectors, v x w. It’s this funny thing where you write a matrix,

whose second column has the coordinates of v, whose third column has the coordinates of

w, but the entries of that first column, weirdly,

are the symbols i-hat, j-hat and k-hat where you just pretend like those guys are

numbers for the sake of computations. Then with that funky matrix in hand, you compute its determinant. If you just chug along with those computations,

ignoring the weirdness, you get some constant times i-hat + some constant

times j-hat + some constant times k-hat. How specifically you think about computing

that determinant is kind of beside the point. All that really matters here is that you’ll

end up with three different numbers that are interpreted as the coordinates of

some resulting vector. From here, students are typically told to

just believe that the resulting vector has the following geometric

properties. Its length equals the area of the parallelogram

defined by v and w. It points in a direction perpendicular to

both of v and w. And this direction obeys the right hand rule in the sense that if you point your forefinger

along v and your middle finger along w then when you stick up your thumb it’ll point in the direction of the new vector. There are some brute force computations that you could do to confirm these facts. But I want to share with you a really elegant

line of reasoning. It leverages a bit of background, though. So for this video I’m assuming that everybody

has watched chapter 5 on the determinant and chapter 7 where I introduce the idea of

duality. As a quick reminder, the idea of duality is

that anytime you have a linear transformation from

some space to the number line, it’s associated with a unique vector in

that space in the sense that performing the linear transformation is the same as taking a dot product with that

vector. Numerically, this is because one of those

transformations is described by a matrix with just one row where each column tells you the number that

each basis vector lands on. And multiplying this matrix by some vector

v is computationally identical to taking the dot product between v and the vector

you get by turning that matrix on its side. The takeaway is that whenever you’re out in

the mathematical wild and you find a linear transformation to the

number line you will be able to match it to some vector which is called the “dual vector” of that

transformation so that performing the linear transformation is the same as taking a dot product with that

vector. The cross product gives us a really slick

example of this process in action. It takes some effort, but it’s definitely

worth it. What I’m going to do is to define a certain

linear transformation from three dimensions to the number line. And it will be defined in terms of the two

vectors v and w. Then, when we associate that transformation

with its “dual vector” in 3D space that “dual vector” is going to be the

cross product of v and w. The reason for doing this will be that understanding

that transformation is going to make clear the connection between

the computation and the geometry of the cross product. So to back up a bit, remember in two dimensions what it meant to

compute the 2D version of the cross product? When you have two vectors v and w, you put the coordinates of v as the first

column of the matrix and the coordinates of w is the second column

of matrix then you just compute the determinant. There’s no nonsense with basis vectors stuck

in a matrix or anything like that. Just an ordinary determinant returning a number. Geometrically, this gives us the area of a

parallelogram spanned out by those two vectors with the possibility of being negative, depending

on the orientation of the vectors. Now, if you didn’t already know the 3D cross

product and you’re trying to extrapolate you might imagine that it involves taking

three separate 3D vectors u, v and w. And making their coordinates the columns of

a 3×3 matrix then computing the determinant of that matrix. And, as you know from chapter 5 geometrically, this would give you the volume

of a parallelepiped spanned out by those three vectors with the plus or minus sign depending on the right-hand rule orientation

of those three vectors. Of course, you all know that this is not the

3D cross product. The actual 3D cross product takes in two vectors

and spits out a vector. It doesn’t take in three vectors and spit

out a number. But this idea actually gets us really close

to what the real cross product is. Consider that first vector u to be a variable say, with variable entries x, y and z while v and w remain fixed. What we have then is a function from three

dimensions to the number line. You input some vector x, y, z and you get

out a number by taking the determinant of a matrix whose

first column is x, y, z and whose other two columns are the coordinates

of the constant vectors v and w. Geometrically, the meaning of this function

is that for any input vector x, y, z, you consider

the parallelepiped defined by this vector v and w then you return its volume with the plus or

minus sign depending on orientations. Now, this might feel like kind of a random

thing to do. I mean, where does this function come from? Why are we defining it this way? And I’ll admit at this stage of my kind of

feel like it’s coming out of the blue. But if you’re willing to go along with it and play around with the properties that this

guy has it’s the key to understanding the cross product. One really important fact about this function

is that it’s linear. I’ll actually leave it to you to work through

the details of why this is true based on properties of the determinant. But once you know that it’s linear we can start bringing in the idea of “duality”. Once you know that it’s linear you know that there’s some way to describe

this function as matrix multiplication. Specifically, since it’s a function that goes

from three dimensions to one dimension there will be a 1×3 matrix that encodes this

transformation. And the whole idea of duality is that the special thing about transformations

from several dimensions to one dimension is that you can turn that matrix on its side and, instead, interpret the entire transformation

as the dot product with a certain vector. What we’re looking for is the special 3D vector

that I’ll call p such that taking the dot product between p

and any other vector [x, y, z] gives the same result as plugging in [x, y,

z] as the first column of a 3×3 matrix whose other two columns have the coordinates

of v and w then computing the determinant. I’ll get to the geometry of this in just a

moment. But right now, let’s dig in and think about

what this means computationally. Taking the dot product between p and [x, y,

z] will give us something times x + something

times y + something times z where those somethings are the coordinates

of p. But on the right side here, when you compute

the determinant you can organize it to look like some constant

times x + some constant times y + some constant times z where those constants involve certain combinations

of the components of v and w. So, those constants, those particular combinations

of the coordinates of v and w are going to be the coordinates of the vector

p that we’re looking for. But what’s going on the right here should feel very familiar to anyone who’s actually worked through a cross-product

computation. Collecting the constant terms that are multiplied

by x, y and z like this is no different from plugging in the symbols

i-hat, j-hat and k-hat to that first column and seeing which coefficients aggregate on

each one of those terms. It’s just that plugging in i-hat, j-hat and

k-hat is a way of signaling that we should interpret

those coefficients as the coordinates of a vector. So, what all of this is saying is that this funky computation can be thought

of as a way to answer the following question: What vector p has the special property that when you take a dot product between p

and some vector [x, y, z] it gives the same result as plugging in [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? That’s a bit of a mouthful. But it’s an important question to digest for

this video. Now for the cool part which ties all this

together with the geometric understanding of the cross

product that I introduced last video. I’m going to ask the same question again. But this time, we’re going to try to answer

it geometrically instead of computationally. What 3D vector p has the special property that when you take a dot product between p

and some other vector [x, y, z] it gives the same result as if you took the

signed volume of a parallelepiped defined by this vector [x, y, z] along with

v and w? Remember, the geometric interpretation of

a dot product between a vector p and some other vector is to project that other vector onto p then to multiply the length of that projection

by the length of p. With that in mind, let me show a certain way

to think about the volume of the parallelepiped that we care

about. Start by taking the area of the parallelogram

defined by v and w then multiply it, not by the length of [x,

y, z] but by the component of [x, y, z] that’s perpendicular

to that parallelogram. In other words, the way our linear function

works on a given vector is to project that vector onto a line that’s

perpendicular to both v and w then, to multiply the length of that projection

by the area of the parallelogram spanned by v and w. But this is the same thing as taking a dot

product between [x, y, z] and a vector that’s perpendicular

to v and w with a length equal to the area of that parallelogram. What’s more, if you choose the appropriate

direction for that vector the cases where the dot product is negative will line up with the cases where the right

hand rule for the orientation of [x, y, z], v and w is negative. This means that we just found a vector p so that taking a dot product between p and

some vector [x, y, z] is the same thing as computing that determinant

of a 3×3 matrix whose columns are [x, y, z], the coordinates

of v and w. So, the answer that we found earlier, computationally using that special notational trick must correspond geometrically to this vector. This is the fundamental reason why the computation and the geometric interpretation

of the cross product are related. Just to sum up what happened here I started by defining a linear transformation

from 3D space to the number line and it was defined in terms of the vectors

v and w then I went through two separate ways to think about the “dual vector” of this

transformation the vector such that applying the transformation is the same thing as taking a dot product

with that vector. On the one hand, a computational approach will lead you to the trick of plugging in

the symbols i-hat, j-hat and k-hat to the first column of the matrix and computing

the determinant. But, thinking geometrically we can deduce that this duel vector must be

perpendicular to v and w with a length equal to the area of the parallelogram

spanned out by those two vectors. Since both of these approaches give us a dual

vector to the same transformation they must be the same vector. So that wraps up dot products and cross products. And the next video will be a really important

concept for linear algebra “change of basis”

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.

Good🙂

Beautifully Explained ! Thank you for the insight !

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

Please make one on series calculus

This is really the best summary of this topic, thanks a lot!

I really love the way you describe.But as a non native english speaker,sometimes I really challange with some sentences you form.For example there is a 17 seconds of only one sentence between 8.58 and 9.15 which is totally mathematical,and i'm lost…Maybe you think about this makes a little bit harder to comprehend.Anyway,I appreciate your studies deep in my heart…

this explains the concept of why a determinant works to calculate area as well. determinants are dot products between a vector and the perpendicular of the other vector

I'm learning linear algebra at school, I don't have time for linear algebra

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

Could you please explain to me why we use j i and k in the determinant? In the geometric observation part, you never mentioned i j and k.

three years after you posted it, some idiot in a far away land watch. after watching it, he watched it in again, and the 10:23 minute mark, his brain snapped, all the pieces lined up. one cannot describe the joy the idiot felt as he witnessed the beauty of truth at that manifested at that moment. truly I say to you, thank you. you are an artist

it burns as much as "mitochondrion is the power house of the cells" that "the cross product of two vectors is a vector that is perpendicular to both". But apart from that, I know nothing about this magical being…

@58 seconds, should j hat not be + j(v1w3 – v3w1)? Also @8.04 seconds, should y hat not be + (v1w3 – v3w1)?

Incredible! It's a pleasure to help you at Patreon.

i pretty much understood everything, except why do we multiply v2*w3-v3*w2 for the first point of the P vector and same goes for the second and third coordinate. Is it similar to the geometric explanation you gave in the determinant episode? that geometric photo of a (parallelogram?) and how it explains the (a*d-b*c)?

if anyone could reply, it would mean a lot.

Hi Grant, thank-you so much for sharing your insight – I cannot tell you how much I appreciate it!

Can ask whether the natural extension to higher dimensions (e.g. in 4D, a ternary 'product' that returns a single vector) is useful? In particular, I've been told in the past that the cross product has a meaning in 7 dimensions as well as in 3. Is this the correct way to generalise from 3D to 7D and if so, why does it work only in 3D and 7D?

Thank-you for reading!

It took me this long to realise there's 3 blue students pi signs and 1 brown professor pi sign….

My god this was exhausting to follow..

But there's still something I don't get:

First he used a variable vector [x,y,z] as the first column of the matrix, I could (barely) follow all that stuff, but then at the end he replaced it with i, j, and k.

Aren't those vectors themselves?

I don't get what this implies.

————————————-

How do you get from

p→ • [x,y,z]

= det( [ x,y,z / v1,v2,v3 / w1,w2,w3 ] )

to

p→

= det( [ i,j,k / v1,v2,v3 / w1,w2,w3 ] )

(where p→ = v→ x w→)

This is really elegant.

It uses a geometrical property of the cross product and writes it in a linear transformation , then by using a notation trick on the left hand side of plugging i-hat,j-hat and k-hat to let the dot product remain a vector,and thus we get a way of computing the cross product on the other side.

Good video. I swore something was off until i rewatched it 4 times.

Some things were just explained in an order that felt vocally fluent but not logically simplistic. The sequence of words chosen confused me at some points (geometry section, maybe I have an odd thinking order). To anyone confused, really pay attention to every word.

Tomasz Kudłacz comment was somewhat more intuitive to me.

Dot product is defined for vectors of higher dimensioms (like 4 dimensional or 5 dimensional vectors). Is cross product defined for vectors of higher dimensions？If yes, how is it defined and what‘s its geometrical meaning？

My Engineering teacher regurgitates the computation trick from the textbook and gave the geometric proof of vectors in 3D space in a drawing. Later on through 3D engineering design software, I could see that the computation trick works. I suspected then that he had no insight and was jumping from a math formula to a concrete drawing. I have to watch the video many more times to let it sink until I can articulate it with math notations that it becomes second nature. Right now my intuition from what lies deeper like the duality, etc are just bubbling in my idiot box forming visual images. From an engineering perspective, a well developed intuitive visualization means you can now assess and tweak the engineering design tools. Amazing and You have enhanced the skills of yet another unexceptional. BTW – Anyone has been taught better in engineering colleges?

FFFFFFFF it was very hard and painfully but at the end I have it. It takes me more an infinite number of time and i had to watch the other videos of the series many time and taking write note. So don't worry if you don't grasp it fast.

Ow, it's because the area of the parallelogram is equal to the length of the p vector so that both multiplied by the projected vector gives a number line. Oh, that's elegant af.

Can somebody tell me why the function he describes was linear?

How many linear transformations to the number line are there? Duality shouldn't be limited to these two particular functions right? There have to be a ton of dual functions out there.

1:04 shouldn't j^ be (v1w3-v3w1)?

I'd like to praise the video for its quality but also point out a thing I stumbled upon while thinking, hoping that others with the same problem will be able to solve it. The thing is: I was not convinced that the geometric interpretation of the transformation leads to a direct conclusion that p has to be both perpendicular to the base of the v,w parallelogram with the length equal to its base. The reason is: such a vector is a candidate for p, but is it the only possible option? Are there others vectors that could, coincidentally have their dot product with [x y z] be equal to the volume of that parallelepiped. The answer is: no, there are no other possibilities (keep in mind that p has to be able to fulfill the criteria for any given [x y z]), and the reason is: focus on arbitrary [x y z] and vary it's orientation in space while preserving its length. The dot product depends on the angle between [x y z] and p, but at the same time, the volume of the parallelepiped depends on the angle between [x y z] and the normal to the base parallelogram. For the equality to hold for any [x y z], it must be that p is in the same direction as the normal to the base parallelogram so the both dot product and the determinant depend on the same parameter, as they are always equal. Once this is realized, it is easy to reason why the length of p has to be equal to the area of the base parallelogram.

ty

in 4:18 YOU SAID:

V×W= Det

But it is actually

V×W= Det*(k hat)

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.

I quickly understood the mathematical demonstration but I find the geometric demonstration not really convincing because I cannot "see" that the dot product result (a "length") is the same as the DET result (which is a volume).

I have no idea what just happened

This really helped. Went slowly through the vid and understood everything!

Easy,

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!

not helpful. sorry.

3Blue1Brown, u got a bug at 1:44 "Numerical Formula" 3rd line expression is just a copy of 1st one.

Bro, you had me stuck on this video for 4 hours

Thank you!

complex anaylsis plz

plz make series on complex analysis

Aaaa… sorry? Sir? Spotted a mistake… at 6:03 you say ”it seems like coming out of the blue”, but actually… ahem… aaarr.. we see that it was coming out of the brown…

Should've known Gibbs was behind this. He plagued me through chemistry, too.

Not gonna lie, this was pretty damn opaque. But fascinating nonetheless. This might be the first time I will have to re-watch the video several times. There were some "Aha!" moments scattered here and there, but I wouldn't able to explain it to someone else if I had to. So time to watch it again.

I love this video, thanks Grant, but I'm having a problem.

This question assumes we understand why cofactor expansion works to calculate the 3D determinant in the first place, and then rephrases it as a dot product.

But, geometrically, why does cofactor expansion work to calculate the 3D determinant?!?!

Grant doesn't have a video on it 🙁

Thanks!

Yesterday, after watching the previous videos on this topic, I got inspired to find an easier method to calculate de determinant of a 2×2 matrix. Given that it was the area of a parallelogram, I started with that. I took the vectors i(a,b) and j(c,d), between them the angle A, and the vector going from the tip of i to the tip of j (forming a triangle). I ended up with det= || i || * || j || *sin(A). Now, what is A ?

I got from Al Kashi's theorem cos(A)= (|| u ||² – ( || i ||² + || j ||²)) / -2 || i || * || j ||

Therefore, det= sin(arccos((|| u ||² – ( || i ||² + || j ||²)) / -2 || i || * || j ||

)) * || i || * || j ||

= sin(arccos( i . j / || i || * || j ||

)) * || i || * || j ||

(in my notebook it was a lot longer since I wrote everything in terms of vectors coordinates, not vector themselves, idk why, it just seemed easier to go through this thing that way

Then I thought… wow that's really helpful, a lot easier than (ad-bc)

Strangely, in this video I learn that the cross product's magnitude is exactly that, I felt relieved that I didn't screw up and a bit sad because I didn't invent a complicated formula that I'm not gonna use 🙁

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.

Interesting. I could visualize now that the gradient vector is a dual of Jacobian Matrix for scalar functions. Nice!

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

Yay i finally got it after the third rewatch , thank u u beautiful human

extremely good ..keep it up .

im totally lost

Well, nice, but WHAT THE FREAK happend at the end of the video!?

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

Holy shit! I just got it on my second viewing. Thank you so much for this.

That is so good. Thank you so much. I can't believe what i just watched..

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

Actually determinant is in chapter 6

Love the intro/outro music.

I cannot understand 😢😢😔😔😭😭

at 10.32 I have a question: how do you find the component of vector[xyz] which is perpendicular to the other two (which create the parallelogram) if those vectors do not create a flat plane?

Why is the determinant geometrically equivalent to a transformation onto the number line? Or is it not equivalent, but simply a convenient computational trick

tks

Why don't you are changing signs accordingly to changing signs patterns (from cofactor martix if I'm not mistaken)?

There is a vector which we may apply the dot product operation between it and x which gives the volume with x, v, and w. This is the dual, and it is unique (I need to check). By a basic geometric argument, a vector exists identical in the dot product function it induces. But then these are the same, or a linear transform has a non-unique dual. Whoa.

At 1:48, there is a typo in the cross product.

A HACK:

read aloud if you get confused

everything will be sorted ////

What about 4D cross product?

Why height/length = area of parallelogram ??

I just have one question. What is the use of VECTOR P!!!!!!?????? PLS SOMEONE ANSWER

6:25 If [x,y,z] happens to be on the same plane as v and w the volume of the parallelepiped = 0. Does it mean this function isn't injective? And shoudn't linear function be injective?

Simply wonderful!

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

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…

@3blue1brown, thanks a lot for your series! Been a pleasure to learn math concepts more conceptually.

Question- at 1:47 in the numerical formula, shouldn't the last row be v2w1 – v1w2?

I'd like to point out that the abuse of notation of putting i-hat, j-hat and k-hat in the matrix can also be expressed like this, using so well concepts that he presented in previous videos:

The function (which we will call "L") we apply to the [x y z]^t vector (^t stands for transposed, just imagine it vertical) must have a representative matrix [p1 p2 p3]. Of course, because of what we saw in the first videos on linear transformations, p1 is where i-hat lands after the transformation, so it is L(i-hat). In the same fashion, p2 must be L(j-hat) and p3 must be L(k-hat).

This way it turns out that p1 must be the determinant of the matrix (which we will call "A") where the first column is i-hat, the second column is v and the third is w (as that's the definition of L in the video).

But i-hat is [1 0 0]^t, so turns out p1=v2w3 – v3w2 – the other (four) terms you'd get by computing the determinant .tragically die because of the zeroes.

The same applies when we put j-hat as input variable: j-hat is [0 1 0]^t, so four terms die and two survive. Same old story for k-hat, which is [0 0 1]^t.

Putting i-hat in the same place where i-hat's "1" component would be and formally computing, obviously makes i-hat being multiply by the correct (scalar) expression. Same applies to j-hat and k-hat.

So abusing notation in that way is just a way to write down just det(A) and pretend to do just one calculation instead of writing it down three times and carry out three calculations to find out the representative of L in terms of the matrix [p1 p2 p3].

By the way, this man truly

isthe greater gift for who studies algebra from now on. I found all the videos about calculus brilliant exactly the same way, but they actually showed me just a few things I didn't got by myself a few years ago. This series in particular put me in front of a whole world of intuitive concepts I overflew while learning linear algebra. And, according to this comment section, I'm not alone.in essence what's really happening is you're looking for a vector p (perp. in the right direction) such that the volume of v,w,p = surface area (v,w) squared.

computationally it then makes sense for each of the "submatrices" when you do the determinant to just be squared and then the root of that is just p1,p2,p3. I hope i'm making sense because in my head it makes sense….

I love your video series and enjoying it thoroughly but despite many attempts to watch and rewatchs, still could not connect all the dots in this video. Perhaps you can speak with slow pace and emphasize where this area becomes length of this vector. I very much appreciate your deep insight but sometime it seems you jump few steps for average folks to appreciate this magic

I didn’t have time to finish the last five minutes, so I came up with my own finish on the go. Given that I am taking the determinant with the x, v, and w vectors, and that it can be espressed as a a dot product with the mysterious vector, I tried letting x be of unit length and perpendicular to v and w in the right-handed direction; e.g., a positive determinant. This volume is 1 times the +area of the parallelogram. If the unit x vector is wobbled in any direction, the determinant gets smaller, so the dot product vector must point in the direction of x, perpendicular to v and w. Also, we now know that the angle between the two vectors is 0, and the length of x is 1, so the length of the other vector must be the area of the parallelogram to satisfy x dot the vector equals the volume of the parallelipided/area of the parallelogram. Both work, but I thought I’d share what I came up with. Mine is easy to understand, but less likely to be come up with if the cross product was not already known.

If the Mathematicians had a choice, why did they make it so that the cross product value is different for different dimensions?

I mean( and I'm assuming that this is a concept that we invented), if i had the chance to I would make it so that it would spew out a number no matter whether the vectors are 2-D or 3-D.

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!

+3Blue1Brown Something I noticed: if the goal is just to derive a formula for the cross product, you can disregard the 3D determinant and its formula completely. Just think in terms of the volume of the parallelepiped. It's easy enough to show that this gives a linear map from R3 to R. All that remains is to find the images of the basis vectors. Consider k for example. The parallelepiped defined by k (and v and w) has volume equal to its height (1) multiplied by the area of its cross section in the x-y plane. The components of v and w in this plane are just (v1, v2) and (w1, w2), giving an area of v1w2 – v2w1. So this is the third (k) component of the map / the dual vector. The other components follow in the same way.

What I think is that duality and dual vectors is making this concept tough to understand.

I have some difficulties to connect scalars x,y,z with vectors i^, j^, k^ in the two determinant expressions

Finally I got to know that why the cross product is perpendicular to the plane of two vectors.

THANK YOU SO MUCH 💕😊

At 4:36 there he denotes things wrongly rest is fine

8:17 mind-blown!

4:00 I don't understand why there is a different cross product in 2D and in 3D. Last video you said that what you described in 2D "is not actually the cross product", but in this video you do refer to it as a cross product. Why is it different in 2D and 3D? Linear algebra should be the same (as in, it can be generalized) to any number of dimensions, right?

Okay but why is the cross product defined this way, why is it useful, and how do you visualize all of THAT?? These last two videos, which relied heavily on trying to explain the "traditional interpretation," really didn't work for me. I'd have much preferred you explain from scratch why we care to know these values, and how that motivates their definitions and calculations.

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.

I guess noone is chanigng the video speed on this one.

Definitely need to watch this one again

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".

Motivation for this equality: duality. Since the determinant of u and w with x as the first column is a function that takes x (3-dimensional vector) and outputs a number (1-dim) by duality, there exists a 1 x 3 matrix that encapsulates this linear transformation. This 1×3 matrix, when multiplied with x, is computationally equivalent to taking the dot product of the 3×1 vector with x. This 3×1 vector is called p.

Most important point:

To find the cross product of vectors u and w, we seek a vector p that satisfies the following:

The dot product of vector p with a vector x = the determinant of u and w with x as the first column.

Reinterpreted geometrically:

Project vector x onto p and multiply by the length of p = the volume of the parallelpiped formed by u, w and x

From the right:

the volume of the parallelpiped formed by u, w and x = area of parallelogram by u and w * component of vector x along the direction perpendicular to u and w.

From the left:

p is a vector perpendicular to u and w. Project x onto p and we get the component of vector x along the direction perpendicular to u and w. Now, all that's left to scale this value to make it equal to the right is to make the length of the vector p = area of parallelogram by u and w.

Excellent video. Chapter 6 was the determinant lecture, always better the second time, ha ha.

Dang. I keep thinking, I'm gonna have to watch this again. Each time it makes a little more sense, ha ha.

very interesting video and answered some questions that i've had, although this give me another question

what would this look like if we added yet another dimension?

do we need a 3rd vector? so that we're essentially finding a vector that is perpendicular to a box?

or is there another trick to fill in that space of the matrix (similar to using a vector of hats)?

Wow