The meaning of the dot product | Linear algebra makes sense

[@Improbabilities]

Just for the record, I do notice the effort you put into these videos. You're doing great! Keep it up!

[@LookingGlassUniverse]

Thank you, that’s very very kind of you :)

[@minandychoi8597]

Looking Glass Universe yeah, please know it doesn’t go unnoticed by us!

[@LookingGlassUniverse]

I really really appreciate that :)

[@icyminglun5138]

IM SO GRATEFUL FOR THESE textbooks arent doing much for me with numbers and formulas :') so thank you mam

[@KHANPIN]

Honestly, I wasn't expecting those pauses in the video, but turns out, they helped a lot with just grasping the concept and understanding. Thanks!

[@shredda7]

literally the thumbnail alone explained it better than everyone else

[@FortniteMaster-vi6qt]

This is incredible, taking a concept that confused me to no end and helping me understand in 13 min. Thank you so much

[@lucasfreitag9794]

Finally a video on KZbin that gives a really intuitive visualisation of the dot product. Well done. Thank you.

[@LookingGlassUniverse]

Thanks so much!!

[@bijannatividad]

I've been looking at a lot of videos and forums looking for the practical meaning of a dot product. Every video and forum I found only defined what dot product is and they were very technical about it. This video gave me the answer after 5 seconds. Thank you!

[@Anna-qp1fi]

The first 8 seconds literally answered a question I've been trying to solve for the last 20 minutes with multiple websites and textbooks.

[@sanjaisrao484]

I cant even imagine how much work you did for this video , its great

[@tejashreebhat127]

The amount of creativity in this video , the amount of efforts you have put in editing this. Salute. 🔥🔥

[@aion2177]

Wow. Finally i understand whats the point of the dot product. Should be called "similarity product" or something to point to it's meaning. Can't believe i went on all this years not understanding this basic idea.

[@no_360scope]

this is actually so sick, a lot of work went into this and it really did help me

[@roger72715]

Suddenly discovered this channel. Loving it!

[@niranjanarunkshirsagar]

Amazing! Our human brain is evolved such a way that we understand anything very easily if it is intuitive & your techniques are so intuitive and gripping that anyone who is very weak in maths can learn by heart the very concept behind "dot product". In that way, your techniques can be called "compatible with evolution". Richard Feynman also had this ability. Go ahead!

[@pilotomeuepiculiares3017]

WOW. This video is better than any classes I had in school or college. I'll watch again later and do the "youtube homeworks" in paper and not half-way in my mind. Thanks

[@LookingGlassUniverse]

Thank you very much! Did you try the homeworks :)?

[@snehasishsaren8422]

How deeply you explained it!...OMG!...This is the only video in youtube I came across, to know more deeply dot product and it's significance...You are really doing great. 🌈

[@JohnWalz97]

You're videos are amazing! Really helped me understand super daunting subjects. You're ability to break these topics down and explain them in a straightforward way is so awesome!!!

[@colorx6030]

Holy crap I just wanted to understand dot product for my Physics class but I somehow ended up in Linear Algebra.

[@pulse5863]

I honestly got stuck at 3B1B Dot product video, and ended up founding this absolutely brilliant and lucid explaination, thank you so much for the efforts! I am making my first patreon contribution to your channel, i hope there is a link!

[@sanjalisubash6307]

Wow! I just started linear algebra and was already confused as to why dot products matter. The way you explain this is so good and you can really make out the effort in the videos. I can't believe I haven't seen your content before! Subbed right away!!

[@Grevil76]

Wow the production quality on this vid is crazy! Thank you for the great explanation too!

[@f1cti]

Outstanding explanation of the dot product! This is the first video-and I've seen plenty- that's helped me get an intuition for this concept. Thanks so much!

[@usernameisamyth]

Ain't u gonna extend the playlist? I loved ur explanation.

[@acborgia1344]

For Homework 2: If we take v_b as the unit vector on its axis, from the previous exercise we know that x = v_u . v_b / ||v_b|| and y = v_v . v_b / ||v_b|| Now we add v_u and v_v v_u + v_v = (x+y) * v_b + perp We apply the same theorem on x+y as on x and y (x+y) = (v_u + v_v) . v_b / [[v_b|| Now replace x and y by their expressions (the ||v_b||s cancel out) v_u . v_b + v_v . v_b = (v_u + v_v) . v_b (1) This shows the distributivity of the dot product (v_a * k) . (v_b . s) = ||v_a * k|| * ||v_b*s|| * cos(theta) = k * s * ||v_a|| * ||v_b|| * cos(theta) = k * s * v_a . v_s This shows the scalar multiplication property of the dot product (not sure if that's the right term lol) v_u = alpha * v_v1 v_v = beta * v_v2 Now we replace v_v and v_u in the formula (1) (alpha * v_v1 ) . v_b + (beta * v_v2) . v_b = (alpha * v_v1 + beta * v_v2) . v_b From last property this is the same as: alpha * v_v1 . v_b + beta * v_v2 . v_b = (alpha * v_v1 + beta * v_v2) . v_b This shows the dot product is bilinear Homework 2 finished

[@pretzels3273]

Why can you just make v_b the unit vector? Aren’t you ignoring its magnitude then

[@acborgia1344]

​@@pretzels3273 I'm honestly not sure why I wrote that (it was 3 years ago), probably a mistake since I never used that property later

[@raphaelsisson3043]

It's truly awesome (in every aspect) the videos you make. You don't need to feel bad about taking some time off. For me at least your videos are a gift and it is not fair with you that we demand anything. That's just my long way to say thank you!

[@LookingGlassUniverse]

Thanks so much :D!

[@MrEiht]

While studying I listened to an old creepy prof. For hours and hours. And you run that down in less that 14 min AND with your sweet voice...thanks :)

[@harrytresor8654]

Answer for the last question: So we have vectors a and b that can be written as a1x + a2y for vector a and b1x and b2y for b. Given the linearity formula for the dot product. a•b = (a1x + a2y)•(b) = a1x•b + a2y•b Substitute for b and we get: a1x•(b1x + b2y) + a2y•(b1x + b2y) a1x•b1x + a1x•b2y + a2y•b1x + a2y•b2y The dot products of orthogonal vectors is 0 so the equation simplifies to: a1x•b1x + a2y•b2y Since the are facing in the same directions, the is the same as multiplying their magnitudes together. Thus a•b = a1b1 + a2b2 ...

[@LookingGlassUniverse]

Nicely done! Thanks for typing it all out :)

[@pretzels3273]

Aren’t you using what we are trying to prove in the proof? You go from a1x•(b1x + b2y) to a1x•b1x + a1x•b2y. You’re using the idea to prove itself. Are you sure that’s allowed?

[@draganjonceski2639]

finaly i understand the dot product i have been looking for this kind of an explanation for so long but only now i realy understand it thanks thanks thanks!!!

[@RichardDownie79]

I am madly in love with your voice!

[@hadmeinthefirsthalfngl2717]

Every comment section of every female youtuber ever. Very cliche way of simping.

[@anegativecoconut4940]

3:35 No God! No God Please no. No! NO! NOOOOO!

[@roygreen8265]

I cant tell exactly how gladful im after this point of view, i thought about it a lot and now thinga makes more sense!

[@christophealexandre1538]

We have been missing you! Every video you make is a little gem.

[@LookingGlassUniverse]

Thank you so so much- that means the world to me.

[@bbq2024lovely]

I realised the depth of the info you share, not just the rules of it. I love it. Just a recommendation, I think it would be much better if it was less distracted. I wanted to focus on sth, but I just kept looking here and there on the necessary graphs, forgetting the point.

[@vasylarsenii4800]

Thank you so much! You're an extremely talented teacher.

[@rizwansabir2448]

after the 1 miunutes of this video ... i am here to comment " it is the best video i had watch so far" luv ya

[@andresvalera1430]

Yay i found another great math channel!, thank you very much now i can understand this not like those "explanations" that basically say the definition and the properties

[@moosehead13]

Super cute video! Very well broken down and thought out! Think you might get a laugh out of this ...I paused the video at some point and there was a black spot on your dry erase board took me a few mins of wiping my screen to realize it was in the video lol. Keep up the content & thanks for taking the time to make this video!!

[@rohitonam5130]

you are great....the animation is great, the moment trig enters DHAN TAN TAAAAA... i laughed very hard..😂😂😂😂 i'll never forget this for my life ,,,, I m very glad i watched it. thank you for this video, it explains very neatly and clearly the basic. I found exactly what i was looking for

[@FF7EverCrisis]

This is the best explanation to dot product. Thank u

[@evasuser]

With gems like this we can partially forget or tolerate the crap that youtube is filled with. Thank you for the video and please keep making. How about PDEs or probability?

[@malna-malna]

I know nothing about filming but this does look good! :) Also, it's so good to finally learn the meaning behind this formulas I've known for years. Thank you!

[@LookingGlassUniverse]

Yay! Thanks for noticing the 1% (probably lower) increase in video quality.

[@55562009]

dot product between one orthonormal basis vector with itself comes out to be 1 because 1*1*cos 0 =1 and with any other orthonormal basis vector it becomes 0 because they are orthogonal i.e. cos 90. Therefore we get the neat formula to multiply corresponding coordinates to get dot product between 2 vectors. Thanks I really had that doubt why this formula works.

[@alonsorobots]

This is SO good! Very entertaining. Critical in convincing people to learn =P

[@hudahussaini9669]

LOL not the horror flashes for the pi circle. Great Video

[@ElPsyKongroo]

I totally forgot about this and we started using dot products for the first time since calc 3, thank you!

[@LookingGlassUniverse]

Great!! So happy to help :)

[@kirancp4758]

This is the best intuitive explanation for the dot product .Can you please do the same for cross product

[@breadyogurt3957]

Can we reboot the series, starting with the cross product? 👀

[@skidogleb]

New favorite math channel! wow this was the style of explanation I've been searching for, such a helpful conceptual take.

[@henrycraft9541]

Yes, an identical unit vector in the opposite direction should be minus. (And this means the dot product has to be Zero). I think I get it! I really like your question, to what extent are the two vectors going the same way.

[@dibakarbarman4662]

this type of teaching is very good and very useful for beginners .

[@LookingGlassUniverse]

Thanks very much!

[@thevideoride]

This video was awesome. A beautiful lecture that help the viewer better understand a confusing topic.

[@pennyl8773]

For 8:09, if x = u cos theta, then why isn't y = u sin theta? For u.v2 , wouldn't the angle be 90 - theta? Thanks.

[@Psd863]

Yay, she's back!

[@daviddeleon292]

Amazing videos. This really helped me understand the dot product.

[@Camilacpinto]

Excellent video, and I loved the toothpast tip!

[@BogdanTNT]

Underrated video. Thank you

[@Thror251]

Is it usual to write the dot product with the dot at the bottom? Everyone I know always writes it with the dot in the center (in LateX this would be \cdot).

[@duckymomo7935]

Thror251 There’s many ways Some write it like this

[@LookingGlassUniverse]

@Mi Les yup :) I write it down because I use \cdot for multiplication sometimes. But it doesn't matter too much!

[@michaelsommers2356]

@@duckymomo7935 Or .

[@michaelsommers2356]

@@LookingGlassUniverse I use \bdot for the dot product, but I also use bold symbols for vectors. I define it as: \usepackage{amsmath} ewcommand{\bdot}{\boldsymbol\cdot} and for the cross product ewcommand{\cross}{\boldsymbol\times}

[@LookingGlassUniverse]

Oh, that’s a nice way to do it!

[@danielmccarthy4741]

Excellent video, thank you for your hard work.

[@mmoonchild276]

May God bless you! This is beyond great.

[@HamidKarimiDS]

Kudos! You are an amazing teacher! Well done. Keep up the good work.

[@ironpatriot2905]

وَحَدَّثَنَاهُ قُتَيْبَةُ بْنُ سَعِيدٍ، عَنْ مَالِكِ بْنِ أَنَسٍ، عَنْ أَبِي بَكْرِ بْنِ نَافِعٍ، عَنْ أَبِيهِ، عَنِ ابْنِ عُمَرَ، عَنِ النَّبِيِّ صلى الله عليه وسلم أَنَّهُ أَمَرَ بِإِحْفَاءِ الشَّوَارِبِ وَإِعْفَاءِ اللِّحْيَةِ ‏.‏ This hadith is in Sahih Muslim. Growing beard is a wajib act and not growing it is a sin.

[@sergiolucas38]

perfect didactic and editing, youre very talented :)

[@TheViolaBuddy]

HOMEWORK 1 Answer 5, a . b / |b| (where the period is the dot product) HOMEWORK 2 Let's assume we're in 2D and the direction orthogonal to b is called a. Then: - As per question 1, u can be written as (u.b)b + (u.a)a - Likewise, v can be written as (v.b)b + (v.a)a - If we add the two, we get that u + v = [(u.b) + (v.b)] b + [(u.a) + (v.a)]a - But from the question in the video before question1, we know how the coefficient in front of b is related to the dot product of the sum and b: (u.b) + (v.b) = (u + v).b; this is exactly what we're proving. Homework 3: Starting with the basis vectors: - vi.vi (the same basis vector dotted with itself) is 1, because they're in the same direction and are of unit length. - vi.vj where i != j (different basis vectors dotted with each other) is 0, because they're in perpendicular directions. - Correspondingly, vi = [0; 0; ...; 0; 1; 0; ...; 0] with the 1 in the ith position in the vector. The sum over k of vi[k] * vi[k] (where k is the index in the "list of numbers" vector vi - this is equivalent to the given definition we're trying to prove) gives 1, as we expect; the only term in that sum that is nonzero is when k = i. Similarly, the sum over k of vi[k] * vj[k] is zero, because the 1's don't match up. - Now consider a and b, where they can be written as a = sum over i of (a.vi)vi, and b = sum over i of (b.vi)vi - If we dot the two together, we can distribute (which we can do because of linearity from HW2) to get that a.b = sum over i of [sum over j of (a.vi) * (b.vj) * vi.vj] (we can reorder a.vi and b.vj because they're just numbers). - But most of the terms in the double sum is zero: in fact, every term where i != j is zero, as we showed three bullet points ago. So we can just drop these zero terms and only care about when i = j, which collapses our sum down to: a.b = sum over i of (a.vi) * (b.vi) * vi.vi - And of course vi.vi is just 1 as well: a.b = sum over i of (a.vi) * (b.vi) - And of course a.vi is just the ith element in the a column vector. So we can rewrite it as a.b = sum over i of a[i]*b[i], which is the second form of the dot product equation that we're looking for.

[@LookingGlassUniverse]

Well done!! And thank you so so much for putting the effort into answering this :)! I really appreciate it

[@bhawnaklair3681]

@@LookingGlassUniverse Firstly, thanks for the help! It's the first time i took pen and paper to do some demos, other wise i assume it's too easy for me...but it turned out it's not at all!

[@bhawnaklair3681]

But now i'm really stuck at the 2° homework. The answer helped me...but I can't understand the last point, where you talk about something previews to the question 1...can you plz help me...

[@atheequest9264]

The way you write the letter b is killing me 😤😤

[@fermongrays1984]

I swear I spent 2 minutes thinking it was a G

[@Anna-qp1fi]

Definitely subscribing! This is an amazing channel!!! Very well explained :)

[@Sanjaybsnl08]

It was a really nice one :) ,it actually gave me a clear picture of dot product

[@mbappekawani9716]

the reason the dot product increases the more similar two vectors are and is smaller the less similar they are is because: 1. you can breakup each vector into the same basis components that can have opposite magnitude pieces of each other when two vectors point in different directions. 2. those negatives cancel out a lot in the calculation with the positives

[@jeroenw9853]

I never knew I had all these problems. But luckily I can solve (most) of them! 🧠

[@avanishpadmakar5897]

Let vector be represented as |a> Then each vector can be represnted as a matrix of the order N×1 where N is number of basis vectors and each number in the matrix will follow a(j1)=|a>•|v(j)> where 1 is smaller than or equal to j is smaller than or equal to N The matrix for product formula follows. And due to linearity property |a>•|b>=a1• (b1+b2....bN)+ a2• (b1....bN)..aN• (b1..bN) As ai is orthonormal to bj for i not equal to j so 0 for i not equal to j ai•bj ={ } |ai||bj| for i=j Hence |a>•|b>=sigma (|ai||bi|) where i varies from 1 to N

[@sajkdlsjf]

Excellent video and explanation. I applaud your efforts.

[@benharris7601]

Thanks for not explaining homework one now I have no clue how to do the rest

[@77kiki77]

Why don't you write 1/√2 as (√2)/2 ? It's easier to understand, and more flagrant with the remarkable trigonometric value than just 1/√2. Though I appreciated this video, twas really helpful, 'cuz it's not ony about understanding the formulas and applying these, it's also about understanding what does it mean. Thanks to you, I did

[@katiefaery]

So well explained. I appreciate it 🙂

[@k-inquisitive9052]

Are you a theoretical physicist

[@ittooklongtomakethis]

lol the toothpick thing is relatable. I use chalk to represent a 3d vector on a chalkboard sometimes :D

[@LookingGlassUniverse]

Haha! Do you teach vectors?

[@swojnowski453]

put a hat on the vectors so that you don't have to remember you have made them unit vectors ;)

[@hashas6998]

You're a great teacher.

[@goSomewhereElse]

This is really great, once in a while I look at a dot product and I'm like "why the heck sr cos() again?"

[@rosuav]

12:37 Inquiring minds must know: what colour were the roses?

[@LookingGlassUniverse]

White but splattered with red paint for some reason...

[@21ruevictorhugo]

I love your videos but I’m starting from scratch. I’m looking for your first video about vectors and can’t find it. When I tried doing a search for “looking glass vectors “ I get only this one.

[@lightyagami-rk2my]

Thanks for this video. I searched for this

[@Kraflyn]

vectors in 3d space have 3 defining properties. Necessarily. If space is 3d. Line that carries it, length, and orientation along that line. Or, if the coordinate origin is defined, you need 3 coordinates to define the point in 3d space that belongs to the vector's top end. Besides, the scalar product comes from defining i*i=1 and i*j=0. We need this indicator type of a product because of the physical axiom: "Movements in two perpendicular directions are independent". Think of rowing in a boat on a river. So we need to define "perpendicular" and "independent" in a mathematical way. Well, i*i=1 and i*j=0 does the tricks. Both tricks. Directions x and y are independent because they do not mix: i*j=0. Since i and j are mutually orthogonal we are done! The sine and cosine thing kinda comes along naturally then. Cheers! :D

[@LookingGlassUniverse]

Yeah, nice work drawing out the connection between ‘orthogonal’ in Maths and ‘independent’ in physics :)

[@viliml2763]

i*i=1 and i*j=0 generalizes to what is called a metric.

[@Kraflyn]

@@viliml2763 Well, it depends on how you look at it. The metric tensor comes from the tensor product, not from the scalar product. However, the components that enter the tensor product come from the scalar product, so yeah :D Cheers :D

[@duegia44]

So we got: = |a||b|cos(θ)(1) as granted, what we have to do is proving = a1b1+a2b2+a3b3+... (2) As |a> and |b> can be expressed as matrices (a1, a2, a3, ...) and (b1, b2, b3, ...) so it dot product will be: = (a1, a2, a3, ...) . (b1, b2, b3, ...) (3) The matrices themselves is a linear combination of their basis time multiplied by a constant, so the dot product (3) could take the form of: = [ a1(1, 0, 0, ...) + a2(0, 1, 0, ...) + a3(0, 0, 1, ...) + ...] . [ b1(1, 0, 0, ...) + b2(0, 1, 0, ...) + b3(0, 0, 1, ...) + ...] (4) Of every combinations, the ones where the vectors is orthonormal to each others (cos they are orthonormal basis) meaning the angle between them is 90 degree, when using formula (1) it means the products equal to 0, and so can be ignored. The only ones that doesnt equal 0 is those which aren't orthonormal to each others, which are the dot product of themselves, i.e: (1, 0, 0, ...) . (1, 0, 0, ...); (0, 1, 0, ...) . (0, 1, 0, ...); (0, 0, 1, ...) . (0, 0, 1, ...); and so on. Because they're just the same thing and have the same length of 1, using formula (1) again just give the result of 1. So the product (4) could be rewrite as: = a1b1+a2b2+a3b3+... Which has proved the product (2). Woohoo, homeworks are fun, maybe from your video i can even think of other work that you guy maybe interesting in, not something new but it could be good, or maybe you can prove me wrong :D

[@LookingGlassUniverse]

Fantastic work!! Nicely done. What other things that could interest us are you thinking of? I’d be really interested :)

[@duegia44]

​@@LookingGlassUniverse It just that what I noticed about the way people understand dot product, is that a lot of people interpreted it as how much two vectors overlap, which is nice and true, but I didnt see much talking about other potential interpretation. As I understand, a matrix could also be use as a linear transformation. They do it by changing the basis vector from the original to a new one, just like your video pointed out, overlap of the old elemental vectors in the new basis direction and then scale by the new unit length of that new basis vector and them sum it up again, that is a way to see it. For example if we have: |2 0 0| |a| |0 1 0| . |b| |0 0 1| |c| It just mean the former matrix is a linear transformation of the vector (a b c) into a new basis where the basis |i> get its length double, turn into (2 0 0). And then we have the dot product, which kind of doing the same thing, only in here the basis is just a 1D line, but as you said before vectors could lost their dimension no problem, it just cant add dimensions. In here one vector just overlap on a new basis which is the other vector, and then scale up by the other vector length which is the new unit length. We can even see below that the new basis just have the 1D length of d, e and f |a| |d e f| . |b| |c| The reason the dot product give out a number is because in 1D there's only length, so the new vector that get transformed into just need 1 number to write it down. The reason the dot product's meaning is to see how much of one vector pointing out in the other's direction is because that literally how you transform a vector to 1D space. In my oppinion, the dot product is a transformation of a vector of any number of dimensions into 1D space with the other vector is its basis. I think I could prove it more rigorously, this is just intuition, but I kinda want to rethinking if I get something wrong, maybe you can help me?

[@LookingGlassUniverse]

@@duegia44 I think you're getting at an extremely important point! If I've understood correctly, you've realised that dot products are related to the so called 'linear functions' or 'dual space' of a vector space. This is an extremely deep insight! If you want to know more about them, you can look them up. I was thinking about whether I should include it in a video too... and I might. We'll see!

[@duegia44]

@@LookingGlassUniverse Damn that's some intense reading, now that I understood what dual vector space is, it just increasing my question. Why do we even care about dual vector space in the first place? It seem like the dual vector space has the exact same space of the original one, same number of dimensions, same bases? The number that it've been transformed into seem arbitrary , just anything belong to the common field? And because the numbers are arbitrary, the dual vector could literally be any vector possible of the vector space (which are still the same)? Hell, it could just be the exact same vector? The notion of linear transformation seem gotten narrower in these dual vector space cases, but to what purpose? Or rather, what so special about these cases of linear transformation compare to others?

[@jeevanandam.exll-a8867]

Yeah I take some time to get that it is cos but where did you get this concept ,

[@eruyommo]

I'll assume all the laws of arithmetic are already proven. Let _u, v, b_ be arbitrary vectors and _x, y, m, n_ belong to the real numbers. Then: 1. u=x*b+m & v=y*b+n 2. u.b=x*|b|_ from the property of the final poll question and first of this homework ( 9:49 ). Similarly, v.b=y*|b| 3. u.b+v.b=x*|b|+y*|b| 4. u.b+v.b=(x+y)*|b| from distributivity of real numbers. 5. u+v=x*b+y*b+m+n 6. u+v=(x+y)*b+(m+n) 7. (u+v).b=(x+y)*|b| from the property of 9:49 8. (u+v).b=u.b+v.b from 4 in 7. Q.E.D. Since v_i and v_j belong to an orthonormal basis, they are perpendicular. Therefore, their dot product is zero (as can be seen in 2:10). Since they are basis, their expression as linear combinations of the basis' vectors is only themselves. The formula shown in 11:03 asks for the product of the corresponding parts of each linear combination. Since the corresponding element to the i-th element in the expression of v_j is 0 and visceversa, this operation becomes 0*1+1*0=0. Q.E.D. For the general expression, I had some difficulties: Let a=x*v_i+y*v_j and b=m*v_i+n*v_j Let the angle _a_ forms with v_i be called _A_ and the angle b forms with v_i be called _B_ , and form assign a and b such that B>=A, then let's call the angle between _a_ and _b_ is T=B-A 1. From the deffinition of dot product, a.b=|a|*|b|*cos(T) 2. a.b=|a|*|b|*cos(B-A) 3. By identity, a.b=|a|*|b|*(cosB*cosA+sinB*sinA) 5. a.b=|a|*|b|*(x*m+y*n) That's where I got stucked. I see that if I proved that |a|*|b|=1, then I'd have my proof, but I can't figure it out and I think I'm wrong since if that were true, that term would not be needed in the main formula. Help?

[@LookingGlassUniverse]

Oh no! I typed a reply but it seems not to have sent. Let me try again briefly. Very very good work on the first part! As for the part you got stuck at- you’re close! But I think it’s easiest if you change track slight. So: try writing your x and y in terms of the dot product (and a particular length). To do this, look to the previous question :) then it helps, I think to draw it out on paper. Give it a go and let me know how it goes!

[@nnnn65490]

Love the vid as always!!! : ) Will you do the cross product too? I actually don't know why it works...

[@LookingGlassUniverse]

Thank you! I wasn’t planning to, since 1)I don’t know the answer (though it could be fun to figure it out) 2) the cross product doesn’t generalise nicely like the dot product. Sorry!!

[@rosuav]

6:10 Awwww, I like your toothpick vectors!

[@LookingGlassUniverse]

Haha! At least there’s one.

[@ApplepieFTW]

9:40 hold on how does this make sense at all. we know that a•b is the part of a in the direction of b, right? so it's a bit like, where the projection of a on b is, that's the vector a, minus the "orthogonal to b" bit. so why does the length of b matter? if I make b very large, and have a stay the same, wouldn't the projection land in the same spot? and if I make a really big, wouldn't it then land on a different spot instead? and also, why is it logical at all that you have to devide by the length of b? since, how exactly does its length change the projection?

[@gpamob]

Amazing. Thank you very very much.

[@dragonlovesdiamond9512]

I have a question, isn’t the dot product also used for how much two vectors are alligned? How can the formula also use this definition and how much these are pointing the same way

[@natasadjurdjevic3533]

You deserve more views

[@Sofialovesmath]

Amazing, you are incredible!

[@andersloren8020]

My guess on homework #2: Show that: (u+v).b = u.b + v.b "Proof": (u+v).b = |u+v| * |b| * cosΘ (u+v).b = |u| * |b| * cosΘ + |v| * |b| * cosΘ |u| * |b| * cosΘ + |v| * |b| * cosΘ = u.b + v.b (as shown in the previous question)

[@nafizkarim5191]

fuck i finally understand dot products not jus the cloud of the confusion that is dot product is the projection of one vector onto another fuck i love u thank uuuuuuu keep up the good work this dot product has been frustrating me so much im trying to learn about 3d rendering

[@MidnighterClub]

OK here's a weird thought. As you draw a vector in various positions around a circle, I notice that the dot product is not unique. 0 (zero) occurs twice as the vector rotates in a circle, once at 90 degrees and once at 270 degrees. But the vector is in different positions!! So are there different zeros here? Is zero not always equal to zero? It seems like there should be some way to distinguish between those different positions.

[@ejetzer]

Gojira There are, but not with the dot product. We go from two vectors of two or more dimensions, to a single real number, so some information gets lost: we can’t distinguish between an angle and its full circle complement, or between acute and obtuse angles, or between negative and positive angles. The dot products tells us how much the vectors point in the same way, but not where they initially pointed.

[@LookingGlassUniverse]

Really nice observation/ question, and great answer. That’s exactly right, the dot product doesn’t uniquely determine where the other vector is relative to the first. Come to think of it, that’s why when you want to specify a vector wrt a orthonormal basis, you need specify the dot product of it with every basis vector

[@mesutkaraman9677]

To handle with nonuniqueness, assuming all the vectors are in the same quadrant may help. Just a triggered thougt, why nonnegativity is so important for recommendation systems. Since all components have positive value, comparing two client's rankings by using dot product is a well suited approach.

[@wavelet4866]

YaY you are back!!

[@LookingGlassUniverse]

Sorry for disappearing!

[@ollicron7397]

I've never seen a b written like a G that's new to me.

[@kaushikgupta9490]

Awesome video !

[@LookingGlassUniverse]

Thank you!!

[@angelisvegan5826]

I would be so so grateful if you answer my question. 2:56 how is the x ( -1) and not (1) ... I mean.. the magnitude couldn't be negative one .. it makes sense that the direction is neg 1 but then x is denotes magnitude.. I'm confused ..ease help me dear

[@arnaldo8681]

That first matrix at 11:10 should be horizontal, shouldnt it?

[@LookingGlassUniverse]

Oh! You’re thinking of the ‘covector’ (the vector on the side). You’re totally right that you can write a.b= (covector of a) times b. But in the video that dot between the column vectors I used is the dot product- not multiplication! Thanks for the question :)

[@LeenaBora]

Hi. The video started with: We want some way to describe - how much 2 vectors are pointed in same direction? And answer is dot products helps in achieving that. But why can't we use simply angle between 2 vectors for the same thing. (I am not talking about cos(theta) ) simply angle. Reason why we can't use angle in this case is, it will work in 2D but not in higher dimension. So we need some other way to describe it. Is this geometrically correct statement?

[@Basaltq]

What poll? I can't see any poll.

[@pablosantanadeoliveira9284]

Shouldn't the answer be Cos . a? Because cos = x/a, right?