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!
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!!!
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!
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!!
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. 🌈
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.
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
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!
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!
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
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 ...
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!!!
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?
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!!
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.
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
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!
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.
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
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
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.
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).
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.
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!!
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
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.
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
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.
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.
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
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.
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 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!
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...
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?
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
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
Very nice video as always. My only question is that shouldn't the dot product be a BILINEAR map instead of a linear map as you mentioned in the video ?
@@anonymoose3423 Well if you have some fixed vector V we can get the map V.(x) which takes any vector x and puts out a member of the base field. Since the dot product is bilinear we know that V.(x) is a linear map and so a member of the dual space. As V was arbitrary and bilinearity of the dot product we have that the dot product takes any V to V.(x) in a linear way.
Absolutely right! But I only talked about the linearity of fixing one index and having the other bit be a linear combo on purpose so I didn't have to say bilinear :P
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
Great video but I have a question. I like how using unit vectors creates a range of -1 to 1 when rotating one of the two parallel vectors wrt the other. But if the lengths of the two parallel vectors is 5, then the range is -25 to 25. Using your definition of the dot product gives "The dot product shows how much of a vector is in the direction of the other. Therefore 5i dotted with 5i...500% of one vector is in the direction of the other vector. Does that sound a little odd? Thanks!
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?
Just for the record, I do notice the effort you put into these videos. You're doing great! Keep it up!
Thank you, that’s very very kind of you :)
Looking Glass Universe yeah, please know it doesn’t go unnoticed by us!
I really really appreciate that :)
IM SO GRATEFUL FOR THESE textbooks arent doing much for me with numbers and formulas :') so thank you mam
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!
This is incredible, taking a concept that confused me to no end and helping me understand in 13 min. Thank you so much
Finally a video on UA-cam that gives a really intuitive visualisation of the dot product. Well done. Thank you.
Thanks so much!!
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!
The amount of creativity in this video , the amount of efforts you have put in editing this. Salute. 🔥🔥
literally the thumbnail alone explained it better than everyone else
I cant even imagine how much work you did for this video , its great
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!!!
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!
Suddenly discovered this channel. Loving it!
this is actually so sick, a lot of work went into this and it really did help me
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!!
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. 🌈
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.
The first 8 seconds literally answered a question I've been trying to solve for the last 20 minutes with multiple websites and textbooks.
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
Thank you very much! Did you try the homeworks :)?
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!
Thanks so much :D!
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!
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
Ain't u gonna extend the playlist?
I loved ur explanation.
Wow the production quality on this vid is crazy! Thank you for the great explanation too!
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 ...
Nicely done! Thanks for typing it all out :)
I am madly in love with your voice!
Every comment section of every female youtuber ever. Very cliche way of simping.
We have been missing you! Every video you make is a little gem.
Thank you so so much- that means the world to me.
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!!!
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?
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!!
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.
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 :)
LOL not the horror flashes for the pi circle. Great Video
Can we reboot the series, starting with the cross product? 👀
after the 1 miunutes of this video ... i am here to comment " it is the best video i had watch so far" luv ya
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
Holy crap I just wanted to understand dot product for my Physics class but I somehow ended up in Linear Algebra.
New favorite math channel! wow this was the style of explanation I've been searching for, such a helpful conceptual take.
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!
Yay! Thanks for noticing the 1% (probably lower) increase in video quality.
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.
Yay, she's back!
put a hat on the vectors so that you don't have to remember you have made them unit vectors ;)
I cant tell exactly how gladful im after this point of view, i thought about it a lot and now thinga makes more sense!
3:35 No God! No God Please no. No! NO! NOOOOO!
Thank you so much! You're an extremely talented teacher.
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
This channel is actually amazing
This video was awesome. A beautiful lecture that help the viewer better understand a confusing topic.
This is the best explanation to dot product. Thank u
This is the best intuitive explanation for the dot product .Can you please do the same for cross product
Amazing videos. This really helped me understand the dot product.
I totally forgot about this and we started using dot products for the first time since calc 3, thank you!
Great!! So happy to help :)
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
Underrated video. Thank you
Excellent video, and I loved the toothpast tip!
Definitely subscribing! This is an amazing channel!!! Very well explained :)
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.
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).
Thror251
There’s many ways
Some write it like this
@Mi Les yup :) I write it down because I use \cdot for multiplication sometimes. But it doesn't matter too much!
@@duckymomo7935
Or .
@@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}
Oh, that’s a nice way to do it!
6:10 Awwww, I like your toothpick vectors!
Haha! At least there’s one.
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.
Kudos! You are an amazing teacher! Well done. Keep up the good work.
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.
Thanks for not explaining homework one now I have no clue how to do the rest
Love the vid as always!!! : )
Will you do the cross product too? I actually don't know why it works...
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!!
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
Excellent video, thank you for your hard work.
May God bless you!
This is beyond great.
lol the toothpick thing is relatable. I use chalk to represent a 3d vector on a chalkboard sometimes :D
Haha! Do you teach vectors?
It was a really nice one :) ,it actually gave me a clear picture of dot product
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.
this type of teaching is very good and very useful for beginners .
Thanks very much!
perfect didactic and editing, youre very talented :)
I never knew I had all these problems. But luckily I can solve (most) of them! 🧠
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
Excellent video and explanation. I applaud your efforts.
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.
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.
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
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.
You're a great teacher.
So well explained. I appreciate it 🙂
Are you a theoretical physicist
Thanks for this video. I searched for this
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.
Well done!! And thank you so so much for putting the effort into answering this :)! I really appreciate it
@@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!
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...
The way you write the letter b is killing me 😤😤
Shouldn't the answer be Cos . a? Because cos = x/a, right?
Amazing. Thank you very very much.
YaY you are back!!
Sorry for disappearing!
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?
Amazing, you are incredible!
Is this the same way the dot product was defined by the original mathematicians who made linear algebra?
I've never seen a b written like a G that's new to me.
12:37 Inquiring minds must know: what colour were the roses?
White but splattered with red paint for some reason...
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
This is really great, once in a while I look at a dot product and I'm like "why the heck sr cos() again?"
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
For the proof of the linearity of the dot product, isn't it kind of obvious if we are multiplying scalars?
I'm actually scared when you do that... 3:35
Very nice video as always. My only question is that shouldn't the dot product be a BILINEAR map instead of a linear map as you mentioned in the video ?
It also can be used as a linear map to the dual space so if we want to be that pedantic it still works :)
George Smith How is the dot product a linear map to the dual space ?
@@anonymoose3423 Well if you have some fixed vector V we can get the map V.(x) which takes any vector x and puts out a member of the base field. Since the dot product is bilinear we know that V.(x) is a linear map and so a member of the dual space. As V was arbitrary and bilinearity of the dot product we have that the dot product takes any V to V.(x) in a linear way.
Absolutely right! But I only talked about the linearity of fixing one index and having the other bit be a linear combo on purpose so I didn't have to say bilinear :P
1.23 B cap. where is B?.. oh only later realise the G is the B, anyhow.. super easy to understand video on dot product, thank u
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
I’ll be back when I’m sober all the big words
Great video but I have a question.
I like how using unit vectors creates a range of -1 to 1 when rotating one of the two parallel vectors wrt the other. But if the lengths of the two parallel vectors is 5, then the range is -25 to 25. Using your definition of the dot product gives "The dot product shows how much of a vector is in the direction of the other. Therefore 5i dotted with 5i...500% of one vector is in the direction of the other vector. Does that sound a little odd? Thanks!
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?
You deserve more views
X = -1? Am confusion
-1 because both have same magnitude but opposite direction so if we go front it is plus and if we go back it is minus.
Awesome video !
Thank you!!