- Vector dot product and vector length
- Proving vector dot product properties
- Proof of the Cauchy-Schwarz inequality
- Vector triangle inequality
- Defining the angle between vectors
- Defining a plane in R3 with a point and normal vector
- Cross product introduction
- Proof: Relationship between cross product and sin of angle
- Dot and cross product comparison/intuition
- Vector triple product expansion (very optional)
- Normal vector from plane equation
- Point distance to plane
- Distance between planes
Proof: Relationship between the cross product and sin of angle between vectors. Created by Sal Khan.
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- Wait, so we have:
||a x b|| = ||a|| . ||b|| . sin(theta)
Isn't that a NUMBER? The result of that calculation - a length, times a length, times a sine should be a number.
In the last video Sai said that the cross product of two vectors is STILL a vector. It doesn't look like a vector to me.
Is it still a vector?(8 votes)
- Wait, I figured it out.
||a x b|| is THE LENGTH of (a x b), not the vector itself. That is a number.(58 votes)
- For the basic equation |AxB|=|A||B|sin(theta), what if sin(theta) is negative? The lenght of a vector is a positive scalar by definition, so if sin(theta) is negative, what does it mean?(14 votes)
- Good question! Recall that the angle, x, between two vectors can only be between 0 and 180 degrees, inclusive. When x is between 0 and 180, sin(x) is between 0 and 1 (think of the unit circle). Therefore, the sine of the angle between the two vectors will never become negative number, and therefore |AxB| will always be positive, as required.(23 votes)
- At4:50Sal begins to factor out the squared "a" terms of the equation and at7:30Sal begins to take the square of the dot product identity--but how do we know to do this? When my professors pull these tricks out, it's amazing--but how did they know to do them? Obviously, they learned them from someone else, but the people who first "discovered" these relationships--how did they know to perfectly combine these equations to come up with such an elegant result? That level of intuition escapes me.(13 votes)
- Is there any difference between llall and lal? Or do they both refer to the magnitude?(8 votes)
The way most people use the notation, ||a|| typically refers to magnitude of a vector, and |a| refers to the absolute value of a scalar. Since both vectors and scalars are usually denoted by lower case letters, it's important to understand which one represents which in the book you are using or the test you are doing.
By the way, not every book or test uses exactly the same notation. Some books might have a special symbol over the letter to make it clear that it is referring to a vector. Other books might use bold print for vectors and matrices and regular print for scalars. Other books might use only greek letters for scalars, and latin letters for vectors and matrices. So if there is a question about what represents what, it is important to ask.(20 votes)
- While I can understand where this proof is coming from, I always ask myself how anyone found this proof. Is there any sort of intuition you can develop for proof-building?(7 votes)
- The only way to get better at proof-writing is practice. Basically every proof-writing problem is unique, but there are broad techniques that you can learn that will usually give you a good starting point.
This document may be helpful:
- Can someone provide link to the video where Khan explains intuition behind |a x b| etc? (01:09onwards)(4 votes)
- at0:30, Sal says that R3 is the only place where the cross product is defined. I understand that it is not defined in R2, but what about R4?(3 votes)
- Sal was simplifying a little bit -- the cross product is also defined in 0, 1 and 7 dimensions (although in 0 and 1 dimensions it is not at all useful because any cross product is always equal to the zero vector). So you could have a cross product in R^7. However, it doesn't work in R^2, R^4 or any other dimension. The reason for this is extremely complicated, but essentially it boils down to the fact that division only works in 1, 2, 4 and 8 dimensions. You already know 1-dimensional division -- it's just ordinary division. You probably know about 2-dimensional numbers (imaginary numbers), and then 4- and 8-dimensional numbers are called quaternions and octonions, and work in a similar way to the imaginary numbers.(3 votes)
- Which is the physics video explaining "the intuition behind what this really means" he refers to at1:10?(2 votes)
- Here: http://www.khanacademy.org/science/physics/electricity-and-magnetism. There's 3 physics cross product videos listed.(5 votes)
- When we are taking the square root in the end (17:00), shouldn't we consider the minus sign too? (as in √4=±2)(3 votes)
- The negative of either side wouldn't change anything aside from the direction of the vector <1,0> and <-1,0> have the same length , just pointing in opposite directions. Also, you could think o it as having taken the principle root, which just means the positive part of the square root.(2 votes)
- I've watched 3b1b's video on the subject, and I think there is a much more simple explanation to this. The length of the cross product, is by definition, the area of the parallelogram that the two vectors make. θ, is the angle between the two vectors. These two vectors are coplanar. So if we look at this parallelogram in 2d(by making this plane which the vectors lie on—plane A—the whole view), it is easy to calculate the area. The formula for the area of a parallelogram is the height of the parallelogram(the "y-component" of the vector that isn't the base), multiplied by the length of its base(the base is the part of parallelogram which is on the span the "x-axis" of plane A). To obtain the height of the parallelogram, we can multiply the the length of the vector that isn't the base(W), by the sin of the aforementioned θ. Multiplying this by the length the base vector(V), yields the expression |V|·|W|Sin(θ).(2 votes)
The goal of this video is to start with our definition of the cross product and the result that we started off with or that we got to in a different video-- I think it was three videos ago-- where we found out that the dot product of two nonzero vectors, a dot b, is equal to the product of their lengths. So the product of the length of a with the length of b times the cosine of the angle between them. We're going to start with these two things. This definition of a cross product in R3, the only place it really is defined, and then this result. And we want to get to the result that the length of the cross product of two vectors. And so obviously, when you take a cross product you get a vector. But if you take its length you get a number again, you just get a scalar value, is equal to the product of each of the vectors' lengths. It's the product of the length of a times the product of the length of b times the sin of the angle between them. Which is a pretty neat outcome because it kind of shows that they're two sides of the same coin. Dot product has cosine, cross product has sin. I'm sure you've seen this before. Well you definitely have seen it if you've watched my physics playlist. And I even do a whole video where I talk about the intuition behind what this really means. And I encourage you to re-watch that, and I'll probably do that again in the linear algebra context. But the point of this video is to prove this to you. Is to prove that with this and this, I can get to this. Now if you're just believe me, and you just say, oh I've seen that before. And I just think it definitely is the case, then you don't have to watch the rest of this video because I'll tell you right now, it's going to get dirty. It's going to be a hairy, hairy proof. But if you're willing to watch and bear with me, let's start proving this result. So the place I'm going to start is with the idea of taking the length of a cross b squared. That's a cross b right there. So I'm essentially taking the length of this vector squared. And we saw in many videos and I've used this idea multiple times, that if I just have some arbitrary vector, let me just say some arbitrary vector. And I take its length squared, that's just equal to that vector dotted with itself or the square of each of its terms summed up all the way to xn squared. So what will this be equal to? Well this is just equal to that vector. And we only have three components, so it's equal to the sum of the squares of each of these components. Let me write this down. It's equal to this term squared. So let me write that down. a2 b3 minus a3 b2 squared. Plus this term squared. So plus a3 b1 minus a1 b3 squared. And then finally, plus that term squared. So plus a1 b2 minus a2 b1 squared. And what is this equal to? Well let's just expand it out. Let's expand that out. So this term right here, we're just going to have to do our expansion of the square of a binomial. And we've done this multiple times. So this is going to be equal to a2 squared b3 squared. And then we're going to have these two multiplied by each other twice. So minus 2. I'm just multiplying this out. Minus 2 times a2 a3 b2 b3. I'm just rearranging them to get the order right. Plus a3 squared b2 squared. That term squared. Then I have to add this term. So plus a3 squared b1 squared minus 2 times both of these terms multiplied. Minus 2 times a1 a3 b1 b3. Plus that term squared. a1 squared b3 squared. And then finally, this term squared. So plus a1 squared b2 squared minus 2 times a1 a2 b1 b2. Plus a2 squared b1 squared. So there you go. And let's see if we can write this in a form-- well, I'm going to write this in a form that I know will be useful later. So what I'm going to do is I'm going to factor out the a2, a1, a3 squared terms. So I could write this as-- let me pick a new neutral color. So this is equal to, if I just write a1 squared, where's my a1 squared terms? I got that one right there and I have that one right there. So a1 squared times b2 squared plus b3 squared. This would be 3 squared. Good enough. Now where are my a2 squared terms? a plus a2 squared times-- I have that one and that one. So times b1 squared. That's that. Plus b3 squared. And then finally, let me pick another new color. I can go back to yellow. Plus a3 squared times-- well that's that term and that term. So b1 and b2. So b1 squared plus b2 squared. And obviously, I can't forget about all of that mess that I have in the middle, all of this stuff right here. So plus, or maybe I should write minus 2 times all of this stuff. Let me just write it real fast. So it's a2 a3 b2 b3 plus a1 a3 b1 b3. Plus a1 a2 b1 b2. There you go. Now let's put this aside for a little bit. Let me put this on the side for a little bit. We'll let that equation rest for a little while. And remember, this is just an expansion of the length of a cross b squared. That's all this is. So just remember that. And now, let's do another equally hairy and cumbersome computation. Let's take this result up here. We know that the magnitude or the length of a times the length of b times the angle between them is equal to a dot b. Which is the same thing as if we actually do the dot product, a1 times b1 plus a2 times b2 plus a3 times b3. Now, just to kind of make sure that I get to do the hairiest problem possible, let's take the square of both sides. If we square this side, you get a squared b squared cosine squared. Then you got a dot b squared or you get this whole thing squared. So what's this whole thing squared? For me, it's easier to just write out the thing again. Instead of writing a square, just multiply that times a1 b1 plus a2 b2 plus a3 b3. And let's do some polynomial multiplication. So first, let's multiply this guy times each of those guys. So you have a1 b1 times-- well there a1 b1. I'm going to do it right here. You get a1 squared b1 squared plus a1-- plus this guy times this guy. Plus a1 a2 times b1 b2. Plus this guy times that guy. Plus a1 a3 times b1 b3. Fair enough. Now let's do the second term. We have to multiply this guy times each of those guys. So a2 b2 times a1 b1. Well that's this one right here. a2 b2 times a1 b1. I wrote it right here because this is really the same term and eventually, we want to simplify that. So that's that times that guy. Then we have this guy times that over there. So let me write it over here. So that's a2 squared b2 squared. Put a plus right there. And then finally, this middle guy times this third guy. So let me write it over here. Plus-- so a2 a3 b2 b3. Now, we only have one left. Maybe I'll do it in this blue color. I have to multiply this guy times each of those guys. So a3 b3 times a1 b1. That's the same thing as this term right here. Because you have a3. Let me write it right here. You have a3 b3 times a1 b1. Then you have this guy times that guy, which is this because it's a3 b3 times a2 b2. Let me put a little plus sign there. And then finally, you have this guy times himself. So you have a3 squared b3 squared. And so if you add up all of this business here what do you get? I'll switch to another color. You have a1 squared b1 squared. Plus, and I'm doing these colors in a certain way on purpose, plus a2 squared b2 squared. Plus a3 squared b3 squared. Plus, and let me do it in this, I'll do it in white. Plus, what do you have here? You have this term times 2. You have this term times 2. And then you have this term times 2. So plus 2 times a1-- let me write that down. Plus 2 times a1 a2 b1 b2-- that's that term. Plus this one right here. Plus a1 a3 b1 b3. Finally, plus this one. a2 a3 plus b2 b3. And you might have noticed something interesting already. If you compare this term right here, if you compare that guy right there to this guy right there, they're the same thing. You have an a1 a2 b1 b2, a1 a2 b1 b2. This term and that term are the same. Let's look at the other terms. Let me pick a nice color. a1 a3 b1 b3, a1 a3 b1 b3. That term and that term is the same. And then finally, if you compare a2 a3 b2 b3. This shouldn't be a plus. This is just this one. So a2 a3-- they're all multiplied. a2 a3 b2 b3, a2 a3 b2 b3. This term and this term is the same. And in this expression, when we expanded it out, we have 2 times this, positive 2 times this. And this term right here when we expanded it out, we have minus 2 times this. So let's see if we can simplify things a little bit. So what happens if we add this guy to this guy? Let's do it. So it's a little exciting. So we get a cross b, the length of that squared. And we're going to add to that this expression right here, so plus the length of a squared times the length of b squared times the cosine of the angle between them squared. What's that going to equal? It's going to be equal to this thing plus this thing. Let's do a simplification. What's this thing plus this thing? Well we already said this is the minus 2 times this. This is the plus 2 times this. So this guy-- let me be very clear. This right here is going to cancel out. When we add the two terms it's going to cancel out with this guy. These guys are going to cancel out. Thank God. Cancel out. Makes our life a little bit easier. And what are we left with? We're left with this right here plus that right there. Then we see we have an a1 squared term, so we just add the coefficients on the a1 squared. We add the coefficients of the a2 squared. We add the coefficients of the a3 squared, and what do we get? We get a1 squared times this coefficient plus this coefficient. So you get b1 squared plus b2 squared plus b3 squared. Things are starting to look a little bit orderly all of a sudden. And then you have plus a2 squared times all of their coefficients added up. So b1 squared plus b2 squared plus b3 squared. And then finally, in yellow, you have plus a3-- sorry, I was trying to do that in yellow. You have a3 squared and you have that. You have b1 squared, b2 squared and b3 squared. So b1 squared plus b2 squared plus b3 squared. And as you see, we're multiplying all of these things by this b1 squared plus b2 squared plus b3 squared, so we can actually factor that out and we get something very interesting. So this is equal to, if we factor this thing out of all the terms, we get b1 squared plus b2 squared plus b3 squared times my a squared terms. Times a1 squared plus a2 squared plus a-- I'm getting excited, the home stretch is here-- a3 squared. So these two things are equal to each other. But what's this thing? What's another way I could write this? This is the same thing as b dot b or the length of my vector b squared. And what's that? That's the length of my vector a squared. This is my length of my vector a squared. That's just a dot a. Let me rewrite everything. So we have the length of a-- that's a darker green. a cross b squared plus this thing. Plus the length of-- plus, let me actually just copy and paste it. It gets monotonous. Plus that thing right there. Why isn't it? If I control, copy and paste. So it's not working. All right, so plus that thing. The length of a squared times the length of b squared times the cosine of the angle squared between them is equal to that. Now what if we subtract this from both sides? What do we get? We get the length of a cross b squared is equal to this minus this. And we can factor-- so let me write that. Actually, let me just subtract this on this line. So if I subtract it from both sides, I could get that out there and I'll put the minus the length of a squared times the length of b squared times the cosine squared of the angle between them. And we can factor this a squared b squared, the lengths of the two vectors out, right? I'm just switching the order. So this is equal to the lengths of a squared times the length of b squared times-- and this is exciting-- time, when you factor this out of this, you just get a 1, minus cosine squared of theta. And what is 1 minus cosine squared of theta? Well, sin squared of theta plus cosine-- this is the most basic trig identity. Sin squared of theta plus cosine squared of theta is equal to 1. So if you subtract cosine squared from both sides you get sin squared of theta is equal to 1 minus cosine squared of theta. So this is sin squared of theta. And then what happens if you take the square root of both sides? This is really exciting. You get the length of vector a crossed with vector b is equal to the length of vector a times the length of vector b times the sin of the angle between them. I just took the square root of both sides of this. And we finally get our result. I never thought I would get here. And so hopefully you're satisfied. You never have to take this as kind of a leap of faith anymore. And hopefully you're satisfied with this. And I'm going to stop recording this video before I make a careless mistake or the power goes out; that would ruin everything.