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Course: Linear algebra > Unit 1
Lesson 5: Vector dot and cross products- 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
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Proof: Relationship between cross product and sin of angle
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.(59 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)- They probably fooled around with it in any way possible to make it equal. It is hard work not magical thinking.(9 votes)
Is there any difference between llall and lal? Or do they both refer to the magnitude?(8 votes)
Akib,
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:
https://math.berkeley.edu/~hutching/teach/proofs.pdf(5 votes)
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).(3 votes)
4:50
Video transcript
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.