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## Linear algebra

### 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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# Defining the angle between vectors

Introducing the idea of an angle between two vectors. Created by Sal Khan.

## Want to join the conversation?

- I don't see how this proof is valid in dimensional spaces other than R2. He defined the angles using a sketch of a triangle in 2D, and then used the law of cosines which wasn't proved to be valid (or meaningful) outside of 2D and 3D space. Isn't there a whole lot of hand-waiving in this proof?(33 votes)
- The definition of an angle between vectors is the angle between two sides of a triangle in 2D with lengths ||a||,||b||,||a-b||. Such a triangle is 2D by construction even for v∈ℝⁿ because ||v|| is a scalar.(91 votes)

- What is the difference between Orthogonal and Perpendicular ? He said that Zero Vector is Orthogonal to everything but not Perpendicular ? I don't get it. o.O(25 votes)
- Two vectors are perpendicular if they are not the zero vector AND their dot product is zero. They are only orthogonal if one or both of them are the zero vector and their dot product is zero. The definition of perpendicular relies on the angle between the vectors being 90 degrees, and with the zero vector, there's no intuitive way of thinking about the angle. The orthogonal case deals with the zero vector, and it is orthogonal to every vector because the zero vector dotted with anything is zero.(4 votes)

- If we wanted to solve for the angle between two known vectors, would it not be easier to rearrange the equation Sal explains at16:54? Instead you could have it as θ= arccos((a・b) /(||a||*||b||)(17 votes)
- Yes. that's perfectly valid. I'm surprised Sal didn't quickly show that here given the title of the video! For some reason he wanted to focus mainly on the perpendicular case.(12 votes)

- "Find two unit vectors that make an angle of 60 degrees with u = (3; 4)."

I've been trying everything for this question, but I've yet to come across an answer. Would anybody be able to explain how to approach this problem? I know that ||u|| = 5 and that another vector v, ||v|| = 1 , but I can't seem to accomplish anything more than this. Please help!!!(4 votes)- Here's an approach which doesn't require us to know anything besides the angle formula and the length formula for vectors:

You want to find vectors v' = [x, y] (where v' means the transpose of v) such that:

u . v = |u||v|cos(60)

or, in other words,

u . v = 5*|v|*1/2

We also want the vector v to have length one, i.e. |v|=1.

In total, we get two constraints on v:

(i) 3x + 4y = 5/2 (from your u, the dot-product formula, and the above equation)

(ii) x^2 + y^2 = 1 (since the length (the square root of this) should be 1)

Right. Now we have to solve this for x and y.

Equation (i) gives us that

x = 5/6 - 4y/3

Plugging this into Equation (ii) gives us

(5/6 - 4y/3)^2 + y^2 = 1,

or (by multiplying out and simplifying a bit):

25y^2/9 - 20y/9 + 25/36 -1 = 0

We're looking for the roots of this second order polynomial. They are

y = 1/10 (4 - 3sqrt(3)) and y = 1/10 (4+3sqrt(3))

where sqrt(3) means the square root of 3.

Alright, now we know all the possible values for y. What are the corresponding x-values? Use Equation (ii) to get

x = 5/6 - 4/3 (1/10 (4 - 3sqrt(3))) and x = 5/6 - 4/3 (1/10 (4+3sqrt(3)),

which simplifies to

x = 1/10(3 + 4sqrt(3)) and x = 1/10(3-4sqrt(3)).

Our two vectors are:

v = [x,y]

with these values for x and y.

Tell me if anything was unclear. :)(13 votes)

- At23:40Sal tells us that the zero vector is orthogonal to all other vectors.

If the zero vector is orthogonal to all other vectors, doesn't that arise from R0 being orthogonal to R1 in the same way that in constructing the second dimension of R2 one sets the x any y axes orthogonal to each other?(5 votes)- By definition, orthogonal is the name given to the relationship between two vectors described when their dot product is 0. The dot product of the 0 vector with any other vector is 0, so by defintion the 0 vector is orthogonal to every other vector. In other words it just has the right properties for that definition (which someone made up).. there is no more to it than that.(6 votes)

- At7:17why sal changed the direction of vector a-b to b-a ?(5 votes)
- Because that's the vector a-b in the opposite direction. He say this at6:55.

To reverse a vector v, you multiply it by negative one, so reversing (a-b) we do: -(a-b) = -a+b = b-a.(5 votes)

- I don't understand how |a| can be less than or equal to |b| + |a-b|.

I get how it has to be less than, but not how it can be equal to, and still be a triangle. It's completely flat, and has no height whatsoever, so why is that a valid triangle?(4 votes)- It may not be a triangle per-se, but all the rules that apply to triangles still apply if you consider it as having one side of length 0. The only real problem is having a length 0 side be perpendicular to anything, but if we replace "perpendicular" with "orthogonal", the problem goes away.(5 votes)

- What is the difference between 'Orthogonal' and 'Orthonormal' ?(1 vote)
- Actually, they are not synonyms.

A set of vectors is orthonormal iff they are orthogonal AND they have length of 1 (unit length).

So, orthonormal implies orthogonal but orthogonal does not imply orthonormal. There is an extra criterion about length.(8 votes)

- Does the word "orthogonal" mean "dot product = 0"? Does the word "perpendicular" have any usage in R^n, n>3?(2 votes)
- You are exactly correct, orthogonal means the dot product is zero. Orthogonal is the generalization of the notion of perpendicular, which is usually used only in two dimensions.

In the comments Matthew Daly points out the technical distinction: "The distinction between the two is that (by definition) the zero vector is orthogonal to all vectors but perpendicular to none of them."

In higher dimensions there is more than line that is orthogonal to another so you have to be careful about that too.

I have heard physics books refer to a line perpendicular to a plane in three dimensions but once again, orthogonal is the more mathematical word.(5 votes)

- Would this also work with spherical triangles?(1 vote)
- No. These inequalities will only work in rectilinear Euclidean/ Cartesian space. Spherical, as well as hyperbolic, geometries are built on entirely different set of assumptions that included Euclidean. For example...

==> In Euclidean space, it is a truism that parallel lines never meet.

==> In spherical geometry, all parallel lines, called "geodesics" or "great circles", always intersect by definition. These are essentially equivalent to lines that maximize their length, which all have the length of 2*(pi)*r. The set {Great Circle Lines} =

==> In hyperbolic geometry, parallel lines also, by definition, always cross.(6 votes)

## Video transcript

A couple of videos ago we
introduced the idea of the length of a vector. That equals the length. And this was a neat idea because
we're used to the length of things in two- or
three-dimensional space, but it becomes very abstract when
we get to n dimensions. If this has a hundred
components, at least for me, it's hard to visualize a hundred
dimension vector. But we've actually defined
it's notion of length. And we saw that this is actually
a scalar value. It's just a number. In this video, I want to attempt
to define the notion of an angle between vectors. As you can see, we're building
up this mathematics of vectors from the ground up, and we can't
just say, oh, I know what an angle is because
everything we know about angles and even lengths, it
just applies to what we associate with two- or
three-dimensional space. But the whole study of linear
algebra is abstracting these ideas into multi-dimensional
space. And I haven't even defined what
dimension is yet, but I think you understand that idea
to some degree already. When people talk about one or
two or three dimensions. So let's say that I have some
vector-- let's say I have two vectors, vectors a and b. They're nonzero and they're
members of Rn. And I don't have a notion of
the angle between them yet, but let me just draw them out. Let me just draw them as if
I could draw them in two dimensions. So that would be vector
a right there. Maybe that's vector
b right there. And then this vector right
there would be the vector a minus b. And you can verify that just the
way we've learned to add and subtract vectors. Or you know, this is
heads to tails. So b plus a minus b is of
course, going to be vector a. And that all just
works out there. To help us define this notion
of angle, let me construct another triangle that's going
to look a lot like this one. But remember, I'm just doing
this for our simple minds to imagine it in two dimensions. But these aren't necessarily
two-dimensional beasts. These each could have a
hundred components. But let me make another
triangle. Well, it should look similar. Say it looks like that. And I'm going to define the
sides of the triangles to be the lengths of each
of these vectors. Remember, the lengths of each of
these vectors, I don't care how many components there
are, they're just going to be your numbers. So the length of this side right
here is just going to be the length of a. The length of this side right
here is just going to be the length of vector a
minus vector b. And the length of this side
right here is going to be the length of vector b. Now the first thing we want to
make sure is that we can always construct a triangle
like that. And so under what circumstances
could we not construct a triangle
like this? Well we wouldn't be able
construct a triangle like this if this side. if b, if the magnitude-- so
let me write this down. It's kind of a subtle
point, but I want to make this very clear. In order to define an angle, I
want to be comfortable that I can always make this
construction. And I need to make sure that--
let me write reasons why I couldn't make this
construction. Well what if the magnitude of
b was greater than, or the length of vector b was greater
than the length of vector a plus the length of
vector a minus b? In two dimensions, I could never
draw a triangle like that then because you would
have this length plus this length would be shorter than
this thing right here. So you could never
construct it. And I could do with
all the sides. What if this length was larger
than one of these two sides? Or what if that length
was larger than one of those two sides? I could just never draw
a two-dimensional triangle that way. So what I'm going to do is I'm
going to use the triangle-- the vector triangle inequality
to prove that each of these sides is less than or equal to
the sum of the other sides. I could do the same thing. Let me make the point clear. I could show that if a, for
whatever reason, was greater than the other sides plus b,
then I wouldn't be able to create a triangle. And the last one of course is
if a minus b, for whatever reason, was greater than the
other two sides, I just wouldn't be able to draw
a triangle in a plus b. So I need to show that for any
vectors, any real vectors-- nonzero, real vectors that are
members of Rn-- that none of these can ever happen. I need to prove that none
of those can happen. So what does the triangle
inequality tell us? The triangle inequality tells
us that if I have the sum of two vectors, if I take the
length of the sum of two vectors, that that is always
going to be less than-- and these are nonzero vectors. This is always going to be less
than or equal to the sum of each of their individual
lengths. So let's see if we can
apply that to this triangle right here. So what is the magnitude,
the length of a? Well I can rewrite vector a. What is vector a equal to? Vector a is equal to vector
b plus vector a minus b. I mean I'm just rewriting
the vector here. I'm just rewriting a here as a
sum of the other two vectors. Nothing fancy there. I haven't used the triangle
inequality or anything. I've just used my definition
of vector addition. But here now, if I put little
parentheses here, now I can apply the triangle inequality. And I say, well,
you know what? This is going to be, by the
triangle inequality, which we've proved, it's going to be
less than or equal to the lengths of each of
these vectors. Vector b plus the length
of vector a minus b. So we know that the length of a
is less than the sum of that one and that one. So we don't have to worry about
this being our problem. We know that that is not true. Now let's look at b. So is there any way that I can
rewrite b as a sum of two other vectors? Well sure. I can write it as a sum of a
plus, let me put it this way. If that vector right there is a
minus b, the same vector in the reverse direction is going
to be the vector b minus a. So a plus the vector
b minus a. That's the same thing as b. And you can see it right here. The a's would cancel
out and you're just left with the b there. Now by the triangle inequality,
we know that this is less than or equal to the
length of vector a plus the length of vector b minus a. Now you're saying hey, Sal,
you're dealing with b minus a. This is the length
of a minus b. And I can leave this for you
to prove it based on our definition of vector lengths,
but the length of b minus a is equal to minus 1 times
a minus b. And I'll leave it to you
to say that look, these lengths are equal. Because essentially-- I could
leave that, but I think you can take that based on just the
visual depiction of them that they're the exact same
vectors, just in different directions. And I have to be careful with
length because it's not just in two dimensions. But I think you get the idea and
I'll leave that for you to prove that these lengths
are the same thing. So we know that b is less
than the length of those two things. So we don't have to worry about
that one right there. Finally, a minus b. The magnitude or the length
of vector a minus b. Well I can write that as the
length of-- or I can write that as vector a plus
vector minus b. If we just put a minus b right
there and go in the other directions, we could say minus
b, which would be in that direction plus a would give
us our vector a minus b. Actually, I don't even
have to go there. That's obvious from this. I just kind of put the negative
in the parentheses. Well the triangle inequality,
and this might seem a little mundane to you, but it really
shows us that we can always define a regular planar triangle
based on these vectors in this way. It tells us this is less than or
equal to the length of our vector a plus the length
of minus b. And I just said and you could
prove it to yourself, that this is the same thing
as the length of b. So we just saw that
this is definitely less than those two. This is definitely less
than those two. And that is definitely
less than those two. None of the reasons that would
keep us from constructing a triangle are valid. So we can always construct a
triangle in this way from any arbitrary nonzero
vectors in Rn. We can always construct this. Now, to define an angle, let
me redraw it down here. Let me redraw the vectors, maybe
a little bit bigger. That's vector a. This is vector b. And then let me just
draw it this way. This is the vector
right there. That is the vector a minus b. And we said we're going to
define a corresponding regular, run of the mill,
vanilla triangle whose lengths are defined by the lengths
of the vectors, by the vector lengths. So this is the length
of b, that side. This is the length
of a minus b. And then this is the
length of a. Now that I know that I can
always construct a triangle like this, I can attempt to
define-- or actually, I will define my definition of an angle
between two vectors. So we know what an angle
means in this context. This is just a regular, run of
the mill, geometric triangle. Now, my definition of an angle
between two vectors I'm going to say-- so this is what
I'm trying to define. This is what I'm going
to define. These can have arbitrary number
of components, so it's hard to visualize. But I'm going to define this
angle as the corresponding angle in a regular, run of the
mill triangle where the sides of the run of the mill triangle
are the two vectors and then the opposite side is
the subtraction, is the length of the difference between
the two vectors. This is just the definition. I'm defining this, the angle
between two vectors in Rn that could have an arbitrary number
of components, I'm defining this angle to be the same as
this angle, the angle between the two sides, the two lengths
of those vectors in just a regular, run of the
mill triangle. Now, what can I do with this? Well, can we find a relationship
between all of these things right here? Well sure. If you remember from your
trigonometry class, and if you don't, I've proved it
in the playlist. You have the law of cosines. And I'll do it with an arbitrary
triangle right here just because I don't want
to confuse you. So if this is side a, b, and c
and this is theta, the law of cosines tells us that c squared
is equal to a squared plus b squared minus 2ab
cosine of theta. I always think of it as kind
of a broader Pythagorean theorem because this
thing does not have to be a right angle. It accounts for all angles. If this becomes a right angle,
then this term disappears and you're just left with the
Pythagorean theorem. But we've proven this. This applies to just regular,
run of the mill triangles. And lucky for us, we have a
regular, run of the mill triangle here. So let's apply the law
of cosines to this triangle right here. And the way I drew it,
they correspond. The length of this
side squared. So that means the length
of a minus b squared. Length of vector a minus vector
b, that's just the length of that side. So I'm just squaring
that side. It equals the length of vector
b squared plus the length of vector a squared minus 2 times
the length of-- I'll just write two times length of vector
a times the length of vector b times the cosine of
this angle right here. Times the cosine
of that angle. And I'm defining this angle
between these two vectors to be the same as this
angle right there. So if we know this angle, by
definition, we know that angle right there. Well, we know that the square
of our lengths of a vector when we use our factor
definition of length, that this is just the same thing as
a vector dotted with itself. So that's a minus
b dot a minus b. It's all going to be equal to
this whole stuff on the right-hand side. But let me simplify
the left-hand side of this equation. a minus b dot a minus b, this is
the same thing as a dot a-- those two terms--
minus a dot b. And then I have minus b dot a. Those two terms right there. And then you have the
minus b dot minus b. That's the same thing
as a plus b dot b. Remember, this is just a
simplification of the left-hand side. And I can rewrite this. a dot a, we know that's just
the length of a squared. a dot b and b dot a are
the same thing, so we have two of these. So this right here, this term
right there will simplify to minus 2 times a dot b. And then finally, b dot b. We know that that's just the
length of b squared. I just simplified or maybe I
just expanded-- that's a better word. When you go from one term here
to three terms, you can't say you simplified it. But I expanded just the
left-hand side and so this has to be equal to the right-hand
side by the law of cosines. So that is equal to-- I almost
feel like instead of rewriting it, let me just copy
and paste it. What did I just do? Copy, edit. Copy and paste. There you go. I don't know it that
was worth it. But maybe I saved a little
bit of time. So that is equal to
that right there. And then we can simplify. We have a length of a
squared here, length of a squared there. Subtract it from both sides. The length of b squared here,
length of b squared there. Subtract it from both sides. And then, what can we do? We can divide both sides by
minus 2 because everything else has disappeared. And so that term and that term
will both become 1's. And all we're left with is the
vector a dot the vector b. And this is interesting because
all of a sudden we're getting a relationship
between the dot products of two vectors. We've kind of gone away from
their definition by lengths. But the dot product of two
vectors is equal to the product of their lengths,
their vector lengths. And they can have an arbitrary
number of components. Times the cosine of the
angle between them. Remember, this theta, I said
this is the same as when you draw this kind of analogous,
regular triangle. But I'm defining the angle
between them to be the same as that. So I can say that this is
the angle between them. And obviously, the idea of
between two vectors, it's hard to visualize if you go beyond
three dimensions. But now we have it at least,
mathematically defined. So if you give me two vectors we
can now, using this formula that we've proved using this
definition up here, we can now calculate the angle between
any two vectors using this right here. And just to make it clear, what
happens if a is a-- and maybe it's not clear from that
definition, so I'll make it clearer here that by definition,
if a is equal to some scalar multiple of b where
c is greater than 0, we'll define theta
to be equal to 0. And if c is less than 0, so a is
collinear, but goes in the exact opposite direction, we'll
define theta to be equal to 180 degrees. And that's consistent with what
we understand about just two-dimensional vectors. If they're collinear and
kind of the scalar multiples the same. That means a looks something
like that and b looks something like that. So we say oh, that's
a 0 angle. And if they go the other way,
if a looks something like-- this is the case where a is
just going in the other direction from b. a goes like that and b goes like
that, we define the angle between them to be
180 degrees. But everything else is pretty
well defined by the triangle example. I had to make the special case
of these because it's not clear you really get a triangle
in these cases because the triangle
kind of disappears. It flattens out if a and b are
on top of each other or if they're going in the exact
opposite direction. So that's why I wanted to make
a little bit of a side note right there. Now, using this definition of
the angle between the vectors, we can now define the idea
of perpendicular vectors. So we can now say perpendicular
vectors-- this is another definition-- and
this won't be earth shattering, but it kind of is
because we've generalized this to vectors that have an
arbitrary number of components. We're defining perpendicular
to mean the theta between-- two vectors a and b are
perpendicular if the angle between them is 90 degrees. And we can define that. We can take two vectors,
dot them. Take their dot product. Figure out their two lengths and
then you could figure out the angle between them. And if it's 90 degrees. you can say that they are
perpendicular angles. And I want to be very clear here
that this is actually not defined for the 0 vector
right here. So this situation right here,
not defined for the 0 vector. Because if you have the 0
vector, then this quantity right here is going to be 0 and
then this quantity right here is going to be 0. And there's no clear definition
for your angle. If this is 0 right here, you
did 0 is equal to 0 times cosine of theta. And so if you wanted to solve
for theta you'd get cosine of theta is equal to 0/0,
which is undefined. But what we can do is create a
slightly more general word than the word perpendicular. So you have to have a defined
angle to even talk about perpendicular. If the angle between two vectors
is 90 degrees, we're saying by definition, those two
vectors are perpendicular. But what if we made the
statement and we can-- if you look at them, if the angle
between two vectors is 90 degrees, what does that mean? So let's say that theta
is 90 degrees. Let me draw a line here. Let's say that theta
is 90 degrees. Theta is equal to 90 degrees. What does this formula
tell us? It tells us that a dot b is
equal to the length of a times the length of b times cosine
of 90 degrees. What's cosine of 90 degrees? It's 0. You can review your unit circle
if that doesn't make a lot of sense. But that is equal to 0, so this
whole term is going to be equal to 0. So if theta is equal to
90 degrees, then a dot b is equal to 0. And so this is another
interesting takeaway. If a and b are perpendicular,
then their dot product is going to be equal to 0. Now if their dot product is
equal to 0, can we necessarily say that they are
perpendicular? Well what if a or b
is the 0 vector? The 0 vector-- let me call
it z for 0 vector. Or I could just draw. The 0 vector dot anything is
always going to be equal to 0. So does that mean that the 0
vector is perpendicular to everything? Well no. Because the 0 vector I said, we
have to have the notion of an angle between things in
order to use the word perpendicular. So we can't use the 0 vector. We can't say just because two
vectors dot products are equal to 0 that they are
perpendicular. And that's because the 0 vector
would mess that up because the 0 vector
is not defined. But if we say, and we have been
saying, that a and b are nonzero, if they are nonzero
vectors, then we can say that if a and b are nonzero and their
dot product is equal to 0, then a and b are
perpendicular. So now it goes both ways. But what if we just have this
condition right here? What if we just have the
condition that a dot b is equal to 0? It seems like that's kind of
just a simple, pure condition. And we can write a
word for that. And these words are often used
synonymously, but hopefully you understand the
distinction now. We can say that if two vectors
dot product is equal to 0, we will call them orthogonal. As I always say, spelling
isn't my best subject. But this is kind
of a neat idea. This tells us that-- well, that
all perpendicular vectors are orthogonal. And it also tells us that the
0 vector is orthogonal to everything else. To everything, even to itself. The 0 dot 0 vector
you still get 0. So by definition,
it's orthogonal. So for the first time probably
in your mathematical career, you're seeing that the words--
you know, every time you first got exposed to the words
perpendicular and orthogonal in geometry or maybe in physics
or wherever else, they were always kind of
the same words. But now I'm introducing a nice,
little distinction here and you can kind of be a little smart aleck with teachers. Oh, you know, it's perpendicular
only is the vectors aren't-- if neither
of them are 0 vector. Otherwise, if their dot product
is 0, you can only say that they're orthogonal. But if they're nonzero you
could say that they're orthogonal and perpendicular. But anyway, I thought that I
would introduce this little distinction for you in case you
have someone that likes to trip you up with words. But it also I think highlights
that we are building a mathematics from the ground up
and we have to be careful about the words we use. And we have to be very precise
about our definitions. Because if we're not precise
about our definitions and we build up a bunch of mathematics
on top of this and do a bunch of proofs, one day we
might scratch our heads and read some type of
weird ambiguity. And it might have all came out
of the fact that we weren't precise enough in defining what
some of these terms mean. Well anyway, hopefully you
found this useful. We can now take the angle or we
can now determine the angle between vectors with an
arbitrary number of components.