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# The dot product

Introduction to the vector dot product. Created by Sal Khan.

Video transcript

Let's learn a little bit
about the dot product. The dot product, frankly, out of
the two ways of multiplying vectors, I think is
the easier one. So what does the
dot product do? Why don't I give you the
definition, and then I'll give you an intuition. So if I have two vectors; vector
a dot vector b-- that's how I draw my arrows. I can draw my arrows
like that. That is equal to the magnitude
of vector a times the magnitude of vector b
times cosine of the angle between them. Now where does this come from? This might seem a little
arbitrary, but I think with a visual explanation, it will make
a little bit more sense. So let me draw, arbitrarily,
these two vectors. So that is my vector a-- nice
big and fat vector. It's good for showing
the point. And let me draw vector
b like that. Vector b. And then let me draw the cosine,
or let me, at least, draw the angle between them. This is theta. So there's two ways
of view this. Let me label them. This is vector a. I'm trying to be color
consistent. This is vector b. So there's two ways of
viewing this product. You could view it as vector a--
because multiplication is associative, you could
switch the order. So this could also be written
as, the magnitude of vector a times cosine of theta, times--
and I'll do it in color appropriate-- vector b. And this times, this
is the dot product. I almost don't have
to write it. This is just regular
multiplication, because these are all scalar quantities. When you see the dot between
vectors, you're talking about the vector dot product. So if we were to just rearrange
this expression this way, what does it mean? What is a cosine of theta? Let me ask you a question. If I were to drop a right
angle, right here, perpendicular to b-- so let's
just drop a right angle there-- cosine of theta
soh-coh-toa so, cah cosine-- is equal to adjacent of
a hypotenuse, right? Well, what's the adjacent? It's equal to this. And the hypotenuse is equal to
the magnitude of a, right? Let me re-write that. So cosine of theta-- and this
applies to the a vector. Cosine of theta of this angle
is equal to ajacent, which is-- I don't know what you could
call this-- let's call this the projection
of a onto b. It's like if you were to shine
a light perpendicular to b-- if there was a light source
here and the light was straight down, it would be
the shadow of a onto b. Or you could almost think of it
as the part of a that goes in the same direction of b. So this projection, they call
it-- at least the way I get the intuition of what a
projection is, I kind of view it as a shadow. If you had a light source that
came up perpendicular, what would be the shadow of that
vector on to this one? So if you think about it, this
shadow right here-- you could call that, the projection
of a onto b. Or, I don't know. Let's just call it, a sub b. And it's the magnitude
of it, right? It's how much of vector a goes
on vector b over-- that's the adjacent side-- over
the hypotenuse. The hypotenuse is just the
magnitude of vector a. It's just our basic calculus. Or another way you could view
it, just multiply both sides by the magnitude of vector a. You get the projection of a onto
b, which is just a fancy way of saying, this side; the
part of a that goes in the same direction as b-- is another
way to say it-- is equal to just multiplying both
sides times the magnitude of a is equal to the magnitude
of a, cosine of theta. Which is exactly what
we have up here. And the definition of
the dot product. So another way of visualizing
the dot product is, you could replace this term with the
magnitude of the projection of a onto b-- which is just
this-- times the magnitude of b. That's interesting. All the dot product of two
vectors is-- let's just take one vector. Let's figure out how much of
that vector-- what component of it's magnitude-- goes in
the same direction as the other vector, and let's
just multiply them. And where is that useful? Well, think about it. What about work? When we learned work
in physics? Work is force times distance. But it's not just
the total force times the total distance. It's the force going
in the same direction as the distance. You should review the physics
playlist if you're watching this within the calculus
playlist. Let's say I have a 10 newton object. It's sitting on ice, so
there's no friction. We don't want to worry about
fiction right now. And let's say I pull on it. Let's say my force vector--
This is my force vector. Let's say my force vector
is 100 newtons. I'm making the numbers up. 100 newtons. And Let's say I slide it to
the right, so my distance vector is 10 meters parallel
to the ground. And the angle between them is
equal to 60 degrees, which is the same thing is pi over 3. We'll stick to degrees. It's a little bit
more intuitive. It's 60 degrees. This distance right
here is 10 meters. So my question is, by pulling on
this rope, or whatever, at the 60 degree angle, with a
force of 100 newtons, and pulling this block to the right
for 10 meters, how much work am I doing? Well, work is force times the
distance, but not just the total force. The magnitude of the force in
the direction of the distance. So what's the magnitude
of the force in the direction of the distance? It would be the horizontal
component of this force vector, right? So it would be 100
newtons times the cosine of 60 degrees. It will tell you how
much of that 100 newtons goes to the right. Or another way you could
view it if this is the force vector. And this down here is
the distance vector. You could say that the total
work you performed is equal to the force vector dot the
distance vector, using the dot product-- taking the dot
product, to the force and the distance factor. And we know that the definition
is the magnitude of the force vector, which is 100
newtons, times the magnitude of the distance vector, which is
10 meters, times the cosine of the angle between them. Cosine of the angle
is 60 degrees. So that's equal to 1,000
newton meters times cosine of 60. Cosine of 60 is what? It's square root of 3 over 2. Square root of 3 over 2, if
I remember correctly. So times the square
root of 3 over 2. So the 2 becomes 500. So it becomes 500 square roots
of 3 joules, whatever that is. I don't know 700 something,
I'm guessing. Maybe it's 800 something. I'm not quite sure. But the important thing to
realize is that the dot product is useful. It applies to work. It actually calculates what
component of what vector goes in the other direction. Now you could interpret
it the other way. You could say this is
the magnitude of a times b cosine of theta. And that's completely valid. And what's b cosine of theta? Well, if you took b cosine of
theta, and you could work this out as an exercise for yourself,
that's the amount of the magnitude of the
b vector that's going in the a direction. So it doesn't matter
what order you go. So when you take the cross
product, it matters whether you do a cross b,
or b cross a. But when you're doing the dot
product, it doesn't matter what order. So b cosine theta would be the
magnitude of vector b that goes in the direction of a. So if you were to draw a
perpendicular line here, b cosine theta would
be this vector. That would be b cosine theta. The magnitude of
b cosine theta. So you could say how much of
vector b goes in the same direction as a? Then multiply the
two magnitudes. Or you could say how much of
vector a goes in the same direction is vector b? And then multiply the
two magnitudes. And now, this is, I think, a
good time to just make sure you understand the difference
between the dot product and the cross product. The dot product ends up
with just a number. You multiply two vectors and
all you have is a number. You end up with just
a scalar quantity. And why is that interesting? Well, it tells you how much do
these-- you could almost say-- these vectors reinforce
each other. Because you're taking the parts
of their magnitudes that go in the same direction
and multiplying them. The cross product is actually
almost the opposite. You're taking their orthogonal
components, right? The difference was, this
was a a sine of theta. I don't want to mess you up
this picture too much. But you should review the
cross product videos. And I'll do another video where
I actually compare and contrast them. But the cross product is, you're
saying, let's multiply the magnitudes of the vectors
that are perpendicular to each other, that aren't going in the
same direction, that are actually orthogonal
to each other. And then, you have to pick a
direction since you're not saying, well, the same
direction that they're both going in. So you're picking the direction
that's orthogonal to both vectors. And then, that's why the
orientation matters and you have to take the right hand
rule, because there's actually two vectors that are
perpendicular to any other two vectors in three dimensions. Anyway, I'm all out of time. I'll continue this, hopefully
not too confusing, discussion in the next video. I'll compare and contrast
the cross product and the dot product. See you in the next video.