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Let's learn a little bit
about the dot product.

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The dot product, frankly, out of
the two ways of multiplying

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vectors, I think is
the easier one.

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So what does the
dot product do?

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Why don't I give you the
definition, and then I'll give

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you an intuition.

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So if I have two vectors; vector
a dot vector b-- that's

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how I draw my arrows.

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I can draw my arrows
like that.

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That is equal to the magnitude
of vector a times the

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magnitude of vector b
times cosine of the

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angle between them.

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Now where does this come from?

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This might seem a little
arbitrary, but I think with a

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visual explanation, it will make
a little bit more sense.

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So let me draw, arbitrarily,
these two vectors.

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So that is my vector a-- nice
big and fat vector.

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It's good for showing
the point.

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And let me draw vector
b like that.

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Vector b.

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And then let me draw the cosine,
or let me, at least,

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draw the angle between them.

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This is theta.

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So there's two ways
of view this.

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Let me label them.

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This is vector a.

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I'm trying to be color
consistent.

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This is vector b.

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So there's two ways of
viewing this product.

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You could view it as vector a--
because multiplication is

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associative, you could
switch the order.

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So this could also be written
as, the magnitude of vector a

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times cosine of theta, times--
and I'll do it in color

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appropriate-- vector b.

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And this times, this
is the dot product.

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I almost don't have
to write it.

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This is just regular
multiplication, because these

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are all scalar quantities.

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When you see the dot between
vectors, you're talking about

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the vector dot product.

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So if we were to just rearrange
this expression this

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way, what does it mean?

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What is a cosine of theta?

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Let me ask you a question.

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If I were to drop a right
angle, right here,

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perpendicular to b-- so let's
just drop a right angle

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there-- cosine of theta
soh-coh-toa so, cah cosine--

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is equal to adjacent of
a hypotenuse, right?

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Well, what's the adjacent?

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It's equal to this.

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And the hypotenuse is equal to
the magnitude of a, right?

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Let me re-write that.

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So cosine of theta-- and this
applies to the a vector.

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Cosine of theta of this angle
is equal to ajacent, which

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is-- I don't know what you could
call this-- let's call

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this the projection
of a onto b.

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It's like if you were to shine
a light perpendicular to b--

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if there was a light source
here and the light was

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straight down, it would be
the shadow of a onto b.

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Or you could almost think of it
as the part of a that goes

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in the same direction of b.

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So this projection, they call
it-- at least the way I get

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the intuition of what a
projection is, I kind of view

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it as a shadow.

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If you had a light source that
came up perpendicular, what

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would be the shadow of that
vector on to this one?

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So if you think about it, this
shadow right here-- you could

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call that, the projection
of a onto b.

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Or, I don't know.

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Let's just call it, a sub b.

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And it's the magnitude
of it, right?

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It's how much of vector a goes
on vector b over-- that's the

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adjacent side-- over
the hypotenuse.

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The hypotenuse is just the
magnitude of vector a.

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It's just our basic calculus.

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Or another way you could view
it, just multiply both sides

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by the magnitude of vector a.

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You get the projection of a onto
b, which is just a fancy

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way of saying, this side; the
part of a that goes in the

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same direction as b-- is another
way to say it-- is

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equal to just multiplying both
sides times the magnitude of a

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is equal to the magnitude
of a, cosine of theta.

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Which is exactly what
we have up here.

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And the definition of
the dot product.

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So another way of visualizing
the dot product is, you could

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replace this term with the
magnitude of the projection of

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a onto b-- which is just
this-- times the

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magnitude of b.

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That's interesting.

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All the dot product of two
vectors is-- let's just take

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one vector.

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Let's figure out how much of
that vector-- what component

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of it's magnitude-- goes in
the same direction as the

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other vector, and let's
just multiply them.

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And where is that useful?

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Well, think about it.

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What about work?

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When we learned work
in physics?

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Work is force times distance.

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But it's not just
the total force

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times the total distance.

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It's the force going
in the same

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direction as the distance.

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You should review the physics
playlist if you're watching

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this within the calculus
playlist. Let's say I have a

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10 newton object.

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It's sitting on ice, so
there's no friction.

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We don't want to worry about
fiction right now.

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And let's say I pull on it.

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Let's say my force vector--
This is my force vector.

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Let's say my force vector
is 100 newtons.

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I'm making the numbers up.

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100 newtons.

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And Let's say I slide it to
the right, so my distance

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vector is 10 meters parallel
to the ground.

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And the angle between them is
equal to 60 degrees, which is

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the same thing is pi over 3.

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We'll stick to degrees.

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It's a little bit
more intuitive.

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It's 60 degrees.

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This distance right
here is 10 meters.

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So my question is, by pulling on
this rope, or whatever, at

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the 60 degree angle, with a
force of 100 newtons, and

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pulling this block to the right
for 10 meters, how much

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work am I doing?

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Well, work is force times the
distance, but not just the

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total force.

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The magnitude of the force in
the direction of the distance.

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So what's the magnitude
of the force in the

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direction of the distance?

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It would be the horizontal
component of this force

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vector, right?

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So it would be 100
newtons times the

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cosine of 60 degrees.

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It will tell you how
much of that 100

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newtons goes to the right.

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Or another way you could
view it if this

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is the force vector.

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And this down here is
the distance vector.

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You could say that the total
work you performed is equal to

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the force vector dot the
distance vector, using the dot

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product-- taking the dot
product, to the force and the

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distance factor.

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And we know that the definition
is the magnitude of

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the force vector, which is 100
newtons, times the magnitude

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of the distance vector, which is
10 meters, times the cosine

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of the angle between them.

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Cosine of the angle
is 60 degrees.

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So that's equal to 1,000
newton meters

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times cosine of 60.

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Cosine of 60 is what?

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It's square root of 3 over 2.

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Square root of 3 over 2, if
I remember correctly.

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So times the square
root of 3 over 2.

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So the 2 becomes 500.

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So it becomes 500 square roots
of 3 joules, whatever that is.

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I don't know 700 something,
I'm guessing.

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Maybe it's 800 something.

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I'm not quite sure.

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But the important thing to
realize is that the dot

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product is useful.

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It applies to work.

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It actually calculates what
component of what vector goes

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in the other direction.

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Now you could interpret
it the other way.

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You could say this is
the magnitude of a

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times b cosine of theta.

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And that's completely valid.

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And what's b cosine of theta?

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Well, if you took b cosine of
theta, and you could work this

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out as an exercise for yourself,
that's the amount of

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the magnitude of the
b vector that's

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going in the a direction.

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So it doesn't matter
what order you go.

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So when you take the cross
product, it matters whether

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you do a cross b,
or b cross a.

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But when you're doing the dot
product, it doesn't matter

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what order.

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So b cosine theta would be the
magnitude of vector b that

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goes in the direction of a.

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So if you were to draw a
perpendicular line here, b

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cosine theta would
be this vector.

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That would be b cosine theta.

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The magnitude of
b cosine theta.

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So you could say how much of
vector b goes in the same

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direction as a?

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Then multiply the
two magnitudes.

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Or you could say how much of
vector a goes in the same

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direction is vector b?

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And then multiply the
two magnitudes.

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And now, this is, I think, a
good time to just make sure

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you understand the difference
between the dot product and

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the cross product.

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The dot product ends up
with just a number.

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You multiply two vectors and
all you have is a number.

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You end up with just
a scalar quantity.

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And why is that interesting?

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Well, it tells you how much do
these-- you could almost say--

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these vectors reinforce
each other.

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Because you're taking the parts
of their magnitudes that

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go in the same direction
and multiplying them.

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The cross product is actually
almost the opposite.

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You're taking their orthogonal
components, right?

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The difference was, this
was a a sine of theta.

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I don't want to mess you up
this picture too much.

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But you should review the
cross product videos.

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And I'll do another video where
I actually compare and

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contrast them.

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But the cross product is, you're
saying, let's multiply

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the magnitudes of the vectors
that are perpendicular to each

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other, that aren't going in the
same direction, that are

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00:10:01,660 --> 00:10:03,380
actually orthogonal
to each other.

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And then, you have to pick a
direction since you're not

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saying, well, the same
direction that

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they're both going in.

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So you're picking the direction
that's orthogonal to

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both vectors.

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And then, that's why the
orientation matters and you

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have to take the right hand
rule, because there's actually

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two vectors that are
perpendicular to any other two

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vectors in three dimensions.

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Anyway, I'm all out of time.

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I'll continue this, hopefully
not too confusing, discussion

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in the next video.

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I'll compare and contrast
the cross

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product and the dot product.

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See you in the next video.

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