Electricity and magnetism
The dot product Introduction to the vector dot product.
The dot product
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- 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.
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