Vector dot and cross products
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Vector Dot Product and Vector Length
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Proving Vector Dot Product Properties
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Proof of the Cauchy-Schwarz Inequality
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Vector Triangle Inequality
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Defining the angle between vectors
-
Defining a plane in R3 with a point and normal vector
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Cross Product Introduction
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Proof: Relationship between cross product and sin of angle
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Dot and Cross Product Comparison/Intuition
-
Vector Triple Product Expansion (very optional)
-
Normal vector from plane equation
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Point distance to plane
-
Distance Between Planes
Proof of the Cauchy-Schwarz Inequality Proof of the Cauchy-Schwarz Inequality
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- Let's say that I have two nonzero vectors.
- Let's say the first vector is x, the second vector is y.
- They are both a part of our set.
- They're both in the set Rn and they're nonzero.
- It turns out that the absolute value of their-- let me do it
- in a different color.
- This color's nice.
- The absolute value of their dot product of the two
- vectors-- and remember, this is just a scalar quantity-- is
- less than or equal to the product of their lengths.
- And we've defined the dot product and we've defined
- lengths already.
- It's less than or equal to the product of their lengths and
- just to push it even further, the only time that this is
- equal, so the dot product of the two vectors is only going
- to be equal to the lengths of this-- the equal and the less
- than or equal apply only in the situation-- let me write
- that down-- where one of these vectors is a scalar multiple
- of the other.
- Or they're collinear.
- You know, one's just kind of the longer or shorter version
- of the other one.
- So only in the situation where let's just say x is equal to
- some scalar multiple of y.
- These inequalities or I guess the equality of this
- inequality, this is called the Cauchy-Schwarz Inequality.
- Cauchy-Shwarz Inequality
- So let's prove it because you can't take something like this
- just at face value.
- You shouldn't just accept that.
- So let me just construct a somewhat artificial function.
- Let me construct some function of-- that's a function of some
- variables, some scalar t.
- Let me define p of t to be equal to the length of the
- vector t times the vector-- some scalar t times the vector
- y minus the vector x.
- It's the length of this vector.
- This is going to be a vector now.
- That squared.
- Now before I move forward I want to make one
- little point here.
- If I take the length of any vector, I'll do it here.
- Let's say I take the length of some vector v.
- I want you to accept that this is going to be a positive
- number, or it's at least greater than or equal to 0.
- Because this is just going to be each of its terms squared.
- v2 squared all the way to vn squared.
- All of these are real numbers.
- When you square a real number, you get something greater than
- or equal to 0.
- When you sum them up, you're going to have something
- greater than or equal to 0.
- And you take the square root of it, the principal square
- root, the positive square root, you're going to have
- something greater than or equal to 0.
- So the length of any real vector is going to be greater
- than or equal to 0.
- So this is the length of a real vector.
- So this is going to be greater than or equal to 0.
- Now, in the previous video, I think it was two videos ago, I
- also showed that the magnitude or the length of a vector
- squared can also be rewritten as the dot product of that
- vector with itself.
- So let's rewrite this vector that way.
- The length of this vector squared is equal to the dot
- product of that vector with itself.
- So it's ty minus x dot ty minus x.
- In the last video, I showed you that you can treat a
- multiplication or you can treat the dot product very
- similar to regular multiplication when it comes
- to the associative, distributive and commutative
- properties.
- So when you multiplied these, you know, you could kind of
- view this as multiplying these two binomials.
- You can do it the same way as you would just multiply two
- regular algebraic binomials.
- You're essentially just using the distributive property.
- But remember, this isn't just regular multiplication.
- This is the dot product we're doing.
- This is vector multiplication or one version of vector
- multiplication.
- So if we distribute it out, this will become ty dot ty.
- So let me write that out.
- That'll be ty dot ty.
- And then we'll get a minus-- let me do it this way.
- Then we get the minus x times this ty.
- Instead of saying times, I should be very
- careful to say dot.
- So minus x dot ty.
- And then you have this ty times this minus x.
- So then you have minus ty dot x.
- And then finally, you have the x's dot with each other.
- And you can view them as minus 1x dot minus 1x.
- You could say plus minus 1x.
- I could just view this as plus minus 1 or plus minus 1.
- So this is minus 1x dot minus 1x.
- So let's see.
- So this is what my whole expression simplified to or
- expanded to.
- I can't really call this a simplification.
- But we can use the fact that this is commutative and
- associative to rewrite this expression right here.
- This is equal to y dot y times t squared.
- t is just a scalar.
- Minus-- and actually, this is 2.
- These two things are equivalent.
- They're just rearrangements of the same thing and we saw that
- the dot product is associative.
- So this is just equal to 2 times x dot y times t.
- And I should do that in maybe a different color.
- So these two terms result in that term right there.
- And then if you just rearrange these you have a minus 1
- times a minus 1.
- They cancel out, so those will become plus and you're just
- left with plus x dot x.
- And I should do that in a different color as well.
- I'll do that in an orange color.
- So those terms end up with that term.
- Then of course, that term results in that term.
- And remember, all I did is I rewrote this
- thing and said, look.
- This has got to be greater than or equal to 0.
- So I could rewrite that here.
- This thing is still just the same thing.
- I've just rewritten it.
- So this is all going to be greater than or equal to 0.
- Now let's make a little bit of a substitution just to clean
- up our expression a little bit.
- And we'll later back substitute into this.
- Let's define this as a.
- Let's define this piece right here as b.
- So the whole thing minus 2x dot y.
- I'll leave the t there.
- And let's define this or let me just define this
- right here as c.
- X dot x as c.
- So then, what does our expression become?
- It becomes a times t squared minus-- I want to be careful
- with the colors-- b times t plus c.
- And of course, we know that it's going to be greater than
- or equal to 0.
- It's the same thing as this up here, greater
- than or equal to 0.
- I could write p of t here.
- Now this is greater than or equal to 0 for any t that I
- put in here.
- For any real t that I put in there.
- Let me evaluate our function at b over 2a.
- And I can definitely do this because what was a?
- I just have to make sure I'm not dividing by 0 any place.
- So a was this vector dotted with itself.
- And we said this was a nonzero vector.
- So this is the square of its length.
- It's a nonzero vector, so some of these terms up here would
- end up becoming positively when you take its length.
- So this thing right here is nonzero.
- This is a nonzero vector.
- Then 2 times the dot product with itself is also going to
- be nonzero.
- So we can do this.
- We don't worry about dividing by 0, whatever else.
- But what will this be equal to?
- This'll be equal to-- and I'll just stick to the green.
- It takes too long to keep switching between colors.
- This is equal to a times this expression squared.
- So it's b squared over 4a squared.
- I just squared 2a to get the 4a squared.
- Minus b times this.
- So b times-- this is just regular multiplication.
- b times b over 2a.
- Just write regular multiplication there.
- Plus c.
- And we know all of that is greater than or equal to 0.
- Now if we simplify this a little bit, what do we get?
- Well this a cancels out with this exponent there and you
- end up with a b squared right there.
- So we get b squared over 4a minus b squared over 2a.
- That's that term over there.
- Plus c is greater than or equal to 0.
- Let me rewrite this.
- If I multiply the numerator and denominator of this by 2,
- what do I get?
- I get 2b squared over 4a.
- And the whole reason I did that is to get a common
- denominator here.
- So what do you get?
- You get b squared over 4a minus 2b squared over 4a.
- So what do these two terms simplify to?
- Well the numerator is b squared minus 2b squared.
- So that just becomes minus b squared over 4a plus c is
- greater than or equal to 0.
- These two terms add up to this one right here.
- Now if we add this to both sides of the equation, we get
- c is greater than or equal to b squared over 4a.
- It was a negative on the left-hand side.
- If I add it to both sides it's going to be a positive on the
- right-hand side.
- We're approaching something that looks like an inequality,
- so let's back substitute our original substitutions to see
- what we have now.
- So where was my original substitutions that I made?
- It was right here.
- And actually, just to simplify more, let me multiply both
- sides by 4a.
- I said a, not only is it nonzero,
- it's going to be positive.
- This is the square of its length.
- And I already showed you that the length of any real
- vector's going to be positive.
- And the reason why I'm taking great pains to show that a is
- positive is because if I multiply both sides of it I
- don't want to change the inequality sign.
- So let me multiply both sides of this by a before I
- substitute.
- So we get 4ac is greater than or equal to b squared.
- There you go.
- And remember, I took great pains.
- I just said a is definitely a positive number because it is
- essentially the square of the length. y dot y is the square
- of the length of y, and that's a positive value.
- It has to be positive.
- We're dealing with real vectors.
- Now let's back substitute this.
- So 4 times a, 4 times y dot y.
- y dot y is also-- I might as well just write it there.
- y dot y is the same thing as the magnitude of y squared.
- That's y dot y.
- This is a.
- y dot y, I showed you that in the previous video.
- Times c.
- c is x dot x.
- Well x dot x is the same thing as the
- length of vector x squared.
- So this was c.
- So 4 times a times c is going to be greater than
- or equal to b squared.
- Now what was b? b was this thing here.
- So b squared would be 2 times x dot y squared.
- So we've gotten to this result so far.
- And so what can we do with this?
- Oh sorry, and this whole thing is squared.
- This whole thing right here is b.
- So let's see if we can simplify this.
- So we get-- let me switch to a different color.
- 4 times the length of y squared times the length of x
- squared is greater than or equal to-- if we squared this
- quantity right here, we get 4 times x dot y.
- 4 times x dot y times x dot y.
- Actually, even better, let me just write it like this.
- Let me just write 4 times x dot y squared.
- Now we can divide both sides by 4.
- That won't change our inequality.
- So that just cancels out there.
- And now let's take the square root of both
- sides of this equation.
- So the square roots of both sides of this equation-- these
- are positive values, so the square root of this side is
- the square root of each of its terms. That's
- just an exponent property.
- So if you take the square root of both sides you get the
- length of y times the length of x is greater than or equal
- to the square root of this.
- And we're going to take the positive square root.
- We're going to take the positive square root on both
- sides of this equation.
- That keeps us from having to mess with anything on the
- inequality or anything like that.
- So the positive square root is going to be the absolute value
- of x dot y.
- And I want to be very careful to say this is the absolute
- value because it's possible that this thing right here is
- a negative value.
- But when you square it, you want to be careful that when
- you take the square root of it that you
- stay a positive value.
- Because otherwise when we take the principal square root, we
- might mess with the inquality.
- We're taking the positive square root, which will be--
- so if you take the absolute value, you're ensuring that
- it's going to be positive.
- But this is our result.
- The absolute value of the dot product of our vectors is less
- than the product of the two vectors lengths.
- So we got our Cauchy-Schwarz inequality.
- Now the last thing I said is look, what happens if x is
- equal to some scalar multiple of y?
- Well in that case, what's the absolute value?
- The absolute value of x dot y?
- Well that equals-- that equals what?
- If we make the substitution that equals the absolute value
- of c times y.
- That's just x dot y, which is equal to just from the
- associative property.
- It's equal to the absolute value of c times-- we want to
- make sure our absolute value, keep everything positive.
- y dot y.
- Well this is just equal to c times the magnitude of y-- the
- length of y squared.
- Well that just is equal to the magnitude of c times-- or the
- absolute value of our scalar c times our length of y.
- Well this right here, I can rewrite this.
- I mean you can prove this to yourself if you don't believe
- it, but this-- we could put the c inside of the magnitude
- and that could be a good exercise for you to prove.
- But it's pretty straightforward.
- You just do the definition of length.
- And you multiply it by c.
- This is equal to the magnitude of cy times-- let me say the
- length of cy times the length of y.
- I've lost my vector notation someplace over here.
- There you go.
- Now, this is x.
- So this is equal to the length of x times the length of y.
- So I showed you kind of the second part of the
- Cauchy-Schwarz Inequality that this is only equal to each
- other if one of them is a scalar multiple of the other.
- If you're a little uncomfortable with some of
- these steps I took, it might be a good exercise to
- actually prove it.
- For example, to prove that the absolute value of c times the
- length of the vector y is the same thing as the
- length of c times y.
- Anyway, hopefully you found this pretty useful.
- The Cauchy-Schwarz Inequality we'll use a lot when we prove
- other results in linear algebra.
- And in a future video, I'll give you a little more
- intuition about why this makes a lot of sense relative to the
- dot product.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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