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
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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
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Vector Triple Product Expansion (very optional)
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Normal vector from plane equation
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Point distance to plane
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Distance Between Planes
Vector Triple Product Expansion (very optional) A shortcut for having to evaluate the cross product of three vectors
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- What I want to do in this video is cover something called the Triple Product Expansion
- or Lagrange's Formula, sometimes.
- And it's really just a simplification of the cross product of three vectors.
- So if I take the cross product of A and then B cross C.
- Now what we are going to do is express this,
- we can express this really as sum and difference of dot products.
- Well not just dot products but dot products scaling different vectors.
- You're going to see what I mean.
- But it simplifies this expression a bit because cross products are hard to take.
- They are computationally intensive and, at least in my mind, they're confusing.
- Now, this is not something you have to know if you are going to be dealing with vectors.
- But it's useful to know.
- My motivation for doing this video is I saw some problems for the Indian Institute of Technology Entrance Exam
- that seems to expect that you know Lagrange's Formula or the Triple Product Expansion.
- So let's see how we can simplify this.
- So to do that, let's start taking the cross product -
- the cross product of B and C.
- And in all of these situations, I'm just going to assume that we have vector A.
- I'm just going to call that A, the x component of vector A times unit vector i,
- plus the y component of vector A times unit of vector j,
- plus the z component of vector A times unit vector k.
- I can do the same things for B and C.
- So, if I say that B sub y, I'm talking about what's scaling the j component in the B vector.
- So let's first take this cross product over here,
- and as you've seen me take a cross products,
- you know that I like to do these little determinants.
- And what, let me just take it over here.
- So, B cross C, B cross C, is going to be equal to the determinant,
- and I put an i, j, k up here.
- i j k
- This is actually the definition of the cross product.
- So no proof is necessary.
- Let me show you why this is true.
- This is just one way to remember the dot product.
- If you remember to take the determinant of 3 by 3's.
- And we'll put B's x term, B's y coefficient, and B's z component. [term, coefficient and component can be used interchangeably here]
- And you'll do the same thing for the C's.
- Cx, Cy, Cz.
- And then this is going to be equal to -
- So first you have its, the i component,
- so it's going to be the i component times B, so you ignore this column and this row, so
- By [times] Cz, By Cz
- Minus Bz [times] Cy, minus so -
- I'm just ignoring all of this, and I'm looking at this 2 by 2 over here.
- minus Bz Cy, minus Bz Cy.
- And then we want to subtract the j component.
- Remember we alternate signs when we take our determinant.
- Subtract that, and then we take that column and that row. So it's going to be [chuckle]
- It's going to be Bx [times] Cz, Bx Cz
- It's a little monotonous, but it'll have an intertesting result.
- Bx Cz minus Bz [times] Cx, minus Bz, minus Bz Cx.
- And then finally plus the k component.
- k, we are going to have Bx times Cy, Bx Cy minus By Cx, minus By Cx.
- So this is, this, we just did the dot product.
- And now we want to take the - sorry we just did the cross product.
- I don't want to get you confused. We just took the cross product of B and C.
- And now we are going to take the cross product of that with A.
- Or the cross product of A with this thing right over here.
- So let's do that -
- Instead of rewriting the vector, let me just set up another matrix here.
- So let me write my i, j, k up here -
- And then let me write a's components so we have -
- A sub x, A sub y, A sub z
- And then let's clean this up a little bit.
- Let's ignore the, we're just looking at -
- No I want to do that in black.
- Let's do this in black so that we can kind of erase that.
- This is a uh, this is a minus j times that, so what I'm going to do is
- I'm going to get rid of the minus and the j
- but I'm going to rewrite this with the signs swapped
- So this is going to be -
- This is going to be swap the signs
- Bz Cx minus Bx, minus Bx Cz
- So let me delete everything else.
- So I just took the negative and I multiplied it by this.
- I'm not making any careless mistakes here. So let me just, so I can make -
- make the brush size a little bit bigger here so I can erase a little more efficiently
- There you go. And then we also want to get rid of that right over here -
- Let me get my brush back down to normal size. All right!
- So now let's just take this cross product.
- So once again, set it up as, set it up as a determinant.
- And what I'm really going to focus on -
- because it'll take the video, well it'll take me forever if I were to do, if I were to do the i, j and k components,
- I'm just going to focus on i component, just on the x component of this cross product
- and then we can see that we will get the same result for the j and k
- and we can see what hopefully this simplifies down to.
- So if we just focus on the i component here
- This is going to be,
- this is going to be i,
- i times, and we just look at this 2 by 2 matrix right over here
- We ignore i's column, i's row.
- And we have
- Ay times all of this.
- So let me just multiply it out, so it's
- Ay times Bx Cy minus, minus Ay times By, times By Cx, By Cx.
- And then, we're going to want to subtract,
- we're going to want to have minus Az times this.
- So let's just do that.
- So it's minus, or negative Az Bz Cx,
- and then we have a negative Az times this, so it's
- plus Az Bx Cz.
- And now I'm going to do this -
- This is a little bit of a trick for this proof right here.
- Just so that we get the results that I want.
- I'm just going to add and subtract the exact same thing.
- I'm going to add an Ax Bx Cx
- and then I'm going to subtract a Ax Bx Cx
- So clearly I have not changed this expression,
- I have just added and subtracted the same thing.
- And let's see what we can simplify.
- Remember, this is just the x component of our triple product,
- just the x component.
- But to do this, let me factor out -
- I'm going to factor out a B sub x, so let me do this.
- We get the B sub x
- So if I were to factor it out
- And I'm only going to look at,
- I'm only going to factor it out of this term that has a B sub x
- and I'm going to factor it out of this term
- and then I'm going to factor it out of this term.
- So if I take out the B sub x,
- I'm going to have an Ay Cy,
- actually let me write out a little bit differently.
- Let me factor it, let me factor it out of this one first.
- So that it's A, it's going to be an Ax Cx
- A sub x, C sub x
- So I used this one up.
- And then I'm going to have a, I'll do this one now
- plus, if I factor the B sub x out, I get A sub y, C sub y
- I've used that one now, and now I have this one
- I'm going to factor the B sub x out,
- so I'm left with a plus A sub z, C sub z.
- So that's all of those, so I've factored that out.
- And now from these -
- From these right over here, let me factor out a negative C sub x,
- a negative C sub x.
- And so if I do that, let me go to this term right over here
- I'm going to have a Ax Bx
- Cross that out
- And then over here I'm going to have an Ay By
- Remember I'm factoring out a negative C sub x
- So I'm going to have a plus A sub y, B sub y
- And then finally I'm going have a plus A sub z,
- A sub z, B sub z
- And what is this!
- Well this right here in green,
- this is the exact same thing as the dot product of A and C
- This is the dot product of the vectors A and C
- It's the dot product
- of this vector and that vector.
- So that's the
- The dot product of A and C times the x component of B
- times the x component, the x component of B
- minus, I'll do this in the same
- minus, once again this is the dot product of A and B now
- minus A dot B, A dot B
- times the x component of C
- And we can't forget all of this was
- multiplied by the unit vector i.
- We're looking at the x component, or the i component
- of that whole triple product.
- So that's going to be all of this
- it's all of this is being, is times the unit vector i.
- Now, if we do this exact same thing
- I'm not going to do it, because it's, it's computationally intensive
- but I think it would, it won't be a huge leap of faith for you
- This is for the x component
- If I were to do the exact same thing for the y component, for the j component
- so it'll be plus
- If I do the same thing for the j component
- We can really just pattern match.
- We have, we have B sub x, C sub x
- That's for the x component
- And we'll have B sub y, and C sub y for the j component
- And then this is not component specific
- so it'll be A dot, A dot C over here,
- and minus A dot B over here
- You can verify any of these for yourself if you don't believe me.
- But it's the exact same process we just did.
- And then finally for the z component, or the k component
- Let me put parentheses over here
- Same idea!
- B sub z, C sub z
- And then you're going to have A dot B over there
- and A dot C over here!
- Now what is, what does this, what does this become?
- How can simplify this?
- Well this right over here -
- we can expand this out.
- We can factor out a A dot C from all of these terms over here.
- Remember this is going to be multiplied times i.
- Actually let me not skip too many steps.
- because I really want you to believe what I'm doing.
- So this, if we expand the i here.
- Instead of rewriting it, let me just do it like this.
- A little bit messier, but let me just, so I can write this i there and that i there.
- I'm kind of just distributing that x,
- that x unit vector, or i unit vector
- and let me do the same thing for j
- So I can put the j there and I can put the j right over there
- And then I can do the same thing for the k.
- Put the k there, and then put the k there.
- And now what are these!
- Well this part right over there
- This part right over here is exactly the same thing as A dot C
- A dot C times Bx,
- B sub x times i, plus B sub y times j
- plus B sub y times j
- plus B sub z times k!
- And then from that we are going to subtract all of this
- A dot B, we are going to subtract A dot B
- Times the exact same thing!
- And you are going to notice this is the exact same thing as vector B!
- That is vector B, and when you do it over here you are just going to get vector C
- So I'll just write it over here.
- You are just going to get vector C.
- So just like that, we have,
- we have a simplification for our triple product!
- I know it took us a long time to get here, but this is a simplification!
- It might not look like one,
- but computationally it is! It's easier to do!
- If I have, I'll try to color code it, if I have A cross
- A cross B cross, let me do it all different colors, C
- we just saw that this is going to be equivalent to
- and one way to think about it is
- You take the first vector times the dot product of the
- The first vector in this second cross product
- The one that we have the parentheses around, the one we have to do first
- You take the first vector there, so vector B
- and you multiply that times the dot product of the other two vectors
- So A dot C, A dot C
- and from that you subtract
- You subtract the second vector [in the parentheses]
- You subtract the second vector multiplied by the dot product of the other two vectors.
- A dot B.
- And we're done!
- This is our triple product
- This is our triple product expansion!
- Now once again, this isn't
- this isn't something that you really have to know
- You could always obviously multiply
- You could actually just do this, you know
- you don't have to, you could do it by hand,
- you don't have to know this.
- But if you have really hairy vectors
- or if this is from some kind of math competition
- And sometimes it simplifies real fast when you reduce it to dot products
- This is a useful thing to know.
Be specific, and indicate a time in the video:
At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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