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Current time:0:00Total duration:14:25

Vector triple product expansion (very optional)

Video transcript

what I want to do in this video is cover something called the triple product expansion or Lagrange is formula sometimes 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 and what we're going to do is we can express this we can express this really as sum and differences of dot products well not just stop brock's dot products scaling different vectors you're going to see what I means but it simplifies this expression a good bit because cross products are hard to take they're computationally intensive and at least in my mind they're confusing now this isn't something you have to know if you're going to be dealing with vectors but it's a useful to know my motivation for actually doing this video is I saw some problems for the Indian Institute of Technology entrance exam that seems to expect that you know Lagrange is 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 let's start taking the cross product of B and C and and eat it in all of these situations I'm just going to assume let's say I have vector a I'm just going to call that that's going to be a the X component of vector a times the unit vector I plus the Y component not B plus the Y component of vector a times the unit vector J plus the Z component of vector a times unit vector K and I could do the same things for B and C so if I say B sub y I'm talking about what scaling the J component in the B vector so let's first take this cross product over here and if you've seen me take cross products you know that I like to do these little determinants and what let me just take it over here so B B cross C B cross C is going to be equal to the determinant and I put an IJ K up here i JK this is actually the definition of the cross product so no proof necessary to show you why this is true this is just one way to remember the dot product if you remember how to take determinants of three by threes and we'll put B's X term these Y coefficient and B's z component and then you do the same thing for the C's CX CY cz and then this is going to be equal to so first you have it's the I component so it's going to be the I component times B so you ignore this column in this row so b YC z v y c z- b z cy - so I'm just ignoring all this and I'm looking at this two-by-two over here - b zc y - bz c y and then we want to subtract the J component remember we alternate signs when we take our determinant subtract that and then we take out that column and that row so it's going to be it's going to be b XC z b x c z this is a little monotonous but hopefully it'll have an interesting result b XC z minus b zc x minus b z minus b z CX and then finally plus the K component K we're going to be x times c y bx c y minus b YC X minus B Y minus B Y C X so this is this we just did the dot product and now we want to take the socks we just did the cross product I want to get you confused we just took the cross product of B and C and now we need 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 another matrix here so let me write my i j k up here and then let me write a is component 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 this we're just looking at now I want to do that in black let's do this in black so that we can kind of erase that now this is a 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 right this with the signs swap so this is going to be this is going to swap the sign it's actually BZ see X minus minus BX see minus BX see Z 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 I can make the brush size a little bit bigger so I can erase that a little more efficiently there you go and then we also want to get rid of that right over there now let me get my brush size 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 only going to focus on because it'll take the video or 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 the I component just on the X component of this cross product and then we can see that we'll get the same result for the J in the K and then 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 two-by-two matrix right over here we ignored eyes column eyes row and we have a Y times all of this so let me just multiply it out so it's a Y a y times BX cy- - a Y times dy times B Y C X B Y C X and then we're going to want to subtract we're going to have minus AZ times this so let's just do that so it's minus or negative a Z B ZC X and then we have a negative a Z times this so it's plus a Z B X C Z and now what I'm going to do this is a little bit of a trick for this proof right here just so that it just so that we get the results that I want I'm just going to add and subtract the exact same thing so I'm going to add I'm going to add an ax BX CX and then I'm going to subtract an ax BX CX minus ax BX be the X is 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 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 let me get the B sub X so if I were to factor it out and I'm only going to look at I'm going to factor out it of this term that has a B sub X I'm going to factor out of this term and then I'm going to factor it out of this term so if I take the B sub X out I'm going to have an A why see why actually let me write it a little bit differently let me factor it let me factor it out of this one first so then it's an A it's going to have an ax CX a sub X C sub X so I use this one up and then I'm going to have a I'll do this well 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 factor that out and now from these from these right over here let me factor out a negative C sub X negative C sub X and so if I do that let me go to this term right over here I'm going to have an ax BX when i factor it out so an ax BX cross that out and then over here I'm going to have an A Y B Y remember factoring 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 to 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 products 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 a the dot of a and C times the X component of B times the X component times the X component of B - I'll do this in the same - once again this is the dot product of a and B now - 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 the X component of 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 and 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 if this is for the X component if I were to 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 well 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 it will 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 you're going to have B sub Z C sub Z and then you're going to have a dot B a dot B over there and then you're going to have a dot C over here now what is what does this what does this become if what I can we simplify this well this right over here we can expand this out we can factor out an a dot C from all of these terms over here and remember this is going to be multiplied times I actually let me not let me not skip too many steps just because I want to I want you to believe what I'm doing so this if we expand the I here instead of rewriting it let me just let me just do it like this it's a little bit Messier but let me just so I could write this I there and that I there I'm kind of just distributing that X that X unit vector or the I unit vector and let me do the same thing for J so I could put the J there and I could put the J right over there and then I could do the same thing for the K put the K there and then put the K there and now what are these well this this part right over here this part right over here is exactly the same thing as a dot C a dot C times and I'll write it out here 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're going to subtract all of this a dot B we're going to subtract a dot B times the exact same thing and you're going to notice this right here is the same thing as vector B that is vector B when you do it over here you're going to get vector C so I'll just write it over here you're 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 a cross a cross B cross let me do in all different colors see we just saw that this is going to be equivalent to and one way to think about it is it's going to be you take the first vector times the dot product of the the first vector in this in this second dot product the one that we have our parenthesis around the one we'd have to do first you take your first vector there so it's 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 you subtract the second vector multiplied by the dot product of the other two vectors of a 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 it 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 has come some type of math competition and sometimes it simplifies real fast when you reduce it to dot products this is a useful thing to know for like your ranges formula or the triple product expansion