QED Prerequisites Geometric Algebra 21 Rotors

Transcript

[Music] thank you welcome back we are in this lesson going to continue our study of geometric algebra we're trying to get in prep prepared to understand these relative frames and pair vectors we're exploring three-dimensional geometric algebra we're taking a break from space-time algebra and today we are going to talk about rotations now in the last selection lecture we did some preparation which we're going to redundantly go through again today hopefully a little quicker I think last time I said you know we're not going to talk about Reflections but upon reflection I've decided that um I don't think there's a better way of understanding rotations in geometric algebra than going through Reflections so that is going to be in fact our method today and let's begin okay so let's begin with a little bit of a catch up right I never object to going through things again in multiple lectures because these are just things that need to be automatic and they're worth repeating and as long as we're learning it's okay to go over the same ground again in different contexts so here we go we're dealing with uh we're headed towards this understanding of rotation so we're going to start building things up again reminder that two vectors A and B take their geometric product a dot b a wedge B we know that a dot b is this inner product which is the magnitude of a magnitude of B times the cosine of the angle between them and a wedge B we know is going to have a magnitude of the area of the parallelogram between the vectors A and B which is the magnitude of a times the magnitude of B times the sine of the angle between them multiplied by some orthogonal bi-vector w v which is an orthogonal bi-vector that is a unit orthogonal bi-vector and meaning that we can write it as uh as W hat V hat and it is in the plane of A and B so it turns it's it's so this is a bi Vector just like this is then once we've done this well we can take this expression and just rewrite it this way which is very critical so the magnitudes come out in front and you end up with this this essentially magnitude 1 exponential form right because we're obviously going to turn this into an exponential in the next line and we do that right away we take a b we multiply it by the exponential of w hat V hat sine Theta and we can do this because we know that W and V multiply to negative one this is something we kind of showed in the last time wvwv is minus wvvw this becomes V squared which is one then what's left is W Squared which is one and you get negative 1 for that one flip you needed so we're going to call that guy I little I to represent the bi Vector that defines the plane between a and b so ultimately I lost my magnitude of B here and so ultimately we know now that we can write the space-time product of A and B as the magnitude of a times the magnitude of B times e to the I Theta where Theta is the angle between the two vectors and I is a unit a unit by Vector that defines the same plane as a and b okay so with this form available to us we also need to Let's remind ourselves we learned last time the projection of a vector X onto another Vector Y and but this is good for by the way any dimensionality right there's nothing in here about three dimensions of space time right the only thing that would make this uh three dimensional is if we defined A and B precisely to be have three independent components but we haven't even specified that so this is definitely true for so far for every every dimensionality for every dimensionality that's based on product of two vectors can be written in this form then again for every dimensionality you can project one vector onto another and we know that we learned last time that the parallel part of the projection is the magnitude of a times the cosine between the uh the cosine between the angles well I said to a here so this probably should be an a right an a vector notice these are not unit vectors so the magnitude of x times the cosine between the angles of the two vectors times the unit Vector in the direction of a and that of course is a aligned X is the component of X aligned with a of course that this is something we may have learned from regular trigonometry but we know that that's also expressible in this form because the magnitude of of of a unit Vector is one so we can write it this way as x dot a x dot a hat in the direction of a hat and then if I replace a hat with its definition which is a over the magnitude of a which is a simple replacement to make and obviously I can take this a hat out and I will get x dot a Vector right now A is a vector so I put a vector symbol up there just to be absolutely clear although I guess for consistency's sake I probably shouldn't have done that right because I haven't done it before although I did it on X right so why don't I do it on a yeah I should be doing it I should be doing it so I'm going to put in these little Vector symbols the full Vector symbol and that should probably be the full Vector symbol here as well right it's the magnitude of a vector and then uh a hat over the magnitude of a and then this gets to a inverse hell okay this gets to a inverse because this guy here see this is why it's worth doing over again you keep you keep kind of having to remind yourself this is equal to a over the magnitude of a that's what a hat is and then you get another Factor the magnitude of a Down Below which is a over the magnitude the magnitude of a squared which is the definition of a inverse right so we end up with x dot a times a inverse and that equals the projection of X onto a right so that's a nice thing to know yeah see getting from this statement to this statement is something you should be able to think about and do it's just we're plugging in the definition of a hat we get a-hat over another factor of the magnitude of a which is a over the magnitude of a squared which multiplied by a is a squared over the magnitude of a squared which is a inverse now it's worth noticing that this is a place where the three-dimensional geometric algebra in the space-time algebra actually do have a meaningful difference right because if we look at a vector say B right in the space-time algebra well what if B is space like right which would mean that um that the magnitude of B is is is less than zero this this would be then then if if for example if B squared was less than zero say let's say that was the case well then B over the magnitude of B squared right if you multiply that by B you're going to get a number on the top that's less than zero and a number in the bottom that's greater than zero so this is going to equal minus one right so it won't be the inverse it'll be the opposite of the inverse so this this uh this is where studying three-dimensional geometric algebra might actually cause a problem to expand to other geometric algebras so the rule that uh works for four-dimensionals for for the space-time algebra said B inverse is B reverse over B reverse B now a vector is its own inverse but what ends up effectively happening is this becomes B over B squared which actually becomes uh which is in B squared is going to be less than zero so when you multiply it by B you get B squared over B squared which will equal one so there's something neat to keep an eye out is in your whatever geometric algebra you're working in you really have to understand how to calculate the inverse so um let's continue the next thing we did is we calculated X's rejection onto y right so the rejection of X onto y I think the same problem here this y I should replace with an a right and the rejection is simply the vector itself minus the projection right so this is one of those little sort of geometric algebra things tricks that you do in these proofs you can always multiply a geometric product by a a inverse right because that's just one so I can distribute the AAA inverse to get x a inverse and then x dot a a inverse a a inverse because the projection part is x dot a a inverse where I am missing a little arrow there right that this is supposed to be a little arrow right here so now once I've done that distribution I simply what do I do um I guess all I did here was I just used the associative property and I multiplied a inverse by a so I eliminated these two because they become one right but I left these two even though they become one also and what I'm going to now do is I'm going to take the a inverse this little mess right here is supposed to be a inverse right I take the two a inverses and I undistribute it to here and then I recognize that this is the geometric product of X with a this is the uh the the metric contraction of X with a so the geometric product minus symmetric contraction is in fact the wedge product so that uh that is uh very simply reduced so this whole term reduces to the wedge product of X with a multiplied by a inverse on the right and fall out that is your perpendicular Vector so this thing needs to be a vector so this is a bi Vector this is a vector right and whenever you see these a inverses you shouldn't be bugged sometimes they write it like this which I always found a little Annoying I I don't like that form but because just because dividing by a vector is kind of weird but but yeah once you realize dividing by vectors just multiplying by an inverse Vector that weirdness goes away pretty quick but just understanding that a inverse is a over in in our in the case here magnitude of a or uh actually the you know what I'm going to just start writing this as a over a squared right because if you're in the 3D geometric algebra a squared will always be positive uh anyway and if you're in 4D it could be negative and that would matter so anyway so you real the point is you know that this is a number in the denominator so you know that this is a vector in the numerator so you know this is a vector here right and when you look at this you say well immediately you know it has no trivect it's it's a product of a bi Vector with a vector so you've the result has to be a tri-vector and a a regular Vector a grade one object and a Grade Three object but the grade three object is zero because a is this a here is actually co-aligned with that a there so this a is linearly dependent on that a so you know the tri-vector goes away and all that you're left with is the vector and we know how to do that Vector calculation it's going to be a DOT a x minus x a um x dot a a right that's what it's going to be that is it's going to be this thing right here right that's what this is that because that's sort of a memorized uh it's a memorized property that I've got carry around in my head so that's a way we calculate the rejection the rejection is this bivactor the rejection of x on a is the bi-vector of X with a times a inverse right so we have those two formulas in our hit pocket and now we are going to understand the reflection now to make sense of this by going through the literature there's there's actually a couple different ways they think of it the first way is taking a vector X and reflecting it through a which is actually taking the reflection of the vector X and the you take Vector a and you're going to actually push X across a which basically means you're taking the perpendicular component of X right X perpendicular and you're actually taking its negative and you end up with a new Vector on the other side of a reflected through a and this is actually done when people are working a lot with two-dimensional geometric algebra but that is really not the good way I think overall to think about it instead we want to flip this parallel part so we're not really reflecting through a we're kind of collecting over and around a and the right way to think of it is about what you're reflecting through a plane like a plane is more like a mirror anyway right so you're reflecting through a plane that is perpendicular to the vector a and when you think of it that way then of course you're going to take the part that's parallel to a and change its sign and your and of course the perpendicular part will won't change sign at all and that's the turns out to be the correct way of looking at things we're always reflecting through planes but we're identifying the planes through their normal vectors now you might immediately say well you've already mentioned a few times that in higher dimensional spaces a plane does not have a single unique normal Vector there are many normal vectors for example if this was the E1 2 plane right if that was the plane defined by E1 wedge E2 well E3 is normal to that E4 is normal to that E5 is normal to that all the way to the n e n where n is the dimensionality of your geometric algebra every one of these vectors is normal to that plane E12 so this clearly doesn't work but geometric algebra saves its own day by saying well no no what we're just doing this because this is three-dimensional so this line which is actually a plane coming out of the Whiteboard you've got to think of that as a hyperplane and it is the hyperplane normal to a and in any D dimensional space a hyperplane has D minus one dimensions and every hyperplane of D minus one Dimensions does in fact have a unique normal or an identifiable single normal vector so what we're actually doing is we're reflecting X through or across a hyperplane in higher dimensions and this is done using this very simple prescription right when you reflect it through this hyperplane you still get a vector but it's a little bit hard to picture this hyperplane but it works fine in three dimensions for us to use three dimensions to understand it and that's what we're going to do so we can just sort of work through the math of how to do this reflection the reflection is obviously the perpendicular part of X so X reflected in this hyperplane so This R represents the reflection through this hyperplane as identified by the vector a so you could in fact write oh this is r a so the a actually identifies I'll put a hat there to show it's a unit vector that little a hat is telling me what the unit Vector of the hyperplane is which in turn identifies the hyperplane and R tells me I'm reflecting the vector X in that hyperplane and obviously I'm throughout this I'm not putting little arrows on x sometimes I do sometimes I don't I it depends there's only we're only dealing with vectors here so it's kind of silly to do it all the time all right so uh how do we do this well the perpendicular part of X stays the same but we change the sign of the parallel part so we're trying to calculate this that's the definition of XR well a pretty typical trick in geometric algebra is to multiply by the geometric product of a vector with its own inverse because this this always equals one right so we're not really doing anything new there so I just distribute because everything is linear I just distribute it right through to get this expression here now what I'm going to do is I'm going to commute the perpendicular part of X with a now a is by definition this perpendicularity is with a so X perpendicular is by its very definition perpendicular to a so these two commute just fine so they just this just straight up commutes and this a perpendicular is [Music] um wait wait wait wait what did I just say this is the perpendicular part anti-commutes with a it anti-commutes with a right so yeah perpendicular vectors just like unit vectors you know we do this all the time E1 E2 equals minus E2 E1 well a and X are perpendicular to each other just like these unit vectors so they anti-commute so when I reverse their order when I reverse their order like I do here I pick up this minus sign on the other hand in this part that's parallel to a well if a and a are parallel you know that it's by Vector though the the geometric product of a and X parallel the bifector part is zero so all that's left is the um is the scalar part so this thing commutes just fine and that's where you pick up we may we commute these two to get these two here right we take these two and commute right here so now we've got these two minus signs here and those two minus signs we just pull out in front to get this big minus sign right here and then x uh perpendicular and X parallel get some together and we pull the the two A's on the left out to the left and we pull the two well we pull the a inverses on the left to the left and we pull the A's to the right this guy becomes X then you get a and minus a inverse and this is the formula for that equals the reflection so this beautiful formula is a pretty simple formula for reflecting X in the hyperplane that is orthogonal to the vector a the orthogonal to the vector a so now that we've defined or we've learned how to calculate the reflection through a hyperplane as we've just described by this formula the next question is well what about two hyperplanes well you just repeat the process right the reflection R is through the hyperplane orthogonal to the vector A and S is the hyperplane orthogonal to the vector B well here's X this is X reflected through a hyperplane orthogonal to the vector a and then by multiplying by B and minus B inverse that's taking the result of this and reflecting it again so this is a composition of of functions really is what this is and you take the two minus signs out and they they clear and you're left with B inverse a inverse x a b and this is the reflection through two hyperplanes now the thing that's important to appreciate here which is the Obscure but interesting and Ultra super important fact is that if we let's look at this diagram here and see what this next super important fact is if a is given by this Vector right here right this is this is the vector a and then this is the vector B so what we're going to do is we're going to take a vector X it's this blue vector and we are going to reflect X first through the hyperplane orthogonal to a which is here's the hyperplane orthogonal to a and then through the hyperplane orthogonal to B which is this line here so that first reflection takes X and remember we're sort of switching the um we're taking the parallel component of X and changing its sign so X moves from there to here right right so this is x r so this is XR so that's the first reflection and then the second reflection is through this plane B right and so this ultimately here is X uh Sr right our first s second that's what I mean by xsr and I guess my Vector needs a little point on it like that so but what's important to understand about this geometric construction is that the ultimate angle of this whole process that means the angle that X is rotated by these two Reflections is twice the angle between Alpha and beta which means that if I want to control the amount of rotation of the angle X I just pick an alpha and beta that have half the angle that I want to rotate if I want to rotate by pi I pick Theta to be pi over 2. if I want to multiply it by pi over 4 I take Theta to be pi over 8 right but this is a little obscure geometric fact that reflecting between two planes normal to two vectors is equivalent to a rotation of the same vector by twice the angle between these two planes or between these two normals and that little obscure fact although it's you know relatively well known I'm sure for centuries uh never was that important and I certainly never thought about it until geometric algebra kind of forced me to but now it's telling me that if I want to rotate a vector X through an angle of 2 Theta I need to just execute a reflection that um by through two planes that I and I can pick a and b to be any two vectors with an angle Theta between them and execute these rotations or these Reflections that I just did right this reflection right here right this is the reflection I just pick any A and B that has the right angle and this reflection uh this double reflection I guess will be a rotation of x and so taking that to heart I'm going to call the bi-vector of any two vectors I'm going to call it a rotor if a and b are unit vectors right because the length of a and b really doesn't matter because actually if you execute this product right you'll notice that the length cancels out because on this side here on this side here you have a b squared and an a squared in the denominator say A and B are not unit vectors these inverses would be B A with two squares in the denominator like that turns out that when you chug through all this those uh the lengths of A and B turn out not to matter right because this is a reflection right this is a reflection so why should it matter so going forward let's just assume A and B are unit vectors and then we're going to say you take two unit vectors you form their bi-vector or you form their geometric product and we're going to call that a rotor or a spinner right and so now we have an object that we're actually sort of treating in a way as an operator right we're basically saying if I want to rotate an object by Theta degrees right I have a Theta degree rotation and that object is X that is an operation right so R is sort of is an operator acting on x well the method of that operation is I create an appropriate rotor and multiply X geometric multiply X on the right by the rotor and on the left by the inverse of the rotor and that would be the effect of operating on X by an operator that rotates X Theta degrees so this is sort of the operator form and this is the geometric algebra implementation form but what's really important is that this works for any Dimension right any Dimension all I have to do is find two vectors in that dimensional space that have the right angle between them and I can create a rotor now I don't know if absolutely 100 rotors are always defined with unit vectors I I the the paper I'm reading or the various papers I read clearly there's no reason not to use unit vectors here so why wouldn't you anyway they do make it a lot simpler because they want your rotors to be written in this form right a nice clean cosine plus sign and that would only happen if A and B were unit vectors right so we we take our rotor we break it up and you know it's a it's a geometric product of two vectors so we know it has this classic form of the grade Zero part and the grade two part right the grade Zero part for two unit vectors is going to be the cosine of the angle between them as the two magnitudes are are one and the um the grade two-part well that's the area of a parallelogram covered by the vectors A and B right bounded by the vectors A and B well that area is the magnitude of a times the magnitude of a b times the sine of the angle between them but eight magnitudes of A and B are one because we're dealing with unit vectors now so you just have this sine factor and then I is any any bi Vector that is that is made of orthogonal components that defines the same plane as a and b and we know that we can do that because we did that in a previous lesson so though that guy we might call W hat V hat where W hat dot V hat equals zero and I would equal W hat wedge uh wedge V hat right wedge uh wedge V hat like this which equals W hat V hat right that's the the geometric product equals just the wedge product because we've set up W and V so they're dot product to zero so once I have this structure and I know that I squared equals minus one because that's always the case right because W hat V hat W hat V hat since W and V were designed to be orthogonal right I end up with W hat V hat V hat W hat that's one switch so I get a minus sign that's V squared so that's a number so that leaves behind W Squared which is another number those they're unit vectors right so that number is one that number is once this whole thing equals minus one oops equals minus one like that so with that once I know that this this bi Vector I squares to minus 1 then Kaboom this immediately can be written in this Euler form and that's a vector and this part here is a vector because I squared is minus one right so you have beta squared x so that becomes a vector and you end up with Alpha squared plus beta squared x so that's actually the original Vector interestingly and um because Alpha squared and beta squared that's cosine squared plus sine squared so that's one so that's the original vector and now if you look at this part all we want to do right now is just show that it's a vector well this piece here this is a vector times a bi vector so normally a vector times a bi-vector is a it's got a a vector Times by Vector is going to have a vector part and a tri-vector part well the vector part we don't care about because the vector part is what we're trying to show we're trying to show that it has a vector part but we do need to show that the tri-vector part goes away and and that brings up a bit of a subtlety which I did not point out earlier so I'm going to have to go back and point it out now when I drew this picture the part that may have you may have missed is that X of course you missed it because it's literally drawn this way right X is in the plane of A and B already right so it just shifts over in this plane but it doesn't have to be that way X does not have to be in the plane of A and B and it'll still rotate right what will end up happening and I'll show you why in a moment is that this becomes X parallel but now and this is why when you read a lot of different books and they take different approaches you have to be very careful that parallel means the part of X that's in the plane of A and B right it's not parallel to a it's not this piece it's not parallel to B it's not that piece right it's actually the part of X that lies in the plane of A and B and that's what moves and the part that's perpendicular right the part that lives here and comes I guess out of the board right that would be X perpendicular right so you got to sort of Imagine something standing right here coming right out of the Whiteboard that's the perpendicular part that part doesn't change at all that part is absolutely fixed so this was one of the reasons we did some of the work in our last lesson so let me show you how this works we imagine this situation where X is out of the plane of w and V where W and V now that's what I is right I equals W hat V hat right it's that geometric product and so um so X has two pieces right X has a component that is in the plane right I'll make it a little dotted line and that will call X parallel right so this is X parallel then it has this part that's out of the plane right uh let's see let me short next parallel a little bit so this picture looks a little better and this part that's out of the planes like that and that's X perpendicular right so now our claim is that the rotated form of X so X rotated equals the rotor inverse acting on x followed by the rotor well now I break X down I'm going to make a whole bunch of substitutions let's go to Black here this will be e well actually I've done it over here I forgot so this is the the process now X so the rotors can be replaced by these two expressions where Phi is the angle of rotation we want to drive on X and I is the bi-vector of the plane that we're going to do this rotation in or it's actually the plane of the two uh the the plane defined by the two normal vectors right which is where the rotation is happening right the reflections are happening in these perpendicular planes let me show you that remember these these perpendicular planes that's where Reflections are happening so there's no rotations going on in these planes the rotation of this piece is purely in the plane given by uh defined by A and B and so that plane we're calling defined by I so we got those we've got our eyes we create our two r's with these exponentials and then I break X into the parallel part a parallel part and the perpendicular part where now parallel means in the plane of I and perpendicular means not in the plane of I out of the plane of I so everything being linear everything being linear I'm just going to bring everything together just like this right I'm just simply Distributing through with e to the minus five Phi over 2 and e to the plus Phi over 2. so I end up with these two expressions but you'll remember in the last lesson we studied the commutation of these two ideas of these two things right we studied how a vector that's perpendicular to I will commute with an exponential and a vector is parallel to I will which will commute with an exponential and what we learned is if you're perpendicular to the to the bi Vector that's in this argument of the exponential if the vector is perpendicular to it this geometric product here because remember this is a geometric product of two vectors so this is a bi-vector plus a scalar right that's what this is but if it's perpendicular to this plane it's going to just straight up commute this is going to straight up commute so this expression here is going to be e to the minus Phi over 2 I times e to the Phi over 2 i x Purp right and these two are just going to add up to zero so you can end up with e to the 0 which is just X purp so this entire side over here is going to just can't this is all going to cancel to 1. but more importantly the other one we know that if you're parallel to the plane defined by this by Vector I you the this commutation product produces a I don't want to say it Andy commutes because it's not a minus sign here but it makes the exponent anti-commute because we proved that the last lesson we proved that if you're in the plane of this bi-vector you can you can pseudo anti-commute with the exponential but the minus sign appears in the exponential which isn't this isn't a surprise because this is just a Taylor series where you know it's a sum of things so you're just anti-commuting with each piece so this this appears linearly in each piece at some point so that's not a big surprise but um uh but these exponents now add up to Phi right so this guy here is going to be plus e to the minus Phi i x parallel which is also we can play the same game X parallel uh e to the Phi I like that right so we could do it either way left or right if it's on the left you've got a minus sign if it's on the right you get a plus sign but the rotation um is always this number of degrees in the in this sense of I right so if if in our case here if I is WV the rotation will be in that the positive rotation will be from W towards V that would be the positive rotational sense so the answer is it's equal to X perpendicular which doesn't change because that's one plus X parallel e to the Phi I right and so now if you started this if we had started this whole analysis in two Dimensions we would have learned that a rotation of any Vector in the two-dimensional plane would immediately rotate in this fashion right we could study this from the beginning and then we have to say yeah that's good but the plane is too simple uh in three dimensions you really have to go to to this expression to rotate a vector because remember every Vector in two dimensions in two-dimensional geometric algebra every Vector is parallel right every Vector is in the plane of I there's only one I this is only one plane right so that's why two Dimensions I think is a little too simple because you end up showing that you end up basically proving that this is how rotations are done rotations are just done R times the rotor right there's no need to do R inverse in two Dimensions because um because in two Dimensions every Vector commutes with r so if I wrote r R inverse like this um uh I I would just immediately uh well I'm sorry and not every Vector commutes with r every Vector commutes uh in this sense right so you know you would have e to the minus uh Phi over 2 I and then you would immediately for in every case just turn it into e to the minus Phi over 2 i x because every X is parallel there's no such thing as X perp in two Dimensions so it's two Dimensions it's too simple for this analysis in my opinion so that's my little contribution is just just skip two dimensions and go right to three but in the end you do find the the parallel part just rotates very nice and simply but the perpendicular part Remains the Same so what's basically kind of going on is ultimately this piece here right rotates uh time twice of angle Theta so it rotates to there and then which means and this piece here stays the same so the whole thing just sort of moves like that you know where this is now the new X parallel is this part so that's kind of cool there's a lot of different pictures you can see online or or um you know in papers about you know describing this but it's a but basically that is the story the story of these rotations is that you're taking the parallel part the part that's in the plane of the pseudo Vector that's defined by the two vectors that that we're using to define the plane of rotation and now we actually do talk about a plane of rotation right being these defined by these two vectors that are actually the normal vectors associated with the hyperplanes of reflection right so that kind of seems complicated but it's just a way of getting us there right it's a way of getting us there um we just we we Define these two vectors because we're going to talk about two Reflections and that gives us this notion of Phi and 2 Phi right which I didn't prove by the way I didn't show that a reflection from of a vector through this the plane normal to a and the plane normal to B will actually be 2 5 but it is and um and that's really really remarkable okay so um uh so I guess that brings us back to this right I needed to show that this was a vector so I needed to show that the tri vectors um went away and so we showed that okay this this V this Vector here uh X this bi Vector well they're they um uh the perpendicular part we know uh will just straight up well I guess we actually have to work it out right we have to say all right I I is actually W hat um wedge V hat right so now we have okay so this part this perpendicular part with W wedge V well it's X perpendicular wedge W wedge V we know that the that's the inflation part the we already know that the this products uh contraction part will be zero because X perpendicular is perpendicular to W perpendicular to V so all of the dot products involved in that geometric product will go to zero so this guy well if I flip X once I pick up a minus sign and then I flip it over here that minus sign goes away so I get this so so X perp I equals i x so if I flip if I flip the purpose so the perpendicular Parts end up canceling right because you don't lose this minus sign so what about the parallel parts well if I change this perpendicular to parallel right well the definition of parallel is that it's in the plane of w and z so that part goes to zero right away right so the only part of this X parallel geometric product with W wedge V is the contraction part so all of this is zero and well what is the contraction part of X perpendicular times geometric product W wedge V what's the contraction part of that well it's X parallel dot w hat V hat minus minus X parallel dot V hat W hat well that's a vector minus a vector so that's a vector so this whole thing ends up being a vector so you have a vector this whole thing here we call the vector v so you end up with a vector w so that's just to show all that's to show that a rotor times an arbitrary Vector is a rotor and we took a little bit of a Divergence in the digression in the middle of that to show this really important fact at this rotor applies to any Vector right that some of which is in the plane of I and some of which is not now just be careful sometimes we've written a perp to mean in the plane of I and a parallel I'm sorry a parallel to mean in the plane of I right and a perpendicular we meant to mean out of the plane of I but then earlier when we talk about Reflections we're talking about the projection of x on a vector a as a parallel and the projection of the rejection of x on a vector a as a perp so those are two different parallel and perpendiculars right this would be x per so again it's it's all notation and it's all contextual and so then there's one last thing about rotors I want to talk about in this lesson and this to me is really really almost a big selling point for geometric algebra and that is say I took the geometric product of a whole bunch of vectors A B C and D right now every blade right every say k blade is basically a sum of objects that look like this right it's the sum of a bunch of geometric products of orthogonal unit vectors you know the way we've been doing it is you know AK is going to be some number a say one two three E1 E2 E3 or e123 depending on how we're writing it at the time uh I noticed by the way Mathematica tends to do this there's a Mathematica package that I'll try to show off um in a lesson or two that's really fun um and I kind of wish it would do this but it actually uses this notation but the point is is that any k-blade can be written as the sum of terms of these geometric product of orthogonal vectors and any multi-vector a can be written as the sum of K blades right from k equals you know 0 to n in your n-dimensional space so imagine taking just an arbitrary blade which is sort of represented by this geometric product and sticking it between two rotors well that would if this was just a vector we know that that would rotate the vector by an amount given by the definition of RR which is e to the I Phi over 2. it would rotate it by an angle of Phi in the plane defined by whatever pseudo Vector we throw up there in in our definition of r so uh perhaps I should be a little more specific now because we're talking about an arbitrary space an arbitrary rotation so I should probably go Omega V where we assume Omega and V are orthogonal unit vectors Phi over 2.

that way you kind of look at this and you don't forget that that is the pseudo Vector this guy here is the pseudo Vector in a plane of a certain plane sitting out there in n-dimensional space right any dimensionality so once I've chosen this and I've got r and r r inverse of course is totally trivial you just throw a minus sign on it right and but look what I can do I can stick r r inverse everywhere in these blades right because RR inverse is just one so this is exactly the same thing as this but then I just regroup like this right I just regroup like that and all of a sudden I have the rotated R the rotated B the rotated C and the rotated D multiplied together so if I want the rotated version of any blade right and by linearity any multi-vector all I have to do is multiply by R inverse on the back and R inverse on the front and I suddenly get the whole thing rotated so it doesn't matter what you throw what multi-vector you throw in between R inverse and R so everything I proved up here for a vector is true actually for any multi-vector in the space you can rotate anything in the vector space you can multiply a bivector a tri-vector a pseudoscaler you can multiply a 10 dimensional thingy whatever just by making one pre and post multiplication in geometric algebra and that I'm is a big selling point because if you're doing this in normal Vector analysis for every dimensionality of thing for every strange thing you threw in here you'd have to come up with a new representation for how that thing rotates and that's not true in geometric algebra and geometric algebra it's very clean all you have to do is take this rotor and execute this multiplication so anyway I'm going to stop there and we will pick up this subject again I'm not sure what the next topic will be in our next lesson so I'll see you next time