Transcript for:
భూపరిమాణం మరియు రేఖాగణిత పరిచయం

rolling okay so each day we'll start with a mathematician spotlight some person that's doing math some current person who's doing math so today the current person is me I'll tell you a little bit about my research so I studied geometry and dynamical systems and specifically I studied billiards so maybe you're familiar with the game of billiards or pool you use a rectangular pool table you've your ball bounces around those that has been understood for a long time since the 1800s at least so I study a pentagonal pool table so you imagine that you have a contactable pool table and you put your ball down somewhere and you shoot it in some direction and maybe it bounces around for a while and then maybe if you're lucky it comes back to where it started and it repeats this path it turns out that if you draw out the path that the ball would take if it repeated like that they're beautiful check these out so I'll pass some of these around these are some examples of periodic paths on the regular pentagon you'll see that some of them have rotational symmetry and the yellow one does not I also have one here on my shirt so nice and one around my neck and one of my ears so anyway there turned out to be really beautiful I've gotten really excited about them in the past year so and this is some current math that's currently being done so there you go ok so you guys all know linear algebra linear algebra is about flat things lines and planes and here in multivariable calculus we're going to study curved things like surfaces and curves and stuff so as a gentle introduction today we're going to talk about lines and planes it's like the flat versions of surfaces and curves so today we're going to talk about lines and planes and the cross-product so in linear algebra you solve systems of linear equations so for example suppose you want to find the intersection of two lines so so find the intersection point of two lines so for instance maybe your lines are X plus y equals two and X minus y equals zero and you want to find where these intersect I think the best way to do this would be to draw a picture but for the sake of doing linear algebra we're going to do it using linear algebra so first you could put these in a matrix so we've just put the coefficients in one one and then the constant is two so 1x plus 1y equals two and then one X minus 1 y equals zero so now we've got this in this augmented matrix and then you will reduce this is review of what you've done before so if you row reduce it turns out this reduces to 1 0 0 1 and 1 and 1 so if this part meant at 1 X plus 1 y equals to 1 X minus 1 y equals 0 this part means X plus 0 y equals 1 and 0 X plus 1 y equals 1 so it tells you there's just this one solution 1 and 1 if we draw this this line X plus y equals 2 you could also write it as y equals 2 minus X so it looks like this and then this guy x minus y equals 0 that's also x equals y so it looks like this and so you can see here and indeed has one intersection point 1 1 okay okay the purpose of doing this is to think about I want to count degrees of freedom so here we have took two variables so we start with um two variables so in general X&Y could be anything we have two degrees of freedom one for X and one for y they could be anything but then we can strain it and generally each time you have a constraint it drops down your degrees of freedom by one so we have one constraint here and one constraint here so we subtract two constraints so here when I subtract I'm just subtracting the numbers not the words and we get zero zero degrees of freedom so it might be that we have no solutions that would be if our lines are parallel or it might be that our solutions are a collection of points and here's just one point so solution is possibly empty or discrete points and in here in this case one point just this point okay so I'm going to use this to talk about maybe some things yeah some some other kinds of equations so here's our next question so question so what kind of object does the equation X minus 2y plus 3z equals 6 describe that's a question I would ask you all but it's written right on your sheet of paper so I won't ask you I will just tell you a way to figure it out if you didn't know how to figure it out so one way to think about it is you have three variables so general you have three degrees of freedom but then we're constraining at once we have minus one constraint this equation so we have two degrees of freedom so we expect our solution there are our thing to be a two-dimensional object okay it's a two-dimensional object we're familiar with two dimension objects we live on one well it's a I mean we live on a three-dimensional thing but we live on its surface it's two-dimensional surface the sphere also things like your piece of paper is a two-dimensional object also things like a like a halfpipe for for snowboarding or skateboarding two-dimensional surface so many two-dimensional things but this one is linear all the terms are linear there's no x squared there's no cosine of Y okay so it's a two-dimensional thing and it's a linear it's a linear two-dimensional thing object so it's a plane so it's a plane it's a plane we have three-dimensional space and we constrain it once we get a plane pretty neat okay so can you tell me some examples of points on this plane anyone have an idea Jack 3:01 let's see three minus zero plus three is six hey it works yay 3:01 nice another one we need three let's do three yeah one negative one one let's see so we have one plus two plus three is six nice okay one more anyone yeah zero zero two zero zero two yeah so we have zero minus zero plus six is six yeah okay good so we can just just plug things in see if they work get some points out on the plane nice okay hopefully somebody no to those down will use them later good so this is leading up to a question which is how do we write the equation of a line in three dimensional space or r3 so r3 just means like R is the real line R 2 is a plane R 3 is three space so how would you written equation of a line in r3 this is a question well you might think oh I know how to write an equation of a line it's just like y equals MX plus B like over here here is a line so X plus y equals 2 but in three space Z can be anything so here this line was X plus y equals two sitting in the plane but if we now consider this plane to be part of three space we actually get Z can be anything so it's like a plane sitting over this line so in three-dimensional space this equation X plus y equals to be a plane so we can't do that so we're going to have to have some other way of writing equation of a line I could tell you what I'm gonna try to make it organically come out of what you already know from linear algebra maybe is a little bit of a a review and then pushing forward and take into calculus so so so so approach let's find the line that is the intersection of two planes okay so I'm going to write down two planes and we're going to find our intersection first I'll draw a picture so our first plane maybe is this plane I have to stand in front of it so everything stays parallel that's pretty much a plane and then here's our second plane okay okay so there are two planes and any two planes intersect in a line so here's our line that we want okay so any two planes intersect on a line and we want to find this line so let's use some good equations so let's our first plane will be our plane from before X minus 2y plus 3z equals 6 and our second plane will be 3x minus 2 y plus Z equals 2 so we have three three variables and two constraints so we expect a one-dimensional thing that's good because we're hoping for a line that's good okay so if we put this in a matrix we will get the coefficients 1 negative 2 3 and then our constant term 6 and then 3 negative 2 1 and our constant term - okay you could row reduce I won't have you do it I asked Wolfram Alpha to do it for me but if you were reduced you get one zero zero one and then some other things we get negative 1 2 and then our constants are negative 2 and negative 4 ok now we have to figure what that means we just did a bunch of little computations so this means reading across X plus 0y minus Z equals negative 2 and then know X plus y minus 2z equals negative 4 so that's the solution that we got allegedly that's a line it's not looking to line it yet but let's let's make it happen so I'm going to write these over here just copy it again so this says another way of writing this is x equals negative 2 plus Z y equals negative 4 plus 2z negative 4 plus 2z and then Z is just Z so the idea here is that we have a one-dimensional thing because whatever we pick for Z it determines what our x y&z are so Z is sort of like a variable that could be anything so the convention is to call it something else so let's call it t so let's this is called a parameter and let's rename it T so we can write this as our x y and z is we can write this as zero plus Z is negative 2 negative 4 0 plus we could say Z but we're gonna say T because it's just any old parameter times 1 2 1 so this is our solution this describes all of our solutions that are intersection of those planes and I'm going to tell you what what we got here so here this part this purple part here this is a point or I mean it's a vector but I think a good way of thinking about it is as a starting point so this is some point that is at the intersection of these things so maybe this here's this point here negative 2 negative 4 0 and then this guy is the direction vector so that's the direction of the line so that's like 1 2 1 so that's this direction vector 1 2 1 and I think of this time as telling you where you are on the line so it's at time T equals zero this whole orange part goes away and we just get this point so at time T equals zero you're sitting here at this purple point at time T equals 1 you add one of these vectors to these at this point so you get to here T equals 1 if you add another one you would get to here T equals 2 doesn't have to be a whole number you could have like T equals PI it could be a negative number at T equals negative 1 you just subtract this vector off and so on so this is a this is called a parametric equation because it expresses X as a function of T y is a function of T and Z as a function of T so it's basically you think about it as a little particle that's traveling along in this case this line at 101 our goal is here now it's there an hour from now we'll be here to our so now we'll be here and so on you know exactly where it is you plug in any time for T you figure out where it is so that's how you do it that's how you define a one-dimensional thing in three-dimensional space use this time parameter yeah questions or ideas okay okay good yay great all right well I like questions too so you can have them so good let's talk about lines and planes so this is one way of talking about lines but earlier you might have talked about it in some other way [Music] so okay suppose you want to uniquely define a line let's say in three-dimensional space if you have one point that's not enough there are many different lines that go through the point but as soon as you get a second point think your line is determined yeah so two points will do the trick so two distinct not the same points those will determine a line so if you have one point here and you have one point here that determines a line between also over there what we've just been doing a lot of the direction will do the trick to so you could have two distinct points you could have one point and one direction so for instance you could have like your point P so here's P and your Direction V so you have P and then you have a direction vector V that also uniquely determines the line and the way that one a short way of writing it down is that your line which is traditionally written R of T and R is a vector quantity here our F key is P plus time times V so that's exactly what we have here we have our R ft our X of T y ZZ of T that's where we are is our P plus T times our vector on the other hand if you're just given two points you can do the same thing because you have a starting point and then you have a vector between them so here your R of T is your starting point could be a or be either one let's pick a plus T times some vector so let's do let's say this vector from A to B let's see to find that vector from a point to another point you do the tail - the head - the tail so this is like B minus a so T times B minus a that would do the trick okay lines good lines are good now we do planes so let's talk about uniquely defining a plane so to define a plane let's see usually people talk about three points defining a plane as if I have suppose I say oh maybe two points will define a plane maybe the two points that define my plane are the end points of this piece of chalk but then if I try to define a plane with it okay I'm ready let's put the plane on Jake but the plane can rotate on this line so it doesn't really uniquely define it but then as soon as I add an extra point like this pinky point I think only one plane so three points uniquely define a plane so three points are good but maybe we can get this direction thing going so maybe like a point and Oh fine by the way the three points have to be a non collinear because if they're all on a single line you're back to the problem of having only two points basically okay how about a point and a direction I wonder if we can get that to work so if we have a point let's say the end of this chalk and direction let's say the direction of the chalk and we want our plane it to be the plane that has that point in that direction it's not going to work there are so many planes that have this point in this direction not so good but there is a direction that you does uniquely define this plane can you see it the sky yeah so so that's a really great idea um you have a point and then you have the perpendicular direction if I tell you a point in a perpendicular direction then there's really only one place to put this plane right there so a point and a perpendicular direction the it uniquely defines the flame so that's going to turn out actually to be the most convenient information that someone can give you to define a plane because it's gonna make it really easy for us to write down a plane equation amazingly enough so let's try it so what's the idea the right so so the goal let's say is to write the equation for the plane containing our favorite points whatever point that is let's say your favorite point is one two three that is that with a normal vector let's say one negative two three and normal vector that is the name for this perpendicular direction normal means perpendicular so that's this perpendicular direction vector so let's do that okay now there's a big idea that we're going to use and here I'll draw a picture of it okay here's the idea we want to find the plane here's the plane that we want that contains our favorite point here's our favorite point one two three and it has the perpendicular vector one negative two three and what I mean by writing an equation for the plane means describe all possible x y&z that are on the plane so for instance this point over here X comma Y comma Z could be anywhere could have been here could've been here come over there any way I want to describe all possible XY z-- + Z Z on this plane here's the great idea the great idea is that if you have any point on the plane and you draw the vector from your favorite point our point in this case 1 2 3 to any other point on that plane that that vector is in the plane and so it's perpendicular to their own vector check that out so this is the big idea so every vector in the plane is perpendicular to the normal vector pretty good idea pretty good it's pretty good thing to notice so I wouldn't expect you to notice a thing like that but other people have noticed in the past in human history and so we can benefit from that so so let's write an equation for all X Y Z in the plane using this insight so we do as we say ok my normal vector my normal vector is 1 negative 2 3 and I know that if I dot it with any vector it'll be any vector in the plane it'll be 0 that's what that's a condition we can use for things two things to be perpendicular so let's find the coordinates of this thing this that during the plane let's see as before to find the vector you do head - tail so this is X minus 1 y minus 2 Z minus C so I'm going to dot this with my X minus 1 y minus 2 Z minus 3 and then set it equal to 0 just show that those things are perpendicular ok I'm going to do two lines in algebra I apologize for my two lines of algebra but they're gonna happen now so if I dot this out we get 1 times X minus 1 minus 2 times y minus 2 plus 3 times Z minus 3 equals 0 ok and then if we multiply out we get X minus 1 minus 2 y plus 4 plus 3z minus 9 equals 0 oh and then if we collect let's put all the variables on one side and all the numbers on the other side and we get X minus 2y plus 3z okay here we had minus 1 minus 9 that's minus 10 plus 4 that's minus 6 so on the other side we'll have a 6 ok equals 6 so there it is our plane it was our plane from the beginning as well the first plane that we were finding points on so that's the idea that's how we can do it and then I'll just write down what we do in general so in general if our normal vector is a b c and the point that's on our plane which we'll call peanut it's our special point is X naught Y naught Z naught then what we write down is a b c dot X minus X naught y minus y naught Z minus Z naught equals zero so that's exactly the same as this example here this was our ABC and then our X naught Y naught Z naught was 1 2 3 this is the same thing and we love to play this out we get what a X minus a X naught plus B Y minus B y naught plus cz minus cz not equal 0 so if we collect like terms we get X plus B y plus cz equals some constant some constant just figure out what it is but if we actually pushed all the constants this ax naught etc or to the other side we get that it's a X naught plus B y naught plus C Z naught so the big idea here the thing that you can take away to make life easy for yourself is that you take the coefficients of the normal vector and they just become the coefficients of your of your plane so you take these the number the entries you just put them in so you can see that here if our normal vector is ABC it turns out that a B and C are the coefficients of XY and Z and then if you want to that that just defines the direct the the show the tilt of your plane and then if you want to like nail it down in space you put you plug in a point and that tells you what the constant is and that nails it nails it down at some location in space so here are 6 nailed it down you might be wondering what if instead of plugging in our favorite point 1 2 3 we have plugged in some other point I am concerned that we may have gotten a different plane equation perhaps we should try a different point so let's try let's try plugging in a different point can somebody remind me of one of the points not the 0 0 1 that one of the other ones any point in the beginning 3 1 1 1 minus 1 1 1 minus 1 1 let's try that so if we had done 1 negative 2 3 dot okay X minus 1 y minus negative 1 so Y plus 1 Z minus 1 equals 0 I hope we would have gotten the same thing we would have gotten X minus 1 minus 2 times y plus 1 plus 3 times Z minus 1 equals 0 again we would have gotten X minus 2y plus 3z and then I wonder what our constants would have been we would have had minus 1 minus 2 minus 3 equals 0 is it the same it's the same yeah we have minus 6 on the left which comes out 2 plus 6 on the right so it's the same so it worked out so that just sort of defines where your plane is sitting in space questions or ideas hmm okay well I shall bring you things 2:30 when you ask questions the next time you will get positive reinforcement and you will learn to ask questions it will be good for everyone okay so finally good so we have learned that if you're given this information a point in a perpendicular direction your life is great because you just plug in the coefficients and then you plug in the point and you've got your plane great but what if you're given some other description what can we do that this is our next task okay so suppose for instance suppose you're given three points three non collinear points so here they are here's the plane that you wish to find the equation of and here are your three non collinear points well it turns out the directions are really great it's we would love to have a perpendicular Direction one thing we can get from three points is we can have one make one point our favorite and then by subtracting points we can get two vectors that live in the plane so here's vector one that lives in the plane and here's vector two that lives in the plane so these are just directions that are in the plane they're infinite lead in many directions that are in the plane all these directions are in the plane so many directions V 1 V 2 and many of their friends so so we can use this to get a point and two directions in the plane that's not so useful we really want the direction perpendicular to the plane so what we want is to be able to take two directions in the plane and get the normal vector the one that's perpendicular to the plane that is our dream so maybe our two directions in the plane are called like x1 y1 z1 and x2 y2 z2 and our normal vector here is ABC so we want to take in these two vectors whatever they may be just number it bunch of numbers and get this one well you could write this down so you could solve this you would say okay I want a b c dot x1 y1 z1 to be 0 because here's my normal vector it's perpendicular to view 1 and it's perpendicular to V 2 so I could write down the second equation I also want a BC dot x1 y1 is sorry x2 y2 z2 to be 0 so these these ones are numbers and these are variables you could solve this this is um this is uh three variables minus two constraints should give us one degree of freedom maybe it's a little surprising that we get one degree of freedom we just expected to get one normal vector but actually the normal vector could be any length it could be twice as long could be a hundred times as long or a Kapone the opposite section could be negative times as long so actually we have a whole line of solutions so it's not so surprising that we'd have one degree of freedom now you could solve this but I'm just going to tell you the answer so the answer ok and I'm not even going to write out the answer because it's a bit long in algebra I'm going to show you a monic device that you can use to get the answer so here's what you do it's a little bit surprising but once again it uses our knowledge from linear algebra so you make a few by three matrix which has entries x1 y1 z1 and x2 y2 z2 and then the top you cheat a little bit you put I J and K in case so I just means the first standard basis vector 1 0 0 J is the next one 0 1 0 and K is the last one 0 0 1 and then you take the determinant of this thing which is what these these bars mean and you expand across the first row so what you do is you say ok it's I times y 1 Z 2 minus Z 1 Y 2 yep plus J times oh no sorry minus minus minus J times X 1 Z 2 minus Z 1 X 2 plus K times X 1 y 2 minus y 1 X 2 okay so what this means is just this you write out this vector so we have y 1 Z 2 minus C 1 y 2 and then the next component is X 1 Z 2 minus Z 1 X 2 and so on X 1 y 2 minus y 1 X 2 so allegedly this vector is perpendicular to both of the original ones it's perpendicular to this and perpendicular to that and you can take the dot product and you'll see that all the terms cancel with both of them so this this is this is the guy and this is called the cross product it's magic it only works in r3 you take two vectors in r3 you take their cross product which means to do this thing and then you get a vector that's perpendicular to the other two pretty nice okay so so we've talked about the direction of the cross product it's perpendicular together to what a vector this vector is actually it's actually a particular vector it has a particular length in a particular direction so the last thing we're going to do today is I'm going to tell you what those two things mean so the cross-product so for example so it's written like V 1 cross V 2 if you were to put a dot here it would be a dot product so it's important to put the cross at the cross product so here's our here's our plane for example we have two vectors V 1 and V 2 you know whatever you do so we know that when we take V 1 cross V 2 we get something perpendicular to both but there's a particular there's a direction like sort of like up or down because I couldn't have drawn this normal vector pointing up or pointing down they'd both be perpendicular it turns out it satisfies the right-hand rule so if you would all take your right hand and do the following so the cross product of V 1 cross V 2 you take your fingers and you point them in the direction of V 1 now I don't know yet whether my hand should be like this or like this could be either and then you curl your fingers towards V 2 so I better turn my hand over I curl my fingers towards V 2 and then my thumb points in the direction of V 1 plus V 2 so in this case V 1 plus V 2 goes that way how about if we had had it differently what if it had been like this here's V 1 and here's V 2 okay now take your fingers point to the direction of V 1 maybe it's like this maybe it's like this we're not sure yet okay no curl them in the direction of V 2 so now the cross product points down the thumb is the cross product so now here's V 1 cross V 2 so there you go so you can always get get out your hand put it in the direction of the vectors to figure out whether it sort of points up or sort of points down depending on your reference frame okay now finally it has a magnitude and the meaning of the magnitude which is the length is the area of the parallelogram spanned by the two vectors v1 and v2 so what I mean by that is here you make this into a parallelogram from the end of v1 you put a copy of v2 and then from the end of each who you make a copy of v1 and that makes a parallelogram and it has some area and that is the magnitude of e 1 cos v2 gives you the area of this thing that's what it means so the bigger the cross product the bigger the vectors are the bigger the area between them I'll tell you one nice thing you can do that with this so suppose you are given three points in space here's three points in space just these three and I asked you or you needed to for some other reason find the area of the triangle that connects those three points this would be hard it would be a huge pain because these might these were need to have three coordinates what would you even do you might use maybe the herons for mattli something like it would have it would be a little tricky but here you can do it with the cross-product here's the clever trick you make one of the sides into a vector v1 you make the other side into a vector v2 and then you find the area of the parallelogram stand by those guys just by doing the cross-product so that the cross-product V 1 cross V 2 will give you the area of this whole parallelogram but we only want half so we just take half and that will give you the area of the triangle between those three points isn't that magic it's so great so that's the idea the cross-product is super useful for finding the equation of a plane containing some points but also the the direction of the magnitude in the baggage of the cross-product are useful two questions ideas all right thanks guys come back next time all right