Lines and Planes via the Student MultivariateCalculus Package



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The Student MultivariateCalculus package contains sixteen commands for defining and manipulating lines and planes in spaces of two and three dimensions. Additionally, all these functionalities are available interactively and syntax-free through the Context Panel system.

This webinar will give some typical, and not-so-typical examples that might arise in the standard multivariate calculus course. As much as possible, the pedagogical model will be to use Maple to obtain a solution, then to use Maple to implement a stepwise solution.

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you at the beginning of the typical course in calculus of several variables this unit on vectors lines and planes and I think it's there to get students to start thinking about three dimensions visualizing and thinking in three dimensions no maples student multivariate calculus package has some nice functionalities for working with lines and planes and you can do it with a set of 16 commands or better yet via the context panel you can do syntax free computing so I'm gonna illustrate that in this webinar to show these tools and to show how much easier it is to do syntax free computing that they have to learn a bunch of commands and related syntax and I'll illustrate it by just doing some examples that come from the typical textbook exercises let's take a quick overlook overview what's available in this package okay so there are two constructors two objects that can be constructed a line object and a plane object because that's what you're gonna manipulate lines in planes then you could work in either two dimensional space or three dimensional space the Euclidean spaces the default coordinate names would be X Y or X Y Z and this table one lists how these objects are represented in maple points or just lists and that that's pretty standard of maple vectors well these wedge brackets these angle brackets are just inequality symbols and that will make a vector the line object in the plane object those are modules so think of them as a Basque into which all the properties of a line and all the properties of a plane would be tossed then you could rummage around in those baskets and get pieces of information out so for example if you were to construct a line object you had defined the line by various ways so if you're in r2 you would give an equation if you are to you could even give it as an expression maple would assume here that you meant to equal zero you can keep the final line with two points and that's true in either r2 or r3 so you would give two points and maybe we would say okay I and I understand what that means I'm going to make a line a point and a direction and that's true in r2 or r3 so there is your point and there's your direction vector and maple says I understand that I can make a line at that you can give a parametric representation in either r2 or r3 and parametric representation you give a list of expressions and a linear in the in the parameter or list of equations that will be of the form chorded name like x equals in the linear expression or you can have a vector of expressions where each component of the vector is the appropriate expression for parametric representation you give two planes because the intersection of two planes it defines a line so if you give two planes and how do you give a plane be linear expression maple assumes that it's equal to zero defining the plane or could be two equations that define a plane or it could be two plane objects that you've already defined two plane objects and maple says okay there are planes hiding in here and I can intersect those two planes and come up with the line if that's what you want a point and a plane maple will take the normal to the plane as the direction of the line so a point in a plane and you get the wine which is essentially orthogonal to the plane and of course if you've already defined the line object in either in our two or three Mabel knows that it's a line and it can give information about that lock now similarly for the plane you could determine a plane or define a plane by an equation of this form you could rearrange the terms or you can just have the expression where maple assumes equal to zero three points determine a plane to sequence of three points point and a normal vector you get plane for which that vector is normal and the plane contains the given point two points and a vector okay this is a direction in the plane two points determine a direction now you now have two directions if you take the cross product you will get a normal that defines a plane you have two vectors and a point well these two vectors must lie in the plane you take the cross product it gives you a normal maple sees that as a valid representation of a plane and says okay I've defined a plane object so a line and the point not on the line that defines a plane you just think that for a quick second you have a line it's gonna lie in the plane on that line you have a direction vector you have a point so you have a point that's not on the line between that point it's not on the line and any point on the line you can get a vector so you now have two vectors cross product you get a normal ya point and a normal you can get a plane to lines well if the lines are skew and is they don't intersect and they're not parallel then what the plane that maple will construct the plane that contains the first line that would be this one and is parallel to this one if the lines are parallel or intersect then you will get one plane that contains both lines if you give a point and a plane then you get back the plane that contains that point buts but it's parallel to the given plane and then of course if you've already constructed a plane object then maple knows what this plane object now all this functionality is based on a set of sixteen commands that appear in the package remember that all of these computations can be done with these commands but really really think in terms of syntax free computing with these manipulations can be done in a point-and-click fashion get dimension so you can determine the dimension of an object two or three dimensions get a direction so if you have a line and you want to know eight the direction vector in that line that is a command get the direction get the intersection whether you have two lines or two planes and you want to see what the intersection is get intersection get to normal from a plane draw a graph get plot get a point pick returns a point arbitrary point on the object line a plane get representation well this will return a representation equations and so forth for either a line or a plane intersect so well this is really a query do these objects intersect yes or no contains do does this object contain this other object distance well this will give you the distance from a point to a plane point to a line line to a plane are these two items equal are these two items parallel are they skew skew lines are they orthogonal lines planes what is the angle between these lines or planes can you project point onto a line point of a plane so these commands have related syntax if you'd like to use these commands you have to know the syntax now when you ask for the representation of a line you've got to have something in mind how you want this to appear well one way you can represent the line is with an equation there is this alternative equations and then you get the equation in a set if you do that in 3d then you get a pair of equations that define a single line more common representation is a vector representation and here the parameter along the line is taking some letter here it's T these the direction a is a fixed point combined vector takes all of this and makes a single vector out of it so it's here it's two dimensions here it's three dimensions parametric useful two dimensions three dimensions x equals x of T y equals y of T Z Z of T these expressions are linear symmetric maybe we'll return the symmetric form of a line in two or three dimensions I've never seen a whole lot of utility in the symmetric formally you can't really do much computation with the symmetric form for plane well there's only one way to represent the plane you get equation if you say equations you will get the equation in a set brackets now that list of sixteen commands with all of the associated syntax could be quite daunting so keep in mind that the functionalities that's hiding in those commands they are available through the context panel and that's what you'll see as we do some of the examples that will illustrate how to manipulate these objects to solve some of the standard problems that one would find at the beginning of a calculus book so now let's take a look at some examples example that appear in a typical calculus book and we can see via these examples how to do these manipulations and I'll show both in syntax free manner and I'll show you how it works if you use the commands so a first example relatively simple straightforward example you are given a point and a direction get the line through that point that has that vector as its direction okay so here's the syntax free mode and I'll do some of these calculations I'll illustrate some of these calculations although I've done a lot of typing in advance so that you don't have to watch this stuff happening I've loaded the student multivariate calculus package through the tools menu tools load package student multivariate calculus so that loads the package now you make a sequence of the point that's the square brackets and the vector those are the angle brackets you then use the context panel here student multivariate calculus and they redo the calculation you'll see that student multivariate calculus lines and planes you select line well that creates this line object when you create the line object you go back to Sue's multivariate calculus and you have the option to select representation now each of those actions brings up a dialog box the first dialog box lets you set the variables X Y Z that's a default variables and I just simply work and I say ok X Y Z and then for representation you have the option of how you want the representation to come out right this one shows what happens if you pick combined vector with parameter T well then you get a vector representation X of T first component Y of T Z of T okay now the equivalent using commands would look something like this you would load the student multivariate calculus package and since we're not doing syntax recomputing you would need with student multivariate calculus define the line so that you use the line command point vector and make the assignment to some name and I use the l1 as the name then you would use the get representation command apply it to line l1 and you use this option form equals if you give it combined vector you'll get this return if you say form equals parametric then this is the return three parametric equations x equals y equals equals if you say form equals symmetric you will get the symmetric form and if you say form equals vectors then you get this form where you see the parameter the direction vector and the fixed point the typical calculus book provides a vector solution and this is what you would have to do if you would just try to imitate the calculus book approach using vector notation you would need the fixed point as a position vector so you define it as a vector the direction vector is defined as a vector the generic position vector to the point XYZ is given the name R and then if you write R equals fixed point plus T times the direction vector you would get this return X Y Z equals parameter parametric representation for each of the three components and what I did here was I pre typed these equality's use the context menu let's bring up the context panel and used a sign name to make the assignment of this item to that name here's a second example equation of line through two points here's a syntax free solution using these tools in the multivariate calculus package so we've loaded the student package I showed you how to do that you make a sequence of the two points sequence of points list here list here bring up the context panel you saw how to make a line the line object is given this name you bring up the request for representation and here you can see the two dialogues that will come up this dialogue since the variables X Y Z under the falsely I keep those and then you have the choice of a parameter name if you want the parametric representation and this is the same as the first example that we showed earlier now the commands for doing that be pretty much the same as they were in the first example with the package define the line using the line command here are the two points their lists assign it to some name I pick down one then use get representation apply it to the named line l1 and select a form for the dis the display of the line combine vector will look like this parametric will look like this same as it was an example one form equals symmetric look like this four equals vectors will look that and remember if you're going to work with these commands you have to either know the syntax or have a way of looking up the syntax when you need it the typical calculus book will provide a vector solution and here would be an implementation of that the vector P and the vector Q are position vectors to the points P and Q and I showed you how to make that assignment through the context menu R is the position vector to the general point XYZ and if you write R equals P plus T times the direction vector what is the direction along that line well you have two fixed points the difference is a vector from P to Q maybe we'll do that subtraction and display the equation in parametric form for the line for a third problem we have a pair of lines this is a vector representation of a line whose parameter is T and this is the vector representation of a line whose parameter is s a B P and Q those vectors can be extracted from these parametric equations here are the parametric equations in T and here are the parametric equations in s and Part A requires that these lines are to be shown skew that is they don't intersect and they are not parallel now a property of skew lines is that they share a common normal fact skew lines can be made to lie in a pair of parallel planes and that common normal is perpendicular to both planes simultaneously and therefore you can find a distance between the skew lines because it's the distance between these two planes so let's tackle this problem first in a syntax free way well begin by installing the student multivariate calculus package you've seen that don't already then you have to define the two lines so you make a list of the parametric equations for each the context panel here for still multiple is lines in planes you have line okay so you can define the lines then if you go back you can assign to a name and it shows you here how to make those assignments square brackets on a script L subscripts it to verify skewness write the sequence of the two names then either go to the context panel and select evaluate display:inline I always do that from the keyboard using ctrl equal I believe all the enter does the same thing my fingers are trained to do to illegal so that then displays the actual line objects here are the names of those objects and here are the actual objects in which case then you can go to student multivariate calculus lines and planes and you have the option to ask are these skew and maple says yes they are skew do the same thing for parallel parallel do the same thing for intersect not they don't intersect so therefore these are skew lines now to find a common normal again write a sequence of the names of the two lines the context panel gives student multivariate calculus lines and planes make plane then query for the normal there you normal alternatively if you want to fall off make use of a more primitive or fundamental approach this is kind of what you would see in a textbook line name evaluating displaying log in line to have maple look at the actual line object and query it for Direction like the planes direction then simply assign that the name v1 and do the same thing here for the second line called that direction vector v2 and then cross product so v1 Crosby two would give you the normal to the plane now that cross-product I'll show you where that comes from I have here in my stack palettes close that one and let's look here for that item that cross-product symbol it's in my favorites palette I found it here in common symbols right click you have the option add to favorites so you add to favorites and you populate a favorites palette so that's where this cross-product symbol came from and if you write that cross-product symbolically and then do convex panel evaluating display:inline you'll again get a computation and a display of the normal vector now we want to get the distance between these lines okay you make a sequence of names l1 and l2 you evaluate display:inline so maple is looking at the actual line objects and then in the context panel mines in planes you have the alternative or the option of distance so you have a distance given and if you click here on the distance you can approximate to various number of decimals I picked 10 and so there's the distance at ten digits now a command based solution would look like this with to get the package in defined two lines with the line command simply give a list that the parametric equations that is enough to have the line objects defined then you use the R skew command giving it the name of the two lines and here you get true for our parallel here you get 12 volts and here for intersects it's the question do these intersect and you get false so you have skew lines common normal if you use the plane constructor plane command and give it the two lines you will get a plane for which the normal vector is the normal you're looking for and the get normal command applied to that plane and here you can see how you can wrap one command around another and you get the normal vector or alternatively you could get the direction vector on l1 and cross it with the direction vector l2 you can say vector for the distance you see a distance command give it the two lines and maple compute the distance between lines and here I floated to five places and then using the command get plot that command unfortunately has not been added to the context panel in any form whatsoever but another observation here to make is that the get plot command Y is the only one item at a time should really have been made to apply to a list of objects so that you could you would have to only use that command once for a bunch of objects so here I did the line l1 did the line l2 did the plane that that contains the point on the line that's what I did here you have a point on line L one and that normal vector so that defines a plane so I drew the plane then I got a point on line L 2 and a normal vector and I got the plane so determined under a picture and using the plots display command plotting all four objects and what you see here is the two planes a point on each plane line lying in that plane direction vector along those lines so you can obtain graphs of this type using the get plot command now the textbook solution would follow this vector approach and I would begin this with the picture of the parallel planes that contain the two skew lines you see one line is red y line is black they lie in a pair of parallel planes and this green line represents the common normal it's normal it's common all for both planes and at this point you have a common normal between the lines now you want to show that the lines do not intersect this is how you would have to do it the traditional way you take two equations X of T equals x of s y of T equals y of s and you solve for s and T well what's really happening what is the geometry here well imagine that these two skew lines are condensation trails two airliners and you're looking up at these two condensation trails well it's going to appear that the lines cross but you're standing on the ground so your coordinates are just x and y so what you would like to find out is what are the X and y coordinates of the point at which these two lines appear to cross that's what s and T will tell you so you take the T value and you put it into Z of T that gives you a height on the one line then you put s into Z of s and that gives you a height on the other line if these two Heights are different than one line lies above the other if these two Heights were the same then for the point at which you're standing on the ground looking up the point that appears to be an intersection point would indeed being a intersection point but here it's not an intersection point so the law do not intersect then you want to check for parallel okay you would need the direction vector on the one line or the P direction vector they all line call it Q now how do you show that these two directions are not collinear you take the cross-product if you do not get 0 the 0 vector then these two vectors are not collinear and incidentally that calculation gives you the common normal in a victus solution for Part B has already been obtained we have found a common normal and finally effective solution for Part C let's look at a picture here is line 1 here is line 2 on line 1 you have a point a you can always find a point it easy to point to select is the given fixed point on a similarly a fixed point on B you can get a vector from A to B by subtraction V now the Green Line is the common normal to get a distance what you do is you project this blue vector on to line 2 so you get this quantity here which you don't need but the orthogonal component of that projection projection on n is what you are looking for and you get the length of that vector and that's the distance between the lines so to do that you need the point a written as a vector point B written as a vector V is obtained by subtraction and then the scalar projection is V that in over the length of N and that number comes out to be 16 over root 11 the fourth example has us finding a distance from a point to a line the line is defined by the fixed point a given as a vector and the direction vector V which is this vector here a syntax free solution begins with the loading of the student multivariate calculus package make a sequence of this fixed point and this Direction vector so that you can make a line object and give that line object a name so you can easily reference it here's the name of the line object this is the point whose distance from the line you would like to know and simply use the context panel to query for the distance a command version of this calculation begins with loading the student multivariate calculus package and for that you using commands so you see with use the line command to define the line object whose that is assigned the name L we're sewing to the name L so here is the point on the line and here is the direction for that line and you have a line object use the distance command apply to that line and this is the point whose distance from the line you chained to find give that some name I used the name D and then apply the eval F command which is the command that floats symbolic numbers and you get the distance now the typical treatment in a calculus book is based on the use of projection so let's draw a figure help see what this projection approach is all about okay the given line is drawn in black the given vector or the direction vector for that line is this red vector the vector L connects a point on the line to the point P whose distance from the line you're trying to find now if you would project this black vector onto the line that would be one component the projection onto the normal or the orthogonal component for this vector would be along this green line and you get the length of that green line or that vector you would have the distance there is a task template a built-in task template and what are the test templates let's just take a quick look tools tasks browse table of contents for the task templates so in linear algebra under visualizations there are two test templates that you might consider projection plot the 1d because you're projecting onto a wine and so here is the test template it is inserted by selecting one of these to insert buttons default content means you get this extra verbiage minimal content means you get enough to make the thing work now I've got that already inserted and here you can see the path linear algebra visualizations projection plot on to 1d so you have to make some assignments a V P fix point on the line the direction for the line and the point whose distance to the line you're trying to calculate okay so here is the test template you are projecting P minus a this figure up here that's the black vector P minus a you are projecting that onto V this is V well anyways along this line is appropriate and this is the orthogonal component and so here you can see sketch you see the line that's in gray here the green vector is the direction vector for the line this black vector is what you're trying to project and you're trying to get this orthogonal component so if you were to float that you would see it is floating point I guess you have to put this stuff in to make the exact calculations work once once the vectors P a V are defined then this exact computation can take place and then there's a vector solution from first principles the difference P minus a I've used L for the vector the orthogonal component upon projection is L minus L dot V over V dot V V get the length of that vector you get u Euclidean norm it will be this number exactly and if you float it it will be this number and this vector is the vector and we found here and so what we did not do here is get the length of this vector because I knew I was going to get the length of that vector here and there is actually an analytic solution possible that's worth noting that here so this is the line a plus DV it's representation of an arbitrary point on the line if you vary tu getting a different point on the line and this difference is then a vector from an arbitrary point on that line to the point P so here is that vector whose length is this quantity it simplifies to this and we want to minimize this quantity because that's the meaning of a distance point of closest approach so I give it the name F right here is F this is what it looks like context menu a context palette here you have the option to differentiate with respect to with respect to T so I did that you want to set that equal to zero because you're doing a minimization problems a calculus one minimization problem if you don't put in two equal zero maple doesn't mind it just figures that that's what you meant you have a solve and you pick the simplest one solve and you're getting value for t and then you have this option substitute into so you substitute into you type F over here click OK and you get this quantity and if you do an approximate here you'll get the number that you got here six five one seven rounds here two one six five four point six five one eight for a fifth problem let's find the equation of a plane containing three points to do this in a syntax free manner load the student multivariate calculus package make a sequence of the three points using the context panel you can define that plane and get a plane object for which you can query representation there is the plane equation of the plane and you have seen already there is an intermediate pop up here where you can define the coordinate names Varick names of the coordinate variables the default are XY and Z which I typically stick with now the command based version of that calculation would use the plane command well you have to load appropriate package would use the plane command you would give it the same sequence of three points and you can wrap the get representation command around the creation of the plane and all use then see is the equation of the plane now typically in a calculus book in the section on vectors lines and planes they're going to promote a vector solution so the vector solution is most easily understood if you take a look at this picture the pink plane is the plane containing the three points a B and C so this is the origin down here and these are position vectors in blue to the points a B and C now by subtraction here and here you get the red and green vectors which then necessarily lie in the plane the cross-product of the red and green vectors gives you a normal and perpendicular it'd be normal to the plane that's done in gold you've got at least two points normal and point you can get your plane so here's how you would do that in Maple and you can see I've typed a lot of this stuff ahead of time a B and C are entered as vectors their position vectors this is the generic position vector to the point XYZ you need two subtractions B might say C minus a I've called that a B and a C you take the cross-product to get the normal N and I showed earlier where that cross-product operator can be found and then the equation is R minus a so this is a vector that must lie in the plane I've used the point a this is generic so you are demanding that this vector lie in the plane and do that you make it's dot product with the normal to the plane be zero and this will give you the equation of the plane and finally I'll show an algebraic solution to this problem because it will illustrate a useful construct in Maple I'm suggesting that we write this equality this defines a scalar valued function of a vector argument the secret here is that if you use the name of the vector the definition V then you give the components as V 1 V 2 V 3 so I've just dragged that over and I've used a sine as a function so you can see here a sine function that takes care of all the machinery of building this function now when this is set equal to 0 you've got a generic equation of a plane V 1 V 2 V 3 taking the place of X Y C so what you really need are these numbers a B and C those are the components of the normal vector and then there's the parameter D because you're not going through the origin necessarily so there are four numbers to obtain okay but really really there's only three because when you set this equal to zero there's got to be at least one number that's nonzero and you can divide by that number and so there's really really three conditions that can be applied well one condition is that a satisfy this equation second condition is that B satisfy this equation and the third condition is that C satisfying this equation so there are your equations and this is why the ability to define a scalar valued function of a vector argument is so powerful because here you only have to just say a and here B and here C and you get these three equations once you have those three equations you you could I guess you gotta use the context panel here you go to solve okay and in solve you can select the variables you're going to solve for now there are three equations but four letters so you can solve for three of them and this is where you would put a comma B comma C and you would solve those three variables in terms of D and then you do a substitute in to write music you substitute into and what do you substitute into you substitute into F of R but R is the vector whose components are X Y Z so you're substituting in here where v1 is now an X v2 is now y v3 is now Z is substituting into the general equation of the plane and each coefficient as a D in it so now you can factor out the D and divide it out and all that kind of stuff but if you would evaluate at a point at any point and give D a good value and what's a good value it looks like minus 86 negative 86 would be a good value to give D all these denominators would drop out and you would get the equation of the plane in example six we're asked for the equation of a plane passes through a point so you have a point on the plane and the normal for the plane well there is a test template we can use I've shown already where the task templates are linear algebra visualizations plane plot so what do you have to fill in the variables X Y Z and you have choices here we have selected that the plane is defined by normal and point normal point point given us vector normal point now this tool generates floating point answers see the normal vector is given as floats the equation of the plane is given in terms of floats and it gives additional information such as the point on the plane that's nearest to the given point nearest to the origin sorry and basis vectors for the plane so here is the picture you get and the slider here controls the number of digits in these answers so it's since been set to 4 and clicking this button is what generates this information and draws this picture and you see you have a picture of the plane that's the origin down here this vector looks a dark blue or black points to that portion of the plane that point on the plane that is nearest to the origin this reddish vector is from the origin to the point on the plane that is specified it has to be able to plane then these two guess they're blue vectors or basis vectors and this red vector is normal to the plane now to solve this problem using the tool of the multivariate calculus package we first look at ace so a syntax free solution load the package make a sequence of the point and the normal vector make the plane object answer its representation and we've seen how to do this use the context menu a context palette and this equation is the equation that we're looking for we can obtain that solution using commands if we do the following load the package use with create the plane object by using the plane command point normal and assign that plane object the name queue if you apply get representation to queue you get back plane one could also just rapper get representation around this command and the intermediate step would not show the get plot command we'll draw a picture of this plane and what you see is piece of the plane the point and the normal and finally let's take a look at how difficult oculist book would solve this in a vector sense load the package define the point as a position vector they find the normal vector and the generic vector to an arbitrary point XYZ then use the vector equation of the plane R minus P so this vector is to lie in the plane and to do that you force it to be orthogonal to the normal vector this is the equation this is the resulting equation of the plane you

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