 So, Ayush had raises question last time that he was having a hard time visualizing the point view of a three dimensional line with respect to this what I would call virtual box right. So, yesterday even I was having a hard time when I went back I was like how would I do that, but then I could figure a way out and maybe I will share that with you ok. So, hopefully things will become clear in particular with regard to the distance that we were taking to get the point view. So, this figure is clear to you right is it is it yes or no perfectly clear to whom is this figure not clear wonderful. So, projection in the horizontal plane on the horizontal plane projection on the vertical plane on the vertical plane these distances are essentially these distances ok. So, pretty much there right and if you figure the intersection of the green and the blue line here and the green and the blue line here you have two vertices of a three dimensional line which is enclosed within a virtual box a virtual box right draw this plane that would contain this projection as well as this three dimensional line ok. Now, this step is critical project this plane forward in a manner that you are actually getting this guy along with this plane and this guy which is the image of this along with this plane get these two hinge lines and when you are drawing this hinge line with respect to the horizontal plane. So, this hinge line would be parallel over here. So, you would be essentially flipping that plane over in such a way that all these three planes lie on the same plane that is what the basic ideas ok. Now, follow this carefully this is fine I have shifted this entire figure over there ok. Now, yesterday two of you came and I asked how you would be able to locate the plane that would contain the point view of this three dimensional line and you know somebody was looking way up there like that alright. So, let me try to get that plane on this picture ok. So, the idea is that we are going to be looking on this line like so ok. So, perpendicular possibly to this plane. So, I draw two planes or rather I draw this plane parallel to this this plane here parallel to this and the back plane which is parallel to this plane which is containing the three dimensional line find so far with me. So, far it is a little messy. So, I would take away certain information that I do not need at this time ok. So, what I am left with is just the horizontal projection over here the vertical projection over here and the respective planes alright. Now, this view is probably not very comfortable for me to work with ok. So, what I will do is I will enclose this entire thingy within a cube and rotate the cube a little are you with me or you are. So, what I will do is I will essentially transfer all this information onto a cube and rotate the cube. So, what I will do is I will transfer this information this information this distance this distance this distance and this distance ok. Just carefully observe one more time this edge is now going to be one of the edges of the cube. I am going to be measuring the red distances and I am going to be placing this blue line on the horizontal surface. I am going to be measuring these green distances on to the vertical plane and then I am going to be having this red line at the end at the two ends of these green lines watch carefully. Now, it becomes a little easier for me to visualize the point view of this three dimensional line of course, for the point view I will have to be looking at this three dimensional line from here. So, I will have to draw a plane that would be that would be or I have to sketch a plane in such a manner that this guy over here is perpendicular to it right perhaps this is what my plane is it will be an inclined plane it would not be parallel to the vertical nor will it be parallel to horizontal it will be an inclined plane. And let me assume that I am going to be having a hinge line here for now that passes through one of the vertices of this red line. So, the idea is for me to flip this plane about this hinge line like this let me try to capture the point view here follow this way carefully, but before I do that I will ask this question to you is this the actual three dimensional line is just the image of that right. So, I should be considering the actual three dimensional line look at that figure over there the black line is something that I am interested not the red line. Now, for that I am going to be using this distance I am going to be using this distance and this is where my black line is. So, this plane black dash plane is going to be capturing the point view where over here is it not it right not there, but here now your question was how do I correlate these distances with this one now it is clear is it right. Now, that plane need not be passing through this vertex of the red line it can be anywhere. So, long as it is parallel to this plane. So, I can move this plane perpendicular to this line along this line yeah once again it is not really matter where the location that planes now what I would do is I would rotate that plane about that black hinge line like. So, and when I do that this guy would be coming where here clear now is it ok for you to correlate this distance with this distance clear yeah anybody else you are right for your consolation this is not good t it does not taste very good anyhow ok. So, having understood this let me move forward. So, visualization is only one aspect of it ok. So, with practice the more you practice the more it becomes easier for you to follow this mechanically anyhow parallel lines given the projections of a three dimensional line c d in the vertical and horizontal plane. And you also have the projections of a point a the vertical plane projection here the horizontal plane projection over here I am expected to draw a line that passes through a which is parallel to c d how do I do this how do I do this I will have to visualize this line point view or true length I will have to figure out the plane in which this line is in true length yeah hinge line this view is going to be I am not sure which one is going to be common all right. So, this view is going to be common because I am using another hinge line I am going to be making an auxiliary plane draw projections perpendicular to this hinge line measure what distance is measure this distance transferred over there measure that distance transferred over there. And this is my line c 1 d 1 in auxiliary plane 1 remember I can use as many auxiliary planes as I can or as I would want. So, this is the first auxiliary plane that is the reason why I am using the subscript c 1 d 1 and this would be in true length great. So, it is really nice to get the feedback from you. So, this would be in true length. So, now that I have located the plane I have to do the same exercise to get the corresponding projection of a on that plane how do I do that I will draw a projector perpendicular to this hinge line from that point I will measure this distance transferred over there that is where my point a 1 is. And then I know that this line is in true length. So, I can draw a line passing through a 1 which is parallel to c 1 d 1 of any length of any length right. And this is let us say point b 1 I take b 1 back I am going to be using the same rules now I am going to be taking b 1 back where do I locate that b 1 over there now would this be in true length a 1 b 1 ok. So, if one of if these lines are parallel and if one of the lines is in true length the other one has to be in true length now how do I locate b 1 in the horizontal plane over there. How do I do that I do not know the distances right now I do not know the distance, but what I know is if this is in true length it is corresponding projection in the other view has to be parallel to the hinge line it has to be parallel to the hinge line which is what I am going to be using. So, I am going to be ensuring that this guy is in true length I am going to go up and I am going to draw a line segment which is parallel to the hinge line ok. Once I get that now I can use the distances now if I project this b h downward I can measure this distance and transfer it back over here down. And there is the projection of a b in the frontal plane on the vertical plane right right ok. So, I have not mentioned a h b h maybe I will mention that later, but if I want to see now the point views of these two lines because both of these guys are in true lengths on this plane. So, if I look at these two lines from here I need to make a hinge line which is perpendicular to either one of these get the projections from where am I going to be getting the distances from there because now this view is common between this guy and the view that I am going to be making ok. So, I will get that distance transferred over there I will get that distance transferred over here so far so good now what ok yeah yeah yeah yeah yeah yeah yeah. So, my mistake was so nice right you guys realize that. So, this guy should be here right ok. So, this distance should actually have been here and that distance should actually have been here ok. So, just flip these guys over and essentially you are going to be getting this point view over here this point view over here. So, point view of c d sorry point view of a b will be here point view of c d will be there yeah ok. Next perpendicular to a line from a point you got the projections c d in the horizontal plane vertical plane and you also got the projection a v and a h in both planes I am doing a Dhoni here. So, I am trying to make runs fast ok. Hinge line it would be a good idea for you to sketch alongside this lecture give you a nice practice and hopefully you will be. So, two things will happen number one your eyes will not get heavy number two you know it will be nice I mean you will get some practice ok. So, I will start with the hinge line so remember whatever exercise I am doing with respect to this view I can do exactly the same thing starting from here nothing stops me ok. So, I will draw a new hinge line parallel to c h d h here take the projections get this distance transferred over there get that distance transferred over there I get c 1 d 1 n true length 1 full and then I will do the same thing for a I will take projection which is perpendicular to this hinge line I will measure this distance transferred over there and I will get my a 1 there ok. So, the question is to figure out a perpendicular from point a in 3 d on to c 1 d 1 or c d the three dimensional line c d this is what the question is alright. So, at this time maybe I will just move forward I will get the I will try to get the point view of c 1 d 1 hinge line perpendicular to this projection I hope I am not making that mistake again get that distance measured over there this is my point view of c d and then likewise I will measure that distance draw a projector perpendicular to that hinge line from a and transferred over there this is my a 2. So, this is my first auxiliary plane that contains the true length of c d this is my second auxiliary plane that contains the point view of c d yeah ok. Now, is it ok for me to draw a perpendicular from a 1 to c 1 d 1 and say that this would be perpendicular why not why not why not or why both questions this would not be the actual length, but this would be perpendicular ok. Do you agree the plane in which this line is in true length I draw a perpendicular this will remain perpendicular ok, but this guy may not be in true length it is. So, on this auxiliary plane it is only the projection of this line that you see to get the true length I will probably have to go over there would this be true length why is that because this guy is parallel to this hinge line if one of the projections is parallel to the hinge line the other projection will contain the true length quite mechanical now yeah. So, this would be in true length. So, you have the perpendicular over here you ensure that this guy is perpendicular and you will find the true length of this perpendicular from here get back you figure this point of intersection let us call this b 1 go back ok. Of course, b 1 has to lie on c d. So, you get b h come down ok of course, b has to lie on c d. So, you get b b. So, this is your first or projection of your perpendicular in the horizontal plane and this is the projection of the same perpendicular in the vertical plane with me. Now, better to understand lot easier to understand compared to previous lectures last part from where till this is fine till this is fine that would be true length this is point b 1 ok b 1 has to lie on c 1 d 1 right come up and these projections they are going to be parallel to each other ok. So, the corresponding projection of b in the horizontal plane is going to be b h on c h d h come down the corresponding projection will be b b ok and you join just a h b h over there and a b b b over here no rocket science how is your galaxy going starting ok anyhow. So, this is clear is this clear ok shortest distances between lines it is going to be a long day today. So, bear with me, but I really want you to understand this because if you do not then you will have difficulty starting Monday ok. So, it was important for us to have this class look at this figure and you will realize that the points or the lines are intersecting or not they are not intersecting the reason why because you do not see the corresponding projection of the intersection in the horizontal plane over here ok. So, they cannot be intersecting they are skewed in space somewhere right alright same thing same thing a hinge line try to get the first auxiliary plane projectors perpendicular to that hinge line from everything a b c and d things might be a little messy. So, I want you guys to follow this carefully measure this distance transferred over there measure that distance transferred on the projection from where on the on the projection emanating from d h over there. So, this is your c 1 d 1 and since this projection is parallel to this hinge line this would be in true length do the same thing for the a's and b's take that distance transferred over here on the projection starting from h take that distance transferred over there and you would be getting a 1 b 1 this guy is in true length how about this guy not in true length fine. Now, let me try to visualize these two lines from a plane from where I can see c d as a point get the point view of c d make a line or make a hinge perpendicular to c d draw this projector measure that distance transferred over there and this would give you the point view of c d draw two other projectors from a 1 b 1 perpendicular to this guy this hinge line where are you going to be measuring the distances from here. Measure that distance transferred over here if you have not noticed let me emphasize that these distances are to be measured they are to be measured from the hinge lines it is critical not to be measured from the points, but from the hinge lines every time. So, measure that distance this little one from this hinge line transferred over there and this is your a 2 b 2. So, on the second auxiliary plane you see c d as a point c d as a point and you see the projection of a b would a 2 b 2 be in true length good no problem how do we find the shortest distance between lines I draw a perpendicular right there I draw a perpendicular all right is this going to be in true length what you say yes it is perpendicular to c d also, but you do not know do you as of now yeah, but with this plane over this projection of the perpendicular give you the give you the true length fine. So, you say yes you say yes, but I will retrace my steps back and convince you that it is actually going to be in true length for those who believe great, but for those who do not believe that this is going to be in true length I will convince you that this is going to be in true length at this time I just pose a question whether this is going to be in true length or not anyhow. So, I track back I call this point m 2 I go back m 2 is going to be lying definitely on a 2 b 2. So, m will be part of a b in general everywhere on every plane. So, this would be m 1 if that is the shortest distance if that is the shortest distance it has to be perpendicular somewhere all right. Now, since this guy is parallel to the hinge line why because this thing is perpendicular to c d shortest distance has to be perpendicular to 2 both both lines right. Now, you just see that it is perpendicular to c 1 d 1, but you can verify later whether it is perpendicular to a b or not in some auxiliary plane anyhow. So, once you verify that this is parallel to this that would be in true length yeah. So, all you to do is name this intersection point as n 1 get both m and n back m lies on a b n lies on c d on every plane. So, this is m h n h m h n h and here this point would be what m v n v wait wait wait wait wait wait wait wait wait wait wait wait wait wait let me let me complete this let me complete this till here till here yeah yeah. So, till here so ignore ignore this part of the red line. So, it is going to be till here yeah. So, while justifying that we got the perpendicular in true length in the second auxiliary plane. So, you assume that the point n 1 lies on series such that m 1 n is perpendicular to c 1 d 1. Yeah. So, why is that so why did we take n 1 such that it is perpendicular to c 1 d 1. The shortest distance is going to be perpendicular to both lines yeah, but how often do you agree the shortest distance is going to be perpendicular to both lines. Now, if you are seeing one of the lines in its true length that shortest distance will be perpendicular to it that is how yeah. If I draw this line so you are asking as to why did I draw this perpendicular to a to b 2 I could have drawn like this, but would that have been the shortest distance I could have drawn that like this, but would that have been the shortest distance this would be the shortest distance and that is the only possibility it has to in reality in three dimensions if I have this line and if I have this line if I am going to be computing the shortest distance between these two guys that line would be unique that would not be changing is just that their projections will be different in different planes step up do we have another pointer turn around this is one of the lines in 3 d this would be the other line 3 figure out the shortest distance perpendicular to both perpendicular to the board yeah yeah show that show that show that yeah now what what is your question sir I am asking that suppose we have two lines yeah one is the shortest one yeah somewhat lengthier than this yeah if I take the projection of both the lines any plane is it always true that the projection of the short shortest line will always be shorter than the projection I mean not necessary not necessary what have a seat have a seat but listen listen listen listen so this auxiliary plane is slightly different in this auxiliary plane you are seeing one line like this in the point view and the other line or the projection of the line somewhere okay how would you see the shortest distance absolutely which is what this is yeah so that plane is slightly different but if you want to do this exercise try to get it a b in its true length and figure out what this projection is going to be okay alternatively alternatively yeah but just a moment before I alternatively you know I started by finding the point view of CD okay do the same exercise instead of making this hinge line make a hinge line that is parallel to HBH get the point view of HBH draw that shortest distance and see really if the true length of the corresponding shortest distances are the same you could do that and you could verify yes okay can can you be a little silent because otherwise it becomes a little difficult for me to hear and can you be a little louder and slow down so C1 D1 and A1 B1 yes yeah yeah yeah yeah yeah yeah okay Yeah, this is what I am saying let me try to understand what you are asking. So, you are saying step 1 I drew a perpendicular from here to here step 2 I drew a perpendicular from here to here. Now, let me counter ask a few questions are you convinced with this step not very much it is true length because the shortest distance has to be perpendicular to both lines. If I am seeing one of the lines in it is true length that means, that I have been able to capture that line on that corresponding auxiliary plane the shortest distance will be perpendicular to it agreed. Do you agree with this then if you agree with this then then this guy is parallel to this hinge line because both of these are perpendicular to C 1 D 1 right. Now, if I go from here to here the projection of any line segment which is parallel to one of the hinge lines the corresponding projection in the other view has to be in true length it has to be right do you agree with that. So, if you agree with this and if you agree with this then your question is this is the point view of C D this is my line here it could be anywhere what is the shortest distance that you are going to be dropping from the point view of C D on to this coming forward rather going forward enough of lines now to planes given the projection of a plane given the projection of a plane in the horizontal plane up there and the projection of the corresponding plane the vertical plane. So, these guys are the projections of plane of a plane a v b v c v down there a h b h c h up there look at the corresponding vertical projections they have to correlate. So, a with a b with b and c with c the question is to find the edge view of the plane number 1 and to find the true shape of the plane if I can can I have one of your yeah. So, imagine that this guy is what you see over here on a triangle and imagine that this guy is what you see over there now the edge view. So, this is the this is the true plane for example, the edge view of the plane will be like this for you the edge view of this plane will be like this and if you flip if you flip this plane over if you flip this plane over or rather if you rotate that plane about this edge view you will be essentially getting the true shape right now the question is how to find the edge view of this plane. And then I know that if I flip this plane over by 90 degrees about that hinge I will be getting the true shape fine now it might seem a little difficult, but it is not very difficult as it seems. Now, be very careful be very attentive and follow the following steps I start with the hinge line now what I do is something smart what I do is something smart I locate a point d h I locate a point d h on d h c h in such a way that a h d h is parallel to the hinge line does this part thing that I have done the second thing well not very difficult I project d h down and get that corresponding point here on d v c v. Now, this guy if you just treat this guy as a line segment it is parallel to the hinge line here this would be in true length and if I look at this plane if I look at this plane in such a manner or rather let me let me go back next step I would do is I would draw a hinge line perpendicular to this line a v d v. So, in the process what I am trying to do I am trying to find the point view of a v d v project this line and measure distances. So, that distance will be transferred over here this distance gets transferred over there this distance from this hinge line to c h gets transferred over there and this distance gets transferred over there fine this is what I have done something very simple something very mechanical and then this is my first auxiliary plane I join these three guys these three points horizontal plane vertical plane that is the vertical plane and that is the first auxiliary plane a 1 d v comes now strangely this is the point view of a v d v and this is my c a 1 this is my b a 1 and here I would have d a 1 and a a 1 once again this is my c a 1 this is my b a 1 and here I would have d a 1 and a a 1 strangely strangely I have got all the four points on a plane onto a single line yeah I have done nothing special I have done nothing special all I have done is I have found a length I have found an edge on the plane or a line segment on a plane to length I have just viewed that corresponding projection of the plane perpendicular to this that is all I have done nothing else. So, strangely this entire plane is now represented in this auxiliary plane by this line segment is this the edge view of the plane this is what we wanted this is what we wanted did not we I could do the same thing starting from the vertical plane let me go over this real quick parallel line project d v to d h this would be in to length make a hinge line perpendicular to a h d h shoot the projectors out from these vertices measure this distance transferred over there measure this distance transferred along c h measure this one transferred along b h draw this line segment this is the edge view of the plane that is the horizontal plane that is the auxiliary plane 1 and again you see that all these vertices lie on the line segment. Now, if I draw a hinge line parallel to this edge view of the plane and try to see how this guy looks perpendicular to this edge view same thing shoot the projectors measure distances from where now from where from here this is between a 1 a 2 hinge line measure that distance transferred over there measure this distance transferred over here somewhere. Now, this distance is from here to a h measure this distance transferred over here this is the true shape of the plane that you will be getting the beauty of this construction is nowhere am I using coordinate geometry nowhere I am using mathematics simple lines that is a beautiful. Can I go back to the previous example and get the true shape of the thing yeah real quick real quick and can I try to compare these two shapes no hey half an hour more I have to finish this lecture just two minutes. So, I get this true shape from the previous example I merge that true shape in such a way that one of the vertices coincide then one of the edges coincide and then I flip it. So, both will expectedly give you the same shape I have to finish this lecture number one, but I will give you the freedom to you know go back and do whatever you want. So, here are two proposals proposal one we take a break for five minutes you can wash your face drink water and then decide whether you want to come back or go to your host room, but I will have to stay here I will have to finish even if none of you are here. I think we are almost done let me see how much let me see how much I have line parallel to a plane is it straight forward get the plane in the edge view get the plane the edge view draw a line parallel to it I quickly go through that quickly go through that see listen otherwise see if I do not finish this then you guys are going to be having a problem in the next week I will not you guys will be facing problems. Bear with me for five or ten more minutes and then you guys are free all right horizontal line h d h stay with me stay with me guys you guys are what I mean you guys are so young you guys are so young I mean full of energy come on you know I so much wish that many of you guys come back and start teaching teach as I teach then you realize I mean what teaching is if not all at least at least 10 percent of you guys. All right so this is d v this is going to be in true length draw a hinge line perpendicular to a v d v shoot the projections measure distances transfer distances straight forward this is not very difficult this is the edge view of the plane edge view of the plane get the points not a problem do the same thing for point p h shoot a shoot a projector perpendicular to this hinge line measure that distance transferred over there this is what p a 1 is draw a line parallel to the edge view of the plane any given length let that point be q a 1 trace it back so if you assume that that is going to be in true length which is actually what is going to be true then this guy has to be parallel to that hinge line get this point q v measure that distance shoot that projector transfer distance get q h this what your line is. So, if you just look at this picture and this picture it is not at all clear if this line is parallel to the plane is it is it strange but only this view gives a clear picture yeah how did if this is in true length if this is in true length this has to be parallel to the hinge line basic golden rule yeah p q should not change in ts view of a b c let us see what this is all right. So, if I made the true shape of the plane and what am I doing here measuring the distance transferring over there measuring the distance transferring over there all right. So, what am I saying so this length this length would remain the same is it over here and over here and the only way this is going to be possible is if both these guys are parallel to the hinge line is it all right fine. So, all right. So, now same thing line perpendicular to a plane from a point stay with me just about getting done hinge line a h d h parallel to the hinge line I get the true length I make the edge view of the plane shoot the projectors get distances transfer distances get the edge view of the plane stay with me mark these points I do the same thing with the point measure that distance transfer over there I get that point as p a 1 and if I want to draw a line perpendicular to a plane from a point this guy has to be perpendicular this point has to lie on the plane q a 1 this should be in true length should this be in true length you guys want to go back this ok if this is in true length then this would be parallel to the hinge line and make sure or know this very well that q the corresponding prediction of q over here has to be lying on this plane ok. So, this is q v and measure that distance project q v up there transfer the this guy has to be lying within the plane this is q h and this is the line which is perpendicular to the plane in both horizontal and vertical planes ok all right forget about that verification p q should be in point view wait this is important p q should be in point view in the true shape of a b c would that be the case all right. So, if you get the true shape of the plane and if you get the corresponding projection of p q on to the true shape this would be the point it should be in the point view you know. So, when you are doing this always verify always have those little thingies that you can use to verify whether you have done something correct or