 So, this is a problem on lines and planes, when I was working on this problem, I did not realize that this problem would have become very interesting and very challenging. In particular because we have multiple solutions to this problem and I will talk about them as I move along. So, the question the problem statement is this given are the two planes A B C and B C D and the two points of a triangle P Q R. So, point P is here point R is here in the horizontal plane P Q is parallel to the plane A B C in its edge view in that view P is at a distance of 25 millimeters from A B C. So, remember that A B C is in the edge view the distance P Q is 20 millimeters in the same view with Q oriented towards C. Now, this is important point Q is such that it is oriented towards C in the edge view of plane A B C. Now, in the edge view of plane B C D point R is at 40 millimeters from the same plane and the distance R Q is 15 millimeters. Now, the question is to find the triangle P Q R in all three views also you need to find P Q R in true shape. So, we are talking about two edge views edge view of plane A B C and edge view of plane B C D. Now, if you look at the view in the vertical plane and the horizontal plane it is possible for us to extract the edge view from the vertical plane of both these planes and the edge views of both these planes from the horizontal plane. So, they are going to be two edge views of triangle A B C and two edge views of triangle B C D and the moment I say that I refer to multiple solutions to this problem. Let us try to explore all possible solutions to this problem. So, we start with the hinge line that would separate the horizontal plane with the vertical plane and then we start extracting the edge view of B C D from the vertical plane. So, we take a horizontal line we take the projection of this point in the top view. So, this is in the third angle setup we get this true length of this edge we take the projection this point draw hinge line perpendicular to this projection and then we start projecting these points to get the edge view of B C D. For that we need to measure distance B or we need to measure vertex B from this hinge line and transfer that distance on to this projection here. Likewise we need to measure this distance and transfer this distance here and similarly this distance needs to get transferred. So, we transfer all distances and we get the edge view of the plane B C D we call them B 1 C 1 and D 1 here. Likewise if we extract the edge view of plane A B C starting from the front view very similar procedure project this point upwards get this in true length extend this projection draw a hinge line perpendicular to this projection. And then project A parallel to this projection line B as well as C and then transfer these distances over here to get the edge view of plane A B C we call this A 1 B 1 and C 1. We will have two more edge views one for B C D and the other one from A B C starting from the top view. Let me quickly go through the construction identically similar to what we had seen before. So, we transfer distances now and here we get another edge view of plane A B C and likewise if we draw horizontal line take this projection down get this length in true length extend that draw a hinge which is perpendicular to this projection project the rest of the vertices B and C and transfer distances you would get the edge view of B C D. Now, coming back to the question we are required to work with the edge view of A B C and also the edge view of plane B C D. So, we can choose any of these edge views for A B C and likewise any of these edge views for B C D. For example, if we choose this as A B C edge view of that then we can work with either this and this or if we choose this as an edge view of A B C then again we can work with this and this. So, we have four possible combinations let us try to investigate each combination one by one. So, let us say we start with this combination now let us go back to this line here P Q is parallel to plane A B C in edge view. So, P Q is supposed to be parallel to this line here and in that view P Q is at a distance of 25 millimeters from A B C. So, P could lie either on this side of the edge view of A B C or on this side, but for that or before that if you are working with these two views let me project starting from point P and R these projection lines which are parallel to these projection lines and let me also project a projection line which is parallel to this projection line here starting from P H. Now, as I said before P is at a distance of 25 millimeters from A B C. So, P could either lie on this side of A B C or on the other side of A B C and P will be lying on this blue horizontal line or possibly on this blue horizontal line. Now, P also needs to lie on this projection line. So, we will have two possibilities for P in this edge view. This is the first possibility, this is second possibility. Possibility represented by P 1 and this one represented by P 1 prime. Now, the second statement the distance P Q is 20 millimeters in the same view and Q is oriented towards C. So, if I choose P 1 as this point I will have to move towards C at a distance 20 millimeters and mark point Q. So, this is Q 1 likewise I can either go this way or this way. Now, if I go this way it makes more sense because then I am traversing towards C. So, I can choose this as my point Q 1 prime or if I want to go this way which is slightly contradictory with this statement that P that Q needs to be oriented towards C. Well, I can do that as well, but before that let me now start working with the edge view of B C D. Now, here I need to locate point R, R is at a distance of 40 millimeters from the same plane that is B C D. So, R can either be on this side or on this side of the plane. So, R will be lying either on this blue line and if I consider R to be lying on this blue line if I take a projection starting from R H over here parallel to this projection line the intersection this projection and the projection blue will give me a unique point R 2. Now, I measure this distance assuming that I choose Q 1 prime here. So, notice that this view is common between this view and this view. So, my Q 1 in this view or my Q in this view has to be lying on this line which is at a distance measuring this one here from this same line. So, Q in this view will be lying on this line. If I choose this as point Q 1 again something very similar I will have to transfer this distance over here. So, Q 1 will be lying on this line. Now, notice that if I am to draw a circle of radius 15 millimeters that circle centered at this point over here will not be intersecting either this line or this line. So, solutions with Q 1 prime here and Q 1 are not feasible. Then let me explore this point here Q 1 double prime if I take this distance transferred over here my Q 1 double prime is going to be lying here on this line. And if I draw a circle centered over here with radius 15 millimeters I will get two possibilities one is this and the other one is this for Q to be identified in this view. So, let me identify this as Q 2 and to locate P in this view again something very similar noting that this view is common between this view and this view. I measure this distance and transfer it onto this projection ray here and this would help me identify P 2. So, I have identified the triangle P Q R here and if I project Q back from this view. So, that this projection line in orange is parallel to this ray line here and if I do something very similar project Q 1 double prime backwards noting that this projection line is parallel to this one the intersection of these two projection lines will help me identify Q H in the horizontal plane R the top view. So, I will have this triangle P Q R here and to identify R in this view it is very simple I measure this distance I already have this projection from R I transfer this distance over here. So, this is my R 1 and this is my third triangle as I said before solution is not likely from this lead. Because if I look at this projection over here this projection is not going to be participating with reference to this line and then P Q R in other views can be easily determined. So, once I have P Q R here P Q R here and P Q R here it is also not going to be difficult for me to compute the true shape or identify the true shape of P Q R. Now, let me explore a different solution. So, H view of A B C is here and H view of B C D is here it is a different H view of B C D that I had considered before again projectors from P parallel to this projection line P is either going to be lying on this side or on this side P is either going to be lying on this blue line or the blue line below. This is something that you know there are two possibilities one is P 1 here the other one is P 1 prime here. This is the first possibility for Q this distance is 20 millimeters second possibility for Q is here and the third possibility for Q is on this side. Now, note that of these three possibilities this one is the only feasible possibility and. So, is this one this one will not give you a solution all also although I have discussed the solution in the previous slide with Q 1 double prime as one of the candidate solutions for Q 1 this one ideally will not give you a solution anyways let me proceed. So, I measure this distance I drop a vertical projector from P here and then I identify my point P in this front view or horizontal plane this distance is the same as this distance. If I am to work with P 1 I measure this entire distance extend this vertical projector down and I will have corresponding P V point here. Now, R is at 40 millimeters from the edge view of B C D. So, we have the edge view of B C D here 40 millimeters from here on this side 40 millimeters from here on this side. So, we will have two possibilities and R is going to be lying on one of these possibilities. Now, I already have R positioned in the top view. So, I can measure this distance now this view is common between this view and this view. So, R is going to be lying at this distance away from this hinge line and this distance is same as this one here. So, essentially R is going to be lying on this green line here. So, the two possibilities this point and this point for R. So, let me mark them as R 4 and let me worry about this solution over here. Now, I draw a circle of radius 15 millimeters centered at this point. I still have not been able to locate point Q which is going to be a tricky thing here. Now, if I need to locate R in the front view I need to extend this projector or this projector here and I need to extend this projector from this point R. Now, this projector is going to be parallel to all these projectors here intersection of this projector and this one will give you a point R B and the vertical plane or the front view. I now draw this projector parallel to this one. I already have this distance now identified I transfer this distance over here and identify point R in this view where A B C is in the edge view. I measure this distance over here and I transfer this distance here to identify this point P 4 prime and I can do something very similar with this point. So, again so notice that there are two possible solutions for point P in the front view or vertical plane. So, this is the first possibility and this one would be the second possibility. So, this distance here from the hinge line is the same as this distance here notice that this is point P 4. So, I got P 4 here R 4 here I got these P's here the Q's and the R's alright. Now to get Q which I have not been able to locate as of now let me try to explore possibilities from all these Q's that is Q 1 double prime Q 1 prime and Q 1. I measure this distance now I transfer this distance over here. So, in the front view my Q will be lying on this horizontal green line. So, Q 1 double prime is going to be lying on this green line here I extend this projection over here and to identify Q I do not have a straight forward method. So, what I will do is I will go for the hit and trial approach to determine Q and this is what I would do. So, let us say I would take some distance and I will formalize this method later on. So, let us say I will take this distance and assume Q to be lying on this line here and Q also needs to be lying on this circle. So, there are two possibilities possibility 1 and possibility 2. Now this distance would be the same as this distance and if I draw horizontal line Q in the horizontal plane or top view will be lying on this line and from here Q has to be lying on this projector. So, Q will be lying over here I take this projection down and I make a projection which is parallel to one of these projection lines over here assuming that this is one of the solutions for Q, Q 4 double prime and I will be able to then identify point Q in the front view Q v double prime and of course, the corresponding projection in the top view will be this point here Q h double prime. So, this is the first triangle I am looking for second triangle I am looking for the third one and the fourth one over here quite complex. Now if I take this as an alternate solution Q 1 prime measure this distance my Q 1 prime is going to be lying on this horizontal line in the front view again the hidden trial approach. Now assuming that if I have Q 1 over here I measure this distance go on to my top view Q 1 prime is going to be lying over here somewhere and if it intersects with this projection line which it has to. So, I will be able to identify Q 1 as this point if I take this second alternative solution this distance would correspond to this distance. So, correspondingly Q is going to be lying on this horizontal line here in the top view I take this projection down from here the first possibility I make a projector parallel to this projector over here and ideally I would assume that this projector is going to be intersecting with the circle, but that is not apparently happening. So, with this distance or this solution over here give me a unique position for point p I would say no likewise this point here the intersection between this horizontal line and this projector emanating from Q 1 prime projected down to lie on this position here and from here if I draw a projector parallel to these guys with this projector also intersect with this circle again the answer is no. So, these two are not possibilities or they are not possible solutions for Q. So, this distance and this distance there would be unlikely to be giving me a solution for Q and therefore, the corresponding triangles P Q R will not be possible for us to find again. So, here I have used the hidden trial approach to determine Q let me start with Q double prime over here or rather let me start with Q 1 here. So, I measure this distance and correspondingly in the front view my Q is going to be lying on this horizontal line again let me go for the trial and error approach. So, let me assume this distance let me draw a line which is parallel to the exchange line over here. So, I will have two possibilities to identify Q one is at this point the other one is this point the two intersection points between this line and the circle and since this views common between this one and this one I can transfer this distance over here and realize that my Q H is going to be lying on this horizontal line. Now, noting that I am getting this projector already from here Q 1 the intersection of this projection line this horizontal line will be a solution for Q here and if I drop a perpendicular I will have one solution over here that is going to be nicely related to this scheme of things in this view. So, I have identified Q 4 here correspondingly Q B here Q in the front view correspondingly Q H here Q in the top view and therefore, I have been able to identify the triangles P Q and R in all 4 views. Once I have been able to do that determining the true shape of P Q R should not be difficult once again I have used the hidden trial approach to determine Q. So, the previous two solutions were done with the edge view of A B C somewhere here. Now, we change things now we consider this as the edge view of A B C and we perform something very similar with the edge view of B C D. So, by now the dynamics of the problem is clear to you I will try to briefly explain what is happening this distance is 25 from A B C P will be lying either on this horizontal blue line or the blue line on the other side. So, by the way this is not horizontal this is parallel to this line here A B C. Now, we measure this distance this view is common between sorry this view is common between this one and this one. So, this distance gets transferred over here. So, P will be lying on this edge or on this line and of course, P is either going to be at this point or at this point let me call them P 3 and P 3 prime. Now, it is much clearer that if I identify Q 3 I will have to identify Q 3 in such a way that we traverse towards point C. So, of course, this would be one possibility for Q 3 and this would be the second possibility for Q 3. So, let me call this Q 3 prime of course, Q 3 or point Q cannot be located over here or over here that is moving away from C 3 not towards C 3. Now, in the edge view of B C D R is to be lying at 40 millimeters from the plane. We measure this distance this view is common between sorry this view is common between this view and this view we measure this distance transfer this distance over here. So, R is going to be lying on this red line as well as on this green line of course, R is then going to be lying at the intersection between these two lines. So, we uniquely identify R. So, we call it R 4 project that on to the front view extend this projection over here intersection of these two projections will give me R v point R in the front view of vertical plane. And if I draw a circle with center at R 4 radius 15 millimeters on this circle is where Q is going to be lying. Let us try to identify Q now. So, if I take this as one of the possibilities draw a projector parallel to these projections measure this distance and transfer it over here. Of course, Q is going to be lying on this horizontal line sorry this orange line and since Q also lies on this circle. So, we have two possibilities one over here the other one over here. Let me mark these possibilities as Q 4 1 and Q 4 2. So, appreciate how complex the solution to this problem becomes if I project this point backwards this would be my first candidate of Q in the front view Q v 1 the second candidate for Q in the front view Q v 2. If I project these guys on to the top view and if I measure this distance transfer it over here this is where I will get Q h 1 and likewise if I do something very similar I will have Q h 2. So, plenty of possibilities for Q starting from the same point here identifying P will not be difficult because P would be the intersection between this projector and this projector. And if I extend this projection line parallel to this one this distance is the same as this distance. So, I have identified P 4 here. So, looks like I have been able to identify P Q's and R's in all of these are possibly not. So, we have R 3 here this distance the same as this distance and then we have a bunch of triangles P Q R's. So, you will have to be really careful if you are attempting to solve this problem it is not straight forward. If I look at this as one of my candidate solutions once again this distance is the same as distance. So, Q 3 prime will result in 2 points here Q 4 1 prime and Q 4 2 prime which are going to be the same as Q 4 1 and Q 4 2, but P will be different a projector parallel to these projectors over here from here P 3 prime. And a projector which is vertically downward intersection of these 2 will give me P v prime you know how to get P 4 prime. So, I will not worry about that very much. So, looks like R 4 prime and R 4 will be the same and R 3 prime and R 3 will be the same. A little more exercise will help you appreciate that this yields another set of solutions and those solutions would come through the primes you know Q 3 prime 4 2 prime 4 1 prime. So, you can join these primes to get another set of triangles P Q R. So, this entire setup becomes lot more complex. The bottom line is we want to appreciate how many possible solutions are available to us. Another question. So, we had considered only this line as the line carrying R 4. Now, the question that you would want to answer yourself is R there are solutions possible with point R lying on this line at 40 millimeters on this side of B C D. Final set of solutions H V of A B C here a different H V of B C D P to be lying on this side or on this side at distance 25. Measure the distance transfer the distance you know about this two possibilities for point P P 3 and P 3 prime and of course, Q 3 will be towards C it will be chosen towards C on both these lines. So, we have Q 3 and Q 3 prime and these distances are 20 millimeters. Now, from here I draw projectors parallel to these projectors over here. So, in this view points P and R will be lying on these projectors 25 R. R is at 40 millimeters either on this side or on this side and since R also lies on this projection line R will get determined as the intersection between this blue line and this gray line here. So, this is the circle of radius 15 millimeters centered at R 2. It is in my opinion not likely to find a solution from this lead. So, we will not worry about that. Now, from here I draw a projector parallel to these projectors hoping to identify point Q in other views. Measure this distance and transfer this distance on to the top view realizing that point Q is going to be lying on this horizontal line in the top view. Now, here let me formalize the trial and error method to determine point Q. So, what I will do is I will choose a bunch of vertical lines rather distances assuming that Q would be lying on this ray and then do something about those distances. So, I draw a bunch of those distances assuming that this is one possibility, this is the second possibility, this is the third possibility for us to identify Q in the front view. Now, notice that this view is rather this view is common between this view and this view. So, therefore, these distances they get transferred on to this view which is precisely what I am planning to do. So, I am planning to transfer this distance this one and this one over here. I already transferred one the second one the third one and I will draw these three lines which are parallel to this hinge line. So, realize what is happening. So, I see a bunch of points of intersections between these lines and the circle all of these intersection points giving the impression that these are the possible solutions for point Q on this view or rather in this view. Now, if I need to identify point Q in this view it will have to be lying on a ray which is parallel to this or rather these projection lines. Now, if I choose this as one candidate solution I will have to project this point along this direction and of course, point Q will be then lying over here intersection between this ray line and this hinge line. This intersection would correspond to this distance. So, would this distance be the same as this distance looks like it would and therefore, we have one solution Q v here. Once we have the solution Q v it is straight forward for us to find Q h and then let us try to identify points p. Now, we have measure this distance point p is going to be lying or point p would be the result of the intersection of this vertical projector and this projector over here P v this distance gets transferred over here and this would also lying on this projector line. So, therefore, this would be point p 2 and of course, this is point Q 2. So, we have P Q r available over here P Q r available over here do we have P Q and r available over here looks like we do not have r in this view let us try to identify that. Now, r in this view will be such that it will be lying on this vertical projector and this distance would be the same as this distance. So, this is point r v here and r would also be lying on this ray which is parallel to all of these projectors here and how do we identify r along this ray we measure this distance transfer this distance over here and this is my point r 3. We can do something very similar with Q 3 prime measure the distance and it would be an exercise for you to figure if the solution is possible from this lead or not and the clue is that this distance is not of sufficient length. Now, we can do something very similar with this as the candidate point for Q. Now, what we can do is we can start with the projector and start with measuring this distance and of course, this distance would be the same as this distance over here. So, the corresponding Q H will be again lying on this horizontal line. Now, it is an exercise for you or let this be an exercise for you to find possibilities with this as the candidate Q 3 and this view. In my opinion a solution is quite unlikely from this lead. Now, for that you need to figure what is happening in this view in particular this distance over here. Well, I will let me stop giving clues to you you might want to figure this thing out yourself.