 So, this is a drawing example that demonstrates the intersection between two planes A B C and D E F. So, both these planes are laid down or depicted as shown in the front view or the vertical plane and the top view or the horizontal plane, the front view the top view vertical plane horizontal plane. Again we need to determine number 1, now this time the line of intersection between these two planes A B C and D E F, if there exists such a line. And if you are able to find that line of intersection then the second task that we need to accomplish is to determine the visibility of the planes, which part of the plane will be visible in for example, the front view and the top view and which part of the plane will be hidden behind the other plane in both these views. Then to determine the line of intersection and visibility of these planes we can use either the edge view method or the cutting plane method. So, what I will do is I will try to demonstrate both these methods using the same diagram and this time trying I am not sure about that, but trying to get the line of intersection to be the same using both the auxiliary plane method or the edge view method and the cutting plane method. So, first the cutting plane, now for that what I would do is I would think about the plane A B C being intact and then I would see plane D E F as composed of a set of three lines. Now, when I say that what I primarily imply is that I would treat in fact, I would see this problem as intersection between plane A B C and each of these three lines. Now, to do that let me first imagine D F to be such a line and let me pass an imaginary vertical plane passing through that line D F very similar to what we did in the previous lab example. And of course, this plane will be containing the actual line D F in three dimensions. So, having said that let me consider this as one of the cutting planes and to differentiate this edge with the other two edges let me draw these inch lines. Now, this cutting plane now imaginary vertical cutting plane is going to be intersecting with A B over here and A C over here. Let me take these intersection points down on to the corresponding edges over here. Now, this point to A B which is right there I will have to make sure that I am using two edge pencil and this point to A C which is right here. Now, what I would do is I would join these two points it is important for me to label, but I would join these two points and perhaps represent it using represent this line using very dim line. Of course, I will come back and label this. So, let me call this point as S H this point as T H the corresponding point here on A B as S of V and this would be T of V. Let me consider another imaginary plane passing through now the edge E F. So, this is the vertical plane that passes through edge E F. Of course, this plane is going to be intersecting with this triangle. So, the corresponding intersection points in the top view are these two. So, this would be on A B this is on B C now. So, let me project this point down. So, this lies on A B very gently using dim line. Let me call this point M H. So, that the corresponding point on A B would become M V. Likewise, let me call this point N H projected down onto B C B C is here once again 2 H dim line and let me call this point N V. And let me join just so that this represents the line of intersection between the vertical plane passing through E F and A B C and projection of that in the front. Now, since this is a cutting plane now, let me represent that using the convention of hinge lines or the section planes that same long followed by two short followed by one long and here also long followed by two short followed by one long. Well, let me name this as plane P 1 let me name this as plane P while this is not necessary. What I did was I considered a vertical plane to be passing through D H E H. So, that plane I name it as P 3 and this plane intersects at U H V H intersects A B C. If I project these down U H lies on A H C H correspondingly U V will be lying on A V C V likewise V H lies on B H C H. So, correspondingly V V will be lying on B V C V and then of course, this line of intersection would represent the line of intersection between the vertical plane P 3 and the triangle A B C and the projection of that in the front view. Now, in the cutting plane method that you have seen in the previous drawing example if there is a vertical plane that passes through one of the edges in the top view. If I come down the corresponding line for example, in this case S H T H. So, the corresponding line S V T V has to be intersecting with the corresponding projection D V F V. So, looks like this point of projection is very close to the index or this vertex F V likewise if I consider plane P 2. So, the corresponding intersection segment was M H N H that is M V and V over here. Now, this is to be intersecting with the image of the line E H F H in the front view which is E V F V. So, if I extend this line little and let M V N V intersect with that I would be getting a second intersection point over here. So, the first intersection point is there the second intersection point is here and if I extend a line during these two points it would probably look somewhat like this again a very dim line once again this one here. Now, just for sanity check let me also consider the cutting plane P 3 this would help me verify if this is indeed the line of intersection or if this is indeed the line on which the actual line of intersection between these two planes would lie. So, P 3 cuts A B at U H V H going down to the front views U V V V and this line segment has to be intersecting with the corresponding image D V E V D V E over here. So, looks like I have this intersection point. Now, in first go it does not seem right because these three points I expect them to be collinear they are quite close to being collinear, but they are not exactly collinear. So, I will have to verify if there is any mistake that I have made looks like there is I am not passing right through U V. So, perhaps if I make a little change in the angle that may make a huge difference if I change this slightly my intersection point would come over here somewhere still close, but not worry. Now, if I look at these two points they seem all right I extend E V F V this seems ok. So, this point seems ok and if I verify this projection is accurate may be a little shit may be perhaps and if I change the angle of this slightly I would see that this is intersecting somewhere over here seems kind of just little error, but ideally they should be lying on the same line anyhow. So, this looks like this would be the line of intersection and possibly not this. So, I will erase that I think I will be able to verify this better in the edge view method and the actual line of intersection between these two planes would be the one that would be common to both the planes. So, it would actually be this line here. So, this is the segment that would be lying within A B C and D F. Let me remind myself by nominating these two points or naming these two points as for example, let us say I of V and J of V. Now, I of V is lying on A B the frontal projection of that. So, if I project this thing upward should be lying on A B is right there I will call this I of H and likewise if I project J V up J V is lying on B V C V or B C in the front view this would go up till this point. Let me call this J H and my line of intersection would be the one that joins I H and J H and that would be the line of intersection between these two planes in the top view. Now, before I start with the projection method to figure out the visibility, may be it is perhaps a nice idea for me to consider the edge view of one of the planes and verify this is indeed the line of intersection that I am looking for. Now, for that better that I draw horizontal from here. Let me verify if this aligns well with the margin looks like it does and I will try to make this a little faster. So, this is the horizontal let me mark this is O H corresponding projection down O H lies on B H C H the corresponding projection down there. So, this point here is O V now A V O V would be true length because this thing is horizontal and this things also parallel to the hinge line that separates the two views the front view and the top view. So, I will come back to the hinge line, but first let me align my drafter the longer side of the ruler along A O and shoot a projection out. So, one way the other way is to shoot the projection the other way or the other side perhaps this box over here would be interfering with the edge view right there. Now, what I will do is I will shoot the projections parallel to that one from all the vertices this is one not. So, nice thing about the edge view that it could be quite tedious from C from D from A from F from B and from E maybe I will have to show another projection and that projection is going to be very close to the one that starts from A. Now, you can see that there is a little problem over here I will probably have to be little more careful perhaps that is the one. Now, having done that I will real on the longer ruler of my drafter draw a hinge line that would separate the front view with the top view and this is going to be a complicated figure. Now, I will align my drafter to be parallel to one of these projections this would correspond to the hinge that would separate the front view from the auxiliary plane. What I will do is I will take a pause measure the distances from all the vertices from this hinge line transfer those distances over here and draw the edge view of A B C which is what I am aiming at and also draw the corresponding projection of the triangle or triangular plane D E F. So, while I do that maybe it is a nice idea for my friend Asutosh who is on the camera to take a pause back from the break anyhow. So, we come back in the meantime what I had done was I had made this figure of this assembly of two planes in the auxiliary plane A 1 where my plane A B C it happens to be in the edge view and of course, D E F it is not supposed to be in the edge view it is supposed to still look like a triangle. Now, the idea behind drawing this edge view was to confirm whether I got the intersection points or not or correctly or not using the cutting plane method. So, let us do that. So, A B C is in the edge view and this is intersecting let me take my two edge pencil. So, this is intersecting triangle D E F at this point here and at this point here. So, of course, the intersection points are going to be lying over here and over here on the A B C plane. So, let me try to realign my drafter along this projector and let me go very close I am not sure if this is. So, let me go let me go very close to this intersection point and of course, this intersection point is lying on E F. So, I need to project this point back on E F E F right there. So, I have to figure my drafters alright looks like this. So, if I project this point on to E F this will probably be intersecting E F over here somewhere. So, let me draw this line and let me keep this point mine and the other intersection point between A B C in the edge view and D E F is over here and that intersection point lies on D E H D E. So, I need to project this intersection point back on D E. So, let me do that and if I do that correctly I expect that point to be lying somewhere over here. So, let me draw this projector let me talk on it and if I get to my 30 degree friend 30 60 set square friend which I call and if I try to align these points. So, looks like I am very close to the intersection line of intersection I J very close little deviation which I guess I can expect because of certain errors in the instruments, but I am really quite close. Let me go back again and see if I can do anything with this yeah possibly I can. So, I just need to make sure that my instruments are free of errors anyhow. So, this confirms pretty much that my line of intersection will be very very close to this line I guess and let me draw this line in blue I J and this is going to be a solid line I know and it is going to be a visible line in both views because this is the common line of intersection between two planes and of course, we will come back to visibility in a while and if I back project the points I will have to realign my drafter if I back project my points and most likely a point I would be somewhere over here J would be somewhere over here and let me join I H and J H as I have mentioned again using a blue line just to highlight that this is the common line segment intersection this is much better. After having gotten the intersection line of intersection between the planes D E F and A B C in both views we are now ready to determine the visibility I will have to be very careful because I tend to make many mistakes. So, first I would like to consider the projection method. So, the logic behind the projection method to determine the visibility of different you know planes or lines is not very difficult. Now, imagine that I have got two lines here this is my two H pencil representing one line this is my H pencil representing the second line. Now, imagine that you are seeing the projections of both these lines in this top view and imagine that this point of this pencil and this point of this pencil or the two lines they are common here in the top view. So, one would be just above the other now if you flip this view over like this of course, these two lines are not intersecting but one point will be on top of the other once again one point will be on top of the other. So, primarily what means is that if I take a projection from this common point in the top view and drop it on the front view I would see that this line will be getting hit by that projection before actually the green line gets hit by the projection let me rotate it and therefore, I would conclude that the brown line is going to be visible and this this part is going to be hidden behind the brown line in the top view once again it is quite straight forward of course, you could clearly see that this line is in front of this line and in the top view the vertical projector is going to be hitting this part first and then later this part. So, we will be using this trick to determine the visibility of the two planes in the front view as well as in the top view and I am going to work on this quite slowly because I do not make much mistakes or rather any mistake here alright. So, you would see that both the planes are represented using dark lines. So, it is going to be a little difficult for me to work but I have these two markers here two friends my black sketch pen or marker and my blue sketch pen marker what I will do is I will kind of work on these two planes by these markers and try to depict visibility and invisibility kind of clearly. Now, let us look at this point S H and S H happens to be common to D H F H and A H B H in the top view and if I project this point down if I project this point down from here I would be hitting A V B V first once again. So, if I come down from here I will be hitting A V B V first that means that this part happens to be above this part in the top view. Now, before I come back to that notice that these edges which are not. So, these edges they are all lying outside the other planes region or let me try to rephrase this better. So, for example, if you look at this part this part of A B and A C both these paths are not lying within D F. So, they have to be visible likewise these two paths they have to be visible because they are not lying within this projected D F and likewise these paths are going to be visible. So, let me go back take may be another pause and work with my marker to darken these lines and then come back and re initiate my discussion on visibility alright. So, I am back now and I have darkened all these edges which are going to be visible anyway. So, it is these edges that we need to figure the visibility for and these are quite a few. So, coming back to this point S H. So, if I project this point down and this of course, in the top view is the intersection between D F and A B if I project this down D F seems to be hit after A B. So, looks like A B will be visible and from here to here. So, this part is going to be visible. So, what I will do is I will take my drafter off once again just make sure that what I determined seemed to make sense. So, if I go from here down A B seems to be intersected by this project or intercepted by this projector before D F does. So, this part is going to be visible. So, using my blue pencil I just mark this thing is solid may be perhaps little darker because I am going to be overriding this by the black sketch pen that I have alright coming back. So, if this comes down from here A B is visible and if A B is visible then of course, D F will be hidden behind it. So, D F will be hidden behind that. So, looks like this would be a hidden line therefore, will be shown by dashed. Now, to confirm if that is really the case this edge of A B is lying above D F. Now, for that I need to shoot a projector vertical projector from here. So, this is the intersection between A C and D F. So, I take my drafter and shoot a projector over here vertical projector down and let me try to figure which is head first. So, looks like A C is head first and before D F here once again. So, looks like this is intercepted first or this intercepts that projector first and then this. So, looks like this part of the line is going to be visible and therefore, is going to be shown by a solid line and that seems to make sense because if this part is solid this part of the edge E F will be hidden. That seems to be working alright for us for now and let me show this using again a solid blue line. Now, let us come to this point of intersection between A C and D E. So, before I shoot a projector down from there I already have one I guess and let me locate the corresponding edges A C and D E here. So, apparently if I go down from here I will encounter A C before I encounter D E. So, this part of the line is going to be visible which is quite alright this has to be a solid line and if this is a solid line then definitely this part of D E will be behind the plane A B C. So, this part of D E will have to be hidden which again kind of makes sense. So, this is maybe I will use a 2 H pencil for that alright. So, if this is hidden let me look at this point here point of intersection between D E and B C. I have a vertical projector down once again D E here B C here and if I go down from here D E B C looks like I am going to be hitting B C before I hit D E. So, it is this part that is going to be visible. So, this part is going to be visible. So, let me draw a solid line here alright. Now, if I go to this point for example, now before that we notice the loop I A C J is kind of solid and this loop belongs to the plane A B C and it is this loop that is lying above the corresponding plane D E F and that is the reason why these two lines are hidden that that kind of makes sense physically as well. Now, coming back to this point point of intersection between A B and E F. So, let me locate A B E F right there alright. Now, if I project this point down here looks like I am going to be getting to see E F before I get to see A B. So, E F will be visible and when I say E F is going to be visible then this part of the A B this part of the edge A B will be hidden. So, maybe I will just without using the ruler this time I will just kind of you know make a few dashes alright. So, when this part is hidden this part has to be solid once again if I take this projector down I tend to be hitting E F before I hit A B right. So, this part is going to be solid I say this part is going to be solid of course, used to be shown using a solid line. Now, what remains is to figure the visibility for this part of the edge B C. So, for that I need to go to this intersection point that is the intersection point between B C and E F. Let me look at where B C is where E F is and if I shoot a projector down from there again I am going to be hitting or encountering E F before I encounter B C. So, looks like E F is going to be or this is this part of the plane E F this part of the plane D E F is going to be visible. And of course, this part of B C will be hidden and therefore, I am going to be using dashed to show this. So, notice what these intersection points have done. So, this is where there is a change of state of this part of A B and this part of A B. So, before I H this part of A B was visible and after I H this part of A B is hidden behind D E F and likewise before J of H intersection point this part of B C was visible and after J H this part of B C is hidden behind D E F. So, it looks like A B C is piercing D E F right through there this where the intersection is happening and after this you will actually see a part of D E F before A B C cannot make sense. So, before I try to figure the visibility of the respective portions of the edges of these planes these two planes in the front view. Let me take a little pause and finalize this drawing using the black sketch pen. So, that a lot of things become clearer. So, this is how the final top view showing the two planes looks. Now, let us try to figure out the visibility of the corresponding edges of these planes in the front view. Now, for that I will use the identical method. Now, it is easier for me to figure the visibility of this edge first because the intersection point is not there and likewise for this edge and of course, we are going to be worrying about these two parts later because that is where the intersection point happens to be there. So, let me focus in this intersection point between D F and A B let me get my drafter over here D F A B. So, I have located these two edges I go up and I try to figure what do I encounter first is a D F or A B. So, looks like I encounter D F first that would mean that in this view D F will be visible and when I say that I will have to be using a solid line representation. At this time let me use my blue pencil of course, I will finalize this drawing later. Now, when this guy or this edge D F is visible of course, then this part of the edge A B will be hidden behind this corresponding portion of the plane D F. So, I would expect this portion of A B to be hidden makes sense that way seems more logical that way. So, maybe I will it is kind of you know mark this part as a hidden line and of course, visit that later. Let me make this little more solid little more clear well I said I will come back to this later. So, I will do that let me focus on this intersection point between A C and D E. So, I get my vertical ruler here I identify A C which is here and D E which is here. So, I go up and clearly I will encounter D E before I meet A C. So, D E would be visible. So, I have to have a solid line here. So, notice that I am not using any information from the edge view with regard to determining the visibility of these portions of edges looks like I am in a hurry. So, I should not do that. So, maybe I will go up again D E D E A C A C. So, I hit D E before I hit A C. So, D E is going to be visible and if D E is going to be visible this part of A C will be hidden you know confirm let us do that. So, there is a point of intersection between D F and A C I take it up D F I hit D F before I hit A C right. So, D F is visible good and if I take a look at this point of section between A C and D E again I get to see D E before I get to see A C in the cuff view. So, of course, this part is visible. So, when this part is visible this part is visible of the plane D E F naturally this part of the plane A B C has to lie behind D E F and logically this will be represented by a hidden line segment. So, just back to this part of A B and this part of B C. Now, if I look at this intersection point between A B and E F I go up well this is where A B is this is where E F is. So, I will probably have to go further up. So, I get to be seeing A B before I get to see E F. So, A B here and E F right there. So, looks like I hit A B before I hit E F for that matter A B will be visible here this again seems to be because this intersection point I V would tend to change the nature of this line segment. So, if this line segment is visible solid then this part has to be invisible hidden. So, this part has to be solid all right. Now, to this intersection point between E F E C E F right there B C here take it up I get to see B C before I hit E F. So, this part of B C will be solid yeah B C before E F. So, this part is solid all right and now what you expect this part of B C to be hidden or solid notice that there is an intersection point that is expected to change the state of the line segment. So, if it is solid over here you know I would expect this segment to be dotted or dashed or hidden behind D E F. Now, for that to confirm that let me look at this intersection point between B C and D E. So, I bring my drafter over here D E is where I have over here yeah and B C is this. So, if I take my projector up I get to be hitting D E before I get to hit B C. So, D E would be visible that means B C would be hidden which is what is expected. So, maybe I will just represent this using dotted lines. Let me finalize this online try to ensure that I do not make any mistake because if I do then I am done with. So, this is solid. So, this part is solid and this part is hidden this part is going to be solid anyways this part is solid this part of D E is again solid and this part of A C is hidden. So, shown using dashed lines right. So, did I leave out any edge looks like I left out this edge here this part had to be solid and looks like I left out this part of E F. Now, what do you expect this part to be should this be a hidden segment or should this be a solid segment well it is not very easy to well it is rather very easy to determine that because if you look at the corresponding portion of A B C this part is solid this part is solid. So, of course in the front view this part of A B C has to be lying in front of this part of D E F and you would expect this line to be hidden. So, maybe I will just go ahead and draw this hidden line if I am not sure maybe I will use a pencil there is a point I would like to make after this yeah why do not I go ahead and draw a dashed line segment I got to be sure of myself at some point of time alright. So, this is how the planes will look the intersection line and different portions of the three edges of three or rather two planes in their visible or hidden state I should have marked this also as a dashed segment better. Now, what we can do is we can confirm if we got the visibility right and the front view by comparing this drawing with the edge view over here. So, now look at this loop D A and if I call this point let us say you know A small A and perhaps this point is small b. So, this point small b is going to be lying on D E perhaps here and this point is going to be lying somewhere over here. Now, D small b small a F D. So, D small b small a F D. So, I am looking at this loop this loop is closer to the hinge line ok. That means that relative to the edge view of A B C this part of D F will be visible and let us see if that is really the case. So, D small b. So, so far so good solid line this part is not a part of the intersection line. So, I ignore this part ok from B till perhaps J V would be interesting for me to plot J V over here let us do that. So, J V would be lying on B C maybe that will give us a better perspective B C by there. So, let me mark this as J 1 and let me also go ahead and mark I 1 I 1 is going to be lying on A B. So, here coming back to the point T B J 1 I 1 F D this is lying in front of the edge view of A B C or closer to the hinge line. So, it should be visible let us verify. So, D small b from J V T I V it is a solid line from I V till A perhaps again this part is something that is not a what concern from A to F it is visible from F to D it is visible. So, all these guys or all these edges they happen to be represented using solid edges and of course, B E 1 A 1. So, this part of D F lies behind the edge view of A B C or away from the edge view of A B C with respect to this hinge line. So, let us verify that is the case. So, if I look at E 1 if I go from E 1 towards F. So, this part is lying behind A B C yeah. So, perhaps I was blabbering. So, maybe coming back and take a look at this loop D F all together. If I go from E 1 to A 1 that is behind the edge view of A B C or away from the edge view of A B C with respect to this hinge line. So, if I go from E 1 to A 1 or E 1 to A E 1 towards A a part of E F has to be hidden behind A B C. So, that is ok. So, if I am starting at A from A till F all right. So, this is where D F will be lying in front of the edge view of A B C. So, this is solid all right and F E rather D E. So, from here to here from D to B I expect the edge to be solid and from here to here there is nothing on the plane O B C that kind of hides this part of D E F. So, this has to be solid. So, it looks like it is ok. So, you can find the visibility of a plane using either the projection method that we had discussed in detail or directly using the auxiliary plane or the edge view method which in a way again is projection method if you are comparing these two views for visibility all right. You know you could do this as an exercise you might want to draw an edge view corresponding to this figure of the plane and using the edge view try to verify if the visibility over here is correct or not. I hope I am correct, but I would still have .