 Today, we will draw details on the interaction between planes and lines. Interaction involves two aspects which we would address. Number one, what would be the intersection point between a line and a plane? And number two, which part of the line will be visible within the projection of the plane. So, if you notice these are two identical pictures, we are going to be discussing two methods. One is the edge view method that will be working over here. The second one is the cutting plane method that will be working over here. Now, both these methods will help us determine the intersection point between the plane and line as well as the visibility of the line. So, let us start with the edge view method first or the auxiliary view method first. So, this is the projection of the plane on the frontal view or the frontal plane. This is the projection of the plane in the top view or the horizontal plane likewise for the lines same over here. And the question now is to determine first the intersection between the line and the plane if there exists one. So, what we will try to do is we will try to first figure the edge view of this plane. Now, for that let us start with the hinge line in between these two views. Switch my pencils, it could be anywhere between these two. You know what the code for the hinge line is? Long dash followed by two short dashes. The trick to find the edge view of the plane is to identify an edge in true length within either this projection of the plane or this projection of the plane. Let me try to identify the edge in true length in this projection. I switch my pencil again to work with the two edge. What I will do is I will draw a horizontal line starting from A. Now, it is always a nice idea because there are going to be too many lines in our figures. It is always a nice idea to label the vertices and the edges before you start your drawing. So, this is my horizontal. Let me label it as D H. H is a subscript that represents horizontal plane. I take a vertical projection down. So, D H lies on B H C H. So, correspondingly D V over here should be lying on the edge B C. This point over here is D V. Now, what I would do is I would release my mini drafter ruler or the ruler of the mini drafter and I will align this ruler along the edge A D and then I will tighten the screw here. I join the two vertices A B and D V. B the subscript it means that all these vertices are represented in the vertical plane or the front view. Now, since A D is horizontal and it is parallel to this hinge in the other view A D will be in true length. To remind myself I just write true length here and then what I will do is I will start taking projections from all the vertices and these projections will be parallel to A D. So, I will start from say P V and then A V, Y D times C V and all these projections are going to be construction lines. So, they will be as light as possible that is the reason why I am using a 2 H pencil. Now, there is a reason why I have fixed my drafter. So, that the longer ruler of my drafter is aligned to be parallel to these projections. It is easier for me to now locate the hinge line anywhere over here and this hinge line has to be perpendicular to these projections. So, I will probably locate my hinge line over here somewhere and draw that again long dash followed by two small dashes. Now, I will take my 30 60 square friend here and what I will do is I will measure distances of each of these vertices from this hinge line and I will transfer these distances over here respectively. So, this perhaps is about 3.7. So, I will mark this point here maybe I will just double check just to make sure I am not making any mistake because if I make a mistake then I will probably not getting the plane in the H view here perhaps 3.7 just about there I will just make a little mark here and then let me measure B from this hinge line. This is possibly about 14 millimeters just close to 14 millimeters and that is B here and let me also measure C of H from this hinge line. So, about 4.8 centimeters or 48 millimeters I will measure C B from here 48 and mark it right here. Now, I will bring my 30 60 square and you could see that these 3 points are collinear. So, looks like the measurements are alright. So, what I will do is I will join these points nicely and I will not forget to label them. Now, A V and D V should be lying on the same point here because these distances are identical. In fact, I should not be labeling them as A V and D V despite many reminders that I have given to myself that I should not be using an eraser I end up using one I should be focusing a little more. Let me also thicken or darken these projection lines just a little bit. So, that my grader follows the construction that I have performed and let me denote these points as A 1 and D 1. So, this is my vertical plane horizontal plane my vertical plane and plane number 1 and the triangle A B C is in the edge view here this point is B 1 and this point is C 1. How about P and Q? Maybe I should also mark those points here I will measure P h from this hinge line it is close to about 76 millimeters 76 millimeters from this hinge line. I do not have to worry about collinearity of P Q because P Q happens to lie on 2 points P and Q. So, I will simply go ahead and mark this as P 1 and how about Q I will measure Q from here it is about 4 millimeters close to that and I forgot to shoot a projection from Q V here which I will do right away and this was close to 4 millimeters let me double check just about this should be 4 millimeters from here from this hinge line I will mark Q 1 here or may be let me first draw that line P Q. Now, notice that the plane in the edge view and the line P Q they are intersecting at this point that is the first sign that would suggest that the plane and the line would intersect. So, that is not the sufficient condition could be necessary condition, but possibly not the sufficient condition. So, we have to ensure that this intersection point also lies on the corresponding projections of the line in the frontal view and the top view and also this intersection point has to lie within the corresponding projections of these planes or rather within the corresponding projections of the plane there is a single plane that we are dealing with. So, before I take this point or project this point backward on to the front view let me join P and Q because that is where this point is going to lie and I am going to be using a very thin light line and then I will project this intersection point back let me call this intersection point as I 1 in the first auxiliary plane. Now, this intersection point hits here somewhere let me call this I v as an I v leak and if I take a vertical projection now for that I would need to realign my drafter perhaps parallel to this hinge line go back and check if it is really aligned possibly not. So, maybe I have to make a little adjustment see if the hinge line is looks like nuts and then I project this intersection point up there right there this intersection point has to lie on p h q h I call this I. Now, notice that not only this projection line is lying rather not only this projection this intersection point is lying on the line p q the intersection point is also lying within the corresponding projections of the plane implying that there is intersection between the plane and the line. Now, I could also possibly think of getting these intersection points using an edge view that I get with reference to the top view of this assembly of the plane and the line. Now, for that I will need to draw a horizontal line here in the front view gentle line call this E v take its projection. Now, E v is lying on A B. So, take its projection on A B let me call this E h now it is time for me to align the longer ruler of my drafter and let me join C h and E h maybe let me make it a little more dark. So, this would be in true length E h C h and I would just mention it over here the reason why it is going to be in true length is because the corresponding projection E v C v is parallel to this hinge line. Now, let me shoot projections from the five vertices in the top view all parallel to E h C h. Let me be a little careful in drawing these projections making sure I do not miss any of the vertex you know a slight chance of a mistake and things will go wrong. So, one needs to focus one needs to be a little careful, but there is no harm as I said in making mistakes did I miss out on any projection possibly not. Now, I am going to be drawing a hinge line which is perpendicular to all these projections. So, let me draw the hinge line over here make sure I cover all projections now this is my horizontal plane. So, I will write h over here this is my second auxiliary plane. So, I will write 2 and then what I will do is I will measure distances from here and transfer them over here put my drafter aside for a while this distance is about 5.5 A v from this hinge line this would be 5.5 over here. So, let me just make a little dot b v from this hinge line is about 13 millimeters just about. So, I transfer that distance over here along the projection from where that starts from b v or b h in this case it is about 13 make a little dot and then c v from here is close to 42 just about close to 42 millimeters. So, I will measure and transfer that distance along the projection that starts from c h it is about 42. So, I just mark it over here and I will try to make sure that these 3 align the very close. So, I draw a line joining these 3 points anyways perhaps let me check this distance again c v is it 41 or 42 possibly 41 millimeters just about perhaps it should be a little closer to the hinge line now these 3 points they tend to align better using a 2 h I try to join these 3 points. So, this is the edge view of this plane on the second auxiliary plane let me label the corresponding vertices this would be b 2 always a nice idea to label c 2 comma e 2 and this of course, would be a how about p and q I did miss out this projection. Now, have it I will have to extend this hinge line of mine which is not a problem there you go let me measure p first from this hinge line this is close to well 73 millimeters 73 millimeters it may extend this projection little and let me also label this as p 2 and q from this hinge line is about close to 7 possibly 7 7.5 millimeters. So, I just transfer that distance here perhaps here and let me write let me label this point as q now what I will do is I will join p and q but not draw and section or not draw this line yet I will try to figure where p q intersects with the edge view of a b c perhaps here and if I project this point back onto the line I am missing out on this intersection point. So, looks like I have not done something right this seems to be ok because a b c the plane happens to be in the edge view on this auxiliary plane. Let me go back and measure these distances once again did I say this was 73 millimeters possibly I mean this scale instead maybe this should have been here looks like there is a problem with the 2 scales and 4 millimeters I am sorry this should be about 6.5 to 7 close to 7 millimeters seems measure that from here just about right maybe shift that little to the left. And once again get these 2 points aligned roughly looks like I get the intersection point the same point or maybe little above. So, looks like I am ok I am alright let me draw projection first it is going to be close there will be some errors but it is ok and then I am ready to join these 2 points. So, this is where my intersection point is let me call this I 2 I had already projected that intersection point over here and. So, this is the same point I h this is the same point I v. So, looks like whichever plane you choose as a reference to draw the edge view of a b c whether this auxiliary plane or this auxiliary plane they should give you the same intersection point. So, the intersection point identified over here and over here let me emphasize these points by encircling them here. So, the edge view method is a little tedious requires little bit of work it is quite effective and quite logic and not so very difficult. Now, the second aspect is which part of the line is going to be visible in this area would be this part of the line that is visible or this part of the line that is visible likewise would be this part of the line that is visible in the front view or this part of the line that is visible. Now, that is very easy to answer if you have the edge views over here. Now, if you look at this view a part of the line is in front of the plane or in other words this part of the line is closer to this hinge line. So, it is this part of the line that would be visible or that would be in front of this plane here. Now, if I start from q and come up to the intersection point I 1 starting from q coming up to the intersection point this has to be visible. So, the plane a b c is lying behind this line. So, I will have to be drawing this part of the line as a solid line and of course, from the intersection point I 1 up till p 1 this part of the line is away from the hinge line when compared with the edge view of the plane. So, this part has to be invisible. So, I will be drawing that using dotted lines or dashed lines I could do something very similar in case of the top view, but for that I need to refer to the edge view here. And that is one of the reasons why this edge view helps looking at this part of the plane this part of the plane is lying closer to this hinge line compared to this part of the line. So, this part of the line will be hidden or will be behind the plane. So, starting from q up till I 2 q up till I h this part would be hidden. So, I will have to be drawing that part of the line using dashed lines and the rest of course, from here to here this part of the line will be closer to the hinge line compared to the edge view of the plane. So, this part from I h to p h will be solid. So, using the edge view method or the auxiliary plane method we have been able to address both issues pertaining to the interaction between the line and the plane. Issue 1 to be able to get the intersection point and 2 to be able to find the visibility of the line with respect to the plane which part of the line is in front and which part of the line is behind the plane in both of these views. Now, took me a little while to draw the 2 edge views find the intersection point and then figure out the visibility of the line p q. Here I am going to be demonstrating the same example using the cutting plane method to determine the intersection point and using the projection method to determine the visibility of the lines in both views. Please time me and check whether this method is easier or more efficient to work with compared to the edge view method or the auxiliary view method. So, ready to time me I will first start with the hinge line that would separate the top view from the front view. Now, this is something which would be tricky and this is something that I will try to make myself understand. Now, what I will do is I will imagine a plane that would contain this projection and that would be cutting this plane a b c. So, my friend over here will be representing that imaginary vertical plane. So, remember this is the top view. So, top view would be the horizontal view and this cutting plane will be the vertical plane that would be cutting this top view and it will essentially be looking like that. This is going to go in as going to remain vertical with respect to the sheet. So, of course, this plane and the actual plane a b c they will intersect and they will have a common line of intersection. Now, that is the line that we want to determine to do that and there is another thing that you would want to remember while you are working with the cutting plane method. So, remember that this vertical plane it will be containing the actual line p q. So, keep that in mind you can work it out. So, let me start with this intersection point between that imaginary plane and this projection plane and let me project this point down. So, this point is lying on a h c h correspondingly when I project this point down it would lie here on a v c b. Let me also do the same for this intersection point I will project this down and I will find that this point over here lies on p h c h correspondingly this point here. Let me mark this point as say s h and t h two points here correspondingly s v and t. Now, if I join these two points here this line s v t v is actually the line of intersection between the imaginary vertical plane here and the plane a b c of course, s v t v is actually projection of that intersection line. Now, if I complete my p v q v line here carefully I probably have to do this a little right because if I have not made any mistake in this method I should be getting the same intersection point as what I got in case when I use the edge view method. Perhaps this intersection point is little off compared to this intersection point, but anyhow perhaps I did not take the points carefully or I probably may have made a mistake somewhere over there not to bother should have the intersection point should have been over here somewhere. Let me give this another shot why not I should be getting the same intersection point should not I seems all right probably it is my drafter that was not proper because if I look at this vertical and if I look at this vertical you see this projection over here is not coming properly. So, let me align my drafter probably and work this thing out again I deserve to be getting the same intersection point do not I before I proceed let me finalize these edges of the triangle it is very important for you to have your instruments calibrated properly and as accurate as possible when you are drawing these concepts on a sheet of paper. Now, this vertical over here seems all right this vertical seems all right possibly not aligned properly maybe I will just re adjust that is not a part of the time that I am supposed to be investing on the cutting plane method. I could also see some discrepancy in the position of p v seems little better but this inch line goes well little adjustment here and there I think it should be now once again projecting this s h point down and this should be lying on a c a c. So, I should be projecting this thing further down right there that is minus v little change not very much from here to here possibly that could make a difference and T v how about that T v moves slightly to the right let me raise this part here somewhere now let me complete this line p q for I start worrying about the line s v T v and if I do everything right and this is the point section that I get again it is not very close to I v that I expected but this time I guess I will give up something must not be right but again it is not the of course intent is to be as accurate as possible but it is more to demonstrate a method. So, this is the point section I will encircle that this point has to be lying on p q in the top view I will have to project this up. So, let me label this as I sub v and label this as I sub h encircle this point alright. So, I expected the positions of these two and the positions of these two to be identical in both views but somewhere I may have made a mistake I will probably have to redraw this again but not today. Now, as far as intersection is concerned the cutting plane method that the imaginary plane that passes through this line p h q h that insects the plane a b c and the projection of that corresponding intersection line is s v T v that would intersect with the projection of the line that we have p q this would give us the intersection point I v in the front view if we take it up that is the projection point I h in the top view. Now, coming back to visibility now for that we will be using the projection. Now, look at this intersection point again s h. So, that is the intersection point between p h q h and a h c h now this point represents the point on p q as well as point on the edge a c in the top view. So, in the top view this point would be the same but in the front view if I bring this point down the corresponding point belonging to a c will be here but the corresponding point belonging to p q will be slightly above right. So, it is. So, if I bring this projection down it is p q that I am hitting first in the front view compared to a c. So, that corresponding part of p q will be visible. So, this suggests that p q is lying above the plane in the top view and therefore, this part will be solid let me switch pencils. Now, if you compare this part of the line p q with the corresponding part that we obtained using the edge view method they seem to be consistent going back to the projection method. Now, this point here is the same point on the edge p c of the plane and the line p q in the top view. But if we take the projection down from here the corresponding point on edge p c is above same point well the image of that point on line p q which would be somewhere over here. So, this was suggest that p q is behind the plane and that would be in the top view. So, that is the reason why this part will be hidden and therefore, shown using dashed line. I could not do the same exercise starting from the front view I take this intersection point take it up my drafter must not be very accurate over here because I end up losing the parallelity between the lines of parallelism between the lines whichever is the correct English or use of English. So, anyhow so I take this projection up from here. Now, this point in the front view represents the point on the edge a c and the corresponding point on line p q up there edge a c would be appearing before p q. So, that would mean that this part of p q will be hidden behind the plane and that is the reason why this should be shown using dashed lines. Now, if I look at the corresponding segment of the line this is also shown using dashed. So, apparently the projection method of visibility or determining visibility happens to be consistent with the edge view method of determining the same. So, now this is hidden and if I take this intersection point up there now this point is shared by p q and the corresponding point on p c, but if I take it up p q would be getting hit by this projection line before p c does that would mean that p q would be visible in the front view again consistent with this segment of p q. So, remark this is a wonderful logical method and so is this. This gives you details with regard to the edge view of the plane and the line the intersection projections back and also it helps you determine the visibility of the line p q. This method the cutting plane method and the projection method in combination to determine the visibility seems a little faster if it is understood properly. You know what I was not very happy with the construction over here possibly because my drafter had some errors in this part of the region. So, what I did was I came over here I had some space I redrew everything over here using the cutting plane method and it looks like the intersection points that I found are quite close to the intersection points that I have using the edge view method. Now, ideally if there are no mistakes and if my apparatus are accurate I should be getting these points to be identical to these corresponding points and that is the ideal case and of course, visibility remains the same. So, this part is hidden behind the plane this part of the line is hidden behind the plane whereas, this part is visible and above the plane and this part is visible and above the plane.