Hi, good morning and welcome to the products and focus so after non farm perils and friday there that came in at 285,000 versus 225 Expected even though we had a little take up on the employment right there and we saw big moves Before the weekend on the dollar especially against Japanese yen Whereas most global stock markets took a bit of a slump on the news as obviously they take a Possible and straight hike in the US as a short-term risk adding some uncertainties to the markets UK market in particular looking quite down Germany 30 down and the US 30 and SPS 500 kind of down a little bit there as well. So this is your Friday candle You've followed that through with just a slight bit more further down was action this morning 17 747 is the next potential support level Is mac these crossing the zero line the other technical show further room for for this to move lower? And certainly this looks pressured in the short term So then moving on to the UK 100 you can see now that we've broken that trend line is Broken a lot lower that was a Thursday's candle Then you've got your Friday's candle almost touching 67 71, which is a support level for me that we talked about a large number of times We've had a very very very small bounce this morning, but The down was pressure looks to be firmly in play the next potential support 6686 should that continue on Japan to do five normally would have been a little bit more buoyant with the music in dollar yen, but obviously Most global equity markets have come off slightly Following the job number on Friday This looks like I can have a flag formation or a pendant formation So consolidation before a re-challenge of twenty eight sixty eight Because they'll be in the cards. So I'm looking at dollar yen Dulling getting quite close to potential resistance 126 This is a 13 year high for dollar yen Great, but kind of bullish engulfing pattern again that we've had there on on the Friday session there and decent amount of momentum As stop shine of resistance and slowly ticking down Telecom the care perspective everything's quite overbought the RSI slow to castig I'm going to see a reduction in the MACD histogram Which is indicative of the fact that there is a slow down the momentum But that's not surprising after the moves that we've seen every break one twenty six then a one twenty nine ten as an Ex-potential resistance level moving on to West Texas crude We actually had a positive session on Friday Even though we had a big surge in the green back which now with pressure pressures commodities So crud doing okay The elections over in Turkey as well probably adding a little bit more and stability to that part of the region for the Popular party there only got 40% of the votes, which is still enough But not enough for a majority and the Kurdish Party in Turkey got 10% of the votes the highest that they've ever done, but there's probably gonna rally a lot more Courage to the cause on that side so a little bit more and stability But crude having a decent day hammer formation actually that we've got right here after selling off the previous couple of sessions $57 is the potential support level other technical the relic are pretty flat This has been moving the side views action for a month and a half So you've got potential resistance round about 62 and then obviously you got your short-term potential support round about $67 Moving on to gold gold got her on Friday actually, but off the session There's quite another surprise a move for gold I thought after no after those non-farm payroll figures that the gold would be really feeling the pain a lot more But off the session lows Trying to have a bit of a positive day today Entry day chart wise you can see it's tried to take up slightly You've got a death cross on the moving averages and you can see the MACD is broken below the zero line here as well Whereas the RSI and still to cast it still show there's further rim again to move lower So the next potential support would be 1137 on gold with potential Resistance now 1186 now be broken that support that's now expected to act as potential resistance So moving on to your dollar Lots of news over the weekend about Greece and the head of the ECB Junker who is ECB but of the European Union member presidency and lots of issues between Junker and the Prime Minister of Greece there Usually known as being more staunch friends But a lot of a lot of words said in the weekend over a lot of European newspapers Patience is fast running out And obviously they've got night till the 30th of June to make the IMF payment of 1.2 billion euros And things are certainly heating up a little bit And that's what's causing a little bit more downward pressure on the euro and you combine that together with the dollar strength who got from friday Um It's it could potentially be the perfect storm, especially it looks like that note that they're so far away on reaching a deal That maybe doesn't happen by the 30th So long term potential support one spot zero eight We are bang on a support level right now Which is around about one spot 11 We're just a little bit above that just now at 43 11 43 Bouncing we're in a 21 period SMA Whereas the other technicals are again are pretty neutral So the fact that we had this graveyard doji formation there on thursday with an ability to fall through and then quite an aggressive sell-off the following day Um, if we break one spot 11, then it's probably quite likely that that downwards pressure would gain Momentum and that means that one spot zero eight could be on the cards. So finishing up with gbp usd Hammer formation again sold off strongly on friday after the figure found stuff potential support one spot 51 85 Which also coincides with that 55 period sma. We talked a lot about during these sessions Is in positive territory today almost at the top end of this range short term spike But we are in the middle of two core ranges now one spot 54 24 and one spot 51 85 So I come in data wise disappointing data out of china Especially trade balance imports Down 17.6 percent they're expecting minus 10. So that's that's pretty bad Again probably adding a little bit of extra juice for Stimulus over in that part of the world But the asian stock markets actually took a tumble on the back of that normally bad news is good news in this instance But I think that import figure was particularly bad And if you have a look at german exports, they're Better they came in slightly better than expected month a month for exports Imports minus 1.3 minus versus 0.5 expectations. That's not so great There's not a huge amount extra due today data wise fast forward on to tuesday. You've got chinese cpi Eurozone gdp And then on wednesday You've got critical inventory So not a huge amount of economic data to be honest to get that excited about so you'd be driving quite heavily by technical factors And we'll see how that pans out So keep around the chart for them because michael hueson and jasper lawyer are two uk analysts We'll be posting lots of interesting technical trade setups for you to have a look at Make sure you make insights part of your later going forward to help you navigate the financial markets And join me again tomorrow to find out what will happen next Thank you


Okay, good afternoon. So, myself I am Professor Shobik Banerjee. We are both here right now and for the next hour or so, what we are planning to do is we have covered so many different topics. So, last week we covered a huge number of topics in great details on statics. We started with basics of vector mechanics, equivalent force movement systems, then we did centroid, distributed force systems, 2D equilibrium, machines and frames, then 3D equilibrium, trusses, friction, complex friction, principle of virtual work and principle of minimum potential energy. So, these were the topics which we discussed in great details. Then we also took various quizzes, solved a lot of tutorial problems and then starting this week we discussed a lot of dynamics, particle kinetics, particle kinematics, impulse momentum theorem, work energy theorem, kinematics of rigid bodies, kinetics of rigid bodies. And vibration dynamics and vibrations, we did that today. So, now there are so many different topics, essentially if you think about it then it is almost close to two semesters worth of syllabus that we had essentially covered in 10 days. So, what we will now try to do is if you have any particular queries, any questions, any doubts which were not clear either during the lectures or which some questions were asked on Moodle, some answers were posted and if you want to ask them again. So, our suggestion is that we will follow the best optimal mode because the time is limited. Why do not you send those queries on the chat? We will have a look at them and we will start answering those questions one by one. So, I request all the coordinators to collate the questions from their respective centers, send them our way via chat and we will start to answer them. Yes, so one of the questions that has been posted, what is the effect of intrinsic damping on natural frequency? So, first of all let us understand what is intrinsic damping? It is driven by the material properties, nothing else. So, mostly it is the material damping that is referred to intrinsic damping. So, for example, let us say you take a continuous system. Now, in civil structure remember we talked about that water tank problem. In the water tank problem if you have that tower and tank on the top, then essentially we are trying to do that as a single degree of freedom system. The question is where does damping come into play? Damping will come into play through the material. So, therefore, ultimately we do have still the equation of motion that is mx double dot plus kx plus cx dot equals to 0. So, now that c that is really coming from the material damping now. So, in the sense nothing happens you know that omega d equals to omega n square root of 1 minus xi square. So, that whatever we have derived that damped frequency equals to omega n square root of 1 minus xi square. So, ultimately your xi that for civil structure which is coming through the material damping can be up to 5 percent that is the damping ratio. So, ultimately xi would be at max to be 0.05. So, as such it is going to affect the damped frequency of vibration. So, there are two questions one is from 1010. Please explain inertial and non-inertial frame of rotation. See I think you know we discussed that in terms of particles actually know how D'Alembert principle is coming into play. And we said that that if you look at fixed reference frame let us say x y and the particle is accelerating with respect to that fixed reference frame x y and z. Then remember there is a resultant force and the acceleration direction is along the resultant direction. So, particle actually see an acceleration which is its absolute acceleration. So, we can use sum of force equals to mass times acceleration. Now, in a non-inertial frame what is happening that let us say observer itself is in the moving body. So, that means what we now said we have another system that is small x and small y. Now, if that frame is actually moving with the particle then what happens that it is not just sum of force that is acting on the particle, but we do need to take into account that there is a force which is a fixed reference frame inertia force that we call the inertia force that should enter as negative mass times acceleration. So, as such then we said that that means as an observer you are actually at rest that means you are actually at equilibrium. So, this will be equilibrium and this will be motion. So, that is the difference you know in simple sense between the inertial frame and non-inertial frame. So, these two is going with the delambert and this is going with along with the Newton's second law. Now, other question that is being posed is let us say 1148 1148 I think 1148 the question is what is the basis of number of members of truss that is the number of basis of 2 n minus 3. So, remember that m equals to 2 n minus 3. Now, this one we have really adopted to construct a simple truss. What was a simple truss? Remember to form a truss first of all I need minimum of 3 members and 3 pins. So, therefore if you count now this is satisfied why because number of member is equals to 3 ok number of joint is equals to 3. Therefore, we see that this relationship is satisfied. So, this is called basic truss it is a simple truss. Now, main point is that if I really draw if I extrude 2 members and connect by pin then remember this is again a simple truss. So, what was the basic philosophy behind simple truss? A simple truss is internally rigid internally rigid means under the action of the load if I do apply any load the internal member should not change angle that means the truss will not collapse. So, that is internally rigid. Now remember the number 3 appears therefore what is happening if I want to make this problem statically determinate then what do I have to do? If I want to make it statically determinate and properly constrained then I do need 3 reactions. So, do need at least 3 reactions has to be there, but these 3 reactions has to be in such a way that the entire truss will behave like a rigid body and it should be properly constrained. So, what we said therefore most appropriate would be putting a hinge and let us say putting a roller at this end. So, in this way this truss will be properly constrained as well as it is a statically determinate now why it is determinate? Now m plus 3 equals to 2 n. So, the number 3 that is the additional unknown that is coming from the reactions 3 reactions right. Now remember then what we have discussed that it is not always necessary that we have to always have 3 reactions that is absolutely not true. So, what can also happen most general class could be m plus r equals to 2 n where m is the number of forces r is the number of reactions and 2 n is the n is the number of joints. So, therefore, what can happen is the total number of unknowns should be total number of equilibrium equation. So, that was the main purpose remember when this is coming into play we can think of problem even just think of is this a truss very simple example remember if I make it a hinge here and if I make it a hinge here again I think that is not coming. So, I am just drawing it here just take a look suppose I make it a hinge here and if I make it a hinge here just see whether it is a statically determinate problem or not. So, in this way I cannot really say that m equals to 2 n minus 3 and all that what I need to look at I have to make sure that m plus r equals to 2 n remember there is one hinge here one hinge here 4 reactions are coming into play, but even with the 4 reactions I do have a statically determinate system. So, it can be solved for a loading p all the member forces and all the reactions can be solved clear. So, m equals to 2 n minus 3 that is just taken to define a simple truss and the number 3 is when you have exactly 3 unknowns that is 3 reactions supporting the truss to make it properly constrained and statically determinate, but it is not always necessary that we have to always go through m equals to 2 you know m plus 3 equals to 2 n we can go to more general m plus r equals to 2 n to solve this problem this is also statically determinate properly constrained. So, from that we have actually went and looked into also partially constrained improperly constrained truss and then we looked at when m plus r is basically you know less than 2 n right like that we have studied also that is partially constrained ok. Is that clear now? Can you give some examples for 1, 2, 3, 4 and 5 degrees of freedom systems? Can we give now a small problem to solve 2 degree of freedom system if you would like? So, can you derive in this case what are the equations of motion is it possible to derive using the D'Alembert principle concept? What are the equations of motion for this system? How do you define the degree of freedom for a dynamic system? So, I think if you can simply you know give us the equations of motion for this system that will be a good exercise for all of us to understand how the you know whatever we have discussed how that can be taken up for further you know studies for 2 degree of freedom or even 3 degree of freedom systems ok. So, can we try to can we expect some answers for this? Now, we gave a little bit of task if any of the centers please help your colleagues also those who are asking that will be appreciated ok. It is not that we also you know try to answer all the questions any remote center please go ahead and try to pose the solution of this problem which is now being shown ok. So, can you please one of the centers at least help your other colleagues just the equation of motions for this ok just writing down the answer, but you can give it a try thing one of the participants some of you have given the correct answers I am looking at center number 1100 1100 yes the answer is correct. So, m 1 x 1 double dot plus k 1 plus k 2 x 1 negative k 2 x 2 equals to 0 then another equation of motion will be m 2 x 2 double dot negative k 2 x 1 plus yeah let me go over a couple of doubts ok. So, one doubt has I was meaning to answer it yesterday somehow could could not get around to do that. So, the one doubt is that that yesterday we discussed about instantaneous center of rotation ok. So, suppose we have a rigid body like this what we discussed yesterday is suppose there is a point a which has a velocity a in this direction and a point b which has a velocity in this direction then how do we find out if these two velocities have different directions then we saw that we just draw a perpendicular from this line and draw a perpendicular from this vector and this particular point where those two perpendicular lines intersect we said that this was c or the instantaneous center of rotation. But now a new question but another question is that what if this v a and v b have the same direction. Now, in that case let us take this point a which has the same velocity which has the same direction for the velocity as at point b this is v b and this is v a. In that case the instantaneous center of rotation has to be along this line perpendicular, but a location cannot be immediately obtained. So, wait one way to obtain that location is to join these two outer extremities and this point c will be the instantaneous center of rotation. And now what is immediately clear is that by proportionality theorem that a c by b c should be equal to magnitude of velocity of a divided by magnitude of velocity of b because these are similar triangles. And if you know what is the magnitude of velocity at a magnitude of velocity at b then straight away you can find out what is the instantaneous center of rotation that it has to be perpendicular to this direction and the length should be how much such that this ratio will be satisfied. So, that is the answer.  If v a is equal to v b then this becomes like this and you realize that instantaneous center of rotation is now at infinity or in other words this is pure translational motion and there is no rotational component that is involved here. So, I hope that this point is clear. There is a question that is asked by center number 1298 and the question is in virtual work principle we have the assumption of small deflections. So, I take it that this is actually a question. So, it is very important that the deformation that are provided or the not a deformations, but the virtual displacements that are provided are small. There is a very detailed and a complex answer for this even though for example, if you look at this simple example for example, if you take a beam like this with a roller and a pin support and we apply some force then if you want to apply principle of virtual work to find out what is the reaction at this support b. Then if you recall what we had done that we release this support and this became our active force diagram or a f d and this was replaced by reaction at b. This was p and this was q and what we did is that that we did we provided this a small rotation of delta theta about point a and then what we saw is that that because this rotation was small. And also because of from whatever discussions we had done yesterday about the kinematics of rigid body we said that if this length is l by 2 and this is l by 2 then this vertical displacement at point b is simply equal to l delta theta and the vertical displacement at the midpoint where the load p is applied is simply l by 2 into delta theta.  You will see that if the rotation is large then this assumption about the displacement of this point perpendicular to this line and the magnitude means l delta theta will no longer be true. If this and exerts as I leave it to you you give it a very large rotation of theta then what will happen is that all these points even the point where the horizontal force is applied will get a reasonable amount of displacement in the horizontal direction. And now if you try to apply the principle of virtual world you will see first of all that when the rotation is large this will be arc of a circle this was point p. Now this point p actually moved here. So, rather than moving just upwards it also had a simultaneously a horizontal direction. So, what will happen is that when we use principle of virtual world this load which is applied here q will also do virtual world. Similarly, the virtual displacement of this point will no longer be equal to l times delta theta it will be equal to l sin theta. And now you try to solve this problem you will realize is that you will get complete garbage answers. And there is a very deep reason for that it is that infinite symbol rotation is a vector and you can then express everything in terms of that vector. Whereas, a finite rotation is a tensor and then everything will go wire go hay wire if you use that. So, in the principle of virtual world all the displacements and particularly rotations should be infinite symbol or very small and small means small compared to the overall dimensions of the structure. That is another question why principle of transmissibility cannot be applied on a truss problem. What we do I think can any of the remote centers could you please answer to this question why principle of transmissibility cannot be applied on a truss problem. Any of the remote centers can you physically explain then raise your hand please you can also try to respond to your colleagues. If you raise your hand raise means just say yes on the chat then I am going to go to that center is that center is ok. Let us see also you would like to hear from you as well what is your thought. Yes go principle of transmissibility is applicable only for mechanics not for some not for internal analysis purpose. If you take a rod you take the forces external forces on both ends that rod is under suppose compression. So, if you use principle of transmissibility you use those forces and drag them on the other end what you see as a tension sorry compression will become tension or if it is under compression if you drag them on the other end that will be external vice versa it will change. So, principle of transmissibility is only applicable for external analysis purpose and not for internal deformation or internal forces. Yes so that was a so the main point is that so physically what it is if you treat the entire truss as a rigid body ok. Then principle of transmissibility can be used to get the reactions let us say I want to solve for the reactions by treating the entire truss as a rigid body, but when it comes to the internal members I cannot put that force external force internal analysis when I am performing then I cannot take it to the individual member of the truss and try to solve for the unknown forces. So, everyone listen to the answer so we also physically explain that therefore principle of transmissibility cannot be used ok for truss analysis when we are looking at the internal members right trying to solve for the forces. There are again questions coming in the form of in case of a truss ok. Why do we consider pin joint in truss while in real structure we provide weld or bolt joint which is fixed. Now remember we have explained this and the point is that there are two simultaneous things happening one is that all the members are connected at a joint and they are concurrent ok that is the first thing. Second thing load is also applied at a joint ok. The other issues are of course that we are not applying any lateral load on the members which are very slender in nature. Now again it is a very much it is a assumption that we assume it pin the reason being we want to say that since the load is applied at joint and we have concurrent force system coming into play therefore what happens basically we do not really have any movement in the joint although in real sense there could be some movement in the joint but joints are you know always have a gusset plate right and they are able to take this movement. So it is again a very you know rough assumption why we are assuming the joint to be pin. Remember the majority will be carried by the axial force the major force that we see is the axial force I am not arguing that there is no bending you know part to it there is a bending ok but if you think of the energy point of view most of the energy will be taken by the axial energy and bending energy will be very small. So but concept really is that load is applied at the joint and we have concurrent force system and therefore at the joint if you look at the equilibrium point of view there is no movement ok although you may have bolted or fixed connection over there. Remember there is no lateral load that is being applied at the member ok. Last but not the least since we have covered the structural dynamics and those who want to take this forward let us say I have not discussed this but you know for both civil engineering applications and let us say for mechanical engineering application and so on I will just give two different problems ok and that can be you know you can take this forward what I would urge all of you for ground excitation ok. So I am just keeping two problems ok. Suppose very basic problem in civil engineering SDOF system I have a lumped mass here two columns, columns can be taken as springs so that means you have stiffness for this column ok. Suppose I excite it UGT that is the ground displacement that is equals to let us say it has some form that is you know ground acceleration is coming in the form of let us say U0 omega square sin omega bar t that is one problem. So in other words you do have a ground acceleration in this case ok. The other problem let us say suppose I have a vehicle ok and that vehicle this wheel it is again you have the stiffness KK ok and you have mass alright. Now suppose it moves at a velocity V and suppose this ground undulation that can be described so therefore you have a periodicity let us say of the undulation is L that length is L and also let us say that ground undulation can be described by this D right and L this vehicle is moving at a constant speed can you not attempt to solve the equations of motion ok. So this is something that will go naturally in the next phase of structural dynamics when we talk about forced vibration. In both cases I do have a forced vibration problem how this force is coming very simple example is earthquake is happening in this case let us say so I have ground acceleration in this case right which has this form again a periodic form. In this case the vehicle is moving at a constant velocity again a undulated surface ok. So therefore I do have ground displacement can be derived right so in some sign omega bar t. So question is what would be the equation of motion for this mass or this system that will be a natural part that we should take forward ok. Now I am not going to really look at how the problem will be solved but remember in this case what is important to realize I again repeat what is important to realize the net displacement of the mass in both cases that has to be decomposed into total you know the two parts the net displacement should be the ground displacement ok plus the relative displacement of the mass with respect to the ground that means one component will be just directly from the ground motion right whatever ground displacement form it has so mass is simply going to be moved by that amount plus you have the relative displacement how that is coming that is relative motion that is due to spring ok. So the spring force that enters that will enter through this ok and then you have the rigid motion that rigid motion of the mass will be determined by the ground motion ok. So I think we have now you know plenty of discussions but there are long list of questions and I do not think we can go through all of this but just to say that we can always take these things forward all the structural dynamics part and so on so forth but ultimately these two problems will also give a very very good understanding of how to go about the equations of motion for these class of problems ok and I think now we can have a tea break.


So I'll be starting with the dot product or what we call as the scalar product. Okay, you've already been using dot product in your physics for calculating the work done by a force, which creates a displacement in a body in a certain direction. Okay, so dot product of two vectors. Okay, first of all, please understand here for finding the dot product, both of them should be vectors. When you multiply a scalar with a vector that is called the scalar multiplication. That is not called scalar product. Are you getting my point? This is a wrong use of word that many people do. Well, let's say you multiply us vector quantity with five, you're doing a scalar multiplication of a with five, you don't say I'm doing a scalar product with five. No, scalar product or dot product is a reserved word that we use for doing such kind of an operation on two vectors, right? So what does this operation actually do? So this operation is basically supposed to give the product of magnitude of one of the vectors. Let's say magnitude of vector A into projection of B in the direction of A, B in the direction of A. Okay, or let me write it in capitals or it's the magnitude of B into, this is your normal into, normal into, into projection of A in the direction of B, of A in the direction of B. Now what is magnitude everybody knows, but what is projection of a vector in the direction of some other vector? That is very important. So first let me explain that, then I will come back to this expression once again. Yes, yes, Aditya will take those questions, don't worry, vector is not going to get over very soon. It has got a lot of concepts into it. And while we are doing other topics in vectors, we'll be revisiting our concept of addition of vectors, etc. Pakul will talk about that. Okay, now what is the projection of, let me just take in general here. What is the meaning of projection of a vector, let's say P in direction of Q, in direction of Q. Okay. So it just means that let's say this is my vector P. Okay, this is a vector P. Okay, and there's another vector. Let me just make it in yellow color. Let's say this is another vector. This is your vector Q. Okay. I want to know the projection of P in the direction of Q. So what does it mean? It means that if I start dropping perpendicular from every point of P onto Q. Okay. So I'll just, you know, I'll just go to the last point. Okay. This length, which I'm showing with white here, this length is called the projection of P on Q. Okay, it's a scalar number. It's a scalar quantity. So projection is a scalar quantity. Okay. So this is nothing but as if you have lit a light source over here and that light source is casting image of this vector P onto Q. Okay. So this is called the projection of P in direction of Q. In direction of Q. Is that idea clear? Okay. So from every point, if you drop perpendicular onto the other vector Q here, so that lines that will be created like an image or a shadow that will be called the projection. Okay. Now coming back to this. What is the projection of B in the direction of A and what is the product with the magnitude of A? That is what is your dot product? So let's take a diagram over here. Let's say there is a vector A. Okay. And there is another vector B. Okay. I'll just make them co-initial. Need to not be co-initial, by the way. So let's say this is vector A. This is vector B. Okay. So let's say this angle is theta. I hope everybody knows that the angle between two vectors is the shortest angle between them when the vectors have been made co-initial. Correct. So if I start dropping perpendicular, now please note I have made them co-initial. So there's no point dropping a perpendicular from that point to itself. Okay. So this length, this length, which I'm showing with a zigzag, that length is called the direction of B on A. So what will be this zigzag length? As per the diagram, if this is theta and this length, all of you can figure out that this length is mod of the vector B. So that would be mod B cos theta. Correct.  Okay. In a similar way, if I say what is the projection of A on B? So what you will do, you will just extend this further. That means you will just show the support of that vector. So I hope everybody knows what's the support of a vector. So support of a vector is the infinite line from which that vector has been carved out. Okay. And you drop again perpendicular from every point of this vector A on to B, then that length will be called as the projection of B, sorry projection of A in the direction of B. And that length would be mod A cos theta. As you can see, this angle will be theta. This is your hypotenuse and the base will be mod A cos theta. So as per the second formula, the expression for the dot product would be mod B mod A cos theta. That doesn't make any difference. So either of the two is going to give you the answer. So whether you take the projection of the force in the direction of the displacement and multiply it with the magnitude of the displacement or you take the projection of the displacement in the direction of the force and multiply it with the magnitude of the force, both of them ultimately leads to the work done by the force. Is it fine? So this dot product was required from physical perspective. This entire scenario was given a small name as a dot product. So that when something is used very commonly, we give them a abbreviation. So when they say dot product, it is magnitude of A into magnitude of the other vector in the direction of that given vector. Is it fine? Any question with respect to what is dot product of two vectors? Any questions? Clear. Okay, so let's take some properties of the dot product. I have already, I think, kept a snapshot so that we don't have to write too much. We'll take this snapshot. Okay. Can you all see the screen? Can you all read this or is it too small? Is it legible? Readable, no? Okay. I have kept it. I have taken a snapshot because I wanted to save time. Okay. I normally love to write, but I thought that most of you would be asking me to leave you early. So, okay. So first thing is dot product is commutative. So as you can see, A dot B, B dot A, there is no difference between it. You're only aware of it. So it's more or less like a revision of whatever you have learned in physics. So dot product is commutative. Dot product is distributive over addition and subtraction.  You can change the order also in which it is written. Okay. So it is called less left distributivity because your vector that you are taking is on the left of B plus C. Similarly, you can also have B plus C dot A. So it'll give you B dot A plus C dot A and both of them will be equal. Hence distributive property will hold true. If A and B are perpendicular, now this is something which needs a small discussion here. See A dot B, if it is positive and A and B are non-sensitive, A and B are non-zero vectors. Okay. So let's say A and B are non-zero vectors. Non-zero vectors mean they're not zero vectors. And if A dot B is positive, it means the angle between A and B, the angle between A and B is acute. Okay. Angle between A and B is acute. And if A dot B is zero, that means the angle between A and B, A and B, let me write it like this, A and B is 90 degrees, means they are perpendicular. Okay. And if A dot B is negative, if A dot B is negative, that means the angle between A and B is obtuse. Okay. That means if the work done by a force on a particle is negative, it means the displacement of the body with respect to the force is in the opposite direction.  So I applied a force, but the displacement was opposite to the direction of a force. So there's a negative work done by me. In fact, the wall did a work on me. Okay. So in this case, the dot product would be negative. Okay. So that is what has been explained over here. So that is why if A dot B is zero, it means the two vectors are perpendicular. So dot product is one of the safest means to find the angle between two vectors also. So by the definition of the dot product, we know that dot product is mod A mod B cos theta. So cos theta is mod A, sorry, A dot B by mod A mod B. Okay. So normally to find the angle between two vectors, we use the dot product. Okay. Between two vectors. Now many people say, sir, we could find angle also by using cross product, isn't it? Because you have already done it in physics that mod of A cross B is mod A mod B sin theta. But in that case, there will be an issue. The reason the reason is if there is an angle which is lying in a first quadrant and second quadrant, which are supplementary of each other, sin theta gives the same value for both of them. For example, sin 30, sin 150, same thing. Sin 45, sin 135, same value. So if the angle between two vector is let's say 30 degree and if you use your cross product in cross product, you will get sin theta as a half. So you will not come to know whether it is actually 30 degree or 150 degree. But cost theta will actually help you make that distinction whether it is 30 or 150 because cost 30 will be positive. Let's say positive root three by two and cost 150 will be negative root three by two. So the change in the sign will be indicating you that there is that angle is obtuse in nature. Therefore, dot product is preferred for finding the angle between two vectors, not the cross product. Are you getting my point? So this could be a question to you, Hind. This could be a question to you in the KVP by interview kind of a thing. Why dot product is preferred, not cross product. Okay. So if two vectors are perpendicular, their dot product is zero. So I dot J, J dot I, J dot K, K dot J, K dot I, I dot K they're all zero. So this is very obvious to everybody. Okay. This is one of the most important properties I have seen in dot product. A dot A is mod A square, even though it seems to be very trivial, but one of the very important mechanism to solve many questions related to dot product. So I dot I, J dot J, K dot K, they will all be one. So very useful. These two are very useful when you are finding the dot product of two vectors, which have been resolved into I, J, K components. I think Kiran sir would have also covered resolution of vectors in two-dimension, three-dimension. Okay. So when you write your vectors in terms of I, J and K, and then you are, you know, taking a dot product of two vectors, then these relations will always be helpful to you in finding your end result. Some properties, which are slightly trivial ones. If you have multiplied a vector with a scalar quantity, that means scale, but vector is, you know, multiply, scalar multiplied with M, then the dot product with B is as good as M into A dot B, or A dot MB doesn't make a difference. So the presence of a scalar quantity along with a vector does not influence much. It can always be taken out and the dot product can be performed and multiplied back with the scalar. Okay. So these two properties are quite trivial. I think this is also very trivial. Yeah. So these all properties that you see, you'll keep on using these, you know, in your bigger problem solving. For example, if somebody says mod A plus B square, let me just take a space over here. Mod A plus B square. So if you see mod A plus B square, if you recall that mod A square is A dot A. So you can write this as A plus B dot A plus B. Okay. So this is going to give you A dot A, which is mod A square B dot B, which is mod B square. And you'll end up getting A dot B, B dot A, A dot B, B dot A are equal to each other because as you know, they are commutative. So this gives you this particular expression. Okay. To A dot B. Okay. This is going to be useful in proving certain type of questions, which we'll be taking a little later on. Okay. Similarly, you can also prove all these things. A, in fact, there is a small sign missing here. Yeah. So A minus B. Many, many times they write just a square. A square is as good as saying A dot A, and it is as good as saying mod A square. So here, the term which I have written, this could also be written sometimes as a square by certain books. Don't get confused. A square just means A dot A, or it just means mod A square. So all these are same equivalent expressions. Okay. So all these things can be very easily proved. Okay. Can somebody prove that if mod A plus B is mod A plus mod B, how is A parallel to B? Can somebody prove this? If mod A plus B is mod A mod B, prove that A is parallel to B. Very simple question. Just write it done if you're done. Done. Anusha is done. Okay. So prove this as of now and prove the F1 also as of now. Done. So to prove D, very simple, you just square both the sides. So mod A plus B square will be mod A mod B square. So this is just now we have written the result of it. Mod A square, mod B square to A dot B. And this is also mod A square, mod B square to mod A mod B. Okay. So a lot of things will get cancelled off. Okay. So you'll end up getting A dot B is equal to mod A mod B. Clearly meaning that mod A mod B cos theta is equal to mod B. So they will get cancelled. So cos theta is a one. Okay. So theta is zero degree, which means the two vectors are parallel to each other. The two vectors are parallel to each other. F1 also done. F1 also done. Do I need to do F1 or should I move on to the questions? We can move on. Okay. Okay. Yeah. So let me begin with this question. If A, B, C are mutually perpendicular vectors of equal magnitude, find the angle between the vectors A and A plus B plus C. Okay. Okay. I'm getting Anusha's answer. Both have given different answers. However. Okay. If you see this situation, compare the situation as if there is a cube. At least they like a cube. Okay. So they have given that ABC are of equal magnitudes and not only that, they are perpendicular to each other also. Okay. So you can take this vector to be A. You can take this vector to be B and you can take this vector to be C. Okay. Yes or no. So A plus B plus C will basically be a vector which is connecting this end to this diagonal. Correct. So this will be A plus B plus C. The question is, what is the angle between this diagonal and the vector A? Okay. Now, if you see, you actually don't need this ABC. You can deal with coordinates as well. Okay. So think as if this is a cube of unit dimension. Okay. So this point is one, one, one. Correct. So what is the angle made by, let's say O and I'll call this point as P. What is the angle made by OP with, let's say the positive, this is your X axis, positive X axis. Okay. In short, they're asking you, they're asking you the angle between two vectors, which is one is IJK. Okay. And the other vector is I what is the angle between these two vectors? Okay. So what is the angle? So cos theta is dot product of these two. I'm writing it initially because this is the first class on this topic divided by their respective magnitude. Okay. So this will become one and magnitude of this is root three magnitude of this is one. So theta is going to be, I'm so sorry. Theta is going to be cos inverse of one by root three. Theta is going to be cos inverse of one by root three. Now, let us say I don't want to use this cube and all. Then how do I solve this question? Okay. Even that is not difficult actually. So if you see A and A plus B plus C, if you want to find out the angle, let's say the angle is cos of, let's say the angle is theta. So cos theta will be dot product of these two by magnitude of these two. Correct? Now to find the magnitude of A plus B plus C, this part. Okay. Let me do a small activity here. This square will be what? If you see this square will be A plus B plus C dot A plus B plus C. Okay. So that is going to give you mod A square, mod B square, mod C square and the dot product of every other vector with the other one will be zero. So ideally it should be 2A dot B. Let me write it down. 2A dot B, 2B dot C and 2B dot B dot C plus 2C dot A. And they will all start becoming zero, zero, zero because they are mutually perpendicular. Okay. So let's say each one of them is lambda. So mod of A, mod of B, mod of C, each one of them is let's say lambda. This is going to be three lambda square. That is nothing but mod A plus B plus C is going to be a root three lambda. So here A dot A will be lambda square. This is lambda and this is root three lambda.  So theta would be cos inverse of one by root three. Easy. This way also could be used to figure it out. Okay. So one important thing which I forgot to tell you in the properties I should have written it. Okay. One important inequality is basically related to the fact that cos theta is equal to A dot B by mod A, mod B. Okay. So all of you please pay attention here. Cos theta is a quantity less than one, less than equal to one. So A dot B will be less than equal to mod A, mod B. Okay. So many times we write this expression like this. A dot B square is less than equal to mod A square, mod B square. Okay.  Okay.  Okay. Then A dot B is nothing but summation AIBI whole square, summation or you can just write it as summation A1B1 whole square. Am I right? So left hand side is this and right hand side is summation, summation A square, summation B1 square. Okay. So this is a very important inequality in the field of mathematics. I think people who are prepared for RMO, PRMM would already be knowing that. What is the name? What is the name? Cushy scores inequality. Please note it down. Some questions in JEE has also been asked on the same. Is it fine? Note it down. So many times there are some questions which you will not even realize that oh it is based on Cushy scores. Okay. And that will be helpful. So how did you get the formula? I didn't understand which formula. This formula. Cushy scores. See cos theta is less than one. No. Less than equal to one. So I took this to the other side. Just focus on this part. Correct. A square both the same or for that matter you can do it here itself. Cos square theta is this square, this square and that will be also less than one. Oh, how did I get this? Okay. A dot B will be what Anusha? What will be A dot B? Have you done dot product of two vectors dissolved into two component physics? What do you do? You just literally take their dot products. Okay. Should I do it and show you because I was assuming that you have done it n number of times in physics. So you already know it. Correct. No. So when you take a dot product of this, you only end up getting A1 B1, A2 B2 and A3 B3. Correct. So can I say A dot B square will be square of this term. This term is abbreviated like summation A1 B1. So this is called summation A1 B1. Okay. If you want, I can write it in more detail. Let me write it in more detail if you want me to that. So it just says A1 B1, A2 B2, A3 B3. Whole square is always less than equal to A1 square, A2 square, A3 square times B1 square, B2 square, B3 square. Okay. This is the cushy scores inequality. T2 inequality is many inequalities are interrelated. T2 and there is something called rearrangement inequality. They were all, they're all related to each other. Okay. Fine. So let's have another one now. Let's have another one now. Okay. Let's take a simple question, which really would have done in physics also using victim, using victim method, prove that in any triangle. A square is equal to B square plus C square minus two BC Cossack famously called as the cosine law. How do you prove this using vectors? Yeah. Done. Okay. So fine. So let's, let's do a simple activity here. Let me make a triangle. Okay. And you can assign any vector, any direction vector to this. Okay. So let's say I have ABC. Okay. Let's call this as vector A. Okay. Let's call this as vector B. And let's call this as vector C. By the way, the choice of the direction is completely arbitrary. Okay. And put any direction you want. Okay. Now in the present way, I have shown it. You can see that A plus B plus C is a null vector because they are cyclically closed. Okay. So from here, I can see B plus C is negative way. Okay. Let's do one thing. Let's write a mod on both the sites. So mod of negative way. And that is as good as mod A doesn't make any difference because if you multiply a vector with a negative one, it just changes the direction. It doesn't change the magnitude. And let's say square both the sides. So when you square both the sides, it's as good as saying you're taking a dot product of B plus C with itself. And this is A dot A. So this is mod B square, mod C square plus two B dot C. And this is mod A square. Now remember, mod A, mod B, mod C are nothing but the side lengths. Okay. This is A. Mod B is nothing but B. Mod C is nothing but C. Okay. So this is B square. This is C square. And this is two mod A, mod B, cause of the angle between B and C, which I'm leaving as of now blank. Okay. Because I would like to ask you what is the angle between vector B and vector C? What is the angle between vector B and vector C? No. Okay. It's pi minus A. Please note, angle between two vectors is defined when they have been made co-initial. So this is the angle between vector B and C, which is 180 degree minus A. Okay. So in this gap, I'll have to put 180 degree minus A. Don't write A. A is not the angle between B and C. Okay. It's pi minus A. So this will become B square plus C square minus two B C cause A is equal to A square. And there you go. This is the proof that we wanted to do. Is it clear? Maybe ask in school exam or however. Okay. So please be careful about writing this proof. We'll take the next question immediately without much waste of time. Let's take another one. Can you prove the projection formula? A is equal to B cost C plus C cost B. Just say it done if you're done. Done. Done. Okay. We'll take the same figure. Okay. So we can do one simple activity here. We'll take A as negative B plus C. Okay. And then we'll take a dot product. Take a dot product with A on both the sides. Okay. On both sides. Okay. So what will happen? This will become A dot A. This will become minus A dot B. This will become minus A dot C. Okay. So A dot A is A square because it is mod A square. So mod A is A actually. Okay. Now this is going to be minus A B cost of the angle between A and B. So A and B. If you see the cost of the angle is cost 180 degree minus C. Right. Because this is the angle between A and B. Which is 180 degree minus C. Okay. So angle between two vectors is always the smallest angle between them when they have been made co-initial. Similarly, this will be minus A C cost 180 degree minus B. Okay. This is the angle between. So this is going to be A square minus A B cost 180 minus C is minus cost C. So it will become A B cost C. And this will become AC cost B. Drop the factor of A. So A becomes B cost C plus C cost B. Done. Any questions? Oh, sorry. I didn't see your message. Anusha. Clear? Anyways, I mean, there's so many ways to do it. Express A as some of two vectors such that one is parallel to this vector and the other is perpendicular to the same vector. Yes. Okay. Siddish. Anybody else? See, these are all just warmer problems. So we are yet to start with the, you know, typical questions that are asked on this topic. 30 seconds. Okay. Okay. Kinshuk. Okay. So since alpha is parallel to B, I can say alpha is lambda three I plus K. Can I say that lambda being some scalar quantity.  In short, it becomes five minus three lambda I and it will become plus J minus two plus. Sorry. Minus two minus lambda. Okay. J will not be affected. Sorry. J will be not affected. So this will be K naught J. Okay. So J will live in minus two J and this will become. Okay. Now this vector is perpendicular to B. So that means beta dot B should be zero because beta is perpendicular to B. That means this start product with B vector. So B vector is three I plus K. So this dot product with three I plus K. So I'm just writing it directly the result. So three into five minus three lambda plus one into this. So minus two minus lambda. This should be equal to zero. In other words, 13 and this is going to be minus 10. Am I right? So lambda is going to be 13 by 10. Correct me if I'm wrong. Correct me if I missed out any information. Do let me know. Oh, once again. Oh, that is five lambda. This is five minus lambda. Yeah. Sorry about that. So that's going to be 15, 20 equal to 10 lambda. So lambda is two. Is it fine? Everybody's getting that. Okay. Put it back over here. So when you put it back, your alpha becomes six I plus two K. Okay. And beta is going to be five I minus two J plus five K minus minus three I plus K into two or six times this, whatever. Okay. So that will give you minus I minus two J and plus three K. That's going to be your beta. Any questions, any concerns with respect to alpha and beta value. This value and this value. Okay. Good enough. Next, please copy down. If you want to copy anything or ask anything, please do let me know. Right. So we'll move on. In a tetrahedron way BC, the edges are off length way and BC are equal to age OBN AC are equal to beach. OC and AB are C each if G1 G2 G3. Sorry, G1 and G2 be the centroid of ABC and AC. And OG1 is perpendicular to BG2. Then find the value of a square plus B, a square plus C square by B square. I'll put the poll also on for this. This length is a BC length is. Okay. So you can call this length as position vector the magnitude of. You can say magnitude of these. This is B. This is C. All right. So I've got four responses so far. Last minute I can give you solve most three and a half minutes. Okay. Five, four, three, two, one. So I could see just 14 of you have responded to this question. Most of you have said option number B. Okay. Which option did you choose? You're there. Okay. You chose B. Okay. That's the safe answer to say. Okay. First of all, if you realize that I'm calling this vector as a that means position vector of a is a position vector of B is B position vector of C is C. Okay. They mentioned that G1 is the centroid of ABC. So position vector of G1 will be. Position vector of ABC by three. Okay. I hope you have done the concept of position vector. Just like we do it for coordinates. The same role has been trained by position vectors in vector. Okay. So just like when you are finding the centroid of a triangle, the vertices are specified. What do you do? You add all the X coordinates divided by three. You add all the Y coordinates divided by three. The same thing is done in one shot by vectors. If you know the position vector of the vertices. Okay. Okay. Let me also call. Let me also ask you what is the position vector of G2. That means OG2. OG2 is the centroid of the triangle AOC. Okay. AOC or OAC, whatever you call it. The position vector of the centroid will be A plus C by three. Correct. They're asking BG2 vector. BG2 is position vector of G2 minus position vector of B. Isn't it? So it's A plus C by three minus B. That's nothing but A plus C minus three B by three. Now as for the given question, these two are perpendicular. Okay. These two guys are perpendicular. So if these two are perpendicular, I can say A plus B plus C by three. Dot A minus three B plus C by three. This should be zero. Okay. That clearly means I'm just doing fast forward here. That clearly means A square minus three B square. Plus C square.  So minus two A dot B, if I'm not mistaken. And it'll give you A dot C, two A dot C. Correct. And B dot C will be minus three B dot C, minus two B dot C. Okay. So this will give you zero. Correct. Now, I have still not used, let me call it as one for the timing. I've still not used the fact that OA is equal to or mod OA is equal to mod BC. And that is equal to A. I've not used this three information yet. How did you go? OG2. OG2 is the position vector of G2. That is the centroid of the triangle. Centroid of the triangle is what? Centroid of the triangle is sum of all the position vectors of the vertices D. So O is zero because O is the reference vector for me. Right. So zero plus A plus C by three. Got it, Ruchita. Okay. So yes. So I've still not used the fact that BC is equal to A. Now BC is what? Position vector of C minus position vector of B. This is equal to A. This is given to me. Similarly, AC is also B. So C minus A is given to me as B. Okay. Let's call it as two. And AB is given to me as C. So mod of B minus A is C. Is it fine? Any questions here? Okay. So let's square here both the sides. So if you square here both the sides. Okay. What does it give us? C squared plus B squared minus two B dot C. Is equal to A squared. Right. So at least we now know. At least we now know that negative two B dot C is a square minus. B squared minus C squared. So I could use it over here. You see here minus two B dot C is here. So I'll use that over over here. Okay. So let me let me just box it for the time being. Let's use the second piece of information. So if I square this also, I'll get C squared plus a square minus two A dot C is equal to B squared. So two A dot C is going to be a square minus B squared plus C squared. So this is another piece of important information for me because I can use it over here. Is it not? Okay. Next, let's use the third piece of information. This gives me B squared plus a square minus two A dot B is equal to C squared. In short, I got minus two A dot B as a negative a square negative B squared plus C squared. So this is another piece of information useful for us. Let's put these three square boxed information in place of these gentlemen over here in one. So when I do that. I'll right here itself. So it gets a square minus three B square plus C square minus minus two A dot B will be minus two A dot B will be minus a square minus B square plus C squared two A dot C will be a square minus B squared plus C squared and minus two B dot C will be a square minus B square minus C squared. This is given to me as zero. So now let us start collecting our a square terms. So how many a square terms will be there? This will get cancelled. So two a square term will be left. Okay. How many b squares minus four b square minus six b square. Oh, by the way, cancellation doesn't mean, cancellation doesn't mean I'm actually canceling it. Cancellation just means I'm taking care of those terms. So here, here also gone and this is two c square. Okay. In short, I can see that a square plus c square is three b square. That means a square plus c square by b square is going to be three. Is going to be three. So the answer to this question is option number b. Is it fine? Any question? Anybody who figured out a shorter way to do it, other than this, do let me know. Angle between the vectors, same. Which vectors? a, b, c. Are you finding it the same? I mean, is there any information which says it is the same? Or did you figure out it is the same? Because I don't see such kind of thing. No, there's no such information that tells us. Can we, sorry, can we go to the next question? Let's take another one. All is on four vectors a, b, c and x. Satisfy the relation a dot x. Now this is scalar multiplication. See a dot x times b. Scalar multiplication is equal to c plus x. b dot a is not equal to one. Find the value of x in terms of a, b and c. So which of the following expression is your expression for the vector x, which you need. Why magnitude? Full vector. What without, without, without direction. What is the meaning of a vector? No, no meaning of vector without direction. X vector. How is it a scalar multiplication? A dot x is a scalar quantity. So it's like lambda. Ruchita, we can go back to that question after this. Really. Okay. I can give you 30 more seconds because I've already given you four and a half minutes, but I could see only eight of you have answered this question. That's just 17% of the population attending the class. Guys, have you done vectors in physics? Have you done vectors in school also, right? Okay. I was just taking the confirmation. School year not done. What about Raja Ji Nagar and Kaur Mangala? Not done. Kaur Mangala it isn't. Okay. See, there is a, okay, let me stop the poll first of all. If you want to vote, you can vote it. Five, four, three, two, one, go. Okay. There's 17 of you seem to be active. So most of you have gone with option number B. It's not because you solved it and got B. It's because B seems to be the safest option to choose. I don't know why. Or B for Bajrang Bali or something like that. Okay. B seems to be the favorite of people who guess. See it. There is already A plus X, A dot X over here and there is an X sitting over here. So what I'll do here is I will take dot product with A on both the sides. Okay. So I'll take dot product with A on both the sides. Okay. So when I do that, I got, I get A dot X times A dot B. Please note that this is just as two skill and numbers multiplied. And this is A dot C, A dot X. Okay. Bring this number to the other side. You'll end up getting A dot X. And you get A dot B minus one equal to A dot C. Correct. So A dot X becomes A dot C upon. Okay. Now put this back. Put this back in the given expression itself, which was provided to you. So A dot X into B. Instead of that, I will write A dot C by A dot B minus one. Now, do you now understand why they had given this information that A dot B is not one. Because now you're getting A dot B minus one in the denominator. So everything provided to you in the question is provided for a reason. Right. So this times B vector is C plus X. And there you go. You can easily make X the subject of the formula. So it's A dot C by A dot B minus one times B minus C. Okay. So does it match with anything? And yes, option eight matches. Why not? If you take the LCM, it matches with option A. So it's A dot C times B minus C times A dot B minus one. Okay. So it matches with option, option, option, option A. That means majority people got it wrong because of the guesswork. Only four people got it right. Yes. Put X perpendicular to A. X perpendicular to A. Okay. If X is perpendicular to A, X becomes negative C. Okay. And you put A dot B. Okay. And you take the dot product and see which option is giving you. Okay. Take a dot product with C and see which option is satisfying that. Good. That's, that's a great idea. Is it fine? By the way, later on, we'll talk about this aspect little more. This is basically called solving for a vector. So it is a kind of a vector equation. So we'll talk more about it under vector equation. See what he did was he took A dot X to be zero. Okay. He got X as negative C. Correct. And with that, he took a dot product. So, so see if you see this. All of these options are representing X, right? So here if you take dot product with C. Sorry, dot product with A and put zero and put your. Put all the, all your A dot C is also a zero because for him, X has come out to be negative C. Okay. So he checked which of them gave zero only option one gave zero. Yeah. A dot C will be zero, right? So that will give you only negative. Yeah. That also you don't need to do. Correct. So he's saying that if A dot C is zero, this will go for a toss. You don't have to do this also. So A dot C will be zero. This will become a zero and this, this term and this term will get cancelled giving you X as minus C, which is what he had actually got by assuming X is perpendicular to it. Right. So he has applied a lot of mind in doing scamming in this. Okay. So that is coming early and he proudly claims that he has done it.  You're, I just took a dot product with A on both the sites. So this vector equation. I took a dot product with A on both the sites. Take dot product. With A on both the sites. Okay. So when you take the dot product, remember this is a scalar quantity. The quantity is taken out. Remember, there was a property in that property list, which I told you M A dot B is same as M A dot B. The same thing I did over here. And I brought this A dot X to the left side. I took that as common. So it'll become a dot B minus one. Okay. So I got eight A dot X like this and I put this back in this given expression, the actual expression given to you in the question. Yes. I was trying to remove that A dot X thing. I was just trying to get only X in that expression so that I could easily make that as the subject of my formula. Oh, sorry. I think Ruchita wanted me to discuss something in the previous. Yes, Ruchita. How did I get one, two and three C one, two and three says beast mod BC is a. So I think you would have been already taught that if you know the position vector of end of two points, let's say this is P Q length. And somebody says this is position vector P. This is position vector of Q. Okay. So P is the position vector of P small Q is the position vector of Q. So what is P Q? P Q is Q minus P. So mod P Q will be what mod of this. That's exactly what I did when I was using the fact BC. Okay. So BC vector mod is a or length of BC vector is a. So BC vector is C minus B mod equal to it. And this is what I squared and I got this. From this part, it is clear, right? I think you only wanted me to tell about one, two and three. All right. So let's take this question. If you put the poll, think how that for P Q would have come and why they have squared the vector a thing something and there is an inequality also. So there are a lot of hints that question is giving you. You just have to grab that. Okay. Two minutes. Two people have responded. Okay. So we'll close in another 15, 20 seconds. Let's say half a minute. Five, four, two, one. So most of you have said option number D. Let's check. See, for P Q term, et cetera, they start coming when you use this expression, something like this, or you can say a famous inequality that you have been using since beginning. So can I say this term that is P X minus Q Y.  P Q are constant. Yeah. P Q X dot Y. Can I write it like this? Correct me if I'm wrong. Okay. So now we already know that this should be greater than zero because it is more of a, you know, vector and you're squaring it also. So it is greater than equal to zero. Okay. So this term is an A. This term is a vector. So this is more day square. Correct. So more day square minus four P Q X dot Y is greater than equal to zero. That means more day square by four P Q. Is greater than X dot Y. Correct. But my question seems to say something else. But my question says X dot Y is greater than equal to more day square by four P Q. Okay. That means there is a contradiction here, which can only be resolved if X dot Y is equal to more days. Sorry. More days squared by four P Q. Correct. So both are saying, you know, things which are looking different from each other. So this contradiction can only be resolved when they are equal to each other. Isn't it? If this is equal to zero, which means this is also equal to zero. Which clearly means that the matter comes all the way till mod P X minus Q Y is equal to zero, which clearly means that this is a zero vector, which means X by Y is equal to P by Q or even mod X by mod Y. Oh, it's P not. Oh, sorry. Yes, yes. It's Q by P Q by P not P by Q by P. Okay. So this will also be Q by P only. That means X and Y vectors are actually collinear vectors. Okay. They're collinear vectors. So the ratio of their lens will be just this proportionality Q by Q by P only. So option number D is going to be right. Any questions? Any questions? Okay. With this, we are now going to move towards, sorry, without mod. They ask for the mod thing. Even with mod, they will be Q by P only. Equation just before this. You're talking about this equation? This equation? Yeah. What is the question? See, you can say X is Q by P times Y. That's what it says. Okay. You can treat this as if you are doing this operation. You can remove this if you want. Okay.  So mod X by mod Y is equal to mod Q by P, but they have given PQ is positive. Isn't it? If PQ is positive, then Q by P will also be positive. Am I right? Yes or no? So this is going to get rid of mod and it will become just Q by P. Now happy? All right. So we are now going to move towards the concept of vector product. Vector product also called cross product. So unlike the scalar quantity, unlike the scalar product, this is going to give you a vector quantity. Okay. A vector quantity normally has a magnitude and has got a direction. Okay. It has magnitude as well as a direction. Okay. So let's figure out what is the magnitude and direction of A cross B. Most of you have already used this concept in your physics to find the moment of a force. Correct. Or current experienced by our force experienced by current carrying conductor. Isn't it? So R cross F torque or moment of a force, it is basically related to your concept of vector product. So magnitude of this vector A cross B. Let me write it like this. This is your magnitude. It's basically nothing but it's the area of the triangle, sorry, area of the parallelogram, which is constructed using A and B as the adjacent sites. So this parallelogram area is given by your modulus of A cross B. Okay. So this area of the parallelogram is what is given by your mod of A cross B. So that is the magnitude part of A cross B. Okay. And what is the direction? Direction is basically obtained by using the right hand thumb rule. What is right hand thumb rule? If you take your right hand and stretch your fingers in the direction of A. Okay. And then curl it naturally towards B. So please curl your fingers naturally towards B. Like this. If you keep it on A, so you keep the edge of your hand on A, curl it naturally towards B. Whatever direction of the thumb you are getting, that will be the direction of, that will be the direction of your A cross B. Okay. So if you call this as let's say eta cap, that will be the direction unit vector along this. So let's say I call this vector as eta cap. So eta cap is the unit vector in the direction of, okay, in the direction of A cross B, where eta cap is obtained by the direction shown by the thumb of your right hand. When you curl your fingers, which we are stretched in the direction of A, naturally towards B. The word naturally is very important. When many people will say, what's that if I stretch it backwards then? No, backward is not a natural curl. Natural curl will come in this way only. I hope you're able to see me on the camera. This is the natural curl of the finger. Is it fine? So this is nothing but this times eta cap. Eta cap is obtained by, so let's say unit vector, this is the unit vector in the direction of, in the direction of A cross B, and this unit vector happens to be perpendicular to the plane containing A and B, okay, and obtained by using right hand thumb rule. Obtained by using, by using the right hand thumb rule. So right hand thumb rule everybody has understood. So perpendicular could be two directions up or down, but which direction will you take? Using the right hand thumb rule, so whichever vector you have written first, first stretch your finger in the direction of that vector, curl it towards that vector which you have written next, which is B. The thumb direction, that perpendicular that is shown by the thumb, that will be, in that direction you take a unit vector, that will be your eta cap, or you can say that is the unit vector along A cross B. Is it fine? Any questions? Any concerns? So both the things are clear, what is the magnitude of A cross B, and what is the direction of A cross B? So magnitude is the area of the parallelogram, which is formed by A and B as the adjacent sites, and eta cap is the direction perpendicular to the plane containing A and B, but there could be two perpendiculars, one up and one down, so which one of the two? So the one which is obtained by your direction of the thumb, when you are curling your finger naturally from the first vector to the second vector, which has been written in the cross-producting. Clear? So B cross A will have the same magnitude, but this time when you're curling your finger from B to A naturally, your thumb is going down the plane. Okay? So that direction will be exactly opposite, same magnitude exactly opposite. And that is why sometimes we say A cross B and B cross A are negatives of each other. In short, cross-product of vectors are not commutative. So we'll be talking about those properties a little later on, but first is everybody clear about what is A cross B, how it is obtained, it's a vector first of all, it has a magnitude and a direction, what is the magnitude that is clear, what is the direction that is clear? Is it fine? Okay. Let's take some set of properties. Sir, always only two directions. No, it has only one direction, why two directions? Who said there'll be two directions? A vector has one direction only. If I understood your question properly. A cross B will be when you stretch your finger along A and curl it towards B, that perpendicular which you get, that is A cross B, that direction. That is why I wrote, it is perpendicular to the plane of A and B and obtained by using the right hand thumb root. Perpendicular to a plane can be up the plane, down the plane also, right? But which of the two will be the answer that is obtained, that is clarified by right hand thumb root. So if I just say perpendicular, you will obviously question me, sir, which way, up way or down way? Perpendicular to the plane containing A and B, where does the question of left and right sit? Tell me, Kinshukh. Let's say, this is the writing part which I used to write. I hope you can see it, okay? There are two vectors on it. Perpendicular either it will be like this or it will be like this. Why not up from B to A? I didn't get your question. B to A means what? Up from B to A means what? If you do B cross A, then it will be curl your finger towards in the direction of B, set your finger in the direction of B and roll it towards A. Then thumb will automatically go down the plane. Then it will go down the plane like this, okay? So that will be a B cross A direction. That is how it has been defined, the cross product. Because we follow a right-handed coordinate system. This is how you generate a vector which is, you can say linearly independent of A and B. In fact, your basic unit vector triad, I, J, K, also follow the same thing. I cross J is K actually, right? So K is a vector which is perpendicular to I and J, but it cannot be perpendicular in a random way. It can be, it is perpendicular following the right-handed coordinate system by default. Now again, if the question setter says, I am following a left-handed coordinate system, then A cross B will go down in this case. That means this will go down. In case the question setter says, we are following a right-handed coordinate system. Got it? Sorry, left-handed coordinate system. Sorry about it. So we follow a right-handed coordinate system. That means when we curl our finger from A to B, thumb direction is the direction of A cross B.  Put your finger along A and curl towards B. Thumb is going down, no? So for a left-handed coordinate system, A cross B will go down. But that is against the convention that we normally follow. You know, in maths and science, we follow some conventions, right? If you redefine that convention, that okay, no. I will take the direction from the pole opposite to the direction of light as positive. Then your lens formula will change. Okay? Similarly, you are all, you know, formula of I cross J equal to K, everything will go for a toss. That will change. Okay? All right. So let's take some properties. And if I am not mistaken, I have the list of properties, I think. Yeah, I have the list. So I have saved your and my time by listing it down. By the way, let me tell you in J advance, they had given one question in one of the cases that, assuming we are following a left-handed coordinate system, then they're asked to simplify certain, some expression. So be aware of that. So left-handed coordinate system, you have to follow the left-hand thumb rule. Okay? Properties of vector product. This is already, I told you that it is not commutative because they are opposite in direction, even though their magnitude is the same. So A cross B, B cross A are oppositely directed. Yes, this is very important. Vector product is not associative. Now this is something which I will talk about in more details in VTP, vector triple product. Okay? So this is to be dealt in vector triple product. So let's not bother too much about it. Vector triple product. So just know for as of now that A cross B cross C is not the same as A cross B bracket cross C. Okay? We'll talk about it later. Just to tell you briefly now, A cross B cross C is a vector in the plane of B and C. While this is a vector in the plane of A and B. Next. Your A, M A cross B is same as M times A cross B. It is same as A cross MB. There's no difference. So the scalar quantity can always be pulled out. And it doesn't matter whether the scalar quantity is with A or with B. The scalar quantity can always be removed separately. Okay? Now see the distribution property of cross product over addition. Please note here you cannot play with the order. You cannot say A cross B plus C is B cross A plus C cross A. That will be wrong. The order has to be maintained. Can't switch the positions. Switching position is possible in dot product because dot product is anyways commutative. But here you cannot do that. Same if you're taking right hand distributivity, please maintain the order B will come first and then A and plus C will come first and then A. Same goes with subtraction also. Cross product of any vector with a null vector is a null vector. So A cross 0 or 0 cross A both will give you a null vector only. Apart from this, there are more properties which I'll pull in. I think this is... If vector product of two non-zero vectors is 0, it means the vectors are collinear. This is very interesting property, very important one also. So if you find that two non-zero vectors, cross product is coming out to be 0, it means the angle between them is either 0 or 180 degree. That is why... See, you're trying to say that this is going to be 0, isn't it? So if A cross B is equal to null, the modulus is 0. And if you write it down, in fact, I will just prove it as of now because I have not written the proof for this that it is mod A mod B sin theta. And if these two are non-zero, which clearly implies this has to be 0, so for that theta has to be either 0 degree or 180 degrees. That means either they will be parallel or anti-parallel. Either they will be parallel or they will be anti-parallel. That means they will be collinear anyhow. That is what this fact number 8 says. So if the cross product of two vector is a null vector, it means the two vectors are collinear vectors. Acharnath, this information I had not given to you. So mod A mod... Sorry, mod of A cross B is mod A mod B sin theta. How do you get this information? Very simple. I already told you that mod of A cross B gives you the area of this parallelogram, isn't it? And let's say this angle is theta. This angle is theta. So area of a parallelogram is what? Area of a parallelogram is base into height. So base is what mod A. So let's call this as a base. Base is mod A. And what is the height? Height is mod B sin theta. So mod B sin theta will act like a height. So this will give you the area of this parallelogram. And that is what we had learned was the magnitude of A cross B. Or mod of A cross B. Please, very, very important thing to be kept in mind. I mean, I feel it is irrelevant to say. But many people, when I ask them what is A cross B, they say mod A mod B sin theta. A cross B is a vector. And you're answering the question of a vector with a scalar quantity. How is it possible? So mod of A cross B is mod A mod B sin theta. Not A cross B. A cross B is a vector. Its magnitude will be mod A mod B sin theta. And it will have some direction also, which is given by the right hand thumb rule, which I already told you. So don't make this mistake. It is a trivial thing to point out. But initially, many people do that mistake. Now see, further more properties. By the virtue of the fact that if two vectors are collinear, their cross product will be a null vector. That is why I cross I, J cross, J and K cross K will be null vectors. Don't say zero. They are null vectors. OK, zero vector, not zero scalar. Zero vector. The ninth point is very important. The vector product of orthonormal triad of unit vectors, I, J, K, basically follow this particular rule. Please note this down. I cross J is K. J cross K is I and K cross I is equal to J. But this is when you are following the right hand thumb rule or right hand coordinate system. If you were following, had you been following a left handed coordinate system, then what would happen? Then J cross I would have been K. K cross J would have been I. And I cross K would have been J. Understood? So that minor change will happen in the way you have written this formula. So I, J and K, they are in such a way, normally if you curl your finger naturally from I to J, it should give you K.  It gives you along I. Similarly, from K to I, K to I, K to I, it will go towards inside the plane of the board, which is the direction of J. No, I, J, K is fixed. That position nobody can change. But if you are following a left handed system, then I cross J. I cross J will go down. That is minus K. Assuming this is your I, J, K, and you're following a left handed coordinate system, then I cross J will be called as a minus K. That means J cross I will become K. Got it, Raghav? Is it fine, any question, any concerns? Okay. Now the 10th point is the Langrange's identity. Okay. Actually this Langrange's identity is a very, very, you can say a special case of Langrange's identity. Langrange's identity is slightly broader. Let me write it first here. So Langrange's identity is actually this. I will prove this later or not right now. When I have done the vector triple product with you. Langrange's identity says, please note this now. We will prove this when I do. We will prove it in vector triple product concept. Okay. Wait for the proof. We will talk about it. This is Langrange's identity. Now what this fellow has given here, he has actually taken C and D as A and B itself. Okay. So as a result, what has happened in the ninth property, they have, they have taken in the ninth property that you see, they have taken C as your A and B as your D. Or you can say D as your B. Let's be consistent while writing. Yeah.   Then this becomes A dot B again. Okay. If I'm not mistaken. And this becomes, this becomes B, B dot A. Have I missed out something? A dot D. Yeah. Correct. And this is also B is B and D is, oh, sorry, A dot A it will become, sorry, A dot A. Yeah. And this will become B dot B. Yeah. So when you do that, it will become automatically mod A square, A dot B, B dot A, mod B square. So if you expand it, it automatically becomes mod A square, mod B square, minus A dot B, the whole square. Okay. This is the expression for A cross B dot A cross B. A cross B dot A cross B is actually mod A cross B, the whole square. This is what they have mentioned here in property number nine. So mod A cross B whole square is mod A square, mod B square, minus A dot B square. And that's what we actually got. So what they have given is a special case of, special case of Langrange's identity. Langrange's identity. Is that fine? This can anyways be proved without Langrange's identity also. So another way of looking at it is mod A cross B is what? Mod A cross B is mod A mod B, into sine of theta, correct? So if you square it, this is what happens. And you may write, you may write sine square as one minus cos square. So if you expand it, it becomes mod A square, mod B square, minus mod A square, mod B square, cos square theta, correct? Which is mod A square, mod B square. And this is nothing but A dot B the whole square. And that is what they had given, right? But to prove this, we need to understand, to prove this, we need to understand vector triple product. We cannot prove it now, as of now, because we have no idea about vector triple product officially. Is this fine? Any question, any concerns related to Langrange's identity? Or any of these properties which I have mentioned? Now how to calculate the cross product of two vectors, when these vectors have already been given to me in a resolved format? This, most of you are already using it in physics. So let's say I want to calculate A cross B, okay? What is the way to do it? Will I literally do the multiplication? Initially, yes, I will do it, but then I will tell you a shorter way to achieve that. So let's say you're doing cross product of this with this. So when you do cross product of A1, I cap with B1, I cap, please note that I cap cross I cap will be zero. So please do not do any kind of a multiplication of a unit vector with itself, cross multiplication, because that will waste your time, because anyways, you're going to get a null vector out of it. So no operations like A1, I cap cross B1, I cap, that will be zero vector. A2, I J cap, B2, J cap. I will not do that because that again, give me a zero vector. Again, A3, K cap and B3, K cap. I will not do that operation. In fact, I will do operation of A1, I cap with B2, J cap. So A1, I cap B2, J cap. What will it give me? A1, B2, K cap, right? So let me do one thing.  So let me make the compartment for them already. Okay. So for K, I will end up getting A1, B2, A1, B2. And please note that when this multiplies with this, you will get A2, B1 minus K cap. So that will give you minus A2, B1, correct? Correct me if I'm wrong, if I missed out anything with respect to how many ways you can get K cap expression. Okay. Now, J cap will come when I and K get multiplied. So A1, B3, A1, B3. Let me write it in white. A1, B3. Okay. And this, this I multiplies with this K. So that will give me minus A3, B1, correct? In fact, other way around. Okay. This will be minus and this will be plus. Sorry. This will be minus and this will be plus. So it will be A3, B1 minus A1, B3. Yeah. So remember I, J, K. So K into I or K cross I is plus J, right? And I cross K is minus J.  Similarly, I cap will be obtained when J and K interact. So A2, B3, but remember A2, B3 will come along with J cross K. J cross K will be I, so it'll be A2, B3. And the other one will be negative A3, B2. Okay. Looking at this, it basically reminds me as if I have expanded a determinant. Whose first row is I, J, K. Second row is A1, A2, A3. Third row is B1, B2, B3. Because if you do this expansion of this determinant, you'll automatically realize that you get I cap A2, B3. Minus A3, B2, which matches with this. Okay. Now I'm writing minus J cap for now, but I will just take the minus inside in some time. So minus J cap. So just hide this row, hide this column A3, A1, B3 minus A3, B1. Okay. Remember, if you put the minus inside, it will become A3, B1 minus A1, B3, like what I have written over here. Okay. Similarly, if you expand the K part, so K times A1, B2 minus A2, B1, which is your this part. Okay. So this and this become actually the same expressions. That is why when we are expanding the cross product of two vectors which have been written in a resolved format, we can always use this determinant to do our work faster. Is it okay? Any questions, any concerns? Time to take some questions on this cross product also. Let's take some questions. Let's start with a very easy question. Maybe something which can come in your school exams also. Okay. Poll is on. If A is any vector, then A cross I square. By the way, A cross I square means A cross I mod square. As I told you, this is another way to write things. Oh, did I ask this in your school exam also? Okay. Maybe I might have asked this. Done. 30 seconds more I can give. Polar. No, people are polling happily. But here also you got it wrong. Okay. Five. Four. Three. Two. One. Go. Okay. Most of you have said option number D, D for the lead. Okay. Let's check. Yeah. Let's say I take a vector as a one I a two J a three K. Okay. If I take the cross product with I. Okay. What I'll get. See when I take cross product with I. I will not happen. So it'll be a two. It will be a two. J cross I. Right. J cross I will be minus K. A three K cross I K cross I will be J. So it will be a three J. Correct. So please be careful while you are choosing the sign. So J cross I is minus K. But K cross I is plus J. Okay. Now eight cross I square means mod of. This square mod of this square is just a two square a three square. Okay. So mod of this square is a two square plus a three square. So similarly, without much waste of time, you can say a cross J. Square. That will be a one square a three square. And a cross K square will be a cross K square will be a one square a two square. So add all the three. If you add all the three, the required expression becomes. Twice of a one square a two square a three square, which is actually twice mod a square. Option number D is correct. How do I know why are you asking me? Who called him? I never call anybody. Why are you asking me that he has got a call? I don't know. School would have called. Who called something you call me after the class. Okay. Okay. Let's do some more questions. Let's take. Okay, let's take some easier ones. Okay, let's take this question. Show that the perpendicular distance of a point C from the line joining a and b. Sorry, they have missed out the word the line from line joining a and b. Okay, so basically this equation says that there is. A and B here. Okay. And there's a C point over here. Okay. They're saying that this perpendicular distance. This perpendicular distance is going to be this expression. Okay. Would you like to try this out or should I discuss this? Can we solve this as a cross I plus a plus K by two. A cross. This the whole square. Can shook. Did you say that it is going to be a cross I plus J plus K. Model is the square. Okay, got it. Yeah, please go ahead. Right out. Okay. See all of you please understand this fact. Let me make. Let me go back to the fact that mod A cross B basically represents. The area of a parallelogram whose adjacent sites are in the correct. So this area is given by mod A cross. Correct. Now can I say. If I just half it. It will give me the area of a triangle. Who's to adjacent sites are in the. Yes or no. Can I say this area will be given by half mod A cross B everybody agrees with me on that. Now, if let's say I make a triangle. Whose vertices are given by ABC. That means the position vector of the vertices are ABC. Then what would be the area of the triangle. So can I say area of the triangle would be half mod. Take any two sites and take their cross product. Let's say I take AB and AC and take their cross product. Right. Just like this, this formula. But remember in this formula and be where side lens or the vector representing the sites. Okay. But in this case, ABC are position vector of the vertices. That is the only difference between these two scenarios. Right. So here I've taken the cross product of any two of the vectors representing the site. Take the modulus of that and half the answer. So can I say it's half mod. What is AB vector AB vector is B minus A. AC vector is C minus A. Correct. So let us expand this and see what happens. So this is B cross C. Okay.  This is minus A cross C, which is C cross A. And A cross A will anyways be another vector. Is it not. So in short, the area of the triangle whose vertices are represented by position vector ABC respectively is given by this expression. Am I right. Now come back to this question. If I construct a triangle over here. If I construct a triangle over here. Okay. Let me call this triangle as ABC. Can I say the same situation is holding true as what it was holding when I derived this result. So can I say area of this triangle ABC will be half mod A cross B B cross C C cross A. Correct. But area of the triangle is also given by base into height half base into height. Okay. Which is D by the way this into is not a cross. This into is the normal multiplication into. Okay. Better I would not write it because then it will create this thing. Correct. So ultimately these two results should be the same. So one and two are equal results because both represent the same area of the triangle. So half mod B minus AT is half mod A cross B B cross C C cross A. Okay. Half half gone. Okay. Divide by divide by odd B minus A. And that's it. This is what we wanted to show, isn't it? So this is the distance perpendicular distance of C from the line joining A and B. Of course the order in which they have written I have not written in the same order. I have written A cross B first then B cross C then C cross it doesn't matter vector addition is commutative. Okay. Is this fine? Any question? Any concerns here? Okay. How else did you do it? A lot of or you couldn't do it. Okay. Is this fine? So this this problem basically comes. This result basically comes from your understanding of the geometrical significance of mod of cross product of two vectors. Area of other figures can obviously be found out if you are making you're talking about area of other polygons. You can make them in triangles and you know start working with the same same expression. I will take more problems, but I would now give some small break to you all because it's a continuous session. We'll have a five minutes break. Okay. So let's meet at 616 p.m. Just a five minute break. Okay. Then I'll continue with some more problems on cross product. So just five minutes break. Let's be back in five minutes. Okay.