 So when it comes to three vectors, so now we are entering into triple product up till now we were talking about product of two vectors. So which included your dot product and your cross product. However, I could not do much problems on those but never mind, we'll be coming across a lot of operations related to dot product and cross product even in our triple products also. So today I'll be starting with the concept of product of three vectors. Or triple products. So now let's see if there are three vectors involved a b and see let's say first tell me which of the operations are valid operations in vectors. So I'll give you these three operations. The first operation is a dot b. Okay. And let me put some brackets also a dot b dot c. Okay. a dot b cross c a cross b dot c. And finally a cross b cross c. Okay. Now look at these four operations carefully and tell me which of these operations are valid operations. That means there are some operations here which is like, you know, they do not have any meaning to it. So which of them are valid. You just write it down on the chat box. Okay. Okay. Okay. Okay. Okay. Which of these are valid operations. Which operations seem to be a correct operation. Of course, where ABC are vectors here. All right is the first operation valid. No. The reason being a dot b becomes a scalar quantity. Right. There's nothing like dot product of a scalar with a vector. However, there is a scalar multiplication as I told you yesterday, but there's nothing like scalar dot vector. Okay. So such operations are not known in in vectors is the second operation valid. No. For the same reason, you cannot have a scalar cross with a vector. Okay. You can't say five cross a that will be a absurd, you know, operation to perform in vectors. The third operation is a valid one. And this operation is actually called a scalar triple product. What is called scalar triple product because through this product you get a scalar quantity. Because a cross B is a vector C is a vector dot product of two vectors will lead to a scalar quantity, like what we did in our scalar product. So this gives you a scalar answer. This gives you a, you know, numeric figure, which is just, which is a scalar quantity, not a vector quantity. And that is why we call it as a scalar triple product. This is also a valid operation because you can definitely do cross product of two vectors. A cross B is a vector in itself. C is anyways a vector. So vector cross vector can be done. And the result is a vector quantity. And hence this is called the vector triple product. Okay. VTP and STP. So only third and the fourth operations are valid, not the first and the second. Okay. Anyways, I didn't expect anybody to go wrong in this, especially that you are nearing an end of vector algebra and going to start with 3d geometry maybe whenever we have our next class. Okay. So let me begin with our discussion on scalar triple product. NPS Kormangala, was this taught in school or was it skipped in school? STP or they are deleted it? Okay. Never mind. So vector, sorry, scalar triple product is also called mixed product. Okay. And it's also called box product. Now there's a reason for the name to be given like this, why it is called mixed product, why it is called box product. We will learn later on when we learned about the properties of scalar triple product. Okay. So let us analyze first of all what quantity or what scalar quantity comes out of this operation. Okay. So I'll be assuming for the sake of, you know, keeping things quite generic that ABC are three non-coplanar vectors. That means all the three are not on the same plane. So let's say a is like this. Okay. We just assume this in white. Let's take B to be like this. Okay. I'm assuming this to be in red. Okay. And let's say C is like this. I'll make one in yellow. Okay. I hope you can understand from the way I have sketched this three vectors that they are not on the same plane. They are not on the same plane. Okay. So what does A cross B dot C actually represent? So if you start doing this operation one by one, let us first perform this cross product operation, which is within the brackets. We know that from our cross product that A cross B, A cross B gives you a vector whose magnitude is the area of the parallelogram and direction is eta cap. Eta being perpendicular to the plane containing A and B. So let's say I make eta cap like this. Okay. This is my eta cap. I hope I have shown the correct direction of eta cap. So eta cap is supposedly a unit vector, which is perpendicular to the plane containing A and B. And it is obtained by using the right hand thumb rule. That means you stretch your four fingers in the direction of A and curl those four fingers naturally towards B. Your thumb direction will be along eta cap. Okay. So this gives you area of the parallelogram. So what I'm going to do now, instead of A cross B, I'm going to write area of the parallelogram times eta cap that is a scalar multiplication and this whole thing dot with C. So that area of a parallelogram is a scalar quantity, eta cap and C are vector quantities. So I can do this operation. Okay. All of you please pay attention. Eta cap dot C. What does it mean? Let's say, let's say this is the perpendicular drop from the terminal point of C on to eta cap. Okay. So eta cap dot C will actually will give you this length. So very simple. See, eta cap dot C is going to be mod of eta cap, mod of C, cos of the angle between them. And let's say this angle is a fight. Okay. So in short, it is one into mod C cos five. Mod C is this whole length. Right. Mod is what? Mod is the length of the vector, isn't it? So cos into cos five will give you this length. So this length will become eta cap dot C. Ultimately, if you see here, you get area of the parallelogram into this height. Okay. Let me call this as H. Okay. Now, area of this parallelogram into this height to be very frank, gives you something like this. Okay. So it gives you the volume of a structure. Okay. And I'm just drawing that volume for you all. It gives you volume of this structure. I'm trying my best to make the diagram. Okay. I hope you are able to see it with a lot of effort I'm making it. So kindly appreciate it. Okay. So if you see, there is a three dimensional box, which is made, whose base area is your area of the parallelogram and height is the eta cap dot C. This is the height of the structure. By the way, this structure is called a parallel pipette parallel pipette parallel, parallel pipette. It depends upon you how you pronounce it. So parallel pipette is nothing, but it's a kind of a cuboidal structure, but instead of having the faces as rectangles, it has a face of a generic case of a parallelogram. Okay. So all the face, if you see, they are made up of parallelograms. Okay. So the back face, this is also a parallelogram. Okay. So all the faces are parallelogram in this. Or you can think as if there is a box which has been kicked from one side and it has got, you know, bent like this. Okay. So cuboid is, you can say a cuboid is a special form of a parallel pipette where all the faces are made up of rectangles. So cube, cuboids, et cetera, they are all special forms of a parallel pipette. So this expression that you have found out, it actually gives you, it actually gives you the algebraic, it actually gives you the algebraic volume of the parallel pipette formed by, formed by A, B, C as, as you can see, A, B, C are, you know, the edges of this parallelogram. Okay. So if you may, sorry, parallel pipette. So if you make a parallel pipette using A, B, C as the co-terminus edges, the algebraic volume, algebraic means it could be, it could give you positive or negative. That means if you really, if you really want to find the volume, absolute volume, you have to, you may have to take the modulus of this because it can give you a negative answer also. It can give you a negative answer also. Is this right? So what does A cross B whole bracket dot C represent in geometrical sense? It represents in geometrical sense the algebraic volume of the parallel pipette formed by these three vectors as co-terminus edges of that parallel pipette. Please make a note of this. Then we will learn something more interesting about this. Any questions so far in this diagram or in this explanation? Okay. Now many people ask me, sir, when could this volume be negative? So this volume could be negative if this angle phi was obtuse in the sense that if your C vector was somewhat like this, let's say hypothetically, then the angle between eta cap and C would have been this angle, which would have been obtuse angle. In that case, what happens? The volume starts becoming negative. Okay. The volume starts becoming negative. Now, few things that we would like to discuss over here. Number one, if you see this operation A cross B dot C. Okay. First of all, do you really need to show the position of the brackets? What do you think? Is showing bracket important here? Can bracket be put like this? In that case, this operation will be a invalid operation in vectors. Right. So first of all, the use of bracket is a waste. Right. So we don't normally put a bracket. In fact, we don't need a bracket. Okay. And now if you see, if you see this expression A cross B, let me, let me just write A as a one I a two J a three K. And let me write B as B one I B two J B three K. Okay. So a cross B, we have all learned in our cross product can be obtained by using this vector. Okay. Don't worry. I will not be expanding it, but I would like you to pay attention over here. Okay. So if you do A cross B dot C, let me write it here. Let me write it here. So what does it give you actually? First of all, let's say hypothetically gives you I something. Okay. Then it gives you, you know, let's say J something. Okay. Then K something. Correct. If you take the dot product of this with C. Okay. Then what happens? You're taking the dot product of something like this. Okay. With. I mean, let me write C also here. I have not mentioned CC being, let's say C one. I cap C two J cap C three K cap. So if I do the dot product with C one, I cap C two, J cap C three K cap, you realize that it, your I will get replaced by C one and you'll have the same bracketed term over here, which I'm not writing J will get replaced with C two and you'll have the same bracketed term and K will get replaced with C three and you'll have the same bracketed term that you had over here. Now, whatever is there, I will, I'm not writing it just to save time. Now what does this indicate? This indicates that the same determinant that you had over here, which you expanded for a cross B in that determinant, had you put your I cap J cap K cap with the components of C vector, then it does the job for you to find out the scalar triple product. So instead of I one, sorry, instead of I cap J cap K cap, I have to put C one, C two, C three and rest row will be same. Okay. Let me write it completely here. Okay. So in short, what are you doing here? You're finding an easier way to calculate the expression A cross B dot C by the use of a determinant. Okay. So this is a big, you know, breakthrough for us because we don't have to calculate a cross B separately and take a dot product of that separately with C. This determinant itself would give you the expression for or answer for A cross B dot C. Okay. By the way, this determinant will give you the algebraic volume of the parallel of people algebraic. Please note it could be negative. Also, we have already seen determinants can be negative positive. Okay. So please note this down and the analysis has not ended yet, but I have to go to the next page. That is why I, I'm giving you some time to just note it down. Then I'll move on to the next page. All right. So let's go to the next page. So we just saw that A cross B dot C. Okay. By the way, since dot product is commutative, you can also write it like this. Okay. You can switch the position of C and A cross B like this. So this is something which we just saw was this determinant. Okay. Why I've written it like this is because you can, you know, now easily remember this because the first vector C is on the first row. Second vector A is in the second row. Third vector B is in the third row. Okay. Similarly, can I do something like this? Can I swap the positions of let's say A1, A2, A3 row with C1, C2, C3 row? And again, swap the position of C1, C2, C3 row with B1, B2, B3 row. Okay. So let me do this operation. I'm swapping the position of R1 and R2. When I do that, C1, C2, C3 will come in the second row. A1, A2, A3 will come in the first row. And again, I'm swapping the position of R2 and R3. So I'm doing these two operations. When I do these two operations, you would realize that the same A dot, sorry, C dot A cross B will now become A1, A2, A3, B1, B2, B3 and C1, C2, C3. Okay. But we also know that I could also write it as A dot B cross C also. Yes or no? Isn't it? So what I learned from here is that C dot A cross B is same as A dot B cross C. Is it fine? Okay. Now, one more operation we can do. We can do this operation. Let me swap the position of the first and the second row. Okay. So I'll just, let me, let me do it in one go instead of writing it twice. Let me do it in one go. So I wish to perform this operation. R1 swapped with R2 and then R2 swapped with R3. So when I do that, it'll become something like this. Correct me if I'm wrong. And making two swaps doesn't change a determinant. Right. So now this is also the same as what we had written in the previous. That means B dot C cross A also gives you the same result. Right. So what does this tell you? In short, the conclusion here is that A dot B cross C is same as B dot C cross A is same as C dot A cross B. Let me not put packets. Right. So these three are called equivalent characterization of the vector of the result. Okay. So these are equivalent characterization, equivalent characterization of the scalar triple product. So whether you do this operation or whether you do this operation or whether you do this operation by the virtue of the fact that they're all related to that determinant. And in determinant, we can do two saplings of two rows and it doesn't change the determinant because of that, these three expressions will be one and the same. Okay. So there is no difference between these three activities. Okay. Yes or no. Correct. Now, so if you see this and this result, let's say and this result. Okay, let's focus on this is it. So you have said A dot B cross C. I'm just putting that result again. So A dot B cross C, you're trying to say it is A cross B dot C. Please note this and this I can stop the positions. Okay. Because it is commutative dot product is commutative. So what does this tell you? It tells you that the dot and the cross positions can be swapped positions can be swapped and it doesn't affect the result. Right. So if the dot and the sub positions can be swapped and it doesn't affect our result, then is there any point even showing the dot and the cross? Let's save time writing the dot and the cross. So many a times this expression is sometimes written as box A, B, C. That means we just put this under in a box means wherever you want, you can put one dot and wherever you want, you can put a cross and that expression is not going to change. It is still going to stand for the scalar triple product of ABC. So whether you interpret it as A cross B dot C or whether you interpret it as A dot B cross C, this result and this result would be the same expression, exactly the same expression, not even differing in sign, exactly same expression. So the position of the dot and the cross in a scalar triple product doesn't matter. And because of that we just write the vector ABC and put it in a box kind of a structure. So since there is a mixing of dot and cross and since we write it as a box kind of a structure or many people say since it gives us the volume of a box kind of a structure, we call scalar triple product also as box product and mix product. The name mix and box has come because of that. Is this fine? So let me further conclude here. So box ABC is equal to box B, C, A is equal to box C, A, B. Why would you have a math exam on Monday? I have not been told. I've been told to give you one paper, but not as an exam point of view, maybe for a practice or something. We didn't connect yesterday, right? So today you give me a call. Okay. Okay. All right. So please, everybody please pay attention over here. So please note this down. Box A, B, C, box B, C, A, B. They're all the same things. Okay. But to be very, very, you know, careful here. If you change the order of any two of them, the answer is going to become negative. For example, box ABC and box ACB, they would be negatives of each other. So this will be negative of this. Okay. So only these three. So you can see that they have been written in a cyclic fashion. A, B, C, B, C, A, C, A, B. If this cyclic fashion is disturbed while writing box product of any three vectors, then the answer is going to become exactly negative. However, magnitude wise, same, but exactly negative. Right. For example, if I give you A, C, B. Okay. Let's say I give you this box product. Okay. Unless I write, or if I write this as C, B, A, or if I write this as B, C, A, they will all be equal. Okay. So a cyclic conversion of this box product is not going to affect the result. ACB, C, B, A, B, sorry, B, AC, I'm sorry, B, AC. They will all be the same. Okay. But if you try to do something like this, ACB and try to just swap first in the second. Let's say you do B, C, A, then it would become negative sign over here. Is the idea clear? What I'm trying to say. So in any box product, if you do a cyclic change like this, it is going to be same. But if you're going to do something like this, then if you do a cyclic change like this, it is going to be same. But if you're going to just take two of them and swap their positions. Okay. Then it would become negative just like what happens in case of a determinant and determinant. If you take any two rows and swap it, you get a negative of the determinant and that's what is going to happen in box product also because box product is basically obtained by writing in a determinant format the components of these three vectors. What's the point? Any questions, any concerns here? Okay. So let's look into further more properties of box products. So I'll be just going to the next slide. If you're done with this, do let me know. Of course, if you talk about position vectors, let's say you're talking about a three-dimensional vector, a position vector. Then in that case, it would give you a parallel pipette which is connecting origin to those points. So you connect origin to these three points. Then construct a parallel pipette by using them as the core terminus edges. That parallel pipette volume will be obtained. Coordinates are what? Position vectors and they will have a common point which is your reference position. Okay. So that will give you a position vector. Sorry. That will give you the volume of the parallel pipette. So you need to understand that position vectors are vectors starting from origin to that point. So let's say origin to that point A, one vector, origin to a point B, another vector, origin to a point C, another vector. From these three vectors as the core terminus edges, you are constructing a parallel pipette. That volume, algebraic volume will be obtained. Would the two parallel pipettes be opposite of each other if you just changed two of the vectors? I mean, volume wise, see what will happen? Why does the volume become negative? Volume becomes negative because the three vectors start behaving as a left-hand coordinate system rather than a right-hand coordinate system. In a right-handed coordinate system, if let's say A, B, C form a right-handed coordinate system in the sense that if I move from A to B, it is more or less in the direction of C or maybe it is making an angle which is acute with C. In those cases, volumes are positive. But if A, B, C are such vectors which are forming or which are very close to a left-handed coordinate system, that means if you curl your finger from A to B, its direction makes obtuse angle with C. Then in those cases, your volume becomes negative in the sense that the volume becomes flipped downwards. Okay, all right. We'll show you more properties of box product. I've already noted it down just to save your and my time. So we'll take that up. So these are the properties of skeletal product. This I think I already discussed with you that I hope it is legible. Everybody can read it. Yeah. If A, B, C are basically given to you in a resolved format, A cross B dot C or A dot B cross C, it doesn't matter. So the expression is given by this determinant. We have already discussed it. We have already discussed it that the position of the dot in the cross can be interchanged without altering it. Hence, many a times, this expression is also written as box A, B, C. Okay, so box A, B, C can stand for two things. Whether you write A dot B cross C, or whether you write A cross B dot C, it doesn't make a difference. Okay, bracket is definitely not required. And as I already told you, they are cyclic, box A, B, C, B, C, A, C, A, B, they're all same. But if you switch the position of any two of them, that means they're not cyclic anymore, cyclicly written anymore, then it will become negative. Okay, few other properties is, let's say if you multiply one of the vectors with K, with a scalar quantity K, then K can come out of the box product. Now can somebody justify this property in light of determinant? In light of determinant? Do you remember one of the determinant properties that if you multiply a given row, right, if you multiply a given row with a common factor, elements of a given row with a common factor, that factor can always be taken common out. That is what exactly is happening over here. And the second part is a scaled up version of the same. So as you can see, this is a scaled up version of the same. That means if you have each of these vectors multiplied to different scalar quantities, K1, K2, K3, then you can always bring K1, K2, K3 out. Look at the six properties. Six properties will also remind you of some property in determinants. As you can see that the first vector is written as sum of two vectors. So you are writing A plus B, C and D. You are taking a box of this. So this could be written as ACD plus BCD. Which property was this? If you recall, there was a property where I told you that if the elements of a given row can be written as a sum of multiple items, then you can break the determinants by breaking it with respect to that particular row. The same property is used over here. Are you getting my point? Yeah. Property number seven is this is what I was trying to tell you. If ABC is in that order that it forms a right-handed system, then box ABC will be positive. So many times a question also is asked on this, whether are the three vectors forming a right-handed system or a left-handed system? So all you need to do is find out your box ABC whether the answer is coming out to be positive answer or a negative answer. If it's a positive answer, yes, they form a right-handed system. If they give you a negative answer, then they form a left-handed system. Property number eight is very important. I'm sure you would have learned about the concept of coplanarity of three vectors. I think Kiran sir would have spoken about it. So when three vectors are a coplanar, let's say vector ABC are coplanar. So can I say that one of the vectors could be expressed as a linear combination of the other two. And this combination is a unique one. Now, this is one of the tests of coplanarity. If I use the same fact in our scalar triple product. If I write box ABC, then what is going to happen? All of you, please pay attention. So when we have written box ABC, we normally write the components of these three vectors and put it in a determinant format like this. Why didn't I write C2 all of a sudden? Now, if you do this operation on this vector, R1 transform to R1 minus XR2 minus YR2. What are you doing then? What are you doing then? Aren't you producing zero, zero, zero on the first row in light of this? Please look at this. Because of this operation, there will be zero, zero, zero produced in the first row. Everybody agrees with me on that or not? As a result, the scalar triple product of ABC becomes zero. So another test for coplanarity of three vectors. If you realize that the scalar triple product of three vectors is giving you zero. Either the base has collapsed or the height of the vector has collapsed or the height of the parallel pipette has collapsed. Isn't it? So if let's say the base collapses, then that A and B which used to be shown by two vectors would not be shown by a single vector. So that single vector and the C vector would not be in the same plane because two vectors are always known to be coplanar. So overall, the three vectors become coplanar. If the height collapses, then the C vector also lies in the plane of A and B. So in each of the two cases, the three vectors ABC have become coplanar. So this becomes a very important and widely used test. Nobody uses this. Okay. After people know about the scalar triple product test, nobody uses this concept for coplanarity. However, this is good to know. But the practical use, the practical way of solving a question whether three vectors are coplanar or not is to just find out their box product. If it becomes zero, then the three vectors are coplanar and that is what has been written over here. If ABC are coplanar, then their box product will be zero. Please note this down. Is it fine? Any questions? Any questions? Any concerns? All right. Look at the ninth property. Ninth property is a very widely used one. However, many people are not aware of it. So can anybody prove this? Can anybody prove this? That if you're three vectors which are involved in the box product, the first vector is made up of a linear combination of ABC. Second vector is also made up of some linear combination of ABC different from the first one. And the third is also a linear combination of ABC. Then you can write it like this. How do you prove this? How do you prove this? Which concept is going to help you prove this? At least tell me that so that we save our time. Which concept in determinants that we have learned is going to match with this property. Anybody in determinants. We did something. Okay. So when you are expressing this as a determinant, let me write that this again. So let's say I have this as my one vector. Y2B, Z2C and X3A. Right. It's related to the product of the determinants. Okay. We'll discuss it out. So when you write it as a determinant, how do you write it? You write the, let me make a bigger one by the way. You write the elements of this particular vector or the components, IJK components in the first row. So if I'm not mistaken, can I say IJK components will be X1A1, Y1B1, Z1C1 and X1A2, Y1B2, Z1C2 and X1A3, Y1B3 and Z1C3. So I'm assuming my A vector is A1I, A2J, A2J, A3K. And similarly our B vector and C vector also. I'm not writing it down. Okay. Similarly. Okay. So if you see this term that you have. Okay. This term is obtained. If you multiply X1, Y1, Z1. Let's say let me write all the components here. A1, A2, A3, B1, B2, B3, C1, C2, C3. I'm not writing everything down. I'm just trying to make you recall what you had done in your determinant multiplication. So when we learned determinant multiplication, we had learned that when two determinants are multiplied, you can multiply them row with row, row with column, column with row, column with column. Right. So let's take this first row and multiply it with this first row over here. So X1A1, Y1A2. Okay. I think it is not written in this fashion. It is written in a row multiplied with column here. Sorry. Yeah. So if you multiply X1A1, Y1B1, Z1C1, you end up getting this. Okay. Similarly, the same row multiplied with the second column will give you X1A2, Y1B2, Z1C2, which is your this. And again, the first row multiplied with the third column will give you X1A3, Y1B3, Z1C3. Correct. That means if I start filling in all the places over here, which I'm not writing by the way, if I start filling in all the places over here, can I say this will be ideally be X2, Y2, Z2, X3, Y3, Z3. In short, this is equal to this. And that is what this property also actually says. So this is X1, Y1, Z1, X2, Y2, Z2, X3, Y3, Z3. And this is box A, B, C. Correct. That is what the property also says. Is this fine? Any question? Any concern in the ninth property? Anybody. See, I could have explained it by writing all the nine elements here, but it'll just, you know, waste our time. That is why I thought I would just explain how do we get that first row from that determinant multiplication. The other rows will similarly be obtained. Okay. Don't worry about it. Is this fine? Any questions? Any concerns? Let's now start taking some questions. Let's start with some questions. Okay. Let me begin with a very simple one. Find the volume of the parallel pippin whose edges are represented by weight need to finish copying. Okay. Thank you. I have come to know that few of you are becoming sluggish towards the end of your preparation. Guys and girls, be very, very careful at this part of your journey. Picking up from here on and dropping it from here on. Both are going to show you drastic results. Okay. I have got calls from few of the parents that so and so has stopped studying. Okay. I would not like to, you know, give that, give off the names, but remember at this point, at this juncture, you are going to make a lot of change in your, you know, career prospects. People who are going to give off or give up at this frame are going to fall drastically and people are going to, you know, double up their effort from this time are going to go higher drastically. So both the things are possible because now the syllabus is almost completed. Right. So the more you start practicing, testing yourself critically, analyzing your weaknesses, you know, the more you're going to grow, the more you're going to relax. I'm tired of two years of preparation. So many chapters. I kill. That is still teaching. Okay. So if that thought comes in, then you're going to go down. Choice is yours. Okay. So step up your efforts at this time. Don't get exhausted. Having a break and rejuvenating is fine. But don't be like, you know, giving up. I don't like, okay, now I'm not going to prepare anymore. So all those things, please let it not come in your mind. That is going to be very, very detrimental towards any of the exams that you're going to write, but that's a board exam or whether it's the competitive exam that you're going to write. If you feel you want to speak to me one on one. And, you know, take my opinion with respect to what to do in the coming few months, you can definitely give me a call. Of course you message me. We'll find out a time to talk. Okay. All right. Find the volume of the parallel preparedness. When I say, when they say volume, they actually mean absolute volume, not the algebraic volume. Just like when we say area, area under the curve, we mean absolute area, not the algebraic area. Indefinite integral gives you algebraic area just like determinant here will give you algebraic volume. Okay. I'm getting four different answers. Somebody's saying zero. Somebody's saying 35. Somebody's saying seven. Somebody's saying three also. Why? One basic question and five different four different answers are coming here. Okay. So determinant. I think many of you have calculated incorrectly. Let's check. Okay. Magnitude. Gives you the volume. Okay. So this outside parallel lines is a magnitude pack. Yeah. Yeah. Please calculate. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. And then plus four minus one, minus six. So that becomes three into two, six. This is a five. Five into three is 15. And this is going to be a minus 28. Okay. That gives you minus seven, but we have to take a mod of it. If you want the volume and that gives us seven cubic units. Okay. Seven cubic units. Okay. That's the answer. Yeah. So this is going to be a minus 28. So that gives me the volume. What determinant, you people are giving me so many options to me. For any three vectors ABC. By the way, this is actually a important property also. Okay. This result is very, very important. I have seen that this is used in several places to solve questions. So box ABC square. a dot c, b dot a, b dot b, b dot c, c dot a, c dot b, c dot c, determinant. So basically this is a question of determinant multiplication. So two determinants are multiplied. Of course, both are same determinants and from the multiplication, you need to prove that the resultant multiplication gives you this, which is quite easy. I think all of you will be able to do it. Done. So don't, don't write everything. Okay. So we'll just do it in a brief manner. So box a, b, c is what? Box a, b, c is a1, a2, a3, b1, b2, b3, c1, c2, c3, right? Where a1, a2, a3, et cetera, b1, b2, b3 and c1, c2, c3, they are all components of vector a, b, c. Okay. So box a, b, c square will be what? This determinant multiplied to itself, better to write it here. I'll save time. Okay. Now pick, do row, row operations. So take the row here, take the row here. So first row with the first row, will it not give you a1 square, a2 square, a3 square? Okay. Then first row with the second row will give you a1, b1, a2, b2, a3, b3. And then the first row with the third will give you a1, c1, a2, c2, a3, c3. Okay. Now I can keep doing it, but I'm just saving my time here because, you know, we all know what to do next. So isn't this a.a or mod a square? Okay. Isn't this a.b? And isn't the last element of this row is, isn't it a.c? And that's how you obtain the first row of your answer. Okay. Similarly, you can prove for the second, third and the, so similarly here also you can do it. Okay. I'm not going to write everything down. So what I wanted to show you here is that this is a very, very important property that is required to solve many questions. In fact, I think I've also taken a scaled up version of this property where instead of a, b, c box square, you have two boxes. Let me just pull out the question. Yeah, this one. No, not this one. Oh, I think I have not written it. Okay. Anyways, let's take more and more questions here. Kind of, yeah. Many things are very similar to what we have learned in our determinants. Okay. This question. If vector a is, I hope you are able to read it. I'm making it big. Yeah. If vector a is given by this, vector b is given by this, vector c is given by this. If the ordered set b, c, a is left-handed, find the values of x. Now, such questions are normally not framed in a JEE main exam and JEE advanced exam. Maybe it could be in a regional entrance exam that they could ask because many a times this left-handed word is explained by the JEE people. They don't expect people to know this beforehand. Okay. So this could, this could not be a direct question in any national level exam, but it could definitely be asked in regional entrance exam where they expect students to have a knowledge of these technical terms. So find the values of x for which these three vectors form a left-handed coordinate system. Give me a response on the chat box. Okay. Arshit, Aditya, Pradyan, Rohan, Kinshuk. What's the surprise to see different answers coming from so many people? There should be a consistency in the answer now. Okay. So b, c, a, b, c, a would be 2, 2x, 1, 1, 0, 1, and x 12 minus 1. This determinant should be negative in order to form a left-handed system. So let's expand it. So this will be 2, negative 12. Okay. So minus 2x, it will be minus 1, minus x and plus 1 into 12 less than 0. Okay. On simplification, this will give you minus 24 plus 2x plus 2x square plus 12 less than 0. If I drop our two factor, it will become x square plus x minus 6 less than 0. Clearly factorizable like this. Okay. Make a wavy curve minus 3, 2 plus minus plus negative means this zone. So x should be in the open interval minus 3 to 2. Okay. So those are given open interval minus 3 to 2. Absolutely right. Is it fine? Any questions? Any concerns here? Simple question. Can I move on? Next. Okay. Again, another easy question coming your way. If a, b, c, r, 3 non-coplanar vector, find the value of value. Value means a numeric value will come for this. Find the value of this expression. Oh, it just looks complicated. It's just touch and go. Sure, whenever when can shook. Yeah. I hope you didn't rush through it. Okay. Now again, three different answers. I'm, I'm getting one, zero and three. Okay. Good one Aditi. Okay. See this top one is box a, b, c. Bottom one is box, b, c, a. Okay. Then second one is box, b, c, a by box, c, a, b. And this is box, c, b, a. And this is box a, b, c. Okay. Please note. Recall that box a, b, c, b, c, a and c, a, b, they will be equal. If you mess with the order of any two of them, it will become negative. Right. So this will become a clear cut one. This will also become a clear cut one. But here there is c, b, a not c, a, b. So this will become a minus one. So the answer for this question is two minus one, which is a one is assigned to those who were rushing through with three answer. Wrong. Okay. Next is a very interesting result. By the way, it is not a question. I can treat it as a question also. Let us say there is a tetrahedron. Let's say there is a tetrahedron. Let's say there is a tetrahedron. I'm just, everybody knows a tetrahedron. So a tetrahedron is basically a three-dimension geometrical figure, which is made up of triangles on all the faces. Okay. So let us say there is a tetrahedron, which is like this. Okay. It's a three-dimension structure. Now a tetrahedron is a special case of a pyramid. Okay. And volume of a pyramid. Volume of a pyramid. What is volume of a pyramid? It is one-third the area of the base into height. We, everybody knows it. It is equally true for any type of a pyramid, whether the base is circular, whether the base is square, whether the base is pentagon, whether the base is a triangle. So any pyramid kind of a structure, if you know the area of the base, if you know the height of the pyramid, one-third is area of the base into height. We'll give you the volume of the pyramid. A cone itself is a pyramid. So the cone volume is what? One-third pi r square h. So one-third is this one-third. Volume of the base is pi r square h. Okay. Now knowing this fact, given we know this fact, tell me in terms of box ABC, what will be the volume of this pyramid or tetrahedral structure, volume of the tetrahedral, where this vector is A, this vector is B, and this vector is C. Okay. So find the volume of the tetrahedral in terms of box ABC. No, sir, it is not that answer. Kinshukh, how is pyramid in tetrahedral different? That's what I said. They're not tetrahedral. Tetrahedral is a type of a pyramid. Cone, a tetrahedral, a prism with a square base. They're all pyramids. Just to cover you, a lot of things are there. Very good, Aditya. That's absolutely right. Aditya has already got the answer. Pradhan also has got the answer. Kinshukh, again, you're wrong. Correct shatih, correct Atharv. So Aditya, Athar, Pradhan, and shatih, right answer. Okay. Anybody who needs a little bit more time, I can wait. But if you have given up, we can start solving it. Anybody wants me to wait? So as I already told you, as I already told you that volume of a pyramid and a tetrahedral, sorry, tetrahedron, a tetrahedron is anyways a pyramid. Okay. So in a tetrahedron, I can use this formula to calculate its area. Okay. So first of all, what is area of, let's say I consider this to be the base. However, you can consider any of the face to be the base, it is your call. So I'll consider the base to be your triangle ABC. Okay. So what is the area of a triangle whose vertices position vector are ABC respectively? Yesterday we did it. I hope you remember it. Half mod A cross B, B cross C, C cross A, correct? Everybody remembers it. This was the result which we have derived in the last class. Correct? Okay. Anyways, so if this is the area of the base, now what is the height of the base? How would you calculate the height of the base? Now, all of you, please pay attention. Very important. Consider this diagram like this. Let's say this is my ABC base. Okay. And this is the perpendicular to the base. And let's say this is your, this is your A vector. Can I say height of this tetrahedron will be the projection of A vector in the direction of the normal to the base. Am I right? Yes, no, maybe, no, sir, you are not right or I don't think so. That is true. If at all, see these kind of feelings is coming, please let me know. Agreed? Height is the projection. So you can say, sir, could have taken any of the three vectors, why only A? Yes, you could have taken OC also, or you could have taken OB also, not an issue. Correct? So it is the height of the projection of it. The height is nothing but projection of any one of the three vectors. You can take, you can take this also. Let's say this is C vector, no, C in the direction of this. Now, first of all, what is the normal to the base vector? Let's say I call this vector to be n cap or n vector. You can call it as n cap also, no issues. So normal to the base, let us say it is n vector. So what is normal to the base? You will say, sir, any vector which you form by taking the cross product of any of the two sides of the triangle, if you take the cross product, you get a vector perpendicular to the plane containing both of them. So it will be along the normal to the, this thing also. Okay, so let's say this is your position vector A, B, and this is C. So all I need to do is just take a cross product of any of the two vectors that you see on the face. So let's say I take this, let me take, it doesn't matter. The order in which you are taking doesn't matter because we are anyway is going to take a mod of it. So let's say B minus A, B minus A cross C minus A. Okay, can I say this will go all perpendicular to the plane containing it? Okay, so whatever this vector comes out, we have to take the projection of A in the direction of that normal. So what I'm going to do, I'm going to do A dot, B minus A cross C minus A. Okay, can I say this will go all perpendicular to the plane containing it? B minus A cross C minus A, okay, divided by the magnitude of B minus A cross C minus A. Yes or no? In the last session, I told you projection of P in the direction of Q. In the direction of Q. What is it? It is P dot Q cap, right? So this Q cap is now this vector divided by its magnitude that gives us the unit vector in that direction. Correct, okay. Anyways, it's good to expand it first, then we'll talk about it. So A dot, let's say inside here, what happens? Let's try to see this. So inside here, it will get B cross C, okay? You will get B cross minus A, which is actually A cross B, okay? And you get minus A cross C, which is actually C cross A. And remember A cross A is anyways a null vector, correct? So it will be magnitude of the same vector down over here, since I've already figured it out. I don't have to redo it again for the denominator separately, okay? If you see this very carefully, you would realize that you have done A dot B cross C, which is box ABC, am I right? Okay, and what about A dot A cross B? Remember this was one of the questions that came yesterday in one of the problems. A dot A cross B, and you told that isn't this, that isn't this, zero. Why? Because A cross B itself is perpendicular to A and B both. So dot product of A with a vector perpendicular to it will be zero. In fact, I would like to now tell you something very generic. In box product of XYZ, if any of the two vectors are same, if any two vector, any two vector, if three vector, then definitely, if any two vector vectors XYZ become same or become equal, then this will become zero. Please remember this. So X, XY, zero, XY, Y, zero, XY, X, zero. Any two of them become equal, they'll be zero, right? Because they will become coplanar then. All the three will become coplanar. Okay, so that's a special situation for coplanarity. So this will be zero. So this and this will be zero. Even this and this will be zero. So need not write them down. Just write zero, zero. Okay, and here you have box. So not box, modulus of, so I've been saying box box so many times that even for modulus, I am saying box. Okay, so this becomes your height. Okay, so now let's use this formula. One third area of the base into height. So finally, coming to our volume of the tetrahedron, volume of the tetrahedron, that becomes one third area of the base. Area of the base is half. We have already seen it. A cross B, B cross C, C cross A. Okay. Into height. Height is box A, B, C by, by this same expression. As you can see, this expression itself is here. This and this are same thing, right? So they will happily get canceled off. Oh, they will happily get canceled off. Okay. Giving you the answer as one sixth box A, B, C. This is a lot of requests you to remember, even though I've derived it in a very, you know, extensive way. Remember this result because it's going to be asked. It's going to be asked directly in some of the comparative exams. Anusha, you answered it as one sixth A, B, C. Where did you answer? I don't see your name here. No, in my chat box, I don't see your name coming. Okay, now that you have brought it up, I'll definitely figure it out. Where is it? No, it is not there. Are you sure you are not missing out on these chats? Okay. Anyways, don't worry too much about it. Okay. Maybe because you got disconnected and your type while you are disconnected, so I could not register it. It is not registered on my chat. Is it fine? Any questions, any concerns? Any questions, any concerns? Okay. Now I have few questions related to tetrahedron, especially related to regular tetrahedron. Some important results related to a regular tetrahedron. Some results for regular tetrahedron. Area vector always perpendicular to the base, yes. Area, if you see A cross B, that vector is having a magnitude of the area and its direction will always be perpendicular to the surface, which contains the two vectors whose cross product was taken. Okay. So first result which I would like everybody to tell me, what is the angle between, what is the angle between any edge of a regular tetrahedron and face not containing that edge and the face not containing that edge. By the way, what is the meaning of a regular tetrahedron? Regular tetrahedron would be a tetrahedron whose all faces are equilateral triangle. Okay. So let's make a base first of all, which is an equilateral triangle. Let me make it like this. Okay, I'm just making a dummy figure. This is an equilateral triangle and these two faces which I have not shown on the face. Okay, so let's say O, A, B, C. So A, B, C, O, A, B, O, B, C, they're all equilateral triangles. So all the four faces are equilateral triangle in case of a regular tetrahedron. Now, what is the question asking you? The question is asking you, what is the angle between an edge and the face not containing that edge? Are you getting my point? That means, let's say I take the projection of this line on this face A, B, C, what is this angle? That is what they're asking us. An edge and the face not containing that edge. Okay, Shatish, are you sure, by the way? Are you just guessing? Okay, anybody else? Yeah, take your time, take your time. Angle between the edge and the face is that, let's say this, I'm talking about the edge O, A, and I'm talking about the face A, B, C. As you can see clearly, face A, B, C doesn't contain that edge, okay? So the angle between them is basically nothing but you can think of it from two perspectives. If you take a perpendicular to the face, right? The complementary of this angle, so this will be 90 minus theta. Let's say theta is the angle, then complementary of the angle would be the angle between the normal and this particular edge, okay? Or if you want to go from a real analysis or a pictorial analysis of the situation, so think as if you're dropping perpendiculars from every point of that edge onto the face which you want to find the angle from. So let's say this is the shadow or this is the projection of that line O, A on the face A, B, C. The angle between that shadow and that edge, that is called the angle between the edge and the face. Use the fact it is regular. Use the fact that it is regular Kinshuk and simplify whatever you have got. Shatish, now you're right. How did you change your answer now? Too early a conclusion. See, very simple. This can be solved by using simple geometry. Do you realize that from, oh, if you drop a perpendicular, that means if you take a triangle like this, I'm just taking a triangle, just all of you watch this triangle, which I'm showing with my white spring, this thing, okay? So in that triangle, so let me make that triangle separately over here, okay? In that triangle, do you realize that this point will be the centroid of A, B, C? Okay. This is anyways the side length of the tetrahedron. Let's say each side, O, A, O, B, O, C, and A, B, whatever. A, B, A, C, B, C, et cetera. They're all equal to lambda. So this is lambda. And this is what? This is the distance of one of the vertex of a triangle from its centroid, correct? So what is this whole distance? What is this whole distance? This whole distance is the length of the altitude of the base space, right? And what is that? What is that? Side length into sine 60 degrees. So root three by two into lambda will be this whole length, AM, correct? So AG will be what? AG will be nothing but two-third of that, correct? In short, it will be lambda by root three. So this length, this length is lambda by root three. And this is lambda. And you're asked for this angle. What is that angle then? So cos theta is base by hypotenuse, which is one by root three. So your answer to this question is cos inverse of one by root three. Is it fine? Any questions? Okay. I think only Shritish got it right and that too in second try. Okay. This result is to be kept in mind because it may be asked directly as a question to you. Second question is what is the angle between two faces of a regular tetrahedron? What is the angle between two faces of a regular tetrahedron? Since it is already written regular tetrahedron, I'll not wait to write it. So what is the angle between two faces? Many people ask me, sir, what is the meaning of angle between the two faces? Angle between two faces is the angle between the normal to the two faces. Okay. And just to avoid confusion, it is not 60 degrees. Those who are thinking 60 degrees, just because those faces are built over two edges, which are at 60 degrees. So please don't have this confusion that since the two faces are on an edge, which are making 60 degrees, even between the faces, there will be 60 degree angle. No, that is not correct. Draw two perpendicular to the two faces. Draw two perpendicular to the two faces, the angle between them. Okay. See, let's say I'm just making a kind of a schematic diagram here. So let's say this is, okay, this is, let me make it in a different way. Okay. Let's say these are two planes. Okay. This is plane one, plane two. When I ask you what is the angle between the two planes, what you will do, you will make perpendicular to these two planes like this. One with this one with this. Okay. So the perpendicular angle, that is the angle between the two planes. So this angle theta, this is the angle between the two planes. Okay. So in a regular tetrahedron, what is the angle between the two faces of a regular tetrahedron? That is simple. I think you can use vectors for that. Maybe if you're not able to use geometry, you can use vectors for that. Anybody who is close to the answer are about to get it. I can afford to wait for some more time. Costs, inverse. Okay. Okay. All right. So let's all figure this out together. See, what is normal to the plane? Oh, AB. What is normal to the plane? Normal to the plane. Oh, AB. So you can say a normal to the plane. Oh, AB could be taken as cross product of a and b. Correct. Similarly, let's take normal to the plane. Normal to the plane. Let's say OAC. Can I not take it as a cross C. Correct. Now what has been asked the angle between the two, two of the faces. So it's just the angle between these two vectors. Let me call this vector as P and let me call this vector as Q. Okay. So angle between the two vectors is their dot product divided by their magnitude. In short, I have to find a cross B dot a cross C upon magnitude of a cross B magnitude of a cross C. Okay. Now the denominator part is quite easy because let's say each of these vectors are off length lambda. Okay. So the denominator part, this is nothing but mod a mod B, lambda square into sine of the angle between A and B vector. Now what is the angle between A and B vector? 60 degree because it's a regular tetrahedron. So that is going to be root three by two. Okay. Sorry for writing in a different color. Okay. Similarly, a cross C modulus will be the same. So lambda square root three by two. Any problem with this so far? Anybody has the denominator part? Anybody has a problem? Any questions? Okay. How would you figure out this part? A cross B, the numerator part. So I'm just writing the numerator separately. So numerator part is a cross B dot a cross C. How will you find this out? Which method will you use? Split. Similarly, I told you a property that started with L. No, but remember A cross B dot C cross D. Languages identity. So it's a dot C, a dot D. Then B dot C, B dot D. Correct. So using, use your languages identity. Correct. So languages identity, this will give you a dot A, A dot C, B dot A, B dot C. Okay. Now hold on. What is A dot A? A dot A is mod of A square, which is lambda square. Correct Anusha. Yeah. Lambda square. What is A dot C? What is A dot C? Who will tell me A dot C? A dot C, A dot C, A dot C. Lambda square by two. Correct. Because A, A root cost 60 is root three by two for you. Okay. Okay. B dot A will be again. Lambda square by two and B dot C will be again. Lambda square by two. Okay. So on simplification, this is going to give you a lambda to the power four by two minus lambda to the power four by four, which is lambda to the power four by four. Okay. So let's write it down now on the numerator. Lambda to the power four into one fourth, lambda square, lambda square, lambda square will go for a toss. So it'll become, if I'm not mistaken, it will become one by root three root three, which is one by three. Okay. So the answer between the angle between two faces of a regular tetrahedron is cost inverse one by three. Now both results look very familiar, very similar. One is one by root three and other is one by three. Don't get confused. Is it fine? Any questions? Any concerns here? Any questions? Any concerns here? All right. So we'll continue with our questions. We have not yet completed our question. Oh, it's only 530. Oh, I thought I would complete scalar product by in one hour. What should I take? Which problem should I take? Okay. Let's take this question. V is the volume of the parallel pipette having three coterminous edges as ABC, then the volume of the parallel pipette having three coterminous edges as alpha, beta, gamma is which of the following? All has been launched. Two minutes on only two responses I've got so far. Okay. Now that I've started getting more five of you last minute. Okay. Five, four, three, two, one. Okay. End of poll. I can see 61% of you have gone for, okay, let me ask somebody. Anusha, which option have you gone with? Anusha says A, but Anusha, people say D. D for donkey. Donkey is maybe a wrong word to use here. D for Delhi. Okay. So please recall, there was one of the properties which I did that is box of X1, Y1, Z1. I'm just writing it down. Okay. If you recall this property, this property gave us determinant X1, Y1, Z1, X2, Y2, Z2, X3, Y3, Z3 times box ABC. Everybody recalls this property that we are done. Now here, we'll try to use this property. See, they're asking us that you want to make a palelopipid whose coterminous edges is alpha, beta, gamma. Now try to treat this as your X1, Y1, Z1, X2, Y2, Z2, X3, Y3, Z3. Isn't this same as saying X1, Y1, Z1, X2, Y2, Z2, X3, Y3, Z3, box ABC. Correct. So this whole thing is this whole thing. Okay. And this whole thing is this whole thing. Yes or no? Right. So now in light of one of the other properties which we had done and I told you that property, that question is actually a property. This thing is what? Box ABC square. Okay. And there's any ways of box ABC. So can I say this whole thing is box ABC cube? Okay. And assuming ABC are forming a right-handed coordinate system and they represent the volume, positive volume of that H, this is actually V cube. So Janta was correct. Janta janardan is correct. Isn't that C supposed to be gamma? I didn't get you. Gamma no, Z3. This is X1, Y1, Z1, X2, Y2, Z2. Oh, alpha beta gamma. Oh, that is a slip of pen. Thank you, sugar. Is it fine? Okay, we'll do one more question and then we'll go to the vector triple product because a lot of things had to be covered. Can I go to the next slide? Okay, let's take this question. Let their vector, let a vector R be A cross B times sine X, B cross C times cos Y and C cross plus C cross A where ABC are non-zero non-copener vectors. Okay. R is orthogonal to this. In the value of secant, secant square Y plus cosecant square X plus C Y cosecant is which of the following pole is on within 49 seconds. Sorry, within 45 seconds, I got one response. Good. Another one. Good. How is the weather now in Chennai? Anybody near and dear ones living there? Are they all fine? Sunny now. Thankfully. Okay, great. Okay. So three people, three different answers. Oh, no, four different people, four different answers. Okay, now seven people and three answers have got equal votes. Are you all planning and voting? Hey, I'm voting A, U, vote B and U, vote C like that. Yeah. No. See, again, two of them are maintaining this thing. So four friends are there. Hey, but if they vote one, they have to tell some other people to vote now. Okay. Maybe I'm thinking too much. So much of plotting should not be going on. Or might not be going on. Okay. Can we conclude this in the next 30 seconds? Almost four minutes. That is the critical time for a problem. Okay. Five, four, three, two, one. Go. Okay. Let me ask somebody. Gurman, which option have you gone for? Gurman, any option? Okay. Gurman says B. By the way, yes, Gurman, most of the people have gone for B also. B and C, most people have voted. Okay. Let's check this out. See, first of all, we know that ABC have been non-coplanar and non-zero vectors. And this entire vector is orthogonal to this. Orthogonal means their dot product is zero. Okay. So let me just write it here. In fact, I will be little fast while I'm doing this calculation. So this dot product with three a plus five B plus two C is a zero. Okay. Now see this dot with a will anyways be zero because a cross B dot a will make two of the vectors same in the box product. It'll become box ABA. Box ABA zero. So I did not write this. Okay. This with five B will again give you a zero. But this will the last term will give you box ABC into two sine X. So it'll become two sine X times box ABC. Correct. Similarly, this guy with the first one will only give you three cause why box BCA BCA and ABC are same. But anyways, initially let me write it as BC only. Yes. The dot product with any one of them is going to give you a zero. And same with this, this fellow with the central one will only give you box five box CAB. Okay. So in short, if you take box ABC common, because anyways, this is also box ABC. And this is also box ABC because they are cyclic. Okay. So if you take box ABC common, you end up getting two sine X plus three cause Y plus five equal to zero. Now, this, this basically brings out two possibilities. Either this could be zero or this could be zero, but this cannot be zero because they have categorically mentioned that they're non-zero and non-coplanar. If they're non-coplanar, they will either be positive or negative zero to me. Correct. Zero means they will be coplanar. So this guy is not zero. So that is clearly implying that this fellow is a zero. That means this is minus five. But now this is an extreme scenario. It can be minus five only when sine X and cause five, both are minus one each. Correct. So what will be six square Y six square Y will be one only. What is Cosec square Y that will also be one and seek wine to Cosec, why that will also be one. So the answer to this question is a three, which is option number a. Okay. By the way, many people went for B and C, including okay. But the answer was option A. Is it fine? Any questions? So this is a question which is beautifully a mix of your trigonometry and a bit of your understanding of box products. And of course understanding of dot products. So this is how a typical question will be framed in competitive exams, trying to test you on various aspects of the chapter. Now I think enough time we have spent on this. We are now going to move towards VTP. I should have actually started with VTP by, by five, 15, five, 20, I'm already 25, 27 minutes late. Okay. Vector triple product. Vector triple product. VTP. So under VTP, we'll first try to find out what is the simplified version of this expression and what does this vector actually represent? How is it related to A, B and C? Okay. Let's try to understand that first. Let's say I call this vector as P as of now. Okay. Now let us look into this fact that B cross C. Let's say that is vector Q. So can I say Q is a vector. Which is perpendicular to the plane containing B and C. Do you all agree with this? So just focus on B cross C. B cross C. I have called it as a new name Q vector. Q vector, not Q, Q vector. So Q vector is perpendicular to the plane containing B and C. Correct. Now A cross Q, which is actually your P is a vector which is perpendicular to plane containing Q. Or you can just write it like it's perpendicular to Q. It's perpendicular to, let me just remove this plane. It's perpendicular to Q. And it's going to be perpendicular to A as well. Am I right? A cross Q gives me that P vector. That P vector will be perpendicular to A also will be perpendicular to Q also. But Q is already perpendicular to the plane containing B and C. That means this vector P is in the plane of B and C and perpendicular to A. Yes or no? Do you agree with me or not? Do you agree with me or not? Everybody agrees to this statement because this is the basis of which we will be deriving our result. So when somebody says A cross bracket B cross C. Now many people will have this question. Sir, if I change the bracket on A cross B, will it give me a different vector? Yes, it will give you a different vector. We'll talk about that also separately. But let us first focus on A cross bracket B cross C. That means B cross C first is evaluated. And then the result is taken a cross product with A. Okay, like that. So A cross B cross C gives you a vector in the plane of these two guys and perpendicular to this guy. Please understand this. This is very important because this look and feel of this explanation itself is asked as a question. Okay, so they'll say there is some unknown vector which is parallel to the plane of so and so vector and perpendicular to so and so vector. Okay, and its magnitude is so and so. Find that vector. So directly you can assume that vector to be lambda times A cross bracket B cross C and get your lambda. That's a simple way to solve that question. So this definition itself is a source of many questions. So A cross bracket B cross C is a vector in the plane of B and C and perpendicular to A. That is please write it down in the back of your head. Sir, how do I write in the back of my head? Okay, anyways, so using this, we are going to now derive this result. So now can I convincingly say that this is going to be a vector which is in the plane of B and C and hence I can write it like this lambda B plus mu C, where lambda and mu are some scalars. Yes or no. Any questions? Okay. And we have also seen that the dot product of this vector with A should be zero because this vector, you have said it is perpendicular to A. So can I say this will actually give you a zero? No, why? It is in the plane of B and C. Okay. And it is perpendicular to A. It is in the plane of B and C and it is perpendicular to A. A itself could be on the plane of B and C. Why not? So why should be A perpendicular to the plane of B and C? Last step. See, using the fact that this whole vector which I call as P, okay, using the fact that this is perpendicular to A vector. I can write this statement. No, Siddhant. Siddhant, that step is clear now. We have already figured out that P which was supposed to be A cross B cross C is in the plane of B and C and perpendicular to S. I'm using, now in the plane of B and C, I've used it. Now I'm using the fact that it is also perpendicular to A. Clear? Okay. Many of you are getting confused how the perpendicular C, let's say this is the plane of B and C. Okay. And let's say this is your A vector. Can I not have a vector like this which is making a 90 degree N? Let's say this is A and this is in the plane of B and C. Let's say this is some lambda B plus mu C. Why A has to be perpendicular to the plane of B and C to be perpendicular to any vector on the plane of B and C? I don't understand that. If any line is perpendicular to the plane of B and C, that is definitely going to be perpendicular to any vector on the plane of ABC. I agree with that. But I don't see that there cannot be any other vector other than the perpendicular to the plane which is perpendicular to a vector, a particular vector on the plane of B and C. Let's say, how is the yellow line? I've made it in the plane of B and C. See, imagine something like this. There is a line on the plane of my writing laptop. Okay. And I take a vector like this. I hope I am able to show you. Okay. See it. My marker and my finger. Can't it make 90 degree? Because there can be so many 90 degree positions made with this. I can make so many 90 degree positions. Are you able to visualize it? See my index finger. 90 degree with my marker. I hope I am, yeah, this is close to the camera. Yeah. 90 degree. And this need not be perpendicular to the plane. All right. We could take so many lines which are perpendicular like this. Of course, normal to the plane is also perpendicular. I'm not denying it. But for this normality between A and the vector on the plane to exist, you need not have a perpendicular to this. You could have some other line and some other plane. The two lines could be two different lines also other than the normal and. Okay. All right. So coming back to this. See my purpose is what? What am I doing here? My end objective is to find out an expression for a cross A cross B cross C in simpler words. Okay. In simpler terms. So I'm in the process of deriving that result. So I've used these two facts. The first fact is my A cross B cross C is in the plane of B and C. And because of that, I have chosen my vector to be some lambda B plus music. And now I'm using the fact that this vector should also be perpendicular to A. Hence the dot product is zero. Okay. Now everybody please pay attention. So I can say. Lambda A dot B is negative mu A dot C. Correct. Let's say I write it like this. A dot B. By mu is. Negative A dot C. By lambda. Correct. Will this be wrong to write it like this? Can I write it like this? Any objection in me representing this expression. Like this anybody has any issues do highlight. But you can write it in the other way around also. Let me just write it like this. Let me write. Lambda by. Let me write it as lambda by A dot C is equal to minus mu by A dot B. Right. Now what I'm going to do, I'm going to call this whole thing as a K. Let me not use K because normally we use K and all in our vectors. So let me call it as give me some different name. Alpha maybe. Okay. Let's call it as alpha. Now what I'm going to do, I'm going to replace lambda with alpha times A dot C. And I'm going to replace mu with negative alpha times A dot B. Okay. In. In this expression. Okay. So that means I'm trying to say that. I'm trying to say that. A cross B cross C is. Alpha A dot C times B. Minus alpha A dot B times C. Now the only thing which is stopping me from getting this complete expression is that alpha fellow. So I will get alpha. Now let us find out the alpha value. So far anybody has any doubt or concern. Please highlight everybody's happy till this step. Now only thing is alpha we have to find out. One thing that you admit here is that alpha will be a constant. Is there anybody who feels alpha is not a constant. See alpha is obtained by a constant divided by a fixed value if ABC are fixed vectors. Okay. Similarly alpha is also obtained by mu divided by A dot B. So these are all fixed vectors. So alpha will be a constant value. Just like your constant of integration. Okay. So now I have to figure out my alpha by using a special value for ABC. So this is a kind of an identity you can say. Where I have to figure out my alpha. Let me take it to the side of my screen. I'll write this down again. So I'll just write this part. I'll just write this part again. So A cross B cross C was alpha A dot C times B minus alpha A dot B times C. Okay. Now since this is an identity, I will use a special value of ABC. So let's choose A as I'll choose simpler ones. Let's say A is I B is also I and let's say C is J. Okay. Now I would request you not to choose I J and K for ABC. Because if you choose I J and K for ABC, you would realize that you will end up getting zero on the right side and zero. Like both the vectors would be zero zero actually. Okay. So in that case, your alpha value will be lost in those zeros. So you'll not be able to figure it out. Hence choose a value by which at least you are not able to get zero on both the sides. So one of the choices I have done is I, I and J. So let's write it over here. So I cap cross I cross J should be equal to alpha. Now remember A dot C will be zero. So this will whole thing will be gone for a toss. And this will be minus alpha A dot B will be one and only C will be left, which is J. So let's see what comes out from the left hand side. So this is I cross I cross J is going to be K. And this is minus alpha. Let me not write dot here as people will think I'm taking dot product of a scalar with a vector. So it is minus alpha J. Now what is I cross K? I cross K is minus J. So minus J is minus alpha J. So alpha is actually a one. Right. So put this back again in this expression. You will end up finally getting the result for this and that is nothing but a dot C times B minus a dot B times C. Okay. So please note this result down. I'll box it over here. So this is the expression or simplified expression for the vector triple product A cross bracket B cross C. Now you have to remember this result. You can't sit and derive it in the examination room. Yes. Alpha will be same for any choices of ABC. Okay. So any choice you can do. I plus J I plus J plus K if you want, alpha will come out to be one anyhow. Right out. But don't put such a vector which will make zero zero on both the sites else you will lose your value of alpha from both the sites. But yes, it's an identity should be true for any three vectors. It's an identity, you know, I'm, I'm trying to find the value of this expression in terms of B and C vector. So I figured out this thing, right? So it should work for, I should be able to find a fixed value of alpha and that value of alpha will be fixed for any ABC combination. And that came out to be one for me. You're trying to claim here and that this vector and this vector are same. No. So alpha should be fixed. It's an identity, you know. So alpha should be some fixed constant value. What is that constant value to figure that out? So I chose some special values for ABC. And I found that it is one. So you choose any ABC you'll get alpha is one only right out. If we have a fixed ABC, you will realize that alpha will also be a fixed value. That is what I said from here. This ratio lambda by a dot C will be fixed. And it will be same as lambda by a dot B always. So many books make a mnemonic out of it. One cross two cross three is 132 minus 123. I mean, it is up to you to make such mnemonics, but I'm not forcing on anybody. So this is how we remember it. Okay. So treat as if this is 123. So one cross two cross three is 132 minus 123. That's how you can keep it in your mind. If at all you are finding it difficult to remember. Okay. Now I will be requesting you all use this result and tell me. What is the expression if I had taken a cross B in the brackets. Yeah, it's an identity. This is true for any ABC. This is the formal. This is an expansion for a cross B cross. Okay. Yes. So now give me the expression for a cross B bracket, then cross C. And we'll try to compare this result with the previous page result. Is it giving me the same thing? Or is it drastically different? And if they are same, then they are same under what occasion? Okay. Okay. Let's discuss this. So first of all, do you all agree that it would be negative of C cross A cross B? Why did I choose to write it like this? Because earlier in that previous page, I had already derived this result. So recall this result in the previous page. We had already arrived that it is A dot C times B minus A dot B times C. Okay. So use this over here. Now A role is being played by C. Okay. And B role is being played by A and C is always being played by B over A. Okay. So this will be negative of C dot B times A. Minus C dot A times B. Okay. On opening the brackets, you will see that you will end up getting A dot C times B minus B dot C times A. Correct. Do you see the difference between this result and this result? They're not the same, by the way. These two results are not the same. Yes or no? Okay. They are different results. So one thing is for sure that A cross B cross C is not the same as A cross B cross C. Obviously one split reasoning here could be, sir, this would be in the plane of A and B, but this will be in the plane of B and C. Okay. Not that they cannot be the same. Now my question is, in general, they are not the same, but when can they be the same? So when is A cross B cross C equal to A cross B cross C? When? When, when, when, when. Okay. When what? Compare it. When A dot B times C is equal to B dot C times A. That means C is some lambda times A or call it some, you know, okay, beta times A where beta is your B dot C by A dot B. Okay. Yes or no? And of course, none of the two should be actually be perpendicular else what will happen, they will become zero equal to zero. That will be true always. So here what you are going to say is that if A and B are, if A and C are, if A and C are collinear, okay, or A and B, C are collinear where A and C are related to each other by this relation. Okay. Then only these two vectors will become, these two vectors will become equal, else they will not be. Are you getting my point? Collinear. I mean, need not be the same. Most of you are saying A should be equal to C, need not be the same, but they should be collinear. Okay. And collinear, the proportionality thing that should be, you should be getting should be your B dot C by A dot B. And of course, neither B dot C nor A dot B should be zero. Okay. That means A and C should not be B and C and A and B should not be perpendicular. Okay. Is this fine? So I think the derivation of the cross product, vector triple product, that is also clear. Okay. Now just to extend this further, there is one small thing that I would like to ask you. If I have four vectors, let's say I have a cross product of four vectors. Okay. Now please understand here. If you treat this vector as a P. Okay. And think this question to be like this. Then this will give you a vector in the plane of, in the plane of C and D, correct. But if you treat the same thing like this, if you treat the same thing like this, A cross B cross Q. This would give you a vector. Let me write it here. This will give you a vector in the plane of A and B. So, reading this and reading this, what is the conclusion that this vector, so from these two, the conclusion is that this vector is a vector, which is along the intersection of, along the intersection of the planes of A and B and plane of C and D. So, this question has been asked in certain competitive exams. So, please note this down. So, they will give you a question like this. They will say A cross B bracket, cross C cross D is a vector in the, in which of the following. So, you will see only in the plane of A and B, only in the plane of C and D, or they will give you in a intersection of the planes of A, B and CD. None of these. So, you have to mark this option, correct, that it is in the intersection of the planes of AB or the plane containing AB and the plane containing CD. Okay. If these two planes are parallel, the cross product will be zero. The cross product will be zero. Why? Because A cross B will be normal to the plane of AB and C cross B will be normal. So, they themselves will be parallel. They themselves will be collinear. So, this cross product will be a null vector in that case. Okay. Correct. All right. So, I think we should go for a break now. Small break. We have a class of 730 today, right? Is anybody who wants to leave early? Because yesterday I asked all of you said, we want to have it till 730. Let's decide it now because if it is still 730, I will give you a break for 10, 12 minutes. If it is 7 o'clock, I'll just give you a five minutes break. Okay. Let's do one thing. I leave you at 715. Okay. Not your, not mine. And I'll give you a eight minutes break. Okay. So, let's meet at, let's meet at 623 p.m. Have some tea, coffee, tea, coffee, sir. Okay. We'll meet in eight minutes. No, no, eight minutes. Two minutes. I mean, two minutes and Maggie will be ready. So, this is another expression which basically results into box ABC square, which is actually box product of A cross B, B cross C and C crossing. Okay. So, let's derive this result. And obviously keep this in mind because it is useful in many questions. Do let me know once you're done with the proof. Okay. So in the interest of time, I'll be also pitching in over here. So this expression is what this expression is a cross B. Okay. Dot the cross product of this with this. Now here comes the role of your vector product. Let's say for the time being called B cross C as a P. So this is P cross C cross A. Okay. So can I say this is going to be, I mean, using my vector triple product expansion, I can say this is P dot A times C minus P dot C times a. Okay. Let's open this bracket up. So this will be now see P dot A is like a scalar quantity. So you're doing a cross B dot scalar times C. So that scalar quantity will not influence that will be separated out and this will give you a cross B dot C. Correct. Similarly, on this side also you can see P dot C times a cross B dot A. Okay. But we know for sure that a cross B dot A would be a zero because in this, you are taking a box product of three such vectors out of which two are equal. Okay. So that only leaves you with P dot A. And this is box ABC. And what is P itself P is B cross C right so B cross C dot a box ABC is as good as saying box BC a box ABC and box BC and box ABC are equal. So that is box ABC square. Any questions any concerns related to this proof. Simple question. Anything that you would like to ask note down please do so. Could you scroll down? Yeah, why not. Done. Okay. Let's take another one. Okay. Let's take this one. If vector A is to I plus J plus K vector B is given to you as following. Okay. A cross B is given a dot B is given find B one plus B two plus B three. Find B one plus B two plus B three. Should I run the poll for this? Do you have a poll? Do you mind having a class on Sunday this coming Sunday? Can we have a class coming Sunday? Morning 8 to 11. I was saying yes sir. Monday math exam. 8 to 10 bio also sir bio also have class. Okay. What time is bio class three to six. Yeah. If you like solving math question. Yes. 8 to 930. 8 to 930 Shraddha will it be okay? One and a half hours like we used to have earlier. At seven o'clock I can't wake up. I mean I definitely wake up that early but seven o'clock you understand. No. A lot of important activities happen during that time. 7 8 to 930 is good. Okay. Chalo one and a half hours is also good. Okay. Three of you have responded. If I'm not able to take up questions related to. I mean vector equation and. Yeah. Vector equation then I'll call for a class. Else there might not be a class also. So we'll see at 715 where are we placed? Depending on that I'll keep a class. Okay. Meanwhile, please focus on this question. Let me know if you have got any results. I can see four of you having responded. Last. 45 seconds. Five. Four. Three. Two. One. Go. All right. Very few of you have responded by the way only seven eight of you have responded. Okay. Let's hear it out from somebody. Okay. Anant. Which option have you gone for? If at all you have gone for some option. Okay. I don't think the answer is B. Okay. I don't think the answer is C. Let's see. Anant versus Janta who is correct. So here. I would request you to do this operation once because I have an information about a dot B. Okay. I have an information completely about a dot A also because a is given. Okay. So I'm barking and basically relying on this information. So a cross a cross B is what? A dot B times a minus a dot a times B. Correct. In short. A cross A cross B, which is actually a determinant that I can form. A cross B J K 211 and five to minus 12. This determinant of course it'll come out as a vector is equal to 11 times a. Okay. Minus a dot a dot a would be two square plus one square plus one square, which is six B. Be something which I don't know. Okay. If I find B, I'll be getting B1 B2 B3 there itself. And hence I will be able to sum that up. Okay. So what I need to do is to get this B thing. Okay. So if you expand this, this will become a minus 12 minus four is minus 14. How is that the LHS? Why not? A is given a cross B is given. So what is the cross? How do you write the cross product of two vectors? Okay. Next is minus J. That will be minus 24 minus five, which is going to be minus 29. Okay. And this is going to be 22 I plus 11 J plus 11 K. Okay. So let's simplify this. So minus 14 I plus 29 J minus K is equal to 22. I plus 11 J plus 11 K minus six B. Okay. So let's do the needful. So six B is going to be 36. Minus 18 J plus 12 K. So B is going to be six I minus three J plus two K. In short, B one is six B two is negative three and B three is two. So what is the sum of what are the sum of these five option number B is correct. Okay. So unant your answer was B only. Yeah. And that is absolutely right. So unant is correct. Janta is wrong. Well done. Good. Is it fine? Any questions? Finding this manually is hard. Yes. Absolutely. That is why we are required to use these kind of tricks. Okay. Let's take another question. Any question, any concerns here? Anything that you would like to copy ask. Please do it. Okay. So in that case, we'll go to the next question. Okay. Let's take this one. This question says box ABC. ABCD B vectors says that box ABC is two. And this, this, this is minus mu D find me value. Okay. Options. Sorry. Paul. I'm putting it on. All right. Three people responded so far. Good. Okay. Let's discuss this question. If you want to give your response. Please do so in the next five seconds. And the count of five. Because I can see only 12 of you having responded so far. Five. Four. Three. Two. One. There is one more question. Four. Three. Two. One. The response has been mostly to in favor of option C. Okay. We'll see which option is right in this case. See this question is actually a very interesting one. First, let us try to figure out what would be the simplified version of this expression. Okay. And that is not difficult to find out. Let's call this as P. Okay. So this is P. Cross C cross D where P is your A cross B. So let P be your A cross B. Okay. So this will give you P dot D times C minus. Minus. P dot C. Times D. Okay. So P dot D times C and P dot C times D. Okay. So P dot C times C. Times A B D. Box A B D. Box A B D times C minus box ABC. Times D. Correct. In a similar way, let us also write down the expression for the others as well. So what will be B cross C. Cross A cross D. I mean, I'll just reduce my effort here. This will be if I'm not mistaken box BC. D. Times A. Minus. Box. B C A. Times A. Oh, sorry. Times D. Not a B. Correct. Am I missing out on anything? Please do let me know. Similarly C cross A. Cross B cross D. I can write it as quickly. C A D times B. Correct. And again, minus. C A B times D. Now, if you add them. If you add them. The required expression, let me name it as E for the time being this whole thing I'm calling as E. Okay. The required expression is going to become. Is going to become. Or let me write it like this summation of. Summation of. A cross B. Cross C cross D instead of calling it as D. I will call it like this. That will become. A B D times C. Plus. BC D times A. Plus. C A D times B. Okay. And correct me. If you see these three terms will be all same, isn't it? So can I say it will be minus three box ABC times D? Will it be wrong to say like that? No. So these three I have added over here. And these three I'm calling it as minus three ABC box C because box ABC box BC and box AB are equal. Okay. No objection so far. Okay. Now see this information is a very vital piece of information for us, which not only tells you the value of this guy. Okay. So now it becomes minus. Six D over here, which I'll change over here minus 60. Because it has been given that box ABC is two. But this also tells you that ABC are non-coplanar vectors. Okay. ABC this also tells you that ABC are non-coplanar vectors. Non-coplanar means they are linearly independent vectors, which means any vector D could be expressed as a linear combination of these three vectors. Right. You can always write a vector in 3D space as a linear combination of three non-coplanar vectors. Does everybody agree with this? This again comes from your concept of linear dependence, independence and the concept of coplanarity and non-coplanarity. Okay. So now this vector D. If I assume it like this. Okay. Let's do some simple activities over here. So let my vector D be this. Let my vector D be this XA plus YB plus ZC. Now you have something called ABD. Right. Okay. ABD. So we'll do one thing. We'll do a small activity here. We will take dot product with A cross B like this. Do this expression. Okay. In short, I'm saying box ABD is equal to, now see here. Everybody please pay attention. What is going to be A cross B dot XA? Zero. A cross B dot YB, that is also zero. But the last term will give you Z box ABC. In short, what do we realize is that Z term is half ABD. Okay. In short, ABD is 2 Z. That means this term that you have written over here, this is actually 2 Z. So it is 2 Z C vector. Okay. Similarly, if I had taken our dot product with D cross C throughout, let me write it here. Similarly, had I taken the dot product with B cross C? Okay. What would I have got over here? So B cross C, I would have just got XBCA. XBCA means to that means box BCD is to X. Okay. That means this term is actually to X. So it's 2 XA. Similarly, this term will become 2 YB. And there's already a minus 6D. In short, what I have written, if you see clearly, I have actually written twice of XA YB ZC, which is actually nothing but your D vector. So I've written 2 D minus 6D, which is actually minus 4D. Okay. So your mu will actually become 4. Option number D is correct. Okay. Because as per the question, this is claimed to be minus mu D. So mu has to be 4. Yes, I'll explain once again, Aditya. See, when you took, in this expression, you take dot product with A cross B. Okay. So you took a dot product with A cross B on both the sides. So this will become box ABD. Isn't it? Isn't it? Yes. This will become this into this will be zero. This into this will be zero. This into this will give you Z times box ABC and box ABC is given to you as 2. No. Okay. So this is 2 Z. So Z will become half of this. Got it now. Is it clear to everybody? So this is a beautiful piece of question actually. Okay. So answer is four, not three. Like most of you gave it. I think majority of you went for option C, which was I think three. Yeah. Answer is four. Any question, any concerns here, please do let me know. All right. So we have not, we don't have much time. So I'll be quickly referring to one of the very interesting concept and that is the concept of reciprocal system of vectors will not take much time. Not a very big concept. I hope you have copied this down. Okay. Should I go to the next page? By the way, box CAD is too wide that you have to figure out. I did not figure it out. I directly wrote it, but it is very easy. You take cross product with C cross A. Sorry. Dot product with C cross A. If you take dot product with C cross A, that will only leave you with Y times C AB. Y times box C AB, which is too wide. So two Y is box CAD. That's why I replaced box CAD with two Y. Okay. Now reciprocal system of vectors. Read this concept like a question because it can be asked like a question to you in the J advance type of exam. Reciprocal system of vectors. So let's say I have a BC as three non-copener non-zero vectors. Okay. So let's say these are non-copener vectors. Non-copener vector triad, you can call it. Then vector A dash B dash and C dash are said to be reciprocal of this system of vectors when they satisfy, when they satisfy the following conditions. Number one condition. Dot product of A with A dash must be one. Same should be dot product of B with B dash. And same should be dot product of C with C dash. Okay. They should all be equal to one each. Let me write one here to be. Okay. And the second thing that they must satisfy is that if you take the dot product of A with B dash, or dot product of A with C dash, they should be zero. Same goes with A dash dot B. Okay. This should be zero. A dash dot C, this should be zero. Okay. That means if you take the dot product of any other other than A dash, BB dash and CD dash, it should give you zero. Okay. Like C dash B, that should give you zero. C dash A, that should give you zero. Okay. So if these two set of conditions are satisfied, then this system and this system will be set to be reciprocal system of vectors. Now note here that IJK system is its own, is its own reciprocal. Okay. They are self reciprocal. That means they all satisfy this condition. As you can see, I dot I, J dot J, K dot K is one, one, one each. And if you take I dot J or J dot K or K dot I, that will give you zero, zero. Okay. So these two are, so if some question comes, write down the reciprocal system for IJK. Then the answer is IJK themselves. They are themselves the reciprocal of each other. Now why the word reciprocal came because of this property. Now they're not literally reciprocal. Don't, don't try to think like, okay, A dash is one by A. There's nothing, you cannot conclude that if A dash is one, then you cannot say A dash is one by that will be literally wrong. How do you write inverse? How do you write reciprocal of a vector? That is something weird. Okay. So because of this property, it is called the reciprocal system of vectors. Now a question arises here that if somebody has given me A, B, C, how do I find their reciprocal system of vectors? That means can I find, can I find A dash in terms of A, B, C? Can I find B dash in terms of A, B, C? Can I find C dash in terms of A, B, and C? Can we do that? Can you tell me what should be my A dash, B dash and C dash? If let's say this is known to us. This is given to us, let's say. Okay. Let's say this is provided to us. Can we find A dash, B dash and C dash? If yes, tell me how. I just know that A dash, B dash, C dash are related to A, B, C by this, by this way. Do it, Aditya. Tell me what do you get? You're on the right track. So Aditya is saying let A dash be a linear combination of A, B, C because it has been given that A, B, C are non-coplanar vector. So I can express any vector in terms of linear combination of A, B, and C. Okay. Carry on. You're on the very, very right track. In fact, I would have also done the same. I'm just giving you time to complete your work. I don't want to play a spoiler here. Everybody, in fact, take this as a clue and begin your work. I have to ultimately get my X, Y, and Z my dear. Once I get that, my job will be done. Any success anybody? So much of shooting. Okay. Anybody with any success? All of you please pay attention. Instead of introducing an extra C, will it be wrong to say that I can also write A dash as X A plus Y B plus Z A cross B? Am I right? Because A cross B is perpendicular to B and C. And that makes A, B, C non-coplanar vectors. Yes or no? No, I can always write it like this. It's wrong in that. I have various opportunities to write it. I can write it like this also. Okay. Since you have objected it to it, I will also write it like this. Y, C, and Z B cross C also for some XYZ combination. Yes. XYZ could be anything. Any vector in the plane of any vector, which is a three-dimensional vector can be expressed as a linear combination of three such vectors which are non-coplanar. Okay. So B, C, and B cross C. Of course, B and C should not be collinear. So B, C, and B cross C should be non-coplanar set of vectors. And any vector can be written in terms of that. Okay. You take any one of the expressions. It's going to work. Don't worry. Can this also be done to solve for my A dash? Yes. Any success here? All right. I'm going to, I'm going to leave this as an exercise for you. I'm just going to, because we're running our sort of time. So I'll just give it as an exercise to you. Please prove that. Please prove that your A dash vector will actually come out to be B cross C by box ABC. Okay. And your B vector, B dash vector will come out to be C cross A again by box ABC and your C dash vector will come out to be A cross B by box ABC. Okay. You can clearly see that with these A dash B dash and C dash, these set of properties will be satisfied. For example, if you do A dot A dash, if you do A dot A dash, you will automatically end up getting box ABC by box ABC, which automatically becomes a one. Okay. And if you do, let's say A dash dot B. That will give you B cross C dot B. Okay. B cross B dot B is box BCB that will anyways be zero. So this will be reduced to a zero. Okay. Right. So with this values of A dash B dash and C dash, you would realize that these all criteria is going to be fulfilled. Now, this itself is asked like a question and hence I'm asking you to do this proof. Many a time they will say there is three set of vectors. They will call them as even E two E three. Another one, they will call it as even dash E two dash E three. And they will say that even E two E three and even dash E two dash E three dash satisfy this set of condition. Then write even dash E two dash and E three dash values in terms of even E two E three. And that is what is the expression that I have used over here, but please prove it. This is not a difficult exercise. You have already learned a lot of your operations already. Now, one thing that I would like you to, you know, establish over here is that the box product of ABC and the box product of A dash B dash C dash will multiply to give you one important point to be noted down. Box of ABC and box of A dash B dash C dash that will always give you a one. Right. Now, this is easy to prove because we already know now what is my A dash B dash and C dash, which is B cross C by box ABC. This is C cross A by box ABC. And this is this is a cross B by box ABC. Okay. If you take out. Remember the box property property that if you have something multiplied, it can be taken out so you can take that out as a cube. And this will simply become box ABC down in the denominator square. Okay. And you have now left with B cross C, C cross A and A cross B, which is as good as saying box ABC square. We have already proven this in earlier questions. Okay. So box ABC box ABC square, they will all cancel off giving you a one. I hope this result you remember we had already proven that the box of these three expressions. What is it? It is box ABC square. And there's only one by box ABC square waiting outside. So they will cancel each other out. Okay. So this is a very important property. Please note this down. So both of these set of vectors should be non-coplanar. Okay. It cannot be coplanar because if anybody is coplanar, then the result would be a zero. So you cannot say zero is equal to one. Okay. So even these vectors. These vectors should be non-coplanar. Of course both are reciprocal of each other. Okay. So they should be non-coplanar. Is it fine? Any questions? Okay. Now, still could anybody figure this out or are you still struggling? How did you get A dash B dash E dash is this? Okay. I will not leave you with any kind of this thing. So let me just solve it. See, this question becomes very easy. If you had taken a dash as a linear combination of linear combination of A cross B, B cross C and C cross A. Okay. This question will become super easy if you take it like this, because if ABC are non-coplanar, so will be their perpendicular. Please note that even these three are non-coplanar. The reason being ABC themselves are non-coplanar. So their cross products will also be non-coplanar. Correct? So now here the answer becomes very easy when you take the dot product with A. So when you take the dot product of A and write this as a one, here if you take a dot product with A, only this middle term will survive, which is, which will become white times box BCA. Rest every term will go to a zero. Okay. In other words, your Y becomes one by box ABC. Yes or no? Yes or no? And what about X? What about your X value and Z value? Now see, if you take dot product with B, it should give you zero. Right? But if you take a dot product of B, what will happen? This will become Z times box CAB. Correct? Rest everything will be zero, zero. So your Z will become zero. And if you take dot product with C, what will happen? It will become X times box ABC and all these terms will become zero, zero. So here X will also become zero. So if you put X as zero, Z as zero and Y as one by box BCA, then this expression comes out clearly that vector A dash is going to be B cross C by box BCA. Okay. That's what I had written over here while I was giving you the, yeah. So this is what I had written. Same. Okay. Anyways, similarly you can prove for B dash and C dash as well by writing it in the same way and finding the XYZ corresponding to that. All right. So one simple question we'll take on this. Yeah. Let's take this question. Find a set of vectors reciprocal to this set. So I have given you three vectors here. You may call them as, you may call them as A vector, B vector, C vector. So find reciprocal set of ABC provided they are, yeah, I think they are non-coplanar. You can check it out. The determinant is non-zero. So find the reciprocal set of vectors for this. It's a manual exercise. Okay. So let's say A dash, A dash will reward B cross C by box ABC. So at least find one of them and tell me so that I'm convinced everybody is understood with this process. Don't find all the three because all the three will take some time. We don't have that much time. So B dash vector would be a C cross A by box ABC and C dash vector would be A cross B by box ABC. Just find one of them and tell me the result. Just find maybe this one only. This you can take for homework. Fast, fast, fast. I don't have much time. I have only three minutes and I have to talk about one more question. Half K minus I. Okay. So this comes out to be half minus I plus K. I hope everybody has got it. Okay. Similarly, when you do it at home, please check the B dash vector that you should be getting. This vector should be half minus J plus K. Okay. Just a manual exercise. You don't have to. And this is I plus J. Okay. Just check it out. Check. All right. So the last thing that I would like to take up is one question related to vector equation. Many a times what happens. And this has become a very common trend in many competitive exams that they will give you some unknown vector. Okay. And they will ask you to find the unknown vector in terms of known vectors. Or maybe I'll take an example to explain that. Okay. Let's say something like this. Now this doesn't require a new set of knowledge. Don't get me wrong. I'm not trying to teach you a new concept here. I'm just trying to teach you how to figure out a unknown vector by using whatever operations you have learned. Okay. Let's say this question. This question has said that if A and B are two given vectors and K is any given scalar quantity. Okay. Then find the vector R with satisfies this condition. Okay. So we have to find R with satisfies this condition. That means you have to find R in terms of A, B and K. Okay. How do you find out? How do you find such things out? Please remember. Vectors are not like your numbers that you can take common and all at your whims and fancies or drop them to the reciprocal, et cetera. They don't follow those kind of rules. Okay. So you already know that vectors have a rigid laws of their own. Okay. So R is some vector, which I don't know. But I know for sure that R cross A plus K scalar times R is equal to B. How do I get R from here? So here there is no standard approach. One of the approaches that many books suggest is that you could take that vector to be a linear combination of the given vectors. The way I used to, the way I used to get those reciprocal system like this. So this is a much, you can say commonly seen approach solving for the unknown vector. So why not take this route to solve for our X, Y and Z? Okay. So I'll substitute this over here. So X A plus Y B plus Z A cross B. Okay. Cross A plus K times X A plus Y B plus Z A cross B. Is given to us as a vector B. Okay. Here you must assume that vector A and B are non-colinear. That means A and B are linearly independent of each other. So A and B are non-colinear. Okay. That means A and B are linearly independent of each other. So if you take a cross product, what do you get? You get A cross A, which is a null vector. Okay. So let me not write it down itself. So I'll get Y B cross A. And this will give me Z. You can say A cross B cross A. Okay. And this is anyways copy paste. I'm just simplifying this part. This is negative of A cross A cross B. Okay. And use your formula. That is going to be A dot B times A minus A dot A. That is mod A square times B. Something happened to my pen. Okay. And this is minus Y. You can say A cross B. And here also just copy paste. K X A, K Y B, just Y B, not K Y B. And Z A cross B. Okay. Now understand here that A B and A cross B are linearly independent of each other. So what you can do is you can club up all A together B together and A cross B together from the left-hand side and compare it to the right-hand side. So now let's talk about A cross B's. How many A cross B's we have. We have a minus Y coming from here. And I think we have KZ coming from here. How many B's we have or how many A's we have. Let's try to collect A's first. So A will be minus Z A dot B. That's coming from here and a K X coming from here. And similarly B vector coefficient will be coming only from this term, which is Z. Oh, sorry. Here also. So Z A dot A or Z mod A square. And from here you will get K Y. Okay. This is equal to B vector. Now from here, few things we get. Number one, since there is no A cross B on the right side, that means this guy should be zero. Correct. Yes or no. Right. That means K Z is equal to Y. That is the first equation. Second equation I'll get is K X is equal to Z times A dot B. Okay. And third equation that I will get is Z mod A square plus K Y is equal to one. So these are the three equations you will get from this condition. Okay. Let's try to solve, let's try to solve for each one of these variables. So I'm putting this first one in the last one. I'll get Z mod A square. K Y will become K square Z equal to one. So for from here, I can simply get Z as one by mod A square plus K square. Okay. So Z is obtained now. One third of the work is done. Okay. So Z is obtained now. X can be obtained from here as A dot B times Z. Z is mod A square plus K square. Okay. Z A dot B divided by K. Okay. So that is your X. Y is obtained from here. So Y is K Z. So K Z is K mod A square plus K square. In short, once you know your X, Y, Z, you know, your R vector can be written as. Okay. This is your X into a vector. Okay. In fact, times a vector is not dot with a vector. It's time a vector. This is a scalar quantity. Remember, this is your X. So I'm just using this formula. I'm just using X, A, Y, B and Z, A cross B. So X, A plus Y. This is Y, B. Okay. And Z, Z, C. Okay. So if you want, you can take things out common and write your answers. So this becomes your answer to the question. Is it fine? Okay. So I think I have touched upon all the aspects of vector algebra, but the onus is on you to practice. Okay. Solve more and more question on vectors. Let me tell you vectors. Sometimes they are very time consuming. You have to think of the right operations to break it through. Okay. So Sunday we will meet. But if Sunday if I meet, I will only be able to finish a little bit of lines. So let it be like, you know, we'll not have a class on Sunday. If A and B were not linearly independent, then probably we cannot write it like this. If they were collinear, then basically we will not be able to write it like this because in that case, R will be collinear to A and B also. Okay. So here they should mention that A and B are non-colinear vectors. Something like this must be mentioned in the question. However, this, this book has not mentioned it in a proper way. There are some authors who don't use all the required piece of information. But yes, I believe they should have mentioned that A and B are non-colinear. Else writing it like this doesn't make any sense. Okay. And yes, so with this, we are done with our linear algebra part, not linear algebra, vector algebra part. Of course vector algebra comes under linear algebra only. We will meet maybe after your semester one to complete lines and 3Ds. Okay. Not a big topic, maybe two classes or three classes is what is good enough. So the last week of your board exam, we will take that up.