 So last class we were talking about linear independence and linear dependence of vectors and how are these concepts used for the tests of co-linearity and co-planarity. I think co-linearity we had already talked about. So I'm going to begin this today's session with the tests for, in fact this is a kind of a revision because I had ended up with talking about tests for co-planarity of three vectors. Okay and I had discussed that when you want to portray three vectors let's say vectors A, B, C as co-planar. So if your vector A, B, C are co-planar, one can be uniquely expressed as a linear combination of the other two. Okay I'm repeating this step. One can be uniquely expressed. So C can be uniquely expressed as xA plus yB. Remember this xy pair is unique. This xy pair is unique. Okay now many people ask me sir what will happen if A, B, C were non-co-planar. Okay see if A, B, C are non-co-planar and you want to express C as a linear combination. This is called a linear combination of A and B. You would realize that there is inconsistency in the value of x and y. Okay there will be an inconsistency in the value of x and y. Let me write it over here. If A, B, C, A, B, C are non-co-planar, if A, B, C are non-co-planar if you try to express C as a linear combination of A and B like this you would realize that the xy value would be inconsistent. These values will be inconsistent or non-consistent in the sense that the xy value that you obtain from comparing the i and the j components will not match with the xy value when you are comparing the k component. Okay don't worry I'll take an example to explain the same. Okay as of now just note this down we'll take a few problems on this particular concept. Is that fine? Now when we talk about hey Preeti good morning everyone good morning Devdas. Physics all done? Still going on? Good morning Shahrakh. Preparation never ends actually I feel that till the moment you enter the examination hall everybody is trying to revise or trying to you know mug up something with the books in their hand. So preparation never ends till you enter the examination hall. Problem solving part of it is done. Okay yeah I think Je has definitely given you an exposure for problem solving. I think you need to work more on derivation presentation of answers following the directives of the question. Right many of us don't distinguish between the directives of the question. What, how, explain, illustrate, describe for everything we write the same answer which is not acceptable actually. If you see the the ones who score very high marks in the board exams they follow the directives of the question properly. Anyways I know I know all the best all I can say is I can pray for you all the best guys. Fine so this is the test for coplanarity of three vectors don't please remember we'll not talk about coplanarity of two vectors two vectors are always coplanar. Okay in a similar way I'll not talk about test for coplanarity of four position vectors. So let me name this now we'll talk about test of coplanarity test for coplanarity of four position vectors or we can say four points position vectors remember they are nothing but four points okay or points remember three points will always be coplanar okay you can always form a plane passing through those three points always okay. Now we say the test is if let's say point A whose position vector is A point B whose position vector is B point C whose position vector is C point D whose position vector of D if these points are coplanar okay remember the previous concept take make any three vectors from these four points let us say let's say I make let's say I make these three vectors AB vector AC vector AD vector okay so what I did I made three vectors by choosing any two sets of points okay from these four position vectors if these four points are coplanar then let me tell you one can be expressed as a linear combination of the other two uniquely uniquely X and Y pair is unique X and Y pair is unique just like we discussed the previous concept of coplanarity of three vectors so what I'm doing I'm just extending that concept of coplanarity of three vectors to coplanarity of four points okay both are the same thing what I'm doing from the four points I am creating three vectors and I can always express one vector uniquely as a linear combination of the other two like this okay if these four points are non coplanar then what will happen you would realize that the X and Y values will become inconsistent or non consistent again I will take a test on this question on this no worries this is the first test okay there's one more test which is slightly more complicated this says says that if ABCD these four points are coplanar then a linear combination of these four vectors let's say XA YB ZC plus let's say UD if you put to a null vector then then X plus Y plus Z plus U would be zero where all XYZ you are not zero okay now this test is slightly complicated okay but it's good to know this in case you're required to solve the problem by this method okay now just an extension of this concept just an extension of this concept let us say I have these four position vectors given to me as this let's say A is A1I A2J A3K let's say B is B1I cap B2J cap B3K cap let's say C is C1I cap C2J cap C3K cap and let's say D is D1I cap D2J cap D3K cap okay now a linear combination of these four position vectors when compared to null vector okay should give you should give you X plus Y plus Z plus U equal to zero where all XYZ and U are not zero okay in other words I'll show you another way of looking at it so when you do X A1I A2J A3K plus Y B1I B2J B3K plus plus Z C1I C2J C3K plus U D1I D2J D3K equal to null okay now please note collect your I terms together collect your J terms together and collect your K terms together so your I term will consist of X A1 Y B1 Z C1 plus U D1 J terms will consist of X A2 Y B2 Z C2 U D2 okay and K terms would consist of X A3 Y B3 Z C3 plus U D3 equal to null now remember IJK are linearly independent they are linearly independent which means if you are comparing such a vector to a null vector the components of each one of them should be 0 0 0 so this should be 0 this should be 0 and this should be 0 getting my point if these three are 0 if these three are 0 let me write it over here you get a system of equation like this you get X A1 Y B1 Z C1 plus U D1 equal to 0 then you get X A2 Y B2 Z C2 plus U D2 equal to 0 you get X A3 Y B3 Z C3 plus U D3 equal to 0 and apart from that you should have X plus Y plus Z plus D should also be 0 now note that this is a system of this is a system of homogeneous this is a system of homogeneous linear equations okay and it should have a non-trivial solution it should have a non-trivial solution why does non-trivial solution see what have I written over here X Y Z U should not be all 0 there should not be all 0 means there has to be a some non-zero values of XYZ also existing which means there's a existence of a non-trivial solution trivial means all of them are 0 which is not acceptable okay if it has to have a non-trivial solution what have we learnt in Kramer's rule we have learned in Kramer's rule that the determinant formed by the coefficients of these variables now here the variables are X Y Z and U okay so the coefficients formed by the determinant formed by the coefficients of these variables okay I'm sorry once again yeah a1 b1 a2 b2 c2 and a3 b3 c3 d3 and 1 1 1 1 this should be equal to 0 okay I'm sure most of you would not like solving a 4 by 4 determinant however it is easy to solve as you can see there is a 1 1 created in the last row so you can take any one column and subtract it from the others to create zeros and start expanding it with respect to that row where zeros have been created but this method is not preferred by students most of the students will prefer this first method okay so this is a slightly complicated a slightly challenging method okay but it's good to know it okay whether you use it or not that is a different matter altogether but you should be aware of this concept is that fine so if three sorry if four points are given to you and they're asked they asked you whether these four points are coplanar all you need to do is quickly make up this determinant and check whether it is coming out to be zero or not if it is not coming out to be zero your conclusion will be they are not coplanar if it is coming out to be zero that means the four position vectors given to you are coplanar that means they all lie on a plate okay let us take some examples to understand this properly by the way any questions with respect to this no sir everyone is clear okay great let me begin with this question if the vectors if the vectors 4i plus 11j plus mk 7i plus 2j plus 6k and i plus 5j plus 4k if these three vectors are coplanar if these three vectors are coplanar find the value of m find the value of m please tie this out for one minute then I'll help you out I'll do this problem in two ways by the way this concept is going to be used again when we are talking about scalar triple product STP shiram says 10 Devdas says 10 please send me your answer in private okay so that others don't read your answer and they're not influenced by it okay now you have sent no problem so that I can respond to your answer I can say right or wrong without others knowing that what you have responded okay so let's give your answer in private to me so that I can say yes or no to your answer by the way let me tell you Devdas and shiram both of you are absolutely correct all of you see the different ways in which I am going to solve this question Adwiyath is also correct the first method that I'm going to use I'm going to use the methodology which I just now explained to you if these three vectors are coplanar one can be uniquely expressed as a linear combination of the other two so let's say I express the first vector you can do it for any vector it's not like I have to take the first vector always take any vector as a linear combination of the other two so let's say I express the first vector as a linear combination of the other two if I do that there must be a unique value of x and y we should satisfy it okay so for that what I'll do is first I will compare the I coefficients so I is 4 on the left hand side on the right hand side it is 7x plus y it is 11 on the left hand side and it is 2x plus 5y on the right okay and it is m on the left hand side and it is 6x plus 4y on the right hand side okay now this system of equation must be consistent consistent means the solution that you obtain from the first two whatever solution you obtain from this must satisfy the third one it must satisfy the third one getting my point so for the first two we can use our simultaneous equation okay so let's multiply the top equation with 5 so 20 is equal to 35x plus 5y and 11 is equal to 2x plus 5y let's subtract it 9 is equal to 33x which means x is equal to 9 by 33 that's nothing but 3 by 11 okay put it in any one of them so 4 is equal to 7 that is 21 by 11 plus y so 4 minus 21 by 11 is equal to y that is 44 minus 21 by 11 is y that's nothing but 23 23 by 11 correct now these two must satisfy the third one so put this value of y put this value of y and x in the third equation so when you put that you get m is equal to 6x 6x means 18 by 11 plus 4y 4y will be equal to 92 by 11 okay that's nothing but 110 by 11 that's 10 so yes m value is going to be 10 okay this is one way to solve it okay other way to solve this would be okay try to remember try to recall try to recall the condition for try to recall the condition for concurrency of three lines concurrency of three lines what is the condition for concurrency of three lines the condition for concurrency of three lines let me write down the equation of the line in the general form like this. So let's say these three lines if these three lines are concurrent, if these three lines are concurrent, what are the meaning of concurrent? Concurrent means they will all pass through the same point. So let's say line number one, this is line number two and this is line number three. All of them pass through the same point. That means there is a consistency in the solution of the three lines. In that case, the determinant formed by these coefficients, the determinant formed by these coefficients of the variable and constant term, this is going to be zero. Just try to recall this. You have already done this before. Right? Now in a similar way, if you look at this situation, okay, if you look at this situation, you are trying to make these three set of lines concurrent, right? You are trying to make them concurrent. Isn't it? That means they have to have a common solution. That's what we mean that it should be consistent. Correct? So if you want to make these three line concurrent, that means they should their coefficient must satisfy the fact that this determinant must be zero. Correct? So if you form a determinant, let me write it over here. If you form a determinant, it'll be 7, 1. Okay. Many people will write it minus 4. Not a problem. Let me write it as minus 4 itself. Okay. This is 6, 4 and minus m. Now you are equating it to zero anyhow. Okay. So does these minus sign matter? Do these minus sign matter? No, they don't. Okay. So these negative signs they don't matter. Correct? So just remove them. Okay. So this should be zero. Right? If you further reshape it, if you further reshape it, you could write it like this. 7, 2, 6, 1, 5, 4, 4, 11 m. Remember, you can always transpose a determinant without changing its value. Correct? Guys, there's something very important that we observe over here. These just become the i, j, k components of the vectors that we are dealing with. Right? So isn't this a much faster and this is advisable to use this method in order to prove the concurrence in order to prove the co-planarity of three vectors. Okay. Rather than finding this x and y value, etc. So if somebody decides to use this method, let's see what result he obtains. Let me expand with respect to the third row. Plus, minus, plus. So 4. This will become 8 minus 30. 8 minus 30 is minus 22, minus 11. This and this goes 28 minus 6. Okay. And plus m. This will be 35 minus 2, which is 33. This is equal to 0. You can drop the factor of 11, I guess, from everywhere. So 4 into minus 2, this will be minus 11 into 2 plus 3 m. Okay. So 3 m is going to be 30 m is going to be 10. Straight shot. We got our answer. Isn't it a much faster and efficient way to solve this problem? Okay. Anybody having any doubt about it? Guys, let me tell you that these concepts are not taken very well by students in the first go. Okay. So in case you have any concerns, any doubts, please stop me right here and get it clarified. Is that fine? Well, now we'll go on and take a problem where we are talking about coplanarity of four points. Okay. So let's take a question on coplanarity of four points. So here goes the question. Find the value of lambda. Find the value of lambda for which these four points, for which these four points are coplanar. Okay. Question is find lambda. Please respond privately to me so that others are not influenced by your answer. Remember, these are points. These are position vectors. They are not vectors. These points are position vectors. They are all position vectors. Position vector means points. Perjuel is not correct. Perjuel, please check your working. Arka is correct. Yesh, no, it is wrong. Arka is your correct. Your answer is absolutely correct. Adhashiram, Devdas, Adwet, please respond. So you're responding privately, so don't hesitate in answering. Let it be wrong. That's fine. No Kavya, that is wrong. Ishika, you are absolutely correct. Shorak, no, it is wrong. Shiramya, no, wrong. Just check your working once again. I'm sure you must have missed out a sign here and there. Not a rocket-sized concept. Okay, glad to see that. Many of you have responded. No Lakshya, that is wrong. Okay, shall we take this up? See, two methods we have discussed so far. Okay, I'll do, I'll take both the methods and explain you how does the answer come. Okay, so let me call these, let me call these points as point A, B, C, D. Okay, formulate any three vectors. Let's, let's formulate A, B vector, AC vector and AD vector. Okay, we know how to formulate vectors when we know their position vectors, right? For example, for AB vector, it's position vector of B minus position vector of A, correct? So we know that AB vector is OB minus OC, sorry, OA. Remember, OB means position vector of B, which is what has been stated to you. Okay, great. So AB vector would be at this vector minus this, which is going to be minus I minus J and plus 4K. Let's find AC vector now. AC vector is position vector of C minus position vector of A. Okay, let's see what it is. This is going to be I plus J minus K. Okay, AD vector will be position vector of D minus position vector of A. So minus I minus lambda minus 3, or you can say minus lambda plus 3J and plus 7K. Okay, now you just use the condition for the coplanarity of these three vectors. Remember the method which I told you, which was the faster of all the methods, make a determinant with the IJK components of these vectors. So minus 1, minus 1, 4, 1, 1, minus 1, minus 1, minus lambda plus 3, 7. Okay, this should be equal to 0. If your vectors AB, AC, and AD are coplanar. So these three vectors are coplanar. Automatically, these four position vectors will also become coplanar. No Shiramya, no Devdas, that's also incorrect. Okay, so just quickly check whether we have done everything correctly. So far, AB vector is minus I minus J plus 4, that's correct. AC is 1, 1 minus 1, that's correct. This is going to be minus 1, minus lambda, that's correct. Okay, let me expand this. Let's expand with respect to the third row. So minus 1 times, hide this, hide this, 1 minus 4 is going to be minus 3. Okay, then lambda plus 3 times. Okay, hide this, hide this. So 1 minus 4 is again a minus 3 plus 7 times. Hide this, hide this. So it becomes minus 1, minus of minus 1, which is going to be 0. This clearly implies minus 3, lambda plus 3 is equal to minus 3. Correct? That means lambda plus 3 is minus 1. Oh sorry, 1, not minus 1. So lambda value is minus 2, lambda value is minus 2. Arka was the first one to give this answer correctly. Is that fine, guys? Did you figure out your mistakes, everyone? That is more important. Where you went wrong? Was it a silly mistake? Was it a conceptual mistake? What type of error did you make? Okay, so always analyze your mistakes. Now this is one method. The other method is what we had discussed in the second method of finding the coplanarity of four points. So let's discuss that. Sign error, see? Be very, very careful with vectors, matrices, determinants because you deal with a lot of elements. You deal with a lot of numbers. Okay? Okay, no worries. So the next method that we are going to take is that non-trivial solution method. So in non-trivial solution method, what did we discuss? We discussed that. We discussed that the determinant made by the coefficients of i, j, k like this. This must be 0, right? Now, ultimately, it's a 4 by 4 determinant and we need to solve it. Okay, so what I'm going to do is, all of you please see the operation that I'm going to perform. I'm going to do r2 as transformed as r2 minus r1 plus r4, correct? So I'm going to add these two and subtract it from the second row. Okay, see, beautifully what will happen? So 1 plus 1, 1 plus 2 is 3, 3 minus 3 will become a 0. By the way, let me copy everything as such because only the second row is getting transformed. Other row are as such, so let me copy it. So 3 minus 2 plus 1, 0, 2 minus 1 plus 1, 0, 4 minus 3 plus 1, 0 and minus lambda, this should be 0, correct? So minus lambda minus 2 should be 0, correct? Is that fine? Now, it is very obvious from here itself to say that if this determinant is 0, okay? One possibility is that this term should also be 0 because it will make the entire row 0. That means minus lambda minus 2 is 0. That means lambda could be minus 2 and straight away we got our answer. Okay, so sometimes this method also is fast. So don't think that, oh, this is a complicated method. I will never follow it. That's why I told you keep your options open. You never know which method is going to give you a faster answer. Please type clear on your chat box if it is clear to you all. Clear, okay, awesome. Now we are going to talk about dot product guys. So today is going to be the class which is revolving around dot product, okay? I'm sure you would have done it umpteen number of times in finding the work done by a force, correct? So today I'm going to begin my discussion with dot product. Dot also called scalar product. So when we say we are finding the dot product or scalar product of two vectors a and b, right? This operation is defined as mod a mod b times cos of the angle between the two vectors. I'm sure everybody knows what is how do we define the angle between the two vectors angle between two vectors is defined as the shortest angle when these two vectors have been made co initial. So when these two vectors have been made co initial, the shortest angle between the two vectors is defined as the angle between the two vectors. Remember, many of us have a very wrong notion about angle between two vectors. Are you getting my point? Okay, so this theta can never exceed this interval. It has to be somewhere between zero to pi only. Okay, because the shortest angle. Okay, note that there was a situation where I asked the student, what is the angle between these two vectors? Okay, and he told me this is the answer. Is this correct? Is this correct? I'm asking you guys, is alpha the angle between these two vectors a and b? Is this correct? Vinda is saying yes. Vinda, no, this is not correct. Okay, because they have been not made co initial, co initial, we have to make it like this. Okay, then this angle beta is the right answer, not alpha. Okay, anyways, just a simple thing that but many people have a wrong notion about it. Okay, now first of all, what is the interpretation of scalar product? What does scalar product actually give you? Okay, if you look at this diagram, let me draw it again. Let's say this is your vector a and this is your vector b. Okay, vector a vector b. Let's say this is the angle theta between these two vectors. Okay, if you look at this term carefully, mod b cos theta. Okay, mod b is nothing but the length of this line. Right? So the length of this line is mod of b, isn't it? Correct? If you do mod b cos theta, it is just like saying you are finding this length. Okay, so let me name it o b a m. So basically your o m length is mod b cos theta. Okay, this is read as projection of b in the direction of a. Okay, so mod b cos theta is nothing but it is the length of the projection of b in the direction of a. Correct? So when you're doing a dot b, you know what you're doing? You're multiplying the magnitude of a. So this is the magnitude of a with mod b cos theta, which is nothing but projection of b in the direction of a. Are you getting it? That's why this concept is used for finding the work done by a force. So when there's a force, let's say this is my force, okay, and this is my displacement. What do you do? You try to find out the displacement of the object in the direction of force to get the work done. Right? You could also see this formula in an alternate way. If you see this formula could also be seen as mod a cos theta times mod b. What is this? This is nothing but you are projecting a in the direction of b. Okay, so this length over here, I'm showing it with yellow, this length over here, this length over here is mod a cos theta. Okay? So what you're doing, you are multiplying the magnitude of b with the projection of a in the direction of b. So it doesn't matter whether you are finding the work done by taking the projection of the force in the direction of displacement or projection of the displacement in the direction of the force and multiplying it with the magnitude of the other. Correct? And for the very same reason, we have a very important property that dot products are commutative in nature. So it doesn't matter whether you do a dot b or b dot a, the result is the work done by, you know, let's say a is a force and b is a displacement. Are you getting my point? Okay? Many a times our direct question will come in your exam, what is the projection of a vector in the direction of another vector? Can somebody tell me if a and b are known to you, how would you directly find the projection of one vector in the direction of other? So let's say I want to find out, find the projection of, okay, let me take this as a question for you. Let me pull out a question on this. Let's take this question. The question is saying that if the scalar projection of this vector on this vector is one by root 30, then find the value of x. Okay? How do we solve this question? So if you're trying to find out, let me call this, this as a. Okay, let me call this as b. Hope you can read the yellow written on white. Okay, so you're trying to find out the projection of a on b. How do you find this out? You just now figured it out that it was mod a cos theta. Correct? Remember, a dot b is nothing but mod a mod b cos theta. Correct? So mod a cos theta, that is what we are looking out for, which is my right hand side, could be directly written as a dot b divided by b divided by mod b. Correct? If you look at it carefully, it is a dot b cap. Correct? So basically the projection of a on b is nothing but it is the dot product of the vector with the unit vector along b. Right? Just remember this formula. Similarly, if you ask what is the projection of b on a, what you will say dot product of b with the unit vector along a. Okay? A direct question can come just like I have given you this question. Okay? In school, especially in UTs and all, they can directly ask a question, find a projection of a vector along a given vector. Okay? So let us quickly complete this. Not a difficult question. Please answer privately. Find the value of x. Okay? I've started writing the answer. Okay, Prajwal has given an answer. Prajwal, you are correct. Shiram also you are correct. So only two people out of so many attendees. Guys, come on. Good morning, everyone. Wake up. Yes, one more response I need before I start solving it. Just a 30 second problem, not more than that. So sorry. Absolutely correct Ishika. Very good. All right. So all you need to do is take a dot the unit vector along b. Right? So what is the unit vector along b? So let me write xi minus j plus k dot 2i minus j plus 5k by root of 30. If I'm not wrong, this is given to us one by root 30. Okay, root 30 root 30 gone. Now guys, I'm sure I have not told you officially how to find dot product when two vectors have been given to you in resolve form. But I'm sure in physics you have already done it, right? So when two vectors are given in resolve form, what do we do is we only multiply the i with i j with jk with k because this other would be zero. I'll show you how it comes. Okay, so 2x plus one plus five is equal to one. Correct? One one gets cancelled. So your 2x becomes negative five. So x becomes negative five by two. Absolutely correct. Those who have answered. Okay, by the way, my main agenda was not to give you a problem right now because I was just in the process of explaining the theory. But since this projection thing came up, I thought I would ask you to solve one. But is this clear? What is the meaning of the dot product of two vectors? Okay, no doubt about it. Great. Now we look into the properties of dot product. That's going to help us solve the question even better. Properties of dot slash kela product. Okay, the first property is a dot a is nothing but mod a square. Okay, this is a very important property. Let me tell you this has helped to solve many questions that will be asked to you in school exams. Right? And I don't need, I don't need to give you the proof for this site because it's a dot a is mod a mod a cos zero, cos zero is one. That's why it is this. Okay, now few important results i dot i will be one j dot j will be one. And of course k dot k will be one. Okay, second property. If a and b are let's say non zero vectors, a dot b is positive. That means the angle between these two vectors is acute. Okay, if a dot b is equal to zero, that means the angle between them is 90 degree. That means a is perpendicular to b. That's why when the work is asked to be done by a force which is causing a perpendicular displacement to it on a body, we say that the work done by that force is zero. Okay, and if your a dot b is negative, remember the angle between them is obtuse. That means instead of doing work, work is done on it. Is that fine? Property number three. I already discussed that dot product is commutative. So I'm just repeating the same thing, just to have a note at one place. Dot product is distributive over addition and subtraction. So you can write it as a dot b, a dot c. And you can do it in any order also. You can do it this also. So it's b dot a plus b dot c dot a. And both results would be the same. Both results would be the same. Fifth property, if two vectors have already been multiplied with a scalar quantity, you can always pull out the scalar quantity separately. Okay guys, one important thing which I again missed out. When we say this property, it implies i dot j is zero, j dot k would be zero, i dot k would be zero. So if you remember i, j, k are orthogonal triads. So dot product of any two of them would be zero. That's why when I was explaining you the dot product of two vectors which have been resolved into i, j, k components, I never bothered to multiply the i component with the j component of the other because I know at the end of the day, i dot j would be zero, j dot k would be zero, k dot i would be zero. So I only multiplied i with i because that will give you a one, j with j because that will give you a one, and k that k, k with k that will also give you a one. So if you have a1, i, b1, j, c1, k, and you're finding the dot product with a2, i, b2, j and c2, k, all you need to do is just do a1, a2 plus b1, b2 plus c1, c2. No other multiplication is required because ultimately they'll give you zero because of this property. You've already done this in physics, guys. I'm not teaching anything new to you here. I'm sure you have done that in physics. Sixth property, when you say a plus b square, it is implied as you're doing mod a plus b square. Both mean the same thing, so don't get confused. Mod a plus b square means you're taking dot product of a plus b with a plus b, which is nothing but mod a square mod b square plus 2 a dot b. Similarly, if somebody says a minus b square, it is as good as saying mod a minus b square which is a minus b dot a minus b, which is nothing but mod a square mod b square minus 2 a dot b. Is that fine? Simple? Okay. Last property, which is actually very simple one, but many people are not able to interpret it. Any vector r could be expressed as r dot i i r dot j j r dot k k. How? How? It's very simple. It's very simple, actually. If you have any vector r, which is nothing but let's say x i plus y j plus z k. You know that r dot i will actually give you an x, r dot j will actually give you a y and r dot k will actually give you a z. So your r could be written instead of x, you could write r dot i itself. So r dot i i, remember this is scalar multiplication. r dot i is a scalar quantity. i cap is a vector quantity. So this is a scalar multiplication plus r dot j j plus r dot k k. Remember this also, sometimes it is useful in solving few questions. Any doubt, any questions, please type n o q if there is no questions. Hey, Andrew is also there. Good morning, Andrew. Okay. Now what are the uses? What are the applications of dot product? Where do we all use it? The very first application is to find the angle between the two vectors. It helps us to find the angle between two vectors. Remember dot product is the most efficient method. Preeti is asking for the previous page. Okay, Preeti, I'll go to the previous page. I don't worry about taking the notes. Anyways, PDF will be sent to you. Do you want me to scroll up, down something? Okay, you got it. Great. Go back to seven. Yeah. So dot product is a very efficient method to find out the angle between two vectors. So as you can see from the definition itself, this is mod a mod b cos theta, okay, which is nothing but cos theta is a dot b by mod a mod b. So we normally use dot product to find out the angle between any two vectors. Now many people will ask why not vector product? Okay, those who have studied vector product while studying moment of a force, you know that mod a cross b is mod a mod b sin theta. So from here theta also can be found out. So why this is not a preferred method? Why dot product is a preferred method? Can anybody answer? You can unmute yourself and talk. This is preferred over cross product. Of course, one reason is you have to find the cross product. Okay, that is fine. Let's say I'm able to find it, but why don't we normally use cross product to find the angle between two vectors? Why dot product is used to find the angle between two vectors? Anyone? Common sense. Why do we use dot product to find the angle between two vectors and not the cross product to find the angle perpendicular? How perpendicular can be found out from both the method, right? It's common sense. Why don't why don't we use cross product to find the angle between two vectors? So does it have something to do with the fact that dot products are scalars and cross products are vectors? See, I'm saying mod a got mod a cross b. This is also a scalar quantity because this signifies. I didn't see that. I didn't see that. Sorry. Sorry. Sorry. The answer for this is very simple. Cross product has a sense of direction. Cross product has a sense of direction. I'm talking about the magnitude of the cross production. Okay, sign zero is zero. So what? Oh, you see, very, very simple answer here. If let's say you found out the angle by using cross product and you got sign theta is equal to half, right? How would you know theta is 30 degree or theta is 150 degree? How would you know this, right? You will not be able to know it, isn't it? But when you're talking about the dot product, the moment it goes in the second quadrant, the moment you exceed 90 degree, your answer will be a negative so that you know that what is the exact value of the angle. Are you getting it? So when you say cost theta is half, you know, oh, for sure it is 60 degree. It cannot be any other value. Okay. Had it been 120 degree, it would have been minus half. But in case of sign, since sign remains positive between zero to 180 degree, we will not be able to exactly know what is the angle. Are you getting my point? It's common sense. So let's say sign theta is 1 by root 2. You will never be able to figure out whether it is 45 degree or whether it's 135 degree. Okay. But cost will tell you know through the change of sign, whether it is obtuse or whether it is accurate. Okay. Fine. That's why we don't prefer cross product to find out the angle. Okay. Great. So if your vectors have already been resolved, let's say a is already been resolved as a1i, a2j, a3k. b is already been resolved as b1i, b2j, b3k. When we say cost theta, it's actually a dot b, a dot b is nothing but a1, b1, a2, b2, a3, b3 by under root of a1 square, a2 square, a3 square under root of b1 square, b2 square, b3 square. I'm just writing the expansion of the very same thing. But this is important because we are going to use this at so many places to find the angle between two lines, to find the angle between a line and a plane, to find the angle between planes later on while we are doing 3D geometry. Okay. Next is it is used in some geometrical proofs like the cosine rule. You can prove the cosine rule using dot product. Okay. I'm sure all of you know cosine rule. Right? So let me just draw a triangle for you. So let's say this is a triangle a, b, c. Okay. We know how to name these sites. Okay. We have learned in the cosine rule. Okay. One of the cosine rule says a square is equal to b square plus c square minus two bc cos a. Can you guys prove this? Prove this by using dot product. I'll just give you three minutes. Okay. Time starts now. And once you're done, please send the snapshot of your working privately to my WhatsApp. Once done, please tell me you're done. Any idea how to do this? Anybody who's trying, please type trying if you're trying. Okay. Oh, okay. Okay. No problem, Shiram. Trying, trying, trying, trying, trying. Many of you are trying. Okay. Can we have a minute more to finish up this time? Okay. Time up. So guys, first we'll assign vectors to these sites. So first we'll call them through some name of vectors. So let's say this is vector a, this is a vector b, and let's say this is a vector c. Now the sign is the sign, the direction of these vectors is your call. Okay. I'll actually make it cyclically closed. That means I have said that a plus b plus c ends up giving you a null vector. Okay. Okay. Never mind. Now what we can do is we can say a plus b is negative of c. In fact, I can do since I want to prove this, I'll do other way around. I'll say a is negative of b plus c. Okay. Now remember, these two are negative vectors, negative vectors of each other. That means the magnitude is still the same. Correct. Yes or no? Negative vectors have the same magnitude is just that their sense is opposite. So can I say the square of their magnitude will also be the same? Right. Now try to recall, try to recall the property which we did. And a mod square is nothing but dot product of a with itself. Okay. Very important property I'm telling you, it will help you solve so many problems. So this is nothing but a dot a, this is nothing but b plus c dot b plus c. Okay. So a dot a is mod a square. Okay guys, let me tell you, since you're calling this as vectors. Okay. Remember, mod a is your side length itself. Mod b is the side b itself. Mod c is the length of the side c itself. Okay. Now, if you use the distributive property, it will be b dot b, which is mod b square, b dot c and c dot b will make two b dot c. Remember, b dot c and c dot b are same. They are commutative in nature, that's why. And the c dot c will give you mod c square. Okay. Now replacing your mod a square with a square, mod b square with b square, mod c square with c square. Two b dot c is nothing but two mod b, mod c, cos of the angle between b and c. Now guys, all of you please pay attention here. This is a mistake which everybody would have done. What is the angle between b and c vector? Can somebody write it down on the screen on your chat box? What is the angle between b and c vector? Absolutely correct Ishika. It's 180 minus a. Shiram, you're wrong. It is not a. It is 180 minus a. Remember, to find the angle, you must have made these vectors co-initial. So this is the angle 180 minus a. 180 minus a. Correct. Don't make this mistake. I'm repeating this endless number of times. When you are writing the angle between two vectors, you should have made those two vectors co-initial and then take the shortest angle between them. Okay. So now this is going to be 180 minus a. Okay. So again, I'm writing it back in terms of simplified expression. So this is two b c cos 180 minus a is minus cos a. Okay. So your expression becomes b square plus c square minus two b c cos a. Okay. Hence cosine rule proved. Yeah, you can do that. No worries. Prajol and Shiram, you can do that. Okay. Next is another question coming your way. Okay. I want to see your application. Let's say this is your a vector. Okay. This is your b vector. My question to all of you is, let us say I name this O, B, A. I drop a perpendicular from B on to O A. Let me call it as M. Okay. Find in terms of A and B only, find in terms of A and B only MB vector. Yeah, sure. Please send me the snapshot, Ramya. Find MB vector only in terms of A and B. If you want to share your solution, you can send me the snapshot and please write done if you're done. Absolutely right. Seems to be right. I think just you missed out on A somewhere. Oh my God, cross product and all you don't need. You don't have to go to cross product. Okay. See here, it's very simple, guys. Can I say O B vector is nothing but O M vector. Let me call this vector as O M vector plus MB vector. Okay. O B vector is known to me. O B vector is B. Okay. O M vector, can I find out? Can I find O M vector? You'll say, yes, we can easily find out O M vector. O M vector is nothing but the magnitude of O M or the length of O M. Let me write it just as O M. Okay. Into a unit vector along A. Yes, Prajwal, you are correct. Prajwal is correct. Absolutely. Now, O M is nothing but mod B cos of this angle theta. Okay. I know I'm not supposed to use theta. I'll pull it out. Don't worry. Correct. And A cap is nothing but a vector by mod A. Yes or no. Correct. Now, what is cos theta? Cos theta is A dot B by mod A mod B. So mod B cos theta, I can replace it with A dot B by mod A. So I can make this replacement at this stage. Correct. So the resulting expression will now become A dot B by mod A. Correct. And there is an A over here. And again, there's a mod A so I can just say mod A square. Okay. Now, guys, let me clarify what have I written over here. I have written, I have written A dot B, which is a scalar quantity times of vector A. Okay. I'm not doing A dot B dot A. That would be an incorrect operation in vectors. Okay. Vector algebra doesn't allow such operation A dot B dot A. Okay. A dot B times A, you can say is something like doing lambda A. So this lambda is your A dot B. Okay. Don't misconstrue this as A dot B dot A. That's invalid operation. Okay. So once I've got this, I can use over here, MB is equal to OB minus OM. OB is nothing but B vector. OM is just now we figured out A dot B by mod A square times A vector. This is going to be your answer. Okay. So some of you, some of you have sent the right answer. Very good. Well done. Let's say, take up questions on dot product. I'm going to start with few basic ones first. Everybody please respond privately to me. Okay. So first question coming your way. A, B and C are mutually perpendicular vectors of equal magnitude. Then find the angle between A and A plus B plus C. Okay. We have an answer. Five by four. Oh, I'm sorry. I should not have spoken it. That's not correct. No, that is also not correct. Just check. You're very close. Yes. Yes. Correct. Kavya is correct. Good Kavya. That's correct. Kavya is correct. Others, please don't hesitate to respond. You're responding privately. Don't worry. Nobody will get to know your answer. One by that or just that because that is an invalid operation. Route three is greater than one. That's correct. Adwet. Yes, Lakshya. That is correct. No, that's not correct. Sigosh, Ramya, Brinda, please share your answers. That's correct. That's correct. That's correct, Brinda. Okay. It's time to discuss this. See here. A, B, C are mutually perpendicular vectors of equal magnitude. So first of all, let's say all of them have the magnitude of K. Okay. K, K, K. And they're all mutually perpendicular means the dot product of any two of them is going to give you a zero. Okay. Now, in order to find the angle between A and A plus B plus C, let's say the angle is theta, you need to do a dot product of A with A plus B plus C. Correct. And if you divide with the magnitude of these vectors, right? Now, everything is fine. The only thing that I don't know over here is what is this guy? Mod of A plus B plus C. Okay. Numerator is also fine. A dot A will be your answer. A dot A is nothing but K square. Okay. This is going to be K. But what is this fellow I need to figure out first? Okay. So nothing to worry. When you say A plus B plus C mod, okay, let's say I square it up. Okay. Square means again the use of that property. That's why I told you that property is so, so important. Okay. So when you expand it, you get mod A square, mod B square, mod C square and you get two times A dot B, B dot C, C dot A. And remember, each one of them would be zero. Remember, each one of them is going to be zero because we have been, we have been already provided with this information that these vectors are mutually perpendicular. So mod A plus B plus C is nothing but three K square. So A plus B plus C mod is root of three K. Okay. So all you need to do is just place it over here. So it's going to be K square by K into three root three K, K, K, K gone. So your answer is theta is cos inverse one by root three. Getting the point. Simple now after solving, after me solving this, look simple. Any questions? Any clarifications? Please let me know. So without much ado, let's move on to the next question. If theta is the angle between the unit vectors A and B, then prove that cos theta by two is half mod A plus B and sine theta by two is half mod A minus B. In fact, you may ask this question directly in your exam, prove that tan theta by two is mod A minus B by mod A plus B. This question actually is the combination of the first two. So this can be a potential question in school. Guys, two minutes to do this. Very simple question. Got it. Okay. Just any one of them you can prove. It's understood from the other. Just type done if you're done. Done. They've done. Good. Charlotte also done. Very good. So I'll just prove one of them. Let's talk about mod A plus B. Okay. Since mod A plus B is appearing over here, let's talk of mod A plus B. If you square it, we know that it's nothing but A plus B dot A plus B. Correct. See, so many places this property is getting used. Good, Vinda and Shiram. Let's discuss this. So this is mod A square mod B square plus two A dot B. Okay. Let's say the angle between them is theta and the unit vectors. Unit vectors mean this is one, this is one. This is two into one into one into cos of the angle between them. Okay. This is mod A plus B square. Okay. Now one plus one is going to be two plus one plus cos theta. We know one plus cos theta is is two cos square theta by two. Right. That means mod A plus B square is four cos square theta by two, which means cos square theta by two is one fourth of mod A plus B square. Okay. Which clearly implies cos of theta by two is going to be half A plus B square. Sorry, half mod A plus B. Why did I take plus minus? Why did I take plus minus? Why only plus? Because theta by two is a cute angle. If the angle between two vectors cannot exceed pi, it has to be between zero to pi. So theta by two has to be an acute angle or max to max. It can become 90 degrees. Correct. So that's why I didn't take a negative value. It's just half of mod A plus B. Similarly, you can also prove the second part. For second part, you start with A minus B mod whole square and you can do it. And if you divide these two results, you can get the third part, which may come as a question to you in school. So I'm not doing it guys just to save your and my time. We can instead take a next problem. By the way, is it clear? Any question? Any question? Any concern? Please ask. Yes, Prajwal. Prajwal, any question? Because it is given that the unit vectors, no, Prajwal, read the question here. A and B are unit vectors. Okay. Great. Let's move on. Here goes the question. If vector A is given by this, vector B is given by this, make an acute angle, find the values of A. These two vectors make an acute angle between them. Find the values of A. No, that's not correct, Preeti. You'll get a range of values. Read the question. Find the values of A. No, no, no, no. You'll get a value of A. You'll not get it in terms of X. See guys, it's very simple. We have already discussed that if the dot product of two vectors is make is positive, then only the angle between them is acute, isn't it? So all you need to use is that A dot B should be positive. Right? What is A dot B? A dot B is X, X plus one plus X minus one plus A. This should be always be positive. That means it should be positive for all X belonging to real numbers. Getting my point. So if you expand this, it becomes X square plus two X plus A minus one is always positive for all X belonging to real numbers. Now, when does the quadratic equation is always positive? Always positive will only happen when the coefficient of X square is positive that is already there because it is one and it has no real roots. Why? Because if it is positive, it is hanging up in the air like this, then only it is always positive. Right? When will this expression will always be positive when it doesn't touch the X axis? Correct? So ultimately, this is why? Correct? Right? So why should always be positive for this particular graph? Correct? So if it is always positive, its discriminant must be negative. B square minus four AC must be negative. Right? So what is B square? B square is four minus four AC must be negative. Correct? In other words, let me expand this four minus four A plus four should always be negative. That means four A should always be greater than eight. That means A should always be greater than two. A should always be greater than two. Absolutely correct Nikhil. This is going to be your answer. Getting this point. So a bit of mixture of quadratic equation along with vectors. Okay? That is how typical JEMain problems are framed. Any question? Any concern? Please ask me. This is clear because the angle should be acute. Angle between A and B should be acute. So this led to a quadratic equation. If a quadratic equation is always positive, remember, this is something which I have discussed with you so many times in past also. If a quadratic equation is always positive, it can only happen when these two conditions are simultaneously met. Okay? This should be simultaneously met. Means take the intersection. Clear? If it is clear, let's move on to the next question without much waste of time. Okay? A very simple little question. Prove by vector method your compound angle identity. Cos A plus B is cos A cos B minus A. Devdas, you need to mute your mic. Just type done once you're done. Just two and a half minutes for this. Time starts now. All of you can see the screen properly. You can hear me out. Yes, anyone? Any success so far? Still trying? Oh, ideally no denominator should show up. Just a minute more I'll give you and then I'll start solving this. Okay? Let's consider that we have two unit vectors. Okay? One making an angle of A with the x-axis. Okay? So let's say I name it as OP vector. It's a unit vector. Okay? So OP vector is a unit vector. Unit vector means its length is unity. Okay? If it is unit vector, can I say the position vector of P would be nothing but cos A i cap plus sin A j cap. Right? In other words, OP vector is cos A i plus sin A j. Okay? Let me take another unit vector, OQ. This time making an angle B in this direction. Correct? Since this angle is minus B, this point position vector would be cos B i comma, in fact plus, in fact minus, sin B j. Okay? Why minus? Because this angle is taken in a clockwise sense. So cos minus B plus cos minus B i plus sin minus B j will give you cos B i cap minus sin B j cap. Okay? So in other words, OQ vector is known to us now, which is cos B i cap minus sin B j cap. Okay? Awesome. Now we'll do the dot product of these two vectors. So OP dot OQ should ideally give you OP mod, OQ mod into cos of the angle between them. Remember, the angle between them is A plus B. This is the angle between these two vectors. Okay? This is the direction for your reference. Okay? Now this is one one each and this is cos A plus B. Now what will be the dot product of OP and OQ? So let's take the dot product of these. So dot product of OP and OQ would be nothing but cos A cos B and sin A minus sin B, which is nothing but minus sin A sin B, which clearly brings me to the proof that cos A plus B could be written as cos A cos B minus sin A sin B. Okay? Hence proved. This may also come as a potential school question. Clear? Any doubt? Any concern? Next. In any triangle, prove the projection formula. In any triangle, prove the projection formula. Hope you know what is the projection formula? A is B cos C plus C cos B using vector method. Done. Shorak is done. Well done. Shiram also done. I at least need two more responses. Then only I'll start solving. Until then please give it, okay Nikhil is also done. Please try it honestly guys. Done. Okay. Let's begin. Again let me make a triangle. Okay? Let me make a triangle like this. And let me assign vectors to these sides. Again I'm taking them as cyclically closed. Now please note mod of A itself is your side A, mod of B itself is your side B and mod of C itself is your side C. Okay? Now what I'm going to do is I'm going to take the dot product with A on both the sides. I'm going to take the dot product off with A on both the sides. Okay. Let me check it out. I'm sure you have taken the angle incorrectly. So when you open this, it becomes A dot A, A dot B, A dot C is equal to zero. Okay? A dot A is A square. Now A dot B very, very carefully. Please find the angle between A and B. The angle between A and B guys, let me tell you it's 180 minus C angle. Okay? So this is going to be A B cos pi minus C. Getting my point. Similarly, the angle between A and C, angle between A and C will be pi minus B. So A dot C would be A, C cos pi minus B equal to zero. How many of you had made this mistake? I'm sure some of you would have made this mistake. Hence the sign was not coming proper. Correct? Now drop the factor of A throughout. So A, this will become minus B cos C. Let me just not write a minus here. And this will become minus C cos B. Right? Take it on the other side and here comes your desired result. Hence proved. Okay? This again, I'm repeating be very, very careful. Angle between two vectors is defined as the angle when they have been made co-initiated. Okay? Is that fine? Next question. Do that the angle inscribed in a semi-circle is a right angle using vector method. Just a two-minute problem. Time starts now. So be very careful. Using vectors, we have to prove that angle subtended by the diameter on the circumference is 90 degrees. Anyone any idea? Done? Very close. Okay. Fine. Fine. I'll give you some time. Done. Great. Okay guys, time to discuss this. Let's say, please follow the diagram which I've drawn for you here. Let's say O A vector is A vector. Okay? And let's say O B vector is B vector. Correct. So O C will automatically become a minus A vector because remember O C is of the same magnitude as O A but reverse direction. Not only that, mod A will be equal to mod B because they signify the radius because they both show the radius of the circle. Okay? Okay. Now let us find out these two vectors. Let me find out this vector BA and let me find out this vector BC. Okay? You can treat O to be your origin also or reference point. You can treat this to be a reference point. So you can treat this to be the position vector of A, this to be the position vector of B and this to be a position vector of C. Okay? So when you say BA vector, it's position vector of A minus position vector of B. Okay? When you say BC vector, it's position vector of C minus position vector of B. You don't have to apply addition and all because your position vector takes care of that. Okay? Now let us find out what is BA dot BC? Let's do the dot product of BA and BC. That means A minus B dot with minus A minus B. Please note that minus one could be taken out common as we have already discussed in one of the properties of dot product that if there is a scalar quantity, you can pull that scalar quantity out. Okay? So this will become mod A square minus B dot A plus A dot B minus mod B square. All right? A dot B, B dot A will cancel each other out. Okay? Now remember one more thing that mod A is equal to mod B because both signify the radius. So which implies mod A square will also be equal to mod B square. That means this and this will also cancel each other out giving you the answer as zero. Now remember here A and B were not zero vectors. They were non-zero vectors as you can see from the figure and if the dot product of this is giving you zero, okay, dot product of this is giving you zero, it implies BA vector is perpendicular to BC vector. That means the angle between these two vectors here is a 90 degree and hence we have been successfully able to prove that the angle subtended by the diameter on the circumference is going to be 90 degree. Clear everyone? Please type clear. Next one. I don't know whether this has anything to do with vectors because the moment you see this question, where is vector in this, right? But to your surprise, yes, vector is involved here. If A plus 2B plus 3C is four, find the least value of A square plus B square plus E square. Find the least value of A square plus B square plus C square. If we take the vectors to be cyclic, can we prove using projection formula? Yeah, see, any method should work fine, but you just have to take care of the fact that you are using all your rules properly. Sure. Sure, Vinda, I'll just go to the previous page for a second everyone. Is that fine, Vinda? Let me know so that I can go. Okay. Any idea how to do this because I'm sure most of you are clueless. Anybody who's trying, if nobody's trying, I can start solving this. Fine. Okay. Sure. Sure. Please. Please go ahead. Yeah, Shana, don't worry. No problem. Take care of your health first. That is more important. The session is already been getting recorded. Don't worry. Okay. That's a good approach, Kavya. Let me tell you, you are on the right track. Go ahead. Go ahead. Kavya, you are on the right track. That's correct. Absolutely correct, Arka. By the way, you can further simplify that. That's absolutely correct. So one of you has only found the answer very well done. Good. Yes, that's correct, Arka. Correct. I'm waiting for few of you to answer. Fine. So let me discuss this. So yes, absolutely correct, Lakshya. That's correct. Okay. See here, most of you started by assuming these two vectors. One was, let's say I called it as P, which is AI plus BJ plus CK. Okay. And other vector, you took it as I plus 2J plus 3K. Okay. So what this information told you was that P dot Q is actually four. Absolutely correct. No doubt about that. Right. Now, if you find cause of the angle between them, you know that P dot Q by mod P mod Q is the angle between them. Correct. Which is nothing but four divided by mod P is under root of A square plus B square plus C square. Okay. And this term is going to be under root of one square plus two square plus three square. Isn't it? That's going to be 14 by the way. Correct. So I can write this as four by under root of A square plus B square plus C square under root of 14. Okay. Now, cos theta. Okay. We all know that it lies between minus one to one. So cos square theta will always be less, will always die between zero and one. Correct. In other words, that will always be less than one. So 16 by A square plus B square plus C square into 14 will always be less than one. That's correct. That's correct, Vinda. Okay. So can I say A square plus B square plus C square will always just take this term on this site will always be greater than equal to 16 by 14. That means A square plus B square plus C square will always be greater than eight by seven. That means the least value of A square plus B square plus C square is eight by seven. This is your answer. Okay. So where vectors was least expected, you actually ended up using vectors to solve it. Okay. So reading this question, nobody will even know in this wildest of thoughts would have thought that okay, we have to use vectors to solve this. So good learning. Okay. Let's move on to the next one. If vector A and B, as given to you on your screen, this and this, find vectors C satisfying the following conditions. It should be co-planar with A and B and it should be perpendicular to B and A dot C should be zero, seven. Please note, all these three conditions are simultaneously satisfied. It's not that you have to solve this as three different questions. They have been simultaneously satisfied by the vector C. Anyone, any success? No, Nikhil, that's not correct. No success. Okay. Pajwal wants a second. Ishika is absolutely correct. Awesome. Very good. Perfect. Ishika is right. Anybody else? All right. So let's try to solve this. No, Preeti, that's not correct. Okay. Let's try to solve it. First of all, let's take the condition number one. First of all, let me call the C vector as XI plus YJ plus ZK. Okay. Let's assume that this is my C vector. Okay. Let me check. No, Pajwal, that's not right. Yes. Yes, that's correct. Preeti, that's the approach. Right. So we'll do that only. The first thing that we have to use is the fact that the vector C is co-planar with A and B. Right? So we can directly use the co-planarity condition that the determinant, which is made by the IJK components of these three vectors must be zero. Okay. So if you expand it, it becomes X into one minus Y into minus three and Z into minus two. So the very first condition that you end up getting is X plus three Y minus two Z is equal to zero. Okay. Next is use the fact that C is perpendicular to B. C is perpendicular to B means dot product of C with B must be zero. That means two X plus Z must be zero. That's your condition number two. That's correct, that's correct. Okay. Third condition is A dot C seven. A dot C seven means A dot C is equal to seven implies minus X plus Y plus Z is equal to seven. This is your third condition. From here, we need to solve for X, Y and Z simple. Right. What we can do is since this is an X and Z, we can eliminate Y from these two. Okay. So what we can do is we can multiply this equation with a three and subtract it. So X plus three Y minus two Z equal to zero. And if you multiply this with a three minus three X plus three Y plus three Z is equal to 21. Subtract these two, you will get four X minus five Z is equal to minus 21. Okay. And we already have two X plus Z is equal to zero. Okay. Multiply this with a five. So 10 X plus five Z is also equal to zero. Add these two. Okay. So when you add these two, you get 14 X is equal to minus 21. Correct. So X is equal to minus three by two. That's absolutely correct. Now put it put this in just this equation to get the value of Z. Z is going to be minus two X. So Z is going to be three. Okay. That is going to be three. Now put it in the last one. Minus X means three by two. Y is unknown. Z is three. This is equal to seven. Okay. So Y is equal to four minus three by two, which is nothing but five by two. So your vector C will now be as Ishika rightly and also Shiramya rightly said minus three by two I plus five by two J plus three K. This is your answer. Okay. I don't think so. It was a difficult question. Silly mistakes is what you must have made. Yeah. Correct. Okay. So the base of a pyramid A O B C is a equilateral triangle. So please read this question carefully. Hope the question is clear to you. First make a diagram. Always make a diagram in such case so that you can refer to that diagram. Okay. Once again, Arka is asking for the previous page. So I'll just go back to the previous page and again come back quickly to it. And of course, I'll draw the diagram pretty. Arka, do you want me to scroll up or scroll down? Let me know once you're done. Done. Okay. Okay. So as requested by you, cost inverse half. That's not correct. Okay. Let's let's let me draw a diagram. Let me first draw a right-handed coordinate system. This time for a change, I'll take this as the X, this as the Y, and this as your Z axis. Okay. Now, basically you're read these couple of lines. O A B C each has side four root two. Okay. So it is an equilateral triangle. Each of side four root two. Correct. O is the origin for sure. A O is perpendicular to the plane O B C. Okay. So now pyramid, we all know pyramid is such a way that each side is a triangle. Now base is a triangle, which is equilateral. Okay. But the fourth vertex is basically a point A, which is perpendicular. So can I say the structure could be imagined something like this. I'm just, this is your pyramid. Okay. So please note that it is in such a way that its base is on the XY. This is a base. Okay. I've kept it on the XY plane. Okay. O is on your, sorry, A is on your Z axis. So that A O is perpendicular. So this criteria should be met. A O is perpendicular to the plane O B C. Okay. Now mod A O is two. That means this point is two K. Okay. Now without the loss of generality, can I say this point, let me name these points B C. Okay. Can I say B will be four root two I because it is completely on the X axis and it is having a length of four root two. Okay. Can somebody tell me what would be the position vector of C? Remember this is 60 degree from here. And this length is four root two. You'll see very simple. Four root two cos 60 I by the way four root two cos 60 is nothing but two root two I. So this is two root two I. Okay. Plus four root two sine 60j. Four root two sine 60 is root three by two. If I'm not wrong, this is going to be two root six J. Is that clear everyone? So at least I know the position vectors of each of the three vertices of these pyramid. Okay. Now question is find the cosine of the angle between the skew straight line one passing through a and the midpoint of OB midpoint of OB here is two root two I. Okay. So let me name this point as D for the timing. So can I say AD vector would be two root two I minus two K destination minus source that is a position vector of D minus position vector of A. Next is and the other passing through O and the midpoint of BC. Okay. What is midpoint of BC? Midpoint of BC will be three root two I plus root six J. So O let me name it as E. So O E vector would be nothing but three root two I plus root six J. You want to find the cost of the angle between them. Correct. That is what the question is asking you. Oh my God. The answer is much, much simpler. Shiram. Okay. Let me finish it off. So cost of the angle between them will be AD dot OE by mod AD mod OE. No, not pi by four. Oh yes, yes, yes, pi by four. Angle is pi by four. That means answer is one by root two. Yeah, because they're asking you cost of the angle. Okay. So dot product of these two will be six into which is 12. 12 minus two root six. Oh, sorry. This is INJ component, right? So yeah. So this is just 12. Okay. divided by under root of square of this, which is eight plus four under root of square of this, which is going to be 18 plus six. Okay. So the answer that we end up getting here is 12 by root 12 root 24 root 24. You can write it as root 12 into root two into root 12 root 12 root 12 and 12 will go off. Answer is one by root two. Remember the question was asking you the cost of the angle cost inverse half you said, then you said cost inverse root two. Oh, this is a J main question. This is a J main question. Don't get perturbed by so much of wording. It is just the angle between two vectors, nothing like that. And for that angle, you just have been provided with information to get their position vectors. So using position vectors, you need it to get the vectors. Once your vectors are known, you just have to use your dot product to get the angles between them. Okay, pretty done. Let's move on to the next one. So the vector always given turns to a right angle passing through the positive x axis on the way. Show that it's new position is this. It's a show that question. So once you're done, let me know you're done or not. Sure, Preeti, I'll move on to the next page, previous page. Sorry. You're asking for coordinates of E, right? Coordinates of E is three root two i plus root six J. Done. Thanks. Okay. Anyone, any idea? Who are trying? Trying. Okay. Okay. Shall I give you two more minutes? Okay. So let me try this out. See a couple of information that has been provided to you in the question and we have to make use of those information. One, if your OB, okay, moves to the position, sorry, OA moves to the position of OB. Okay. This is 90 degree and in between it is crossing the x axis also. Okay. You know, three information has been provided to you here. What are those information? Let me show you first. Let me take the OB vector to be xi plus yj plus zk. First piece of information provided is the length of OA should be equal to the length of OB. That means both these vectors should have the same magnitude, right? Of course, of course, you'll come to that, okay. Number two information is OA, x axis and OB are coplanar. That's what Arka is pointing out. They are coplanar, getting my point. Yeah. Thirdly, OA dot OB is zero because they are making 90 degrees with each other. Correct? Aren't these three information sufficient for us to find x, y and z? I think so they are sufficient. So first let me make use of the first condition that mod OA is equal to mod OB. So mod of OA is nothing but 1 square plus 2 square plus 2 square under root, which is equal to mod OB. Mod OB is x square plus y square plus z square, which means x square plus y square plus z square is equal to 90. That's the first piece of information. Correct? Second piece of information, let us use that OA, x axis and OB are coplanar. That means the determinant formed by the coefficients of i, j, k for these vectors must be zero. Correct? So x, y, z, one, two, two, and let's take x axis to be i cap. Doesn't matter. Any vector you take doesn't matter. It'll give you zero anyways. Okay? So when you expand it with respect to the third row, it becomes only y, 2y minus 2z will appear and that is going to be zero. That is y is equal to z. That's your second information that we have. y should be equal to z. Third piece of information, AB dot, sorry, OA dot OB should be zero. OA dot OB is zero means what? Which means dot product is x, 2y, 2z should be zero. This is your third piece of information. Correct? Now we have to use these three to solve for x, y, and z. Now since y is equal to z, I can say x plus, let me put y as, let me put z as y itself plus 4y is equal to zero. Okay? So that means x is minus 16y. Let's use this information in the first equation. So x square means, I'm so sorry. I'm so sorry. This is 4y. x square means 16y square. y square, let it be y square. z square is also y square. This is going to give you a 9, which means 18y square is equal to 9. That means y square is equal to half. That means y is equal to plus minus 1 by root 2. y is equal to plus minus 1 by root 2. Getting the point? Okay? So z is also plus minus 1 by root 2. Correct? So x is minus plus 4 by root 2. Correct? Yes or no? Correct? Now there are two possibilities. One being 4 by root 2. In fact, 4 by root 2 minus 1 by root 2j minus 1 by root 2k or minus 4 by root 2 plus 1 by root 2j plus 1 by root 2k. Okay? But remember if you're turning 90 degrees, the new position, let's say this is your x-axis and this is your OB. Okay? This was your OA. This must be an acute angle. If this is an acute angle, that means dot product of with i should be positive. So this cannot be your answer. Only this is your acceptable answer. So this is your answer to this question. Is that fine? Very interesting question. So basically it was testing you on various fronts. Coplanarity, it was testing you. It was testing you on your idea of angle. Okay? Then it was testing you whether you know how to interpret the acute angle between the two vectors. So various concepts got tested in this question. Let's take another one. One second. Let me just look out for... Okay, I have a question from my side. Okay? The question goes like this. There's an arc AC of a circle. Ask arc AC of a circle which subtends the right angle at center O which subtends 90 degree at the center O. Okay? Now there's a point B which divides the arc in the ratio 1 is to 2. Okay? If OA is given to be A vector and OB is given to be B vector, then find OC in terms of A and B. Let me make the diagram for you so that things are quite obvious to you. Okay? So this is the arc. Let's say AC is the arc which subtends a 90 degree over here. Okay? And there's a point B which divides this in the ratio of 1 is to 2. This vector is A. This vector is B. You have to find OC vector in terms of A and B. OC vector in terms of A and B. Nikhil, your answer is too complicated. 2B minus root 3A, that's absolutely correct. Shiram, can you send me the picture of your working to my WhatsApp? Others? Others, please respond. Please respond. Okay? Let others first send their response or at least try to answer this. Then only we can discuss this. Anyone? Please respond, guys. I'll give you one more minute. Then we'll start solving it. Great. That's good. That's a good approach, Shiram. Okay? So let me take over. Now, first of all, if this ratio is 1 is to 2, would you all agree with me that this is 30 degree and this is 60 degree? The angle division will also be in the same ratio as the length of these two arcs. Correct? Now, we all know that these lengths are nothing, but they are the radius or they are the radii of this circle. So all of them, let me say it's r. Is that fine? Okay? Now, another important thing that is worth noting that A, B and C are in the same plane. So I can always write, I can always write C as a linear combination of A and B. Is that fine? Is that fine, everyone? Right? Okay. Now, if I take a dot product with A throughout, let's say I take a dot product with A throughout, this is what I will end up getting. Now, C dot A. C dot A, do you see that the angle between C and A is 90 degrees? So this would be 0. Yes or no? Okay? What about the angle? What about A dot A? A dot A would be, now, can I say A dot A is nothing but mod A square. And mod A square is like r square. Correct? Yes or no? Oh, I'm sorry, I should have written a B over it. Sorry, A over it, by mistake I wrote down. Now, what is B dot A? B dot A would be mod A mod B into cos of the angle between A and B, which is 30 degrees. So can I say it is r square cos 30 is root 3 by 2. So this is B y square root 30 root 3 by 2. Yes or no? Correct? So in other words, I can say 0 is equal to 0 is equal to x plus root 3 y by 2. Okay, so what do I conclude from this? Remember, x and y both are not 0. So what do I conclude from this? Nothing, no conclusion can be drawn from here. Okay? Now, what about the magnitude of C vector? Magnitude of C vector is also r. Correct? Can I use that? Can I use that? So magnitude of C is equal to r vector. Can I use this information? So magnitude of x A plus y B is equal to r. That means square of this is square of this. That is nothing but x A plus y B dot x A plus y B is r square. Correct? What does this give me? x square A dot A. A dot A will be r square. y square B dot B that is also r square. Then we'll have x y A dot B two times. So two x y A dot B. Okay? So this will be x square r square y square r square plus two x y A dot B will again be r square into root 3 by 2. Okay? Now drop the r square factor from everywhere x square plus y square and this will be just root 3 x y is equal to 1. Okay? Can we solve for x and y from these two? Okay? From here I can say root 3 y is negative to x or y is equal to or I can put everything in terms of x only no problem. So x is going to be negative root 3 y by 2. So this will give you 3 by 4 y square plus y square minus root 3 into minus root 3 by 2 y square is equal to 1. So this will give you 7 by 4 minus 3 by 2 y square is equal to 1. Correct? So this is nothing but 6 by 4. So 1 by 4 y square is equal to 1. So y is equal to plus minus 2. If y is equal to plus minus 2, x will be equal to plus minus root 3. In fact minus plus root 3 minus plus root 3 not plus minus minus plus root 3. Okay? So I have two possibilities for c. One is minus root 3 a plus 2 b and other is plus root 3 a minus 2 b. Correct? Now can we accept both the answers or one of the answer has to be rejected? Anyone? You can unmute yourself and speak out. Can I say the first one has to be rejected? Why? Why this answer has to be rejected? Look at the angle between b and c. The angle between b and c that they want is actually an acute angle. So if you take the c value as minus root 3 a plus 2 b what will you get? Oh I'm sorry. Let's take here. If you take your c as in fact first one is accepted, second one is rejected. This one is accepted. This one should be rejected because if you take the dot product of this with b this should give you a positive answer but this will end up giving you 3 a dot b minus 2 b dot b. b dot b is r square. a dot b is going to be r square root 3 by 2. Correct? So that's going to be 3 by 2 r square minus 2 r square which is a negative value. Okay but here you can see in the figure that the angle between b vector and c vector this angle must be acute 60 degree. Correct? So therefore root 3 a minus 2 b is rejected and only this becomes the answer. Very good. Very good Shriram. Only two of you could solve this. Fine then guys. We'll bring this session to an end and all the best for Tuesday's physics exam. Do well. Avoid silly mistakes. Write a mock paper today and tomorrow if possible. Analyze your mistakes. Check your speed. Okay do a lot of derivations and go. Present your answer neatly and it should be as detailed as possible. Okay thank you so much everyone. Bye bye. Have a good day.