 Hello everyone. This is Shankar. Let us study some physics. Okay. So this video is a problem solving video. Okay. On the chapters, scalars and vectors. Okay. How to use this video is very, very simple. What you have to do is I'll do some questions with you along with you. But first what you can do is you can pause the video as soon as you see the question and you can start attempting the question. And if you don't know the answer to it, you can do it with me along with the video. Okay. So this is, this is also can be used like an assignment sort of thing where you can just, you know, watch the questions, try it on your own. In case you're not able to do that is when you look, okay, look through the solutions that I am describing. The next thing is who are, you know, the people that I made this video for. Okay. Who are the people who are made this video for? Are those people who understand the theory in class, but for some reason are not able to approach the question. So there are a lot of people like that who are able to understand everything in class, but when they go home and try some questions, they're unable to approach the question. So this is aimed at those kids. Plus also this video is not going to be, you know, directly high level question solving. Just like my teaching, I'll always start at zero level. That means I'm going to start with NCRT level questions. Then I go to J mains level question. Then I go to J advanced level question and then beyond that. Okay. So that is the order that I'm going to follow. So the first five, six questions are going to be NCRT level, then the next 10 questions will be J mains level and the next 10 questions will be J advanced level. Okay. So if you don't have an idea of how to approach question, this video is going to immensely help you. Also it is for those people, this NCRT section is specially, especially for those people who are preparing for competitive coaching. Okay. So competitive in this competitive coaching, what exactly happens is we are so in focus with the competitive side of things that we forget how to write the answers. And that creates a problem for us in UTS and other things. Okay. So if you want to excel in school exams as well. So the first question, five questions that I'm solving for NCRT, I'm going to solve it in the board manner. Okay. So I also have experience teaching board kids. So I know exactly how to prepare for boards as well. Okay. So and you cannot give up on boards as well because now suddenly if you start having, you know, if you want to go to a university outside or if you want to, you know, just get some good marks in boards, this method of writing answers is also going to help you. Okay. So in NCRT, there's a, there's a R2 writing answers and in UTS, what happened? You start arguing with your school teacher. So the answer is correct. Why didn't I get marks? Okay. But I'll show you why you're not getting marks because there's a pattern of writing answers. Okay. In CBC and we are going to follow that whenever we are doing the NCRT question for the first five questions, I'm going to do it how I would personally do in a board's exam. Okay. So please pay attention. Have fun. Okay. And it's not necessary that you have to complete a video in one go. So I'll do probably 20, 30 questions. You should also do it over a period of time. Let's say, let's one or two days, you just do five, six, any number of questions that you are comfortable with. Okay. So let us start our discussion. This is the first question from NCRT. Very easy question. Find out the magnitude and direction of resultant of two vectors, A vector and B vector in terms of their magnitudes and the angle between them. Okay. So this essentially they're asking us to derive the parallelogram law of addition. Okay. Parallelogram law of addition, which we have already derived in class. Okay. So let me show you how to, how to write the answer if this, if this question is asked in board or in UTS or whatever school exams you have. Okay. So let me show you how it is done. So the question gives us three data here, A vector, B vector and theta, theta angle between them. Okay. So let me draw them. Let me draw them. This is the B vector. Okay. This is the A vector. This is the A vector. Okay. This is the A vector and the angle between them is theta. Okay. So I'm going to name this because this is a parallelogram law. I'm going to name this. Let's call this O. Let's call this B. Let's call this Q. And you know what is the next step, right? We parallel transpose A to the top like this. Okay. Let's call this A vector. Okay. And we parallel transpose B to the right side, making, completing what we call a parallelogram. Okay. So this parallelograms completed when I, when I transpose A to top and B to the right side. And then what we have to do is we have to extend A vector to this side and we have to drop a perpendicular. Okay. We also have to name this. Let's call this S. Okay. And P, Q, R, S, T. Let's call this point T. Okay. And parallelogram law states that whatever the length of this, this diagonal, this diagonal, the length of this diagonal is going to be the resultant we are looking for. Okay. So parallelogram laws states that this value of this R is going to be the addition of A vector plus B vector. So A vector plus B vector is essentially R vector. But finding out the length of R vector is your job. Okay. So let me show you how to write this. The first thing you will write here is this. Let O, Q and let O, Q, B, A, B vector and let O, P, B, A vector. Okay. And let theta B, D angle between them. So the first statement that you're writing in any board's exam is you have to define the variables that you took. Okay. So most of the people miss this, but you should not. Okay. If you want to score 100%, if you, if you, if you want to be sure you're doing really good in board's exams or in school exams, this is what you have to start with. Whatever variables you took in the question, you mentioned them in the beginning. Okay. Let this is this, let this is this, let this is this. Okay. This is how you write. Next thing you'll write is that parallelogram law states that parallelogram law states that, states that R vector is equal to A vector plus B vector. Okay. So then you write, then you write in triangle SPT. Okay. So this is a triangle SPT. This is a 90 degree triangle. ST is going this length ST will be equal to how much I have derived this in class. This angle is going to be theta. Why? Because corresponding angles and this length is B, this, this B vector is B. So this is perpendicular in this triangle. So this perpendicular is going to be equal to B sine theta. I have derived this in class. Okay. I have derived this in class. The value of PT would be again from trigonometry. This is, this is the adjacent side, the hypotenuse we know. So PT upon B is how much? This is actually adjacent side upon hypotenuse is cos theta. So from here you can get PT is equal to B cos theta. I'm going to write this here. Okay. PT is equal to cos theta. So notice how I'm writing this in triangle SPT, ST is equal to B sine theta, PT is equal to B cos theta. You can also add this from trigonometry. Okay. From trigonometry. I got this from analyzing this triangle SPT trigonometry. Okay. Now, in the next step is in triangle OST, this big triangle OST, R is the one that we are looking for OS, OS, this length OS is what we are looking for. OS square is actually equal to ST square plus OT square. This is what, this is what Pythagoras theorem. Okay. You can write this here. This is Pythagoras. I don't know if that is correct Pythagoras theorem. Okay. I don't know if Pythagoras is, I think this is more correct Pythagoras theorem. Okay. Pythagoras theorem. I don't know if the spelling is correct. You have to check that. Okay. Please check that for me and let me know in the comments if I wrote that correctly. Okay. So Pythagoras theorem, you can just mention this is what I got from Pythagoras theorem and then just plug in the values. What is OS? OS is what we have to find the value of ST is what ST is B sin theta, B sin theta square. Okay. And OT square is what OT is a combination of OP and PT. OP is what A and PT is what B cos theta. Okay. This is what I have to square. Now, when I square these two terms, when I square these two terms, this will become B square sin square theta and this one will become, I'm opening this using A square plus B square plus twice AB. Okay. So this will become A square plus B square cos square theta plus twice of AB cos theta twice of AB cos theta. Now, B square sin square theta plus B square cos square theta, when you add them together, this will become B square and this will become A square plus twice AB cos theta. Okay. So from here, this OS or R square will be equal to B square. So this resultant is how much under root of A square plus B square plus twice AB cos theta. Okay. But that gives us only the magnitude. Okay. But I told you that a vector is made up of both magnitude as well as direction. So for the direction part, I told you what you can do is find out this angle, the angle between R vector and, you know, AA vector. We call this phi. This angle is phi. So in this triangle, OST, again, in this triangle, OST. So again, you will write like this in triangle OST. 10 phi is going to be equal to perpendicular upon base. Okay. So perpendicular, perpendicular upon base. Okay. So what is the perpendicular here? Perpendicular is ST and the base is OT. So ST by OT. Okay. ST by OT. Again, I plug in the values of ST and OT. ST is B sin theta and OT is A plus B cos theta. Okay. So from here, this whatever this fancy angle phi we took that is equal to tan inverse B sin theta upon A plus B cos theta. So if you write like this, I'm pretty sure your school teacher will have no way of cutting your marks and you'll get full marks. Okay. But again, people who are from competitive coaching, they will underestimate, there's a tendency for them to underestimate this and be like, what is this sorcery that I don't understand why I have to write this? I perfectly understand that statement. Okay. There is actually no need to, if you understand this, what is the point of writing all this? The answer is there's a set pattern to writing things in CBC. If you want to perform good in UTS, this is the way you have to write so that marks of yours is cut by your school teacher. Okay. And again, if you're only focused on advanced coaching, okay, that and you don't care about your good marks at all, then you can pretty much just skip this and just, I don't know. But one thing you have to understand, if you just write this formula, okay, this is the formula, just take it. Okay. Just take it, madam. Madam is not going to give you marks for this. Okay. If you just write the formula for this. Okay. You have to do some sort of derivation, which includes steps like these. Okay. So this is it. Okay. And let's move on to the next question. Okay. So this is the next question that we are going to do. Okay. Rain is falling vertically with the speed of 35 meter per second when it starts blowing after sometime at a speed of 12 meter per second in east to west direction, in which direction should the boy waiting at a bus stop should hold his umbrella. So this is again a vector question. It's a little bit of application. So whatever vector addition we have studied, we have to apply this. So pause the video and try this question and let and then you can try it with me. Okay. If you are unable to do so, try it yourself first, pause the video right here and then try it. Okay. So let me start solving. So if I make a diagram for this, remember as physics students, you are expected to make good diagrams. Okay. You should be able to make good diagrams. Okay. So the rain is falling vertically with a speed of 35 meter per second. Okay. So I'm going to make a vector like this that shows the rain. So this is a vector that is representing the rain. The magnitude of this is 35 meter per second. Okay. This is this is the rain. This is the rain vector. Okay. Then if I consider usually we consider east on the right hand side, west on the left hand side. So wind starts to flow in the from from east to west. So that means this vector is going like this from let's consider this as east, this as west. So this wind, wind here is actually flowing from east to west direction with what velocity? 12 meter per second, 12 meter per second wind. Okay. If I want to find out the resultant, if I want to find out the resultant, I have to find out this resultant. You what is the resultant between 12 meter per second and 35 meter per second? The angle between them is 90 degrees. Okay. Obviously, if something is falling down and something is going horizontally, that means the angle between them is what? It's 90 degree. Okay. So what you can do is use parallelogram law here. Okay. Actually, they are just 90 degrees. So you can simply use Pythagoras theorem also. But if you're confused, these are how to do it. So the best way whenever you have two vectors, the easiest way to do is do this is with parallelogram theorem. Okay. Parallogram law of addition, because in that, you don't have to think too much. You have a formula, set formula to finding out the magnitude and the direction. Okay. So let's let's let's do this. Let's say R is equal to under root a square plus b square plus twice a b cos of 90 degree. Now, as soon as you put 90 degree, this whole term will become zero. Okay. This whole term will become zero. So that means the the resultant is just under root a square plus b square. And this is something that you are familiar with. If two vectors are perpendicular to each other, the resultant will be under root a square plus b square. Okay. This is something that you can remember, or you can also use the cosine law or this parallelogram law of addition to do it. Okay. And once you put cos 90 cos 90 becomes zero, it will automatically come out. Okay. So here, all you have to do is under root of 35 square plus 12 square. Okay. And let me do 35 square. I don't know how much 35 square is. So 55 to 25, 2 comes out 15 plus 15 is 32. So 3 comes out. It is a 9. 9 plus 3 is 12. Okay. So this is 1225 plus 12 square is 144. Okay. So if I take an under root here, this will be under root 9. This is going to be 6. The screen is going. So 13 69. This, I think is the square for 37. Okay. Let's check. Let's check. 77. Yeah. 49. 4 comes out. 21. 42 plus 4 is 46. 4 comes out. 3, 3 and 9 plus 4 is 13. Yes. 13 69. So this is the square root of 37. So the resultant is going to be 37 meter per second. Okay. It is going to be 37 meter per second. So the resultant of 35 meter per second and 12 meter per second is going to be 37 meter per second. Okay. Now. I need to find out what is this direction. Okay. At what angle from the vertical or the horizontal, whichever one you are comfortable with, what angle is 37 meter per second, uh, flowing at. We have to give a direction also. Right. We cannot just give the magnitude and be over with it. So to find out this angle. So what I can do is if I want to find out from the vertical, let's call this angle alpha cause alpha would be what cause alpha would be based upon hypotenuse 37 being the hypotenuse. If you imagine a right triangle like this, imagine a right triangle like this, because you know all these vectors are from trigonometry only. So anytime you are confused, just make a right triangle and figure out what you are trying to find. Okay. So cause alpha would be equal to how much cause alpha would be 35 by 37 base upon hypotenuse. Okay. So from here alpha can come out to be cause inverse of 35 by 37. This answer is absolutely correct. Okay. You can also, uh, you know, do it in terms of sign also sign. If you want to use tan, you can also do tan also. But if you just define this right angle, right angle triangle, after that you can use any trigonometry formula to find out the angle between, uh, this 37 meter per second and the vertical or the horizontal. Okay. All you have to is just define the triangle. You can use any trigonometric, uh, you know, ratio that you're comfortable with and that is how you do. So the final answer, uh, for this particular question is, uh, there's, uh, the, the, you know, the rain is actually falling at 37 meter per second, um, at, at, at an angle which is equal to this. This I can actually put it in the calculator and find out, but no need, no need. Okay. Uh, CBC is not expecting you to know the value of cause inverse 35 by 37. Okay. If you write it like this, you'll get the full marks. Okay. So remember to just underline your answers before you exit the question. Okay. So let's go to the next question then. Pretty easy stuff. Okay. Let's do this one. Okay. A motorboat is racing towards north at 25 kilometers per hour and the water in the current, the water current in that region is 10 kilometers per hour in the direction of 60 degree east of south. Okay. So let us understand this. Let us make a diagram. So north, I will usually point upwards. Okay. So for me, generally, this is the convention I use north upwards east west and this is south. This is what I usually follow. Okay. So based on this, the motorboat was going from going towards north with 25 kilometers per hour. So I have 25 kilometers per hour pointing upwards. Okay. And then this, uh, this this water current is actually going in 60 degree east of west, east of south. That means so here actually I'll teach you how to understand this. Okay. East of south means you will start at south. I'll start at south. From south, I will go 60 degree east. Okay. So whenever you get a convention like this, always look for the last, last word. South. Okay. East of south means I start at south. From south, I go 60 degree towards east. Okay. So that means, uh, the vector, the vector of this water is actually going to, if I extend this like a, uh, perpendicular. So this is the vertical. This is the south direction from this. I have to make this at 60 degrees. Okay. So this is like 60 degrees. Okay. And, uh, this magnitude of this is going to be 10 kilometer per hour. Okay. So find out the resultant velocity. Find out the resultant velocity. So again, this can be done by two ways. Okay. It can be done, uh, since we don't have anything, the question doesn't say you have to do it graphically or component wise. So I always prefer to do it in component wise. So I'll do it in component wise, but you can also do it using graphical metal. Okay. All you have to do is find out a square plus b square plus twice a b cos of theta where theta is the angle between 10 kilometer per hour and 25 kilometer per hour, which in this case is going to be 120 degree. I taught this in class. Okay. It's not 60 degree. The angle between 25 kilometer per hour vector and 10 kilometer per hour vector is 120 degree. So you'll have to find out cos 120 degree. Okay. And cos of 120 degree, uh, it's actually equal to, uh, minus half cos 120 degree is equal to minus half. Okay. So you can try using that process also, but I'm going to take components because in component method, you'll never have any mistakes. Okay. So if I want to take components of this orange vector, I told you towards the theta is cos theta. Okay. So this 10 kilometer can be broken down into two vectors, one vector pointing downwards, which would be 10 cos 60. Uh, so that means five kilometer per hour downwards. Okay. And one in this direction. Okay. One component in the right hand side and this would be 10 sine 60. Okay. 10 sine 60 would essentially mean, uh, five root three. Five root three kilometer per hour. Okay. So essentially you have three vectors now. So I, I got rid of this 10 kilometer per hour. I broke it. I broke it into two vectors, one vector pointing right side, one vector pointing downwards. Okay. Now, what is the use of this? What is the fun part is because 25 and now five kilometer per hour are exactly opposite to each other. I can say the net resultant of these two vectors will be 20 kilometers in the upwards direction. Let me stay, stay put that statement one more time. Since 25 kilometers is upwards and five kilometer per hour is exactly downwards in the minus y and this is plus y. Okay. Uh, when they are just opposite of each other, I can simply subtract them. I can simply subtract them. If they're in the same direction, I can simply add them. If they're in opposite direction, I can simply subtract them. Okay. So 25 kilometer per hour minus five kilometer per hour. So that means the net upward will be 20 kilometer per hour. Okay. And then don't forget this five root three in this direction. So I have five root three in this direction. Okay. So if I want to find out the resultant from this, so this can be written how this can be written as five root three i cap plus 20 j gap. Okay. This is in kilo. This is how I did component form if you remember. Okay. This is how I broke things into components. Now they're asking what is this magnitude? I already told you how to find out the magnitude from from the component form always just square. Okay. Five root three square plus 20 square. And this will give you what under root of five square 25 into three. This is 75. This is 400. Okay. So under root of 475 is the answer we are looking for. Okay. This is the answer we are looking for. So I think 375 is not a perfect square. Is it 25 is 625. Yes. It's not a perfect square. Okay. So no need to worry. You can keep the answer like this only. No need to worry. You'll get find the resultant velocity of the book. Yes. This is all you have to find out. You don't even have to find out the direction. Okay. So root 475 is the answer to this question. Okay. Okay. So let us do this question. In this question, it's saying that there are two forces each equal to P by two that are acting at right angles. Okay. And their effect may be neutralized by a third force acting along their bisector in the opposite direction with a magnitude of how much. So if I try to make the diagram for this, then the diagram would look something like this. So there are two forces P by two and P by two, which are acting at 90 degrees. Okay. These are acting at 90 degrees. These are acting at 90 degrees. This is what the first statement says. Okay. Then I have to find out a force in this direction or a vector in this direction, which will essentially just cancel out all of this. All of this. Okay. So if I think about this, let's call this this x vector. Let's call this a x vector. Okay. So how will this, how will I proceed with this question? The answer is very simple. So first I have to find out what is the resultant of P by two and P by two. Okay. So first I have to do that. And since they are in the perpendicular direction, so if I want to find out the resultant, I can use the parallel graph law. But in the previous question you saw, if it is in 90 degrees, I can simply find out the resultant by just doing under root of P by two whole square plus P by two whole square. Okay. So this will give me under root P square by four plus P square by four. So if you, oh sorry. So this would mean, okay, this answer is equal to under root of two P square by four under root. So two and two will cancel. So your final answer will be P by root two. This is, this is the addition of P by two and P by two that will be in this direction. Okay. Obviously I have to find out whether how much this angle is. How much is this angle? So for this finding out this angle, you can either use one of the formulas that was given in the course course of, you know, based upon happiness or you can complete this triangle. If I complete this triangle, this will be P by two, this will be P by two and this angle is how much? Let's call this angle alpha. 10 alpha is what perpendicular, which is opposite side P by two divided by P by two, which means one. So this angle is nothing but 45 degree. If 10 alpha is one then alpha is 45 degree. So that means this P by two and P by two, they are given separately to us. But if I add them together, if I add them together, I'll get a vector whose magnitude is what P by root two, which is pointing in the opposite direction at 45 degrees. Now they're asking me in the question, they're asking me what should be the magnitude of this X vector? What should be the magnitude of this X vector so that it cancels out this entire thing? So that means the addition of P by two, P by two and this X vector needs to be zero. So in other words, they're asking me what should be the value of this X vector so that if I add X vector to this P root two, this will become zero. The answer is the X vector should also be equal to P by root two. When X is equal to P by root two and they're both in opposite direction, they will just cancel out and make this zero. Okay, and that is exactly what is asked in the question. This word neutralized, this fancy word neutralized just means the sum of all the vectors should be equal to zero. Okay, so the answer, how you find it? The answer is equal to P by root two. Some people will be doubtful. Okay sir, why does that need to find out this? Why do I need to find out this 45 degree? Because if this angle we do not know if this is not 45 degree, this is not exactly going through the middle then these two will not be in the opposite direction. These are in opposite directions because this vector that we find out found out P by root two is also in exactly in the middle is the bisector of this 90 degree, like 45 degree and 45 degree is going exactly through the middle. And in the question it is said this X vector is going through the middle, the bisector. So that's why they are in the same direction. Otherwise, I won't be sure whether X vector and this P root P by root two vector, it's not P root two, it's P by root two. So this P by root two vector and this X vector will not be in the same direction if that angle is not equal to 45. So that warranted the need for me to find what is that value 45 degree. So that is why I found out this. Okay, now let us do this next question. Okay, so let us do this next question. In this next question they are saying we have the A vector is given to us. i cap, 2j cap, minus 3 cap. When a vector B is added to this A vector, we get a unit vector along the X axis. This is very easy to understand. As soon as they say unit vector along the X axis, I exactly know what this is. Okay, so what is a unit vector along the X axis? i cap. Okay, i cap is a unit vector. One i cap is the unit vector pointing towards the X axis. One j cap is the unit vector pointing towards the Y axis and one k cap is the unit vector pointing towards the Z axis. So here we have, so they are saying that A vector plus B vector, okay, A vector plus B vector should be equal to 1 i cap. Okay, so A vector is how much? It is 1, it is i cap plus 2 j cap minus 3 k cap. So plus B vector is going to be equal to 1 i cap. So what I have to do is I have to rearrange the terms. So B vector will be 1 i cap and 1 i cap will cancel. If I take this side, so this will be minus 2 j cap plus 3 k cap. So I think the first option is the correct one. Okay, let's move on to the next question. Okay, another easy question right here. Okay, if the magnitude of if the magnitude of X and Y components of A vector are 7 and 6. So from the first statement, they are saying A vector A vector has 7 and 6 as the X and Y components. So how will I write this? A vector I will write as 7 i cap plus 6 j cap. So from the first statement only, I can get this. Also the magnitude of X and Y components of A plus B. So if you do A plus B vector, your components will be how much? It will be 11 i cap plus 9 j cap. This is what is given in the second statement. The magnitude of X and Y components of A vector plus B vector is 11 and 9 respectively. What is the magnitude of B? So what do I need to do? I just need to simply subtract A vector from A plus B vector. So if I do A plus B minus A, so that means I'm just subtracting these two. So 11 i minus 7 i plus 6 j. Oh, sorry, plus 9 j minus 6 j. So what did I do? I subtracted the i from the i, subtracted the j from the j. So from here, 11 minus 7 is 4 i cap. 9 minus 3 is 9 minus 6 is 3 j cap. Okay, so this is what this is the B vector. But they don't want the B vector. The question is asking what is the magnitude of B? So remember, I told you how to find out the magnitude if you are given the component form. All you have to do is square the components and take it under root. So the magnitude of B vector would be nothing but under root of 4 square plus under root of 3 square, making this under root of 25, which is 5. So for this option, A is correct. Okay, let's do some more questions. Again, let's do this formula-based question. The angle between z-axis and the vector is asked. Okay, so we have a vector A vector. Let's call this A vector, which is i cap plus j cap plus root 2 k cap. If I want to find out the magnitude of this A vector, how do I find out? I don't already told you. All I have to do is square. I have to square the components. So the first one is 1 square, the second one is 1 square, the next one is root 2 square. Okay, so that means we have what? Under root of 1 plus 1 plus 2, which is under root of 4, and the magnitude of A vector is 2. The magnitude of A vector is 2. Now, if you remember the formula, for if you want any angle, like angle with the x-axis, angle with the y-axis, or angle with the z-axis, all you have to do is find out the cos of that angle. So cos of gamma, where I'm assuming gamma is the angle between this vector and the z-axis, all this is equal to, is the z-component divided by that. If you forgot the formula, you can always refer to the class nodes. Okay, this is also available on the slide. If you want to find out what is the angle between the vector and the z-axis, all you have to do is find out this ratio. The z-component, z-component, so like A z divided by magnitude of A. So in our case, the z-component is root 2, root 2, and divided by 2. So if you do this, you will get 1 by root 2 as your answer. So essentially, we are looking for an angle. So cos gamma is equal to 1, 1 by root 2. So that means gamma is 1 degree 45 degree, right? Because if you know, if you remember the trigonometric ratios, cos 45 is equal to 1 by root 2, okay? So we have solved this. So the angle between z-axis and the vector is 45 degree. Okay, cool. Okay, so in this question, the question is saying that we have a vector like this. We have some B vector, let's call this A vector. A vector we have as 0.5 i-cap plus 0.8 j-cap. And this z-component we do not know. So they are saying that this is equal to c times k-cap. Okay, and they are finding, they're asking us to find out what is this value of c. Now, only thing that is given is that this A vector is a unit vector. So what is the property of unit vector? Unit vector is defined as this. It is a vector whose magnitude is 1, whose magnitude is 1. So all I have to do is find out the magnitude and how do I find out the magnitude? Square all the components, add them and take a square root. So this is nothing but 0.5 square plus 0.8 square plus c square. And this will be equal to how much? This will be equal to 1. So squaring both sides, you will get 0.5 square plus 0.8 square plus c square is equal to 1. Okay, if you square both sides, 1 square is also 1. Okay, so this will be, c square will be nothing but 1 minus 0.5 square minus 0.8 square. So I think 0.5 square is 0.25. Okay, 525 and this is minus 0.64. Okay, so if I do this correctly, I'll have 0.75 minus 0.64, which makes this 0.11. Okay, so obviously this is equal to c square. Okay, then the value of c would be under root 0.11. Let's see if we have the data option. Yes, we do. So option number a is correct for this question. Okay, so let's move on to the next question. Easy stuff. This is easy stuff. Okay, let us do the next question. In this, they are saying there's a plus b equal to c. The magnitude of a is given, magnitude of b is given. And the value of c is also given. And we are asked the value of, what is the angle between a and b? So we'll use the simple, you know, a parallelogram law. R is under root a square plus b square plus twice a b cos theta. Okay. And in this, our resultant is the c1. So I'll plug in all the values. c is going to be 3. a square is going to be 3. b square is going to be 3. 2 times under root of 3 multiplied by under root of 3 and cos of theta. Okay. So if you solve this equation, you are going to get the value of cos theta. So this will become 9. This is 6 plus 2, 3, 6 cos theta. Okay. So I end up with this cos theta equal to half. Okay. So for what angle is cos theta equal to half theta equals to 60 degrees. So the answer to this question is 60 degrees. Let's go. Okay. Let's do a question which is slightly more difficult. This is based on vector product. Okay. So here there is a, there are two vectors given to us. Vector a and vector b. And they're asking which of the following statements are true. So we have to check whether it is perpendicular or parallel. Or also we have to give, we have to check their magnitudes because there are options that are talking about the magnitudes. Okay. So we have studied both the dot product and the vector product, the scalar product and the vector product. So we know that the definition of dot product is what? Dot product is essentially a dot b is defined as magnitude of a, magnitude of b cos of theta. Okay. So although we studied this, there's one thing that we need to know about this, that we are dealing with cos theta. So anytime, anytime a vector is perpendicular to b vector, anytime a vector is perpendicular to b vector, a dot b should come out to be zero. It should come out to be zero because we're dealing with cos theta here. And I also told you in case of, when you do i dot j, those are perpendicular vectors, the answer always comes out to be zero. So here if they're asking if a is perpendicular to b, then all you have to do is find out a dot b. You will have to find out a dot b. If this is numerically equal to zero, then option a would be correct. Okay. So let's check it out. So a vector is how much it is i cap minus j cap plus 2k cap. And how, what is the value of b? b is 3i minus 3j plus 6k. Okay. So if I multiply them together, if I multiply them together, the answer would be 3 ones at 3 plus 3 plus 12. Okay. So the answer is, the answer is it's 18. It's not zero. So I can say for sure a is not perpendicular to b. It's not perpendicular to b. Let's check whether they are parallel to b. So how do you check the concept of parallel? How do you check whether something is parallel? Okay. So if you have two vectors like this, okay, I have i cap plus j cap plus k cap and I multiply this thing by two. So for example, 2i plus 2j plus 2k. Okay. These two, these two vectors will be parallel. So what I'm trying to say is, if the two vectors are such that if you divide one vector by a particular scalar number, then you'll get the same vector. As in, if I try to say it simply, this would be something like this. The components, the individual components are a direct multiple of the other vectors. So here you can see I got this vector by multiplying this with two. Okay. So if I multiply this with minus one, this would become minus i minus j minus k. Okay. So this is again another parallel vector, parallel vector but in the opposite direction. Okay. So this is how we check the parallel thing. Okay. So in this, if you see this a vector and b vector, if I multiply a vector by three, if I multiply a vector by the number three, I'll get the b vector. Yes or no? Do it. Do it. Check it yourself. Like if I multiply a vector with three, this would become, let me take a different pen here. If I multiply a vector with three, this would become three i cap minus three j cap and three two just six k cap. If you check this, this is exactly matches with b. So that means these two vectors are parallel. So how do you check the parallel nature? You just divide the components. Okay. You divide the components. If you're getting the same number, that means they are parallel. So what do you do? For example, x component of b is three. So three divided by one is equal to minus three divided by minus one. And this is equal to six divided by two. So all you have to do is check this ratio bx by ax, b y by ay and bz by az. If the ratios are, all of these ratios are equal, then the lines will be parallel. Okay. If the ratios are negative, that means they are anti-parallel. They're parallel, but in the opposite direction. Okay. So here option b is correct. We can also check the magnitudes here. The magnitude of a would be what? The magnitude of a would be under root one square plus minus one square plus two square, which would mean this is equal to under root of four, five, six. Okay. So under root of six magnitude of b, on the other hand, is under root three square plus minus three square plus six square. Okay. So that would mean under root of nine plus nine. That is 18 plus 36. That makes it, how much? 54. Okay. So they're asking us the ratio. So if I do b vector divided by a vector, then I'd get root 54 by six. So that is nine. So that would mean under root of nine is three. Okay. So I can say that a vector has like one-third the magnitude of the b vector. Or b vector is three times a vector, which is not in any of the options. So option b is the only correct answer. Okay. It is the only correct answer. Let's move on. Okay. So the next question is something like this. If p vector dot q vector is given to us as zero. So what inference do you get from this? Either the vector p is a null vector or the q vector is a null vector or cos theta, because p dot q is defined as magnitude of p magnitude of q into cos theta. This is equal to zero. Zero. Okay. Now unless specified, don't just assume key we are dealing with null vectors because it's not common to get null vectors. Okay. So the magnitude of p and q are there. Anyway, in the next statement, it says find out p cross q. If any of those p vector and q vector are null vectors, then you cannot take a cross product like that. Okay. So what essentially we have to come to a conclusion key cos theta must be zero. Then, okay, cos theta must be zero. If cos theta is zero, then theta is what angle theta is 90 degree because cos 90 is zero. Right. So that means this theta must be 90 degree. Okay. We inferred from that. Now they're asking what is the magnitude of p cross q. We know the magnitude of p cross q is magnitude of q multiplied by magnitude of q. Okay. And sine of theta. But we already know the value of theta. Theta is 90 degrees. So this is magnitude of p magnitude of q sine of 90 degree and sine 90 is one. So the answer is magnitude of p magnitude of q. So option A is the correct answer in this. Okay. Option A is the correct answer in this. Let's go. Okay. So let's do this question again. Like this is one of the other common question they will ask you what is the unit vector. And I already told you the unit, what is the unit vectors formula. So a vector is made up of magnitude as well as direction. So if I divide the vector by its own magnitude, we get what the unit vector. Okay. So we have what i cap plus j cap. Okay. If I want to find out the magnitude of this thing, the magnitude of this would be under root one square plus one square. Because if I write only i cap plus j cap, that means there is one i cap and one j cap. So this, you know, the magnitude is root two. Okay. So if I want to find out what is r cap, I simply divide i plus j divide by root two. Okay. So what I get is one by root two i cap plus one by root two j cap. Okay. So that is how you do this kind of questions. Okay. Let us do this question. So this is a little bit more applied compared to other questions. In this, the force is given to us. Force is given as 10 i cap minus three j cap plus six k cap. And it is said that this displaces from this point to this point. So the displacement vector is not given, but it is given that the point is going from what position to what position. So to find out what is the displacement, all you have to do is final position minus initial position. Okay. That is the formula for displacement. So whatever displacement r vector, I'm presenting by r vector, is given by final position minus initial position. Final position is 10 i minus two j plus seven k. And I have to subtract it from the initial position. So this is the initial position. This is the final position. So initial position is what six i plus five j minus three k. Okay. So everything here is in meters. So if I subtract them, 10 minus six becomes four i cap minus two minus five becomes minus seven j cap and seven minus minus three is plus 10 k cap. So I got the displacement vector from these two. Now, if they are asking, what is the work done? Okay. So I know the formula for work done is force dot displacement. So all I have to do is multiply these two vectors using a dot product. So 10 i minus three j plus six k. I have to dot this with four i minus seven j plus 10 k. Okay. So when I dot it, I simply multiply the coefficient. 10 fours are 40 minus three times minus seven is plus 21 plus six times 10 is 60. So the answer would come out to be 121 joules. Okay. So 121 joules b is the correct option here. Okay. b is the correct option here. Let's move on to the last question. Okay. So this is the question that we are going to do. They're asking what is the angle between a vector and b vector when a and b vector are given. So remember dot product is also useful, very useful in finding out the angle between two vectors because what I why I'm saying that is because a dot b is simply magnitude of a magnitude of b and cos of theta. This was a definition from here cost theta can in most cases cost theta can be found out using by taking this into the denominator and making it like this. Okay. So I took the magnitude of a to the left hand side and magnitude of b to the left hand side from here I got the value of cost theta. So I'm going to do this and find out cost theta. Okay. Now a dot b is very, very simple. I just multiply the coefficients here. So two fours are eight plus three threes are nine plus four twos are eight. So all I did is I found out a dot b from this component form. So I multiply two times four then I multiply three times three m. This four times two I did. So this is the a dot b on the numerator on the denominator we want magnitude of a magnitude of a is just a square addition of all the squares of the component. So two square plus three square plus four square and then under root of four square plus three square plus two square. Okay. So when you do this you're on the numerator you get 16 plus nine is 25. Okay. And in the denominator this first one would be two square is four. So four plus nine plus 16. Okay. So 29 the first one is 29 under root 29 again because this is again four square three square so if I add all of them together the simplified form would be 25 by 29. Okay. So cos theta comes out to be 25 by 29. So theta is what theta is cos inverse of 25 by 29. Okay. No need to find this. Okay. Just write it like this cos inverse 25 by 29. I think it's one of the options. Yes. C is the correct option in this one. Okay. So with that I'll end this video. Okay. So again because not a lot of questions were able to be done on class. So that is why I issued a problem solving video. There are different problems from this also but this should lead a foundation of you know you being able to do other questions from vector. So I'm going to attach a bunch of assignments with this. So this should be used as a foundation and then once you get the hang of it do as many questions as you can do module do assignments and then you should be ready for scalars and vectors and do this for every chapter. Okay. So I'm making this video for scalars and vectors probably it won't it's not possible for me to make for other chapters also. Okay. But we'll see what we can do. I will attach other videos. Okay. And then we can work on it. Anyway, we'll see you next time. Bye. Bye.