 Let's take another one. A man travelling towards east at 8 km per hour finds that the wind seems to blow directly from the north. On doubling the speed, he finds that it appears to come from northeast. Find the velocity of the wind. Sir, does this north mean his north or the normal north? His north. His north and normal north will be the same, right? Turn towards the right and then if he is travelling then the north will be like the other side. North is fixed, north is a fixed thing. My north and your north will be the same, right? But if they are facing opposite then it will be different north. Growing from the north. North direction doesn't change if we are facing other side. If I turn round, north will remain north, right? It is not from his north or something. It is from the north. That's it. Are you all aware of the concept of relative velocity? When you see this word seems, seems means relative to the man. See this word seems. So whatever is the velocity of the man, the wind relative to man appears to come from north. This appears to come from north. Let me just make a diagram like this. So let's say the man is moving in this way. So this is your V man. This is what he sees. He sees that the wind is coming from the north. So this is the V man minus, sorry, V wind minus V man. Since it is coming from the north means it is going towards the negative y axis. So you can call it as minus pj. Sir is it a root 2? Correct. What is the direction? Sir, is it once again? Okay, so let's say the wind velocity is xi plus yj. So what is given to you here is Vm minus, sorry, Vw minus Vm. Why am I writing m every time here? Sorry. Vw minus Vm is minus pj. So xi plus yj minus 8i is minus pj, which is nothing but x minus 8i yj is minus pj. Now, guys, we have not yet studied about the concept of linear dependent and linear independent. But here let me tell you, since there is no i component over here, you can safely say that this is 0, which means x is equal to 8. Okay. Now, second scenario is if he doubles his velocity, let's say he makes it 69. Now this becomes your new velocity of the man. Then the wind appears to come from northeast. Okay. So let me say it is minus qi plus j. Okay. So this time V wind minus V man is going to be minus qi plus j. So xi plus yj minus 16i is minus qi minus qj, which makes it x minus 16i plus yj is minus qi minus qj. Now, x minus 16 should be equal to minus q. x is already 8. So minus 8 is equal to minus q. That means q is equal to 8. And from here y is equal to minus q. That means y is minus 8. Okay. So the velocity of the wind becomes 8i minus 8j. That means the wind is appearing to come from the north. This is 8i minus 8i minus 8j. Correct. That means the wind is coming from northwest. Okay. And the speed of the wind, that is mod of the speed of the wind velocity will be 8 root 2 whatever units. Clear everyone. Please type CLR if it is clear. Okay. Next question. So how important is this chapter like considering the competitive? Very important. Very important. Vector at least two questions. 3D at least two questions. Vector 3D combined three to four questions will come. Find a unit vector C if this bisects the angle between C and 3i plus 4j. So basically there is a vector 3i plus 4j. Okay. There's another vector C which is actually a unit vector. So C is a unit vector. And there is a vector which bisects the angle. This vector is your minus i plus j minus k. Find C vector. So here's a very important concept which I want to talk about via this problem is see if you have any vector A. Okay. And you have another vector B. There is a vector which is bisecting the angle between A and B. Okay. Let me say C. Then let me tell you that C vector can be expressed as lambda times A by mod A plus B by mod B. Now of course my question would be how? How is this possible? Can you at least justify this? So what I'm claiming here is that the C will lie along. Lambda is just saying that there is a vector along the direction of this vector. At least somehow you get this vector. Lambda you can always justify. Lambda means any vector which is collinear or lying parallel to that vector. Then we'll come back to this problem. So as of now park this problem and let's try to focus on this. Anyone? Any idea? Aditya, Shriya, Hruthu, Ardhara, Samyukta. You can unmute yourself and talk. No worries. Don't know sir. You have muted yourself to just say don't know. Sir, could it be like from parallelogram law of vector addition? And if we take the parallelogram, if we take a special case, then we can prove this. But are we allowed to take special cases? Yeah, go ahead. Let's see what is your special case? How special? Tell me through special case only. No problem. Technically I'm not sure whether in parallelogram angles are bisected. That's why I'm taking a special case of a square. So then by parallelogram law of vector addition, c vector will be the resultant vector and that. Absolutely. There's nothing wrong in that. See you actually, you know, you're doing whatever is required. See if I take a unit vector along a, let's say unit vector is like this. Okay. So let's say I take a unit vector along a cap. Okay. A unit vector along BB cap. Can I say since these two vectors are of the same length, one unit. Okay. This parallelogram will have, will be more or less like a rhombus. Isn't it? So if I, if I take a bisector over here, can I say this bisector angle will connect this point to the midpoint of these two? Correct. Yes or no? Do you agree that bisector angle will connect the midpoint? So if let's say this is your O and this position vector is a cap, this position vector is B cap, then this position vector would be a cap plus B cap by two. Correct. Can I say any vector in the direction of, let's say I call this point as C, any vector in the direction of C, can I say it will be some constant? Let me call that constant as lambda. Not lambda. Let me call it as beta. Beta times A cap plus B cap by two. Correct. So even C lies in the direction of this. So C is nothing but some constant times A cap plus B cap by two. Can I just pull out a two out and write A cap and B cap as A by mod A and B by mod B? And beta by two is just another constant which we can call it as lambda also. So can I now say that any, this is a very important result guys. Let me tell you there's some reason why I'm deriving this all because this result is important. This result is directly used in solving many problems. Okay. So any bisector angle vector, which is bisecting the angle between vector A and vector B could be written as lambda times A by mod A plus B by mod B. Getting my point. Now, can you please use this to solve this given problem? Now come back to this question. Now this C is different. Okay. Don't get confused. This is a bisector. This C is the one which is getting bisected. So now don't have the same C carry forward. Okay. Now solve this. So can I say minus 3 by 5 I cap and 4 by 5 J? No, no, no. The answer is slightly more complicated. Can I say this vector is lambda times C vector. Let's say C vector is XI plus YJ plus ZK. So C vector itself is a unit vector. So you don't have to do A by mod A. By the way, let me just make a confirmation here that this will be equal to 1 plus 3I plus 4J by 5. Is that clear? By the same logic that we discussed over here. Okay. This bisector vector is lambda times the unit vector along C, which itself is a unit vector. So I've just written it like this plus unit vector along 3I plus 4J, which is 3I plus 4J by 5. Now let us compare the coefficients. Minus 1 is lambda times X plus 3 by 5. J component is 1 over here. J component here is lambda times Y plus 4 by 5. K component is minus 1 lambda times Z. Is that fine? So now Z is minus 1 by lambda. Y is nothing but minus 4 by 5 minus 1 by lambda plus 1 by... Which you can write it as 5 minus 4 lambda by 5 lambda also. And X is nothing but minus 1 by lambda minus 3 by 5. Which is nothing but minus of... Now how do I get the lambda value by using the fact that X square plus Y square plus Z square is equal to 1. Because X, Y, Z are the components of a unit vector. So 5 plus 3 lambda square by 25 lambda square. 5 minus 4 lambda square by 25 lambda square. And 25 by 25 lambda square should be equal to 1. So let us take 25 lambda square on the other side. So it is 5 plus 3 lambda square. 5 minus 4 lambda square plus 25 is 25 lambda square. Let me expand it. If you expand it, 9 plus 25 lambda square here itself I will get. And I will get 30 lambda minus 40 lambda which is minus 10 lambda. And constant terms would be 25 plus 25 plus 25. That is 75 is equal to 25 lambda square. So 25 lambda square, 25 lambda square gets cancelled off. Giving you lambda value as 75 by 10 which is 15 by 2. That means X... Now put it back over here. Put it back at these values. This value, this value and this value. So X will be nothing but minus 1 by lambda which is minus 2 by 15 minus 3 by 5. Minus 3 by 5. You can write it as minus 2 by 15 minus 9 by 15. That is minus 11 by 15 is your X. Y will be minus 4 by 5 plus 2 by 15. That is minus 12 by 15 plus 2 by 15 which is actually minus 10 by 15. And Z will be minus 1 by lambda which is minus 2 by 15. So your final answer will be your C vector is minus 11 by 15 i minus 10 by 15 j minus 2 by 15 k. Any questions so far? No sir. So this is going to be your answer.