 Let's move on to the next concept, simple question based on section formula. Done? Sir, one minute, just 5 seconds. Okay, Shriya is done. Sir, done. So let me discuss it, we divide the point P and Q. P has position vector i plus 2j minus k. And Q has position vector minus i plus j plus k. Okay. If the division ratio is 2 is to 1, internally. Okay. Then my formula is, okay, I'll just write down the formula for internal division. It is, let's say this is P vector, this is Q vector. So it will be 2Q plus P by 3. For external division, it will be 2Q minus P by 2 minus 1, which is 1. Okay. So just plug in the values 2Q plus P by 3. And this will be, so this will give you a minus i plus 4j plus k by 3. Okay. And external division will give you a minus 3i, 0j plus 3k. Is that fine? Is this what you are getting? All of you? Yes, sir. Fine. So now we are ready to take up the concept of resolution of vectors. First we'll talk about resolution of vectors in 2D. So let's say we have a vector a. Okay. I can resolve these vectors along two perpendicular directions. And normally we choose those two perpendicular directions as your direction of the x-axis and the y-axis. Okay. So this vector can be resolved along x direction and y-axis direction. Okay. Now how do we resolve it? Very simple. See, let's say the magnitude of this vector is mod a. Okay. And let's say this angle is theta. Correct. So if I drop a perpendicular from here, can I say this length would be mod a cos theta? Okay. So let me name it as OAP. So OP length will be mod a cos theta and AP length will be mod a sin theta. Correct. At the same time, we know that if the coordinates of a is x, y, x actually could be written as mod a cos theta, y could be written as mod a sin theta. Okay. Now think as if you want to reach from O to a. Then take a path from O to p and then take a path from p to a. Correct. O to p path would be nothing but mod a cos theta into a unit vector along the x-axis which we know as i cap. Okay. Pa is nothing but the length of pa that is mod a sin theta and a unit vector along AP. So this unit vector is the same as the unit vector along the y-axis which is j cap. Okay. So your OA vector which is nothing but your vector a could be written as OP plus pa and OP plus pa is nothing but mod a cos theta i mod a sin theta j. Okay. And since your x and y are your mod a cos theta and mod a sin theta respectively, you can write this like this. Okay. So one thing is for clear over here is that the position vector of a, the position vector of a that means the vector connecting origin to the point a which is nothing but this vector is actually the coordinates of that point along with i and j unit vectors attached to them. Are you getting my point? So if I say the position vector of a point is 3i plus 4j, what does it mean? That means the vector connecting this point with the origin, this is 3i plus 4j and this coordinate of this point is actually 3 comma 4. Okay. So this is how we can resolve a vector along the direction of x and the y-axis respectively which we call as the resolving of vectors. Is that fine? This concept is very well known to you. Okay. Now the problem comes when we are trying to resolve a vector in 3D. So let's talk about resolution of vectors in 3D. Okay. So let's say this is my right-handed coordinate system. This is your x, this is your y and this is your z. Okay. And let us say we have a vector like this, a vector. This point is a. Okay. Now before we understand the concept of resolution of vectors in 3D, we have to understand something called direction cosines. Have you heard of this term in your school? Yes. All of you have heard this? Yes, sir. How do you define direction cosines? It's the cos of the angle made by the vector with x, y, z. Absolutely correct. So basically let's say this is your vector a and this is your positive x-axis, positive y-axis, positive z-axis. If you draw a plane which is containing oa and ox, I hope you can imagine a plane which passes through ox and oa. Then this angle which we call as angle alpha. Okay. If you draw a plane through oa and oi, let me call this plane as containing this angle beta. And if you draw a plane through oz and oa and that plane contains this angle gamma, then cos of these angles basically are called the direction cosines. These are your direction cosines. Normally we address them by simple alphabets to make our life easy. LMN. Okay. Later on, this will be a very useful concept for understanding the equation of plane and line in three-dimension. Okay. So in 2D, we have the concept of slope. So in 3D, the same concept of slope is now converted to direction cosines concept. So there's nothing called slope of a line in 3D. The equivalent concept of slope in 3D is direction cosines. Are you getting my point? Okay. Now, in order to resolve this vector along 3D, okay, let me again use polygon law of addition. Okay. So let me drop a perpendicular from a on the xy plane. Okay. So let's say, yeah. Let me draw, drop a perpendicular, let's say a dash. Okay. And let me connect a line parallel to a dash with the x-axis. Let me call it as p. Okay. Now, let's say the coordinate of this point is xyz. Okay. So can I say this length is x, can I say this length is y and this length is z? Okay. You will all realize that your length x is nothing but the length of this vector into cos of the angle between the vector and the x-axis. Y will be nothing but mod A into cos of the angle made with the y-axis and z length will be nothing but mod A into cos of the angle made with the z-axis. Correct? So in order to reach from O to A, I have to take a path. You can see the figure over here. From O to A, if I want to reach, I can take a path O to P, P to A dash, A dash to A. Okay. So it's O to P, P to A dash, A dash, A dash to A. Okay. O to P, O to P is nothing but A dash cos alpha into I. Okay. A dash cos alpha times I vector. I is the unit vector along x-axis. P A dash. This is this length over here. You can see the motion of my pen over here. P A dash. Let me slightly put it to the left. Yeah. P A dash will be mod A cos beta in times j cap and A dash A vector would be mod A cos gamma into k cap. Okay. So if you take mod A common, you'll get something cos alpha I cos beta j cos gamma k. And this is nothing but your vector A. This is nothing but your vector A. Okay. So basically you have resolved the vector A along the I, j and k components. So you can write this like this also. Okay. Okay. One thing is very clear from this derivation is that if you look at this term, A by mod A actually represents cos alpha I cos beta j cos gamma k, which is nothing but saying that A cap, remember A by mod A is A cap is li plus mj plus nk. Now this is a very useful relation for us. This relation says that the direction cosines are nothing but the i, j and k components of the unit vector along that given vector A. So if I have a vector A like this, then a unit vector along this A, this is nothing but let's say this, this vector has direction cosines l mn. Then this vector is actually li plus mj plus nk. Very useful information. We'll be using this later on very frequently. Is this clear guys? Direction cosines represent nothing but the i, j and k components of a unit vector of the unit vector along the direction of A. Is this clear? Yes, sir. Now, if this is clear, let me ask you some questions. Hope you can read this question. Let me know once you're done. Okay, Shreya is done. See, very simple problem. Let's say I'm showing my original coordinate axes like these. Now you rotate it 5 by 4 in an anticlockwise direction. So let's say now along the z, that means z doesn't change. So this is 5 by 4, this is also 5 by 4. Now in this rotated coordinate axes, these are the components of a vector. Find the components of these vectors in the original system. Okay, now when you rotate, let's say z axis is not rotated. Let me show you a 2D view. So let's say this was your original x and original y. Okay, and let's say you rotate a vector, you rotate these coordinate axes 45 degrees anticlockwise. Okay, 45 degrees anticlockwise. Okay, now any unit vector in this direction, let's say I call it as i cap dash. Can I say i cap dash is nothing but it is original i plus j by root 2. Yes or no? Do you agree with me on this? Okay, similarly, the new y component, let's say j cap dash. j cap dash could be written as minus i cap plus j cap by root 2. Do you agree with me on these two fronts? Yes or no? Then only I'll move forward. Yes? Okay. Now, if I want to make a vector, remember k will not change. So k cap dash is same as k cap. So if your vector was in the new coordinate system, 2 root 2 i cap dash, 3 root 2 j cap dash and 4k cap dash. Then if I convert it back to i and j normal components that we had in the beginning of the system, this i cap dash is nothing but i cap j cap by root 2. j cap dash will be minus i plus j by root 2. k cap dash is same as k cap. Okay, so this root 2, this root 2 cancels, this root 2, this root 2 cancels. So it's 2 i plus j plus 3 minus i plus j plus 4k. If you expand it, it becomes minus i plus 5j plus 4k. That means the components, these are the components of the vector in the original coordinate system. That is what we are asked to prove over here. So as you can see, this is what we are asked to prove. Is that fine? So basically here it is the concept of knowing how your unit vector, when you are rotating the x y axis is 45 degrees anti-clockwise, how does the unit vector actually gets changed?