 Now why am I all of a sudden talking about these cases is because you are now going to study co-linearity, co-linearity of two vectors. What is the meaning of co-linearity? Co-linearity means they are parallel. So we say vector A and vector B are co-linear when they are linearly dependent. Vector A and B are co-linear when they are linearly dependent. What is the meaning of linearly dependent? That means if you do m1A plus m2B equal to null then m1 and m2 not all will be 0. Are you getting my point? In other words you could express A as minus m2 by m1 times B which is nothing but you can express A as lambda times B. So if you could express one vector where lambda is some scalar quantity. So if you could express one vector as lambda times the other then we say that the vector A and B are co-linear to each other. In other words they are linearly dependent on each other. That's why the concept of co-linearity evolves with the concept of linearly dependent. If they are non-co-linear my dear friends, let me write the reverse case also. If A and B are non-co-linear then they are linearly independent. In other words if you try to do this then it will only result into m1 and m2 both being 0. Let us try to understand the deeper meaning of this. Let's try to understand this through vector addition. Let's try to have another perspective of it. If let's say this is your vector A and this is your vector B. Are they co-linear? According to my diagram let's say they are parallel. Are they co-linear? Now if you see this, this is trying to say that if you manipulate these two vectors in such a way that their resultant becomes 0. That means if you do m1A plus m2B in such a way that the resultant becomes 0 this is possible without m1, m2 being both 0. How? Let's say this is double the length. So what I will do is I will half its length so that it becomes like this. Then I multiply it with a minus 1 so that it reverses its direction. Now if I add these two so 1 into A and minus half into B, 1 into A and minus half into B do you realize that this vector and this vector will close the path and it will become a 0 vector. So m1, m2 here both are non-zero. That is what this theorem actually says. It says that it is possible for you to multiply these two vectors with a certain scalar quantities that means you are fiddling with their direction and magnitude in such a way that they both cancel out each other. That is possible for co-linear vectors and hence these vectors are linearly dependent. Have I made myself clear? Yes sir. But if you have such vectors my dear friends like this, A is like this and B is like this. Can you try making the resultant of these two as a null vector? Can you somehow close this path such that this plus this ends up being a closed path? Can you do something like this in this case? No unless until you make both of them 0 isn't it? There is no way you can close the path. You try fiddling with the direction and length of these two vectors. Do anything you want. Says that they cancel out each other. Can that happen? No it will not happen. It can only happen when you have multiplied both of them with a 0. That means you have made them both dots. Then addition of two dots will give you a dot that is a null vector. But if you have any other non-zero quantities for M1 and M2 it is not possible to close the path between these two vectors or it is not possible to make them cancel each other out. So these cases are called these vectors would be called linearly independent. So two non-colinear vectors are linearly independent. Two collinear vectors are linearly dependent. Are you getting this part? Now a same concept can now be applied to collinearity of three position vectors. Collinearity of three position vectors or points you can say. Remember position vectors and points there is no difference. What is position vectors in vector is points in coordinate geometry. Getting my point. So many points here. So you say position vectors a b c. See guys we never talk about collinearity of two position vectors. Any two position vectors will always be collinear. So it's a futile effort to talk about collinearity of two position vectors. We can always pass a line through two points isn't it? So in order to make sense we will talk about three or more than three points. So three points let's talk about three points let's say point a, point b and point c whose position vectors are as shown. They will be collinear. That means they will all lie on a line. If. Now there are two ways in which I normally explain it. If a, b, c are collinear then construct any two vectors by choosing any two of the points. Let's say I take a, b and a, c. Then one can be expressed as lambda times the other. If this condition is satisfied then point a, b, c will also be collinear. Mind you you can take any two vectors you like. You can take a, b and b, c also. So even if you can express this as lambda times this, okay, any one of these conditions then these three points will be collinear. That's one way of understanding it and this is the easiest way. Okay. Normally I also choose this method to prove whether my three vectors are collinear or not. Okay. Second method is slightly tricky but you should also understand this. Second vector says that if you express the linear combination of these vectors, if you express the linear combination of these vectors as a null vector then m1, m2, m3 not all zero, not all zero must satisfy m1 plus m2 plus m3 equal to zero. This is slightly challenging way but this is easy to understand once you know the logic behind it. So what am I trying to say is that if a linear combination of a, b, c express as a null vector, okay, it should lead to m1 plus m2 plus m3 as zero but all of them should not be zero. That means of course m1 plus m2 plus m3 will be zero if all of them are zero. But here my situation is it should be zero despite m1, m2, m3 not all being zero. Then only point a, b, c would be collinear. Okay. How does this actually come about? It's very simple. It actually comes from the section formula itself. Let's say a is this, b is this and this ratio is let's say, let's say x is to y. Okay. So if I use my section formula, b would be nothing but xc plus ya by x plus y, correct. Now it can be minus also doesn't matter. Okay. If you cross multiply x plus yb is xc plus ya. If you write this in form of this term, if you write, if you express this as this, you write it as ya minus x plus yb plus xc equal to null vector. I can clearly see here that this is playing the role of m1. This is playing the role of m2. This is playing the role of m3. You can see that m1 plus m2 plus m3 will actually give you y minus x plus y plus x which actually is zero. That's what this condition is saying. But each one of them is not zero. You remember m2 and m3, they need not be zero all of them. Are you getting this point? Guys, is this understood? If it is understood, let me take a problem. Hope you can read this question. The vectors 2i plus 3j, 5i plus 6j, 8i plus lambda j have their initial points at 1 comma 1. Find the value of lambda so that these vectors terminate on one straight line. Done? Do you understand the question? First of all, see there's a starting point 1 comma 1. So one vector is like this. Let's say this vector is 2i plus 3j. One vector is like this. Let's say 5i plus 6j. Another vector is like this. Let's say 8i plus lambda j. All of these three vectors, they end on the same straight line. So this is the terminating point. Let's say a, b, c are the terminating point. So what should be the lambda values as that they all terminate at the same point? 11 by 3 is not the right answer. No problem, Aditya. See, let's say the position vector of a is not known. So can I say position vector of a minus i plus j is actually your given vector 2i plus 3j. That means position vector of a is nothing but 3i plus 4j. Similarly, position vector of b would be 6i plus 7j and position vector of c would be 9i plus lambda plus 1j. Now these three vectors, these three points are collinear. That means they are linearly dependent. Take any two of the vectors. The third could be expressed as lambda times the other. Okay, so let me take these two vectors and these two vectors. Initial point 1, 1 comma 1, right? Why does minus 1 comma minus 1 comma 1? I think you misread the question. Okay, so I'll use the formula ab vector is lambda times ac vector. I'll use this formula or the other way around. Ac vector is lambda times ac vector is lambda times ab vector. Anything you can use, not a problem. Now, I don't want to use lambda because lambda is already used. Let me use something like beta. Okay, so ac vector means c minus a. c minus a would be 6i lambda minus 3j. This is beta times ab. Ab will be 3i plus 3j. Correct? Now let's compare the coefficients. So 6 is equal to 3 beta. That means beta is equal to 2. And lambda minus 3 is equal to 3 into beta. 3 into beta is 2. So lambda is equal to 6 plus 3, which is 9. lambda is equal to 9. Okay, all right. So one last question we'll take up and then we'll call it a day. If a and b are two non collinear vectors, then show that these three points, points means position vectors are non collinear if the determinant given by that expression is 0. Okay, let's say I use the second method this time. So if I say these three vectors are collinear, can I say x times, let's say this vector y times or let me call it as, okay, let me use x, y, z no problem. Since I was using m1, m2, m3, so I thought I would use the same, but I'm sure you would not get confused even if I replace m1, m2, m3 with x, y, z. Okay, let's say I equate this to null vector, then this should be true for x, y, z not all zero. Correct? Yes or no? Now let me first collect the terms a and b separately on my first equation. So if I take a term separately, so x, l1, y, l2, z, l3, a. Now let's collect the b terms. So x, m1, y, m2, z, m3. Okay. Now please note a and b are non collinear. Non collinear means they should be linearly independent. If they're linearly independent and you're equating it to a null vector, that means these two terms individually must be zero, zero each. That means, let me write it like this. That means l1, x, l2, y, l3, z should be zero. m1, x, m2, y, m3, z must also be zero. And other than that, we have x plus y plus z also equal to zero. Okay. Now remember, we have to find a situation where x, y, z all are not zero. That means this system of equation, this system of equation must have non trivial solution. Non trivial solution means at least x, y, z should be non zero in some cases. Yes or no. Now remember the condition when a system of equation has a non trivial solution. It can only happen when the determinant formed by their coefficient must be zero. Try to recall your determinants chapter. Okay. From determinants from, yeah. From Kramer's rule for system of homogeneous equations. Okay. Is that clear? And this is what we wanted to prove. The other if you know these three points, let's say I call this as l i cap, m1 j cap, l2 i cap, m2 j cap, and this point as l2, l3 i cap, m3 j cap. Now think as if there's a triangle, think as if there's a triangle whose coordinates are known to you, l1, m1, l2, m2, l3, m3. And you want this to be collinear. That means the area of this triangle must vanish. Area should become zero. What are the area expression? Half l1, m1, 1, l2, m2, 1, l3, m3, 1 should be equal to zero. Okay. So even if you drop this half, it will be zero. Now you just transpose it. If you transpose it, it becomes this. Okay. Another way of understanding the same thing. Is that fine? So next class, when we meet, we will be talking about coplanarity of two three vectors, et cetera. And a similar conditions will now be seen in that cases as well. And then after I'll start with dot product. Okay. So today, mostly our session was oriented towards understanding the introduction part of vectors. Okay. More importantly, the concept of linear combination, linear independent, linear dependent is what you may have learned new in this part today. Okay. Anyways, I'll stop this session here over and out from my side. But yes, all the best for you tomorrow's exam. Do well. Avoid silly mistakes. Okay. Thank you. Thank you, sir. Bye-bye. Have a good day.