In this lecture, we will be looking at components of vectors. So why do we have to look at this particular topic? So the main reason is that although we have discussed about vectors and defined them, and defined various operations like addition on vectors, still we do not know exactly how to perform algebra on them. So this particular topic, that is, to split the vector into components, will actually help us in doing algebra on vectors. Ok, So again let us start with something which we are already family with that is, the three-dimensional right-handed coordinate system. So as I have drawn here, you have the x-axis, you have the y-axis, and if I turn my fingers from X to Y-axis you actually point in the z-axis and hence it is a three-dimensional right-handed coordinate system. And suppose you have a point p with coordinates a, b and c. So, you can clearly see from this diagram that the x-coordinate, which is marked on the x-axis is a, on the y-axis is b, and on the z-axis, it is c. Ok, so this particular point, how can you represent it as a vector? So in order to do this, there is a convention which is followed and the convention is as follows – that the unit vector along the x-axis is generally denoted by ‘i’. ‘i’ with a hat on top of it. So what this means is, this is the unit vector along x-axis. Similarly, we reserve the letter ‘j’ for the unit vector along the y-axis. So ‘j’, it is usually known as a hat, ‘j’ hat is a unit vector along the y-axis. And similarly, ‘k’ is the unit vector along the z-axis. Ok, so now since we have the unit vectors along these three axes, now there is a way in which you can write the point P as a vector, and that is simply as ‘a’ times ‘i’, so P can be written, ‘a’ times ‘i’, the unit vector ‘i’, plus ‘b’ ‘j’, the unit vector ‘j’, plus ‘c’ ‘k’, where i, j, k are unit vectors. So this is a way in which we write a vector into the components in three-dimensional space. Ok, so now what else? So further when one asks you about components of a vector, you have these three scalar components which are a, b, and c, and the vector components are ai, bj and ck. the next thing which we can do with respect to the components is suppose you want to find the length of the vector and that you need to find with respect to the components. So how do we do that? Now again if you look at this figure, it is very clear that when I took the projection of this vector on the XY-plane, what I got is this particular line Ok? And this you can clearly see is, if you look at this particular triangle, it’s a right triangle. and it’s two sides are ‘a’ and and this side is ‘b’. So what do you get is, this particular side will become the root of ‘a’ squared plus ‘b’ squared. The length of this particular side is root of ‘a’ squared plus ‘b’ squared. All right now let us look at this particular triangle this particular triangle. Here you have one side, we already found out that it’s length is equal to root of ‘a’ squared plus ‘b’ squared, and the length of this side is equal to ‘c’ because it’s z-coordinate is equal to ‘c’, and hence again using pythagoras theorem in this particular right triangle, you will get that the length of this particular side, so if you want to name, I will get it as OP is equal to Again using Pythagoras theorem, you’ll get it as square root of the sum of the squares of the two sides. So the square of this side of ‘a’ squared plus ‘b’ squared, and the other side is c, so you will get plus ‘c’ squared. So using Pythagorean theorem we have actually deduced that the length of the particular vector OP, or the position vector of p is equal to the square root of ‘a’ squared plus ‘b’ squared plus ‘c’ squared. So this can be written as, As you can see, I use this particular notation to denote the length of the vector OP, it is equal to square root of ‘a’ squared plus ‘b’ squares plus ‘c’ squared. Yeah, so we know that once you have the components of a particular vector we know how to write down it’s length in terms of its components. Alright, so then coming back to one more thing is that we learnt previously about certain terms which are associated with a vector and these are the direction ratios and the direction cosines. Ok, so suppose I give you a vector p equal to ai + bj + ck then what can you say about its directions ratios and direction cosines? Ok so the first thing is direction ratio. It’s nothing but simply the scalar components of the particular vector. Ok, so p, if i give you this vector, ai + bj + ck and I ask you the direction ratios, they are simply a, b, and c. Ok, so the direction ratios are equal to nothing but the components of the particular vector. Ok, and coming to direction cosines, direction cosines as you remember, the direction ratios and the cosines are proportional, but cosines are actually the direction cosines are actually direction cosines are cosines of the particular angles. Ok, so what you need to do is, direction cosines will be given by the same direction the ratios, but divided by its magnitude because as we have seen in the previous class, that is, this relation between the direction ratios and the direction cosines. Ok so the direction cosines will be given, so there are three direction cosines – the first one will be given by ‘a’ divided by square root of ‘a’ squared plus ‘b’ squared plus ‘c’ squared. The second direction cosine will be given by ‘b’ divided by square root of ‘a’ squared plus ‘b’ squared plus ‘c’ squared, and similarly, you can write down the third direction cosine, and will be equal to ‘c’ divided by square root of ‘a’ squared plus ‘b’ squared plus ‘c’ squared. Ok, so now you have got this. In fact, geometrically, what the direction cosine will give you? It gives you nothing but the unit vector along this particular vector. So as we you see, the unit vector along a particular vector is given by the components divided by the length of the vector. So in this case, as we have seen already, when you took the vector with components a, b, and c, if you look at the magnitude of this vector having components l, m, and n, you will get that ‘l’ squared plus ‘m’ squared plus ‘n’ squared is equal to one. In fact we saw that in the previous lecture. So, what we have found out is that the direction cosines actually give the components of a vector whose direction is same as the given vector but it’s magnitude is equal to 1. So, as I told earlier, splitting the vectors into components actually helps us to do algebra on vectors. So, what are the operations that we generally do on vectors? We do addition, subtraction, and scalar multiplication. So let us see how exactly one can do these operations when they are given in terms of components. Suppose I give you one vector say the first vector is a1 i plus a2 j plus a3 k. So, we are writing it down in the three components, and suppose I give you the second vector b as b1 i plus b2 j plus b3 k And I ask you to add these two vectors ‘a’ and ‘b’. All right, so now once they are given in the components, actually the addition becomes very simple. It’s straightforward, which is the obvious thing to do is to add their components just as you add two real numbers. So you will get a1+b1 as the first component, so (a1+b1) times ‘i’ plus the next one would be a2 + b2 times ‘j’ and the last one would be a3+b3 times ‘k’. Alright, so now you know how to add two vectors when they’re given in terms of components. Similarly you can do subtraction so all that you need to replace is if you want to add ‘a’ minus ‘b’ here, the only changes would be you have to do a1-b1 a2-b2 and a3-b3 The next thing that comes is how do you multiply by scalars? Suppose let us take again the same vector a as a1 i + a2 j + a3 k and suppose I want to multiply by a scalar ‘k’. So, ‘k’ times ‘a’ . Obviously what came to your mind is to it’s nothing but k a1 i + k a2 j + k a3 k I’m sorry for the slightly using the letter k again but I hope it’s clear. Here the k which I meant here is actually a scalar and do not confuse it with the unit vector k with a hat on top of it. So once we have this, so scalar multiplication is also fine. Going next, using components when can we say that two vectors are equal to each other? And that is again straightforward. The two vectors are equal to each other if and only if all their components are equal to each other. For example, here if I ask you when ‘a’ and ‘b’ are equal to each other? The answer is that only if a1=b1 and a2 is equal to b2 and a3 is equal to b3. Even if one component differs, the vectors are not equal. Ok, alright, so with these things done, now we are in a state to solve a few problems to illustrate the concepts which we have discussed so far. And let us start with a simple example. that is, suppose I give you, let us look at the first problem which is to find the direction cosines of a particular vector. So suppose I give you the point whose coordinates are say 1, root 3, and root 5 and I ask you to find the direction cosines of this the position vector of p. As we have learnt already what you need to do is, before going to direction cosines if I ask you direction ratios and it’s pretty straightforward. The answer is 1, root 3, and root 5. Okay but how to find the direction cosines what you have to do is you have to, the first direction cosine will be, ‘l’ will be equal to one divided by the square root of those sum of squares. So square root of 1 squared plus root 3 squares plus root 5 squared. And as we can see the answer comes out to one divided by 1 plus 3 plus 4 plus 5 and squared root of nine is equal to three. So the answer is 1 by 3. Alright, coming to the next thing which is ‘m’. The second direction cosine will be equal to root 3 divided by this remains in the same, denominator remains the same, it is root 9, So it’s 3 and the hence, the answer to this is 1 by root 3. And similarly the last direction cosine will be given by ‘n’=root 5 divided by 3 and this is the last direction cosine. So we have got the direction cosine of the given position vector. It is 1/3, 1/root 3 and root 5/ 3. Okay and you can easily check for yourself that it satisfies the condition for the direction cosines that is this squared plus this squared plus this squared will be equal to 1. Alright! Now let us look into under the problem which makes use of several concepts which we studied in this lecture. Ok, so let me give you to two vectors say a given by say i+j+k and say b, let it be equal to i+2j+3k Yeah so once I have given you these vectors and the question is to find a vector with magnitude equal to say 5 and I need to specify direction also, right? along a+b. Ok, so how to proceed with this problem? So as we have already seen when two vectors are given in component form its straightforward to add them up, you just have to take the components and add them up separately. So in this case it will be a+b will be sum of the components. So it will be 2i + 3j + 4k Ok, now what you need to do is, you need to find a vector whose direction is same as this vector but it’s magnitude is given as 5 Okay, so the way to do it is find a unit vector along this vector and that is pretty simple, you have to just divide this vector by its magnitude So the unit vector along this direction, if i call it as some a+b so let us stick to the notation So, a+ b hat a+b hat is a unit vector along a + b so that it will be equal to the same vector divided by its magnitude so it’s magnitude is squared root of 2i + 3j + 4k divided by its magnitude which is squared root of the sum of squares of the this, so this is 4 + 9 which is 13, 13 + 16 which is 29. So it becomes divided by square root of 29. Ok so what we have got now is, we have got the unit vector along the vector a+b This one. But but this one has magnitude just equal to 1. But the question asks you to find a vector whose magnitude is equal to 5. And that is straightforward you just have to multiply this vector by 5. So the required vector will be this particular vector times 5. Ok so that gives you gives us the required vector.