Introduction to 3D Vectors & Line Equations | Class #09 | A2 Mathematics 9709

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Okay. So, first class for vectors, A2 vectors. Uh initially this was also a topic on in the AS vectors, but they completely removed that topic from the AS and they also So, nerved it a lot in A2 as well. So, now it's much more easier and uh bearable. So, vectors is what? Vectors is uh little bit, you know, connection with physics because vector quantities, you know, vector is something which has magnitude, which has direction. These are the two main symbol. Right now, the vectors that we use, these are very generalized vectors. There is no specific vector quantity that we work with. Then then it

goes into the M1, the physics part. We are working on the mathematical part of the vectors. So, over here, we would only have the magnitude of any vector and the direction. And any vector, we know it can be represented with a arrow with an arrow. So, the body of the arrow is tells you it it denotes the magnitude. So, the longer the vector is, the bigger the magnitude. And the arrow head, that gives you the direction. Otherwise, if if there's a specific vector, for example, force is a vector quantity, displacement is a vector quantity, then you would have the unit as well. Unit of force, unit of uh displacement.

So, but but that would not be you will you won't see any of those over here. These two main things. And you would work with coordinate geometry. You have 2D vectors, you have 3D vectors. 2D vectors are only in the XY plane. Right? The Cartesian coordinate system, 2D. So, you have the horizontal movement, you have the vertical movement, that's it. And when you have 3D vectors, you have the X Y Z axis. XY Z XY axis axis. So, now you would have three coordinates that you would write. X, Y, and Z. How much you have moved in the X, how much you have moved in the Y, and in the Z. So, 2D we know it is a

simple Cartesian coordinate system, vertical and horizontal movement. And these are the simple ones. We did this in the IGCSE as well. You write with how much you have moved in the X, how much you have moved in the Y, and then you write it down. When you have 3D, 3D is when you have a third dimension as well. We live in a 3D world. So, you have the X axis, you have the horizontal, you have the depth perception perception as well. So, now in the world that we live in. And there's a fourth dimension as well. I'm moving on to the physics, but what is the fourth dimension? Does anyone know? Anyone know?

3D is the 3D is the space where we live in, and there's a fourth dimension. Time. Time is the fourth dimension. I don't know if you people know. But, it's okay. That is not relevant over here. So, we will move on forward. So, this is the X, this is the Y, and this this is the Z. And all three of them are perpendicular to each other. Right? And then, if you have a coordinate over here, you would move in the X, you would move in the Y, you would move in the Z. So, all three of them. So, now you will also have equation vector equation of line. 3D line. You would learn how do you write the equation of a 3D line and uh

many more things as well. Now, moving on further, 3D vectors. Now, we are going to work on the 3D vectors because 2D vectors you have seen in the IGCSE as well, all levels as well. So, 3D vectors, the two formats that you write down the 3D vector is and is this column vector form. Uh X Y Z. Column vector form. And you have the other one, which is the IJK form. So, you write X I plus Y J plus uh Z K. This is the IJK format. So, any vector can be written down in these two forms. So, they are This is telling you how much you have moved in the X and the Y and the Z direction. So, and now, furthermore, there are

types of vectors. The first one is position vector. The vector relative to the origin. Connected to the origin. For example, OA or OB or OD. O is the origin. And if you connect your point, your ABCD point, any of the point that you have from the origin O, and say you have to move from the origin to that point, and the distance that you have to move, X Y and Z, that is called the position vector. It is connected to the origin. So, any coordinate, if you have the coordinate as well, for a 3D point, the coordinate can also be said to be the position vector. Because we know coordinate is also what?

Connect It is when you when you plot down that coordinate, it is also relative to the origin. You count the things, how much you move in the X, in the Y, in the Z from the origin. So, both of them, they are interchangeable. You can say that they both are equal to each other. The coordinate of a point and the position vector of that point, it would be the same thing. If you have to move from one point to another. Then you have For example, you have Let me take an example. OA is this, -1, 2, and 0. So, this is telling me that from If I want to move from the origin O to the point A, O to A, I have to move -1 in

the X, 2 in the Y, positive from the Y, and nothing in the Z. This is how it works. Then you have Then you have the the position vector, and then you have the yes, displacement vector. Any other vector is called the displacement For example, AB. If you want to move from the point A to the point B, if you want to go Sorry, if you want to do CD, from the point C to another point D. These are displacement vector. How much you have to move from one point to move in the X, in the Y, in the Z to get to the other point. So, these are called displacement vectors. And how can you open it up in terms of

position vectors like this? So, if you have AB, AB can be opened up like this, OB - OA. So, the position vector of the second letter minus the position vector of the first letter. Similarly, if it is CD I can open it up like OD minus OC. So, position vector of the second letter minus the position vector of the first letter. So, if you have any other displacement vector, if you have LM or XY or anything, you can open it up the same way. So, LM will be opened up like OM minus OL. Okay? That's it. Then you have unit vector. The main purpose of unit vector is purpose is direction. For direction only.

Because magnitude of a unit vector will always be one. So, how do you find unit vector? For example, if you have AB vector and they're asking you to find AB and this when you have this a kind of a hat a vertical arrow you can say this is telling us that they're asking you to find the the symbol vector. So, this is equals to the unit vector of AB will be equal to the vector AB divided by the vector AB magnitude. And one more thing, I forgot. How do you find the magnitude of a vector? The magnitude of a 3D vector will be simply x ^ 2 + y ^ 2 + z ^ 2 and you put square root. This is the magnitude.

Okay, any questions anyone? You can take the screenshot as well. These are the basic parts. I'll share you the notes as well, it's okay. So, the parallel vectors, when two when two when two vectors are parallel they are scalar multiples of each other. For example, if they tell you that AB vector is parallel to CD. So, this would automatically mean that AB is equals to K times a scalar multiple, K times CD. So, the only difference would be the magnitude. One of them would be bigger, the other would be smaller, but the direction would be same. And you can also say that for parallel vectors the unit vector will be same

because unit vector is the direction. So, if the two if the two vectors are parallel then direction is same, so unit vector would be same. And the other thing is that they they would be scalar multiples of each other. Either one would be the double of the other, it would be half or 1/3 or 2/3 or anything. Direction would be same, just the size would be different. These are parallel vectors. >> Calculate the magnitude of a vector, use unit vectors, displacement vectors. Position vectors, you've done this. Understand the significance of all symbols used in the the equation of a line, straight line.

Now, the next thing we would move on to uh doing a simple two, three examples. Adding, subtracting, or anything. And then we'll move on to the equation of a 3D straight line. So, look, they have given you, let's suppose, OA is given 1 3 -5. OB is given -2 7 3 and OC is given 10 -1 4 and OD is given 5 C -2C. Okay, so these are a few vectors given and they're asking you to find, let's suppose, AB or find CB B or find DE such that it is AB plus CB. Okay. Okay, so AB AB is a displacement vector. It is not connected to the origin. And the all others over here that they have given you, these are they are what?

Uh they are position vectors connected to the origin. So if I open up if I want to find AB, it would be equal to I told you the position vector of the second letter minus the position vector of the first letter. So OB minus OA. Now you have OB, you have OA. So OB is minus two seven three minus Uh OA is one three minus five. So you will add them just like the usual way. Minus two minus one minus three seven minus three four three minus minus five eight. This is your AB vector. Similarly, if you have CB, so CB if you open up in terms of position vectors, it would be OB minus OC. So OB you write it down minus two seven

three minus 10 minus 11 sorry, minus one four. So minus two minus 10 minus 12 seven minus minus one eight three minus four minus one. This is your CB. Then they have told us that DE is such that we have to find DE such that it is AB plus CB. So AB is we have found out minus three four eight plus CB is minus 12 eight minus one. And now add it. So minus three minus 12 minus 15 four plus eight is 12 eight minus one is seven. Easy. Just adding subtracting things. They can ask you to find the magnitude of DE. Let's suppose. They can ask you to find find the value of C such that such that OD is parallel to

DE. So, now DE magnitude. Magnitude I told you it is simply it is square root of x squared plus y squared plus z squared. So, this would be square root of -15 squared plus 12 squared plus 7 squared. How much will that be? 20.4 units. Now, this one is a little bit interesting. I told you the last one right now. Find C such that OD is parallel to DE. I told you what when when two lines are when two vectors are parallel, they are scalar multiples of each other. So, OD is K times DE. And now, write down what is OE. O OD OD is 5 C minus 2 C. And it is K times -15, 12, and 7. Now, what? You can equate the X

with the X and the Y with the Y and the Z with the Z. So, you will get 5 K is equals to, sorry, 5 is equals to -15 K. And now, from this thing, your K answer would come out as -1 upon 3. Similarly you would do the same thing. C is equals to 12 K. And if you know the value of K, your C will come out as 12 upon Sorry, 12 multiplied by -1 upon 3, which will be equal to -4. And you can find the -2C as well. I may have not formed this equation right. Uh let me Let me change it. Find C and D. This is D. So now how will you find D? So D is equals to K * 7. Sorry, this is red. So D will be -1 upon 3 *

7. Right? K was your -1 upon 3 * 7. This would be your -7 upon 3. So this means that your OD vector, complete vector OD, it is equals to 5 -4 and -7 upon 3. Okay? Get it? Any questions, anyone? This is the very basic part. This is something that you people have done in your IGCSE. I'm just right now introducing the 3D vectors, the X, Y, and Z, all three of them. So now we can see that we have discussed the first three What do you say this? Points in the syllabus outline. And now we will do the equation of a 3D line and then intersection, scalar product of parallelogram cuboid and Okay. Okay. Okay. Okay.

Angle between two points. Oh, sorry, two vectors. The scalar product. Only maybe we'll do the equation of a 3D line today, not the complete thing. Equation of a 3D line. The general form is R is equals to position vector on the line plus the parameter times the direction vector on the line. R is any arbitrary point 3D point on that 3D line. Position vector is one point that you know. One point that we know is on that line. This is the parameter. The value of this parameter will change and you will get different values for the points on that line. So, parameter lambda you can be it can be lambda or mu

or any other as well as or T. It's just a variable. It's just a parameter. You will change its value 1 2 3 4 it will change and you will get new points on that line. And this is called the direction vector. Direction vector works like the direction which is the works like the gradient you had in your 2D line. This is called the direction vector. And I will teach you how do you find this? So, you would need to find this thing and this thing. You would find this thing and this thing. And you will replace on this form and that would give you the 3D line. So, just like you had in your IGCSE O level, the equation of a 2D

line, they are if you had two any two points, you can find the equation of the 2D line over here as well. It is written, find the equation of line passing through um A which is 0 5 1 and B -2 1 3. So, we will find the equation of a line which is passing through this these two points. Now, I told you the coordinates can also be written down as the position vectors. So, you already have two points which through which the line would pass. So, this thing we already know. We know the point through which the line is passing through. We can use any one of them. The problem is we don't know the direction vector. So, the direction

vector is would be the AB. Right? The direction of the line because you know that if the line is like this and it is passing through A and B, so the vector AB will be the same as the direction of the line. So, direction of the line would be the same as the vector AB or you can use BA as well. Okay, there was one more thing I forgot to tell you. I'll I'll tell you after this thing. So, direction vector over here in this case you will do OB minus OA. So, OB you know, minus two minus sorry one three minus OA is 051. So, direction vector of the line is minus two minus four two. Now, you can replace it. It will be R is

equals to any one of the points you can use 051 plus lambda and the direction vector minus two minus four and two. This is the equation of the line. Equation of the 3D line. Easy. That's it. Simple. Okay, there was one more thing. One more type of not type of but another one of these types of not specifically not the type but an important part. So, if you have a missed point negative vectors. So, if you have a vector AB it will be equal to negative BA. So, if AB is like this A to B it will be the same as negative BA. So, the only difference will be the direction. So, if you multiply it with the minus

sign, the direction would be exactly opposite, exactly the other side. But, you can see the They both are parallel to each other. They are anti-parallel, you can say. The direction is opposite, but they are parallel. So, over here in this case for the direction vector, if you find BA or AB, both of them would work. In this case right now, we had AB, we found AB, right? It was equals to -2 -4 2. So, if I want to find BA, so BA would be just if we multiply it with the minus sign. So, this would become 2 4 and -2. Do you get it? Just the direction is opposite in the other side. Otherwise, everything is

same. You have a lot of possible questions on this basic part, finding the equation of the line. And the next part uh they would do find find the point of intersection between two lines and something other than it. But, right now, this is the basic one, finding the equation of a 3D line. If they've given you two points, so you This is how you find it. Let's find Let's do some Open up some past papers. You would have it over here. Decide whether each of the points are intersecting, parallel, or skew. I'll tell you what this is. Finding the equation of the line. Where is the equation of the line?

Okay, look at this one. Part B, we won't to the part B right now. The first part is right now relevant to us because that's what we have done right now. They have given you with the with respect to the origin O, the points A and B have position vectors. A and B points are given to you. And OA You can write it down. I always like it better in the column vector form. 1 2 2. It's easy to understand. And OB is 3 4 and there is no K, so 0. So, the point P lies Okay, okay, later on. Find the equation of the line uh passing through Sorry, this AB. So, you would need the direction vector, which I told you it can be either you

can use AB or BA, it would be the same thing. The direction would be the same. The purpose is the direction for the direction vector, not the magnitude or the if it is antiparallel or negative or positive. Direction should be same. Which I know that it would be OB minus OA. So, OB 3 4 0 minus 1 2 2 which will be equal to 2 2 minus 2. 3 minus 1, 4 minus 2, and 0 minus 2. Yes. So, now the R will be you can choose any one of the points through which the line is passing through. We know you can choose OA or OB. So, 1 2 2. Plus lambda and 2 2 minus 2. This is your answer for the first part. They can ask you further find the point

on the line when your lambda is 1, when your lambda is two. So, you will just substitute the value of the lambda over here and you will get new points. You know, those will be the points on those on this line. Okay, let's see something else. Whenever you see a question from using the planes, look at the question number five, planes. The plane M is parallel to we the plane was in our syllabus before, but now it's no more. So, you will ignore. These are the more related to the other parts as well, so In the next class, hopefully we will finish everything. Right now, I would don't want to because we don't have much

time. You can try these questions finding the equation of the line passing through this question number three. And you can try this one as well, question number four. And question number four, what do they have given you? Determine whether the point P lies on the line or not. So, this P they have given you a point on the line, which is 514. So, the R R is a point on the line. So, you can replace the R with 514. And then you can form three equations, one for the X, one for the Y, one for the Z and check if you get the same answer for lambda for all three of them. So, if you get the same answer for alpha

for all of them, this means that this satisfies this point satisfies the equation of the line. And that this line lies on the this point lies on the line. Okay? You can try I'm bringing you over here. All right, I'll stop the recording as well. >> There you go. And you people can try this and this. And both of these this is a simple one. Okay, I'm stopping the class or recording.