Vectors in Component Form - Video Tutorials & Practice Problems

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1

concept

Position Vectors & Component Form

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Welcome back everyone. So up to this point, we spent a lot of time talking about vectors and what they look like. Now we're called that visually vectors are these arrows drawn in space. And we've talked about how these arrows can be stretched or shrink or can be combined with other arrows to create result in vectors. Well, what we're gonna be talking about in this video is position vectors and component form. And this might sound a bit random and confusing but don't sweat it because it turns out all we're really going to be learning about in this video is a simple way to write numbers that represent our vector. And I think you're going to find mathematically it's actually very intuitive. So without further ado let's get right into things because this is an important concept to understand. Now, let's say we have this vector which we'll call vector V as you can see this is an arrow or drawn in space. Now, what we can do is represent this as a position vector and a position vector is simply a vector where the initial point is drawn at the origin. So if we want to draw vector V as a position vector, I just need to relocate this. So the initial point is at the origin and that right there is the position vector and that's all there is to it, it's just moving your vector to the origin of the graph. Now, the question becomes, how can we write this vector with some kind of numbers? Well, we can see that we have some numbers here on the graph and we can use these numbers to figure out what our vector is because what we do is we represent these vectors using component form where we have an X component and a Y component. And all these components do is tell you the length of the vector in the X and Y directions. So we can see in the X direction we need to go three units to the right. So we'd have three and then in the Y direction, we need to go two units up. So we'd have two. So this vector is three comma two. And that's all there is to it. As you can see, it's really straightforward. Now it turns out that there are also ways that you can represent these vectors if you don't have a position vector. So say that you were given some initial point of the vector like a point right there. And then a terminal point over here, you could figure out what the vector is in component form by using this equation down here. And to really put this equation to use and understand it rather than just looking at it. Let's actually try an example where we have to do this. So in this example, we are told if a vector has initial 0.23 and a terminal 0.35 without drawing the vector, write the vector in component form. So we're not allowed to just graph this immediately and figure out what it looks like. What we need to do is use this equation to figure out what our vector is. But this equation is actually pretty simple to use. So all I need to do is recognize that our vector V is going to be the difference in the X components common, the difference in the Y components. So we can see that we have the final X minus the initial X and then we're going to have the final Y minus the initial Y. Now I can see what these values are based on the points above. So if I go ahead and go to this first point and see this first point is 23, I can see that the second point is 35. So what I'm going to have for the X points is the difference between three and two. So we're gonna have three minus two. And the reason that I put three first is because there, it's the final X minus the initial X, it's gonna be 0.2 minus 0.1. So we have three minus two and then we're going to subtract the Y values. So the final Y is five and then the initial Y is three. So this is what our vector is going to be. So we're gonna have three minus two, which is one and we're going to have five minus three, which is two. And that right there is the solution that is vector B. So this is how you can solve problems when you can't initially draw them or don't initially have some kind of graph of the vector. Now, if we want to know what this vector looks like, we actually can use this graph just for reference. Well, I can see that our vector is 12. And if we draw this as a position vector starting at the origin of our graph, we're going to go one to the right and we're going to go two up. And that right, there would be our vector V. So as you can see whenever you're dealing with these types of vectors, you're going to have the X component, which is how far we travel in the X direction and the Y component, which is how far we travel in the Y direction. And that's always going to be the case when using component form. So that is how you can represent vectors using numbers and how you can draw position vectors which are at the origin of your graph. So hope you found this video. Helpful. Thanks for watching.

2

Problem

Problem

True or false: If $a ⃗=⟨3,2⟩$ and $b ⃗$ has initial point $(3,-1)$ & terminal point $(6,1)$, then $a ⃗=b ⃗$.

A

True

B

False

C

Cannot be determined with given information

3

Problem

Problem

True or false: If $a ⃗=⟨3,-2⟩$ and $b ⃗$ has initial point $(-10,5)$ & terminal point $(-7,3)$, then $a ⃗=b ⃗$.

A

True

B

False

C

Cannot be determined with the given information

4

example

Position Vectors & Component Form Example 1

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1m

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Let's see if we can solve this example. So in this example, we're told if vector V has initial 0.43 and terminal 0.12 write vector V in component form and sketch its position vector. Now our first step is going to be to find vector V in component form. And we can do that using this equation. It's going to be X two minus X one comma Y two minus Y one. And this will give us the vector we're looking for. Now the X two is going to be the final X or basically the terminal X and I can see the terminal point here one is going to be that X value. So we're going to have one minus the initial X, which we can see is four. Now next we're going to have Y two minus Y one. Now Y two is going to be the Y value for the terminal point which is two and then we're going to subtract off the Y value from the initial point which is three. So this is the vector we end up getting now to solve this. What we can do is we can do the subtractions here. So we're going to have one minus four, which is negative three and we're going to have two minus three, which is negative one. So the vector is negative three, negative one. And that is the vector that we're looking for in component form. Now we're also asked to sketch the position vector and the position vector is just going to be this vector that starts at the origin of the graph. So if I start here at the origin, I can see we're at negative three so that we can go three units to the left. And then we're going to be at negative one, which is one unit down. And this right here would be our vector V. So notice how it goes backwards, it goes towards the negative side of the X axis instead of the positive side because our vector is negative three negative one. So this is how you can find the vector and sketch its corresponding position vector. I hope you found this video helpful. Thanks for watching.

5

concept

Finding Magnitude of a Vector

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4m

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Welcome back everyone. So in the last video, we talked about how to write vectors using component form, which ends up looking something like this where we have these brackets. Now, what we're gonna be talking about in this video is how you can use this component form. We learned about to find the magnitude of a vector. Now, the magnitude is something that we've talked about in previous videos, recall that it describes how much of something you have or basically when it comes to vectors, it describes the total length of the vector itself. And it turns out finding this length, there's actually a pretty straightforward way that you can calculate a number to represent it. So without further ado, let's get right into an example of this because this is an important concept to understand. Now, let's say that we have this vector down here and we want to find the magnitude. Well, our first step should be to figure out what this vector is in component form. And we can do that by just taking a look at our graph. So I can see that we need to find the X and Y components. Now, the X component is going to be four units to the right. So our X component is four and our Y component is going to be three units up. So our vector is four comma three and that's it, that's the vector in component form. But how could we use these components to figure out the length of this vector? Could you think of some sort of way that we could do that? Well, I want you to recall something we learned about called the Pythagorean theorem. The Pythagorean theorem is an equation that we use on right triangles. And what it does is relates all the sides of a triangle together. So we have a squared plus B squared equals C squared where C is the hypotenuse or alongside of the triangle, then you can rearrange this equation to get that C is equal to the square root of A squared plus B squared. And plugging in the two legs of the triangle will allow you to calculate the hypotenuse. It turns out we can actually use this exact same equation and the same strategy to find the magnitude of any vector because notice something these vectors here, they form a right triangle, these two vectors are perpendicular. And then this would just be the hypotenuse. So if we want to find the magnitude or length of this vector, all we need to do is take the square root of the X component squared and the Y component squared and add them together because that's the same thing as having these two legs of the triangle and treating them like the opposite and adjacent side. So let's go ahead and do that. So we have this vector magnitude that we're trying to calculate. And it's going to be the square root of our X component squared, which is four squared plus our Y component squared, which is three squared. Now four squared is 16 and three squared is 9, 16 plus nine is 25 and the square root of 25 is five. So this right here is the magnitude or length of our vector. It's five units long. And that is how you can calculate this using this pythagorean theorem that we've already learned about. So you can just use this equation whenever you have your vector in component form. Now to make sure we understand this. Well, let's actually try an example where there's not a nice graph given to us. So in this example, we're asked to calculate the magnitude of vector PQ if we're given the initial point or some point P 12 and the final 0.53. Now to figure this out, recall that we learned this equation in the previous video which allows us to calculate our vector in component form if we're given two points. Now this is going to be the final point which we'll call 0.2. And this is going to be the initial point point one So let's go ahead and use this equation. We have that our vector V is going to be equal to the final X minus the initial X comma, the final Y minus the initial Y. Now, what I'm going to do is plug the numbers in. So we're going to have five and one which are the ini initial X components. So we'll have five minus one and that's going to be comma and then we're going to have subtracting the Y component. So we're going to have three minus two. So this is what our vector is going to be. Now, five minus one is four and three minus two is one. So this right here is our vector in component form. Now, since we've calculated the vector in component form, recall that we can now use this equation to figure out what the magnitude of our vector is. So the magnitude of our vector is going to be the X component squared plus the Y component squared. And all of that is going to be underneath a square root. So let's go ahead and plug the values in. So we're going to have the square root of our X component squared, which is four plus our Y component squared, which is one now four squared that comes out to 16 and one squared is just 1, 16 plus one is 17. And we can't simplify the square root down any further. So the magnitude of our vector is 17. And that is the solution to this problem. So this is how you can find the magnitude of any vector. And sometimes you will be given the vector in component form. Other times you need to figure out what the vector is in component form. But you could always use this equation when you're trying to calculate it. So hope you found this video helpful. Thanks for watching.

6

Problem

Problem

If vector $v ⃗=⟨-4,-10⟩$, calculate the magnitude $|v ⃗ |$.

A

$4\surd29$

B

$2\surd116$

C

$2√29$

D

$\surd-116$

7

Problem

Problem

If vector $v ⃗$ has initial point $(-1,2)$ and terminal point $(9,5)$, calculate the magnitude $|v ⃗ |$.

A

$2√29$

B

$\surd109$

C

$\surd73$

D

$\surd113$

8

example

Finding Magnitude of a Vector Example 1

Video duration:

2m

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Let's give this problem a try. So in this problem, we're told if vector V has initial 0.12 and terminal 0.44 sketch V is a position vector and calculate its magnitude. Now to solve this problem, the first thing I'm going to do is figure out what vector V is in component form. This is always a good first step when you're dealing with these types of problems where you're given the initial and terminal point of a vector. Now to solve this, define this component form. What I can do is take the final X and subtract off the initial X and then take the final Y and subtract off the initial Y and that's going to give us our vector. Now, first off, we can see that for X two, this is gonna be the X value for the terminal point which is four. So what I can first do is plug in four to this equation. Now, next, I'm going to subtract off the initial X and the initial X is one. So we're going to have four minus one for the difference in the X's. Now, as for the Ys, I can see that we have Y two and Y two is going to be this Y value for the terminal point which is four. And then this is going to be minus the initial Y and the initial Y is two. So we're going to have four minus one, which is three comma four minus two, which is two. So this right here is the vector. So now that we have our vector in component form, what we can do is we can now sketch the position vector. And the position vector just means that our vector is going to start at the origin. So if I go here at the origin of our graph, I can see that our vector is three. So that means on the X axis, we're going to go over here 123 and I can see that our Y value is two. So that means we're going to go up two. So this is what our vector is going to look like. And that's vector V sketched as a position vector. So we've now found the component form and we've sketched our position vector. And our last step is going to be to calculate the magnitude of this vector and to calculate the magnitude, we can use this equation. The magnitude of V is equal to the square root of VX squared plus VY squared where VX is the X component and VY is the Y component. So we can see that we have these values that we calculated already. So we're going to have a VX which is three. So we have three squared plus two squared now three squared is nine. So we're gonna have the square root of nine plus two squared, which is four and nine plus four is 13. So the magnitude of our vector V is equal to the square root of 13. And this is the magnitude of our vector as well as our vector sketched as a position vector. And that is the solution to this problem. So hope you found this video helpful. Thanks for watching.

9

concept

Algebraic Operations on Vectors

Video duration:

4m

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Welcome back everyone. So in previous videos, we've talked about how you can add or subtract vectors using the tip to tail method. So recall if we add one vector, we could add the tip of that vector to the tail of a second vector and then the resulted vector that connected the two points would give you the sum of those vectors. This should be reviewed from previous videos. Now it turns out there is also a way that you can add or subtract vectors if you are given numbers as opposed to just a graph. And that's what we're gonna be talking about in this video. This is a very important skill to have in this course as well as likely future math or science courses. So let's just go ahead and get right into things. Now, if you have two vectors and you're given them in this component form, the way that you can add or subtract these vectors is by adding or subtracting the individual X and Y components. So to understand this, let's say that I have vector V and I want to add or subtract it to vector U. Well, what I just need to do is add or subtract the individual components. So I can add or subtract the X components. So we'll have VX plus or minus UX and then likewise, I can add or subtract the individual Y components. And that's all there really is too adding or subtracting vectors. So let's go ahead and try it with this example down here. So here we have vector 23 and we wish to add it to vector three negative one. Well, to do this, I'm first going to add the X components which are two and three. And then I'm going to add the Y components which are three and negative 12 plus three is five and three plus negative one is positive two. So this right here is the vector V plus U and that's all there is to it, that's the solution. Now it turns out if you actually use the tip to tail method on these vectors, you're going to find that you actually will get this result. So let's say I have vector 23, which I can draw on this graph right here. And then let's also say I have this vector three negative one. So I can draw this right about there. So we have vector V and here we have vector U. And if I go ahead and use the tip to tail method where I take the initial point of this, of the first vector and connect it to the terminal point of the second vector, notice how this is the vector that we get for V plus U. But notice how this vector ended up being +12345 units to the right and it ended up being +12 units up. So we get the exact same result if we use this tip to tail method. So as you can see the math actually checks out here when we this visually. Now something else I do want to touch on is this idea of multiplying a vector by a scalar, which is something else we've discussed, you can do this with the numbers too. But I think the best way to understand this is going to be to try an example where we have to do this. So let's say we have this situation where we have two vectors V and U and we're asked to find the vector V minus three U. Now to solve this, what I'm first going to do is write out what we have. So we can see that we have vector V and that's given to us it's 85. Now, what I also see is that we have vector U, but what I need to do is figure out what vector three U is to define vector three U. Well, that means I need to take our vector U which is 24 and I need to multiply it by three. But how exactly do I do this? Well, it turns out if you want to multiply a vector by a scalar, all you need to do is distribute the scalar into each of the individual X and Y components. So let's say that we have some scalar constant K. All I need to do is multiply this by the X and the Y component. So we'll have K times the X component and then K times the Y component. So going down here, I can just distribute this three into each component. So we're going to end up with three times two, comma three times four. Well, three times two is six and three times four is 12. So this is the vector three U. So we have vector three U and we have vector V. Now, all I need to do from here is figure out what vector V minus three U is and to find V minus three U. Well, I just figured out what V and three U are. So I just subtract the two vectors. We said that V is 85 and we said that U is 6, 12 or three U I should say. So we end up with these two vector and I just need to subtract them. So what we're going to do is subtract the individual components. So we'll have eight minus six, which comes out to two and then we're going to have five minus 12 which comes out to negative seven. So this here is the solution to the problem. And this is how you can do these basic operations with vectors like adding or subtracting vectors or multiplying vectors by scalers when you're dealing with numbers rather than a graph. So hope you found this video helpful. Thanks for watching.

10

Problem

Problem

If vectors $v ⃗=⟨2,1⟩$, $u ⃗=⟨3,4⟩$, and $w ⃗=⟨-1,1⟩$, calculate $v ⃗+u ⃗-w ⃗$.

A

$\langle4,4\rangle$

B

$\langle5,4\rangle$

C

$\langle6,6\rangle$

D

$\langle6,4\rangle$

11

Problem

Problem

If vectors $v ⃗=⟨4,1⟩$, $u ⃗=⟨-8,3⟩$, and $w ⃗=⟨-2,-1⟩$, calculate $w ⃗-3(v ⃗+u ⃗)$.

A

$\langle10,13\rangle$

B

$\langle-14,13\rangle$

C

$\langle-14,-13\rangle$

D

$\langle10,-13\rangle$

12

example

Algebraic Operations on Vectors Example 1

Video duration:

3m

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Let's see if we can solve this example. So in this example, we're told if vector A equals 31 and vector B equals negative 49 and vector C is equal to five times vector A minus two times vector B, we're asked to calculate the magnitude of vector C. Now, what I can see here is that we're trying to ultimately find the magnitude of this vector. But this vector depends on the other two vectors we have in this problem. So we definitely have our work cut out for us here. But whenever solving these problems where we have multiple vector operations, I personally like to start small with the problem and to work my way to solving the whole thing. So let's actually do that. Now, the first thing I'm gonna do is figure out what five times vector A is because I'm gonna really be focusing on this vector C here and to find five times a well, five times vector A is just going to be five times this vector. So we won't have five times 31. Now five times three, we can multiply this scalar into each of the components. So we have five times three, which is 15 and then we'll have five times five, which is five. So this is the vector five A. Now next, I'm going to figure out what vector two B is. So vector two B is going to be two times vector B. Well, I can see here that vector B is negative 49. So that's gonna be our vector. And then what I need to do is distribute the scalar into this vector as well. So we're gonna have two times negative four, which is negative eight. And then we're going to have two times nine, which is 18. So this is vector five A, five times A and vector two times B. And what I need to do from here, they need to subtract these two vectors. So to find vector C, it's going to be five A minus two B. So we can see here that five A is 15 5. And then we can see that vector two B is negative 818. And then what I can do is I can subtract these two vectors. So if I subtract them, we're going to have 15 minus negative eight and 15 minus negative eight is the same thing as 15 plus eight, which is 23. And then what I'll have is I'll have five minus 18 which comes out to negative 13. So this right here is vector C and now that I've found vector C, our last step is to just find the magnitude of C, we can use the Pythagorean theorem. And the Pythagorean theorem for these vectors is going to be the square root of the X component of C square plus the Y component of C square. Now, what I can see here is that the X component is going to be 23. So we're going to have 23 squared plus the Y component squared, which is negative 13 squared. So this is what we get now 23 squared that turns out to be 529 and negative 13 squared. Where if you, well, if you square a negative number, you're gonna get a positive number and negative 13 squared comes out to positive 100 69. So we're going to have 500 29 plus 169 which comes out to 698. And it turns out this is actually not a radical that you can simplify down any farther. So the magnitude of C is the square root of 698. And that is the solution to this problem. So this is how you can handle multiple vector operations as well as finding the magnitude of a vector once you've already done operations on that vector. So I hope you found this video helpful. Thanks for watching.