In this video, we're going to talk about vectors. But what is a vector, really? And let's distinguish it from something called a scalar quantity.
So what's the difference between a vector quantity and a scalar quantity? Well, a scalar quantity only has a magnitude. A vector quantity has both a magnitude, but it also has... direction. So for instance, speed is considered to be a scalar quantity.
However, velocity is a vector quantity. So let's say if I'm driving a car going 40 meters per second. What I told you is my speed.
I only gave you a magnitude, just a number. But if I say I'm going 40 meters per second north, now I've given you my velocity. I gave you both speed and direction. So the 40 meters per second represents the magnitude.
It tells me, I mean tells you how fast I'm going and north tells you the direction of where I'm going. And so that's basically a vector. It gives you, it combines magnitude with direction.
Another example, would be force. Let's say if I have a box. I can apply a force of 300 N directed east, or I can apply a force of 200 N directed north.
So let's say the first example, 300 N, that's the magnitude of the force, and the direction east, that's the second component. which makes it a vector. So make sure you understand that a vector has both magnitude and direction. Another scalar quantity would be temperature. You can't apply direction to temperature.
If I ask someone, hey what is the temperature today? Well let's see, it is 80 degrees east. What? That would make no sense.
You can't say, you can't apply a temperature with a direction. It doesn't make sense. So temperature would be considered a scalar quantity.
You can't say 80 degrees Fahrenheit north or something like that. It just doesn't, it doesn't apply. And so that's a scalar quantity.
Another one would be mass. So let's say you want to find out how much matter is in the object. So this box has a mass of, let's say, 50 kilograms.
You can't apply a direction to that. You can't say it's 50 kilograms north or 80 kilograms south. It just doesn't work. And so mass is a scalar quantity. So if you want to determine if something is scalar or a vector, ask yourself, can I apply direction to this, whatever it is that I'm considering?
So like volume, is volume a scalar quantity or is it a vector quantity? What would you say? So let's say this container can hold 50 milliliters of fluid.
Can I apply direction to that? Can I say it has 50 milliliters of fluid north? That wouldn't make sense.
So therefore, volume cannot be a vector. It's a scalar quantity. And so hopefully that highlights the difference between what is scalar and what can be a vector.
Another way to think of a vector is to think of a directed line segment. So let's say if we have two points, point A and point B. A is the initial point, B is the terminal point. So we can draw a vector from A to B that looks like this.
And so you can write it this way, with the arrow pointed at B. So that's vector AB. The left of the arrow indicates the magnitude.
Now, where the arrow is pointed tells you the direction. So let's try another one. So let's say this is point C, and this is point D.
So based on this arrow, which one is the initial point, and which one is the terminal point? So the initial point is where the arrow starts and where it ends, so to speak, that is the terminal point. So we can call this vector, vector CD, with the arrow pointed at the letter D, or towards D.
Now there's two common ways in which we could describe a vector. And that is, we can give it a magnitude and its angle. So let's say this is the x-axis, this is the y-axis.
And let's say this vector has a length of 5 and it's directed at a 40 degree angle. Let's call it vector v. So the magnitude of vector v is 5. The magnitude represents the length of the vector, how long it is. So this vector would have a magnitude of 10 because it's longer.
And this vector would probably have a magnitude of 2 or 3 because it's shorter. And so the length of the directed line segment tells you the magnitude. So that's one way in which you can describe a vector, using its magnitude and an angle. Another way in which you could describe a vector is using components. Let's say we have the vector a, and we have 2, 3. So this tells us that the x component of the vector, you can say a sub x, That's 2, and the y component is 3. So let's represent it graphically.
So here's the y-axis, here's the x-axis. So starting from the origin, we're going to travel 2 units to the right, and 3 units up. So here is the x component.
It's 2 units along the x-axis. And the y component... It's three units parallel to the y-axis, and that will give us this vector, vector a. So you can describe a vector using the x and y components of that vector. Now there's something that you need to be able to do, and that is you need to be able to distinguish a point from a vector.
A point... has parentheses. For example, this is the point 3 comma 4. A vector, let's say vector b, has these inequality symbols like this one. Let's say this is, it has the components 4 and 5. So this would be a vector.
You wouldn't use parentheses to represent a vector. So if you want to represent a vector in component form, you need to use those arrows instead of a set of parentheses. Keep that in mind. So if we were to plot it, a point would be just a point.
So point P is right here. It's at 3, 4, and that's it. It's not a directed line segment.
It's simply a dot in space. Now, a vector, on the other hand, is different. Let's graph vector B on this graph. So we're going to start from the origin, and then we're going to travel four units to the right, because that's the x component, and then we're going to go up five units.
So let me use a different color. So this is four, this is five, and vector B starts from the initial point, which we just based it from the origin. and it ends at the terminal point, which in this case is the point, but the initial point is. And so that's a vector B.
It has an x component of 4 and a y component of 5. Now let's work on a practice problem. Vector V has the initial point and terminal point. Part A, find the component form of vector v. So what we want to be able to do is we want to find the component form of vector v. We need to determine its x component and its y component. How can we do so given the initial point and the terminal point? What we need to do in order to find it is take the difference between the x values of the terminal point and the initial point.
So vx is going to be 5 minus 1. To find vy, It's simply the difference between the y-coordinates of the terminal point and the initial point. So think of final minus initial. We need to subtract the terminal point minus the initial point. So 1 minus negative 2. 5 minus 1 is 4. 1 minus negative 2, that's basically 1 plus 2, which is 3. So the component form of vector V is...
4 negative 3 let's illustrate this with a graph so the initial point is at 1, negative 2. So here it is. And the terminal point is 5, 1. I was going to plot 6, 1. Here's 5, 1. So this is the initial point and this is the terminal point. Now to go from the initial point to the terminal point, how many units do we need to travel along? the x-direction or parallel to the x-axis.
The initial point has an x value of 1 and the terminal point has an x value of 5. So we need to travel from 1 to 5 parallel to the x-axis. So we have to travel 4 units to the right and that gives us the x component. As you can see, it's 4. So that's vx. Now the initial point has a y value. of, let's see if I can fit it here, negative 2, and the terminal point has a y value of 1. So to go from negative 2 to 1, we need to travel up 3 units.
And so this is the y component, this is the x component. Now how can we determine the magnitude of vector v? Remember, the vector is right here. I'm going to highlight it in red.
How can we determine the length of that vector? Since we have a right triangle, we could use the Pythagorean theorem to determine the length of the hypotenuse of the right triangle. Now, if you recall, the hypotenuse is c and the legs are a and b.
And so we have the formula c squared is equal to a squared plus b squared. So c is the square root of a squared plus b squared. Therefore, if we wish to find the magnitude v, which is the length of the vector, which is the same as c, we could use this formula. It's the square root of vx squared plus vy squared.
In this example, vx corresponds to a, vy corresponds to b. So to finish part B, the magnitude of vector V is going to be the square root of 4 squared plus 3 squared. And 4 squared is 16, 3 squared is 9, 16 plus 9 is 25, and the square root of 25 is 5. So basically we have the 3, 4, 5 right triangle. And so that's how you can find the magnitude.
and the component form of the vector v, if you're given the initial point and the terminal point. Let's try this problem. Vector v has initial point a, negative 4, 1, and terminal point b, 8, 6. And vector u has initial point c, negative 7, negative 15, and terminal point d, 3, 9. Determine if the two vectors are equivalent.
So how can we determine if two vectors are equivalent? Well, they need to have the same magnitude and the same direction. So let's begin by finding the component forms of vector v and u.
So the component form for vector v, let's start with vx, is going to be the difference between the x values of the terminal point and the initial point. So it's going to be 8 minus negative 4. And Vy, the y component of V, is going to be the y coordinate of the terminal point minus the y coordinate of the initial point. And so that's going to be 6 minus 1. And so we have 8 plus 4, which is 12, and 6 minus 1, which is 5. So this is the component form of vector V. Now let's do the same for vector U.
The x component of the terminal point is 3 and the x component of the initial point is negative 7. So we're going to subtract 3 and a negative 7 together. Now the y component of the terminal point is 9. And the y component of the initial point is negative 15. So it's 9 minus negative 15. So we're going to have 3 plus 7 and 9 plus 15. 3 plus 7 is 10. 9 plus 15 is 24. So this is the component form of vector u. Here is a question for you. What is another way in which we can describe vectors v and u?
based on the information that we were given. Another way in which we could describe it is using the initial and terminal points. Vector U has the initial point C and the terminal point D. Therefore, vector U can be called vector CD and the vector V has the initial point A and the terminal point B.
So vector V is also vector AB, which is equal to 12,5. So now that we have this information, let me get rid of a few things. Now that we have the component form of vector V and vector U, can we say that these two vectors are equivalent? So looking at them, the numbers are different, which means that the vectors will be different. But if we want to determine if two vectors are equivalent, we need to determine the magnitude and the direction.
Let's start with the magnitude first. Let's find the magnitudes of vector v and vector u. The magnitude of vector v is going to be the square root of 12 squared plus 5 squared.
12 squared is 144, 5 squared is 29, and 144 plus, I think I said 29, 5 squared is 25. 144 plus 25 is 169, and the square root of 169 is 13. So that's the magnitude of vector v. Let's do the same for vector u. It's going to be the square root of 10 squared. plus 24 squared.
Now 10 squared is a hundred and 24 squared that's 576 I believe so 100 plus 576 is 676 which I'm going to use a calculator at this point. The square root of 676 is 26. So the magnitudes of these two vectors are different because the components are different. But what about the direction?
Do they have the same direction? A quick way to determine if the two vectors have the same direction is to determine the slope of each vector. So if you recall, the slope is the change in y values divided by the change in the x values.
Now for vector v, the change in the y values, looking at the initial point and the terminal point, it's 6 minus 1, and that basically gave us vy, which was 5. Now the change in the x values for vector v, it's 8 minus negative 4. So you could say x2 is 8, x1 is negative 4. And so that gave us 12, which is the x component of v. So basically the slope of vector v, it's simply vy over vx. So I'm going to write m sub v, the slope of vector v. So it's vy over vx, which is 5 over 12. The slope for vector u. It's going to be uy over ux. So the y component of vector u is 24, and the x component of vector u is 10. Now, we can reduce 24 and 10. They're both even numbers, so if we divide 24 by 2, that's 12. If we divide 10 by 2, it's 5. And so these two vectors, they don't have the same slope.
This is 5 over 12, this is 12 over 5. So they're completely different. In order for two vectors to be equivalent, they need to have, number one, the same magnitude, which these two vectors do not have, and at the same time, they need to have the same slope, which these two vectors also do not have. So right from the beginning, you can tell if they're equivalent just by looking at the component forms of the two vectors. But if you don't have that, if you have the magnitude and the direction, then you can look at that.
So if two vectors have the same magnitude and the same slope, they will be equivalent. Or if they have the same components, they will also be equivalent to each other. Let's talk about adding vectors.
So let's say this is vector A and this is vector B. Let's say C is the sum of vectors A and B. How can we add the two vectors together? So what you need to do to graphically add the vectors is connect them from head to tail. So if we draw vector A and then we draw vector B right after A, so basically we want to take the initial point of B and connect it to the terminal point of A.
And to draw vector C, start from the initial point of A and Let me use a different color. Draw the vector towards the terminal point of B. And this is vector C.
That is the sum of vectors A and B. Now, you could start with A or you could start with B. So let's say if we started with B. If we drew B first, and then afterward, we drew A. We would still get the same vector C, which is going in the same direction and has the same length.
So the order in which you do it doesn't matter. Now let's say if we want to draw a vector d, and we're going to define it as being b minus a. How can we draw this vector?
Well let's start with b. We have positive b, which is the original vector, and When you see b minus a, it's best to view it as b plus negative a. So we're going to add negative a to vector b. So if this is a, what's negative a? When you multiply a vector by negative 1, the direction of the vector will change.
It's going to change by 180 degrees. So all you need to do is reverse the vector. So vector A is going to be going this way.
So now to draw vector D, we're going to start from the initial point of vector B to the terminal point of vector A. And so that's going to be vector D. So that's how you can subtract vectors B and A from each other.
Now let's say that vector E is 2a plus b. Go ahead and draw this vector. What's going to happen if we multiply vector a by a scalar quantity, which is just a number, in this case 2. So what is 2a?
All it is, it's basically doubling the length of vector a. So if this is a, 2a it's going to be twice the length. And then we need to add vector b to it.
Now to draw vector E, we are going to draw it from the initial point of the first vector to the terminal point of the second vector. And so this is vector E. It's the sum of 2a and b. So now you know how to graphically add and subtract two vectors together.
And you could do it with three vectors if you want to. So let's say that vector D is the sum of vectors A, B, and C. And we're going to say this is A, this is B, and let's say this is C. So let's start with A, and then connect it to B, and then connect B to C. So starting with the initial point of the first vector, we're going to...
draw a vector to the terminal point of the last one. And so this is vector D. That's how you can add three vectors together.
Now let's work on some vector operations. So let's say if we're given vector A, and it's going to be 4, negative 2. And let's say that vector B in its component form is negative 3 comma 5. So perform the following operations. Let's calculate the value of 2a plus 3b and also let's say 5a minus 4b. Go ahead and try those two problems. Feel free to pause the video.
So let's start with the first one. 2a plus 3b. So the first thing I'm going to do is replace a with what it's equal to.
So a is equal to 4 comma negative 2. And then I'm going to replace a b with the stuff that it's equal to, which is negative 3 comma 5. And then I'm going to multiply a by 2. So I'm going to multiply the x component by 2. 2 times 4 is 8. And I'm going to multiply the y component by 2. 2 times negative 2 is negative 4. Now I'm going to do the same thing with b. So 3 times negative 3, that's negative 9. And 3 times 5 is 15. Now when adding vectors together, you need to add the corresponding x components together. and the corresponding y components together. So 8 plus negative 9 is negative 1, and negative 4 plus 15 is 11. So the answer for the first part, that is part a, is this.
It's negative 1 comma 11. Now let's do the same for part b. So we have 5a minus 4b. So let's begin by replacing A with 4, negative 2, and let's replace B with negative 3, 5. So let's begin by distributing the 5. 5 times 4 is 20, and 5 times negative 2 is negative 10. Now I'm going to distribute the negative sign, so I'm going to put a plus here.
Negative 4 times negative 3 is 12. And negative 4 times 5 is negative 20. If you don't use the negative sign, it will look like this. 4 times negative 3, which is negative 12. And 4 times 5 is 20. So this is what you should have if you don't change the negative sign to a positive sign. But I like to put a positive sign here, so I'm going to distribute the negative to... the stuff that's on the inside. That's just the way I like to do it to avoid making mistakes.
So now let's add the corresponding x components together. 20 plus 12 is 32 and negative 10 plus negative 20 is negative 30. And so this is the answer. This is equal to 5a minus 4b.
That's it. Now let's talk about position vectors. What is a position vector?
How would you describe it? A position vector is any vector that has its initial point placed at the origin. So the initial point has the coordinates 0, 0. Let's call this vector v. Now let's say the terminal point of vector v is 3, 4. So vector v is a position vector.
But notice that vector v, its components, is the same as the coordinates of the terminal point. Because if you subtract the x coordinates, 3 minus 0 will still give you 3. And if you subtract the y coordinates, 4 minus 0 will still give you 4. So when dealing with position vectors, the coordinates of the terminal point will be the same as the components of the position vector, due to the fact that the initial point is placed at the origin. And that's basically what you need to know about position vectors. So to review, remember, a position vector is any vector that has its initial point placed at the origin.
Now what about unit vectors? What is a unit vector? Think of the keyword units. What is the basic unit of any number?
Most numbers are based on a unit of 1. And think of the unit circle. When you hear the word unit circle, what do you think of? The unit circle is basically a circle.
with a radius of 1. Likewise, a unit vector is a vector with a magnitude of 1. So some textbooks will describe a unit vector as just being U. Sometimes you'll see this symbol associated with a unit vector, but what you need to know is that the magnitude of any unit vector is always 1. That's what you want to take from this. Now how do we go about finding unit vectors from other vectors?
So for instance let's say if we have some generic vector, vector v, and let's say it has a magnitude of 4. What do we need to do to find the unit vector of vector v? Now recall that every unit vector has a magnitude of 1. So to find the unit vector of vector v, we just need to divide vector v by 4. Because 4 divided by 4 is 1. And so that will give us a shorter vector that's in the same direction as V, but with a magnitude of 1. And this is going to be called the unit vector of V. So therefore, to find any unit vector, you need to take the vector. Let's say if we want to find the unit vector of V. Take that vector and divide it by its magnitude. And that's how you can find the unit vector of any generic vector.
Now it's important to be familiar with this equation because if we multiply both sides by the magnitude of vector v, we are going to get this. Vector v can be described as the product of the magnitude of vector v times its unit vector. So what does this mean?
Recall that in the beginning of the video we said that vectors can be described using two things. A vector has both magnitude and direction. So this represents the magnitude of the vector.
Therefore, the unit vector describes the direction of vector v. It tells us where the vector is going and we could extend the length of that vector to any other number. We can increase the length, decrease the length. We just need to multiply by some magnitude. So all vectors can be described as the product of the magnitude of the vector times its unit vector.
So it's a combination of magnitude and direction. Now let's work on this problem. Find a unit vector in the same direction of v. So how can we do that?
What's the first thing that we should do? What I recommend doing... is write in the formula. So the unit vector u is going to be the vector v divided by the magnitude of v. And v, in its component form, is 4 comma negative 3. And now the magnitude of v is going to be the square root of 4 squared plus negative 3 squared.
And we know 4 squared is 16, negative 3 squared is positive 9, 16 plus 9 is 25. and the square root of 25 is 5. So this becomes 4, negative 3, all divided by 5. And now what you could do is divide each number by 5. So 4 divided by 5 is just 4 fifths, and then we have negative 3 divided by 5. And so this is the unit vector of vector v. It's 4 over 5 comma negative 3 over 5. And if you want to check it, You need to show that the magnitude of this vector is 1. And let's do that. Let me make some space first. So if I take the square root of 4 over 5 squared plus negative 3 over 5 squared, I need to get 1 in order to prove that this is indeed a unit vector.
4 squared is 16, 5 squared is 25. Negative 3 squared is 9. and 16 plus 9 is 25. So we have 25 over 25 which is 1 and the square root of 1 is 1. So we can confirm that this is indeed is a unit vector. So now that you know how to find a unit vector, I think it's a good time to talk about standard unit vectors. So what exactly is a standard unit vector? There's three of them that you need to be familiar with. Perhaps you've seen them before.
I, J, and K. Each of these are unit vectors with a magnitude of 1. So I has an x value of 1, but a y value and a z value of 0 when dealing with three-dimensional systems. J is a unit vector with a y value of 1, and k is a unit vector with a length of 1 along the z-axis.
So make sure you associate i with the x-coordinate system, j with y, and k with z. And remember that each of these vectors have a length of 1. So for a two-dimensional system, where this is the x-axis and this is the y-axis, I is going to have a length of 1 along the x-axis, so that's I. And j is a unit vector with a length of 1 along the y-axis.
And z is the same thing, but along the z-axis. So how do we apply this? Well, let's say that we have vector v, and in component form, it's 4 comma negative 7. How can we represent this vector using the standard unit vectors? So we can also say that vector v, it's 4 times the unit vector i, and then plus negative 7 times the unit vector j.
Or we could simply say it's 4i minus 7j. So this is the x component of vector v, and this is the y component. So the x component vx is basically 4i, and the y component of vector v is negative 7j. And so you could represent any vector in its component form or using standard unit vectors.
So let's try another one. Let's say we have vector w, and it's 3, negative 5, 4. Go ahead and represent this vector. using the standard unit vectors.
So vector W can be written as 3i minus 5j plus 4k. So wx is 3i, wy is negative 5j, and wz, the z component of vector w, is 4k. But this is the answer though. Now let's say that vector A is 3i minus 5j and vector B is 2i plus 6j.
And let's define vector W as being 4a minus 3b. Go ahead and find vector W using the standard unit vectors. So go ahead and perform the vector operations. So what we're going to do is we're going to replace a with what it's equal to, 3i minus 5j.
And then we're going to replace b with 2i plus 6j. And so we have 4 times 3i, which is 12i. and 4 times negative 5j, which is negative 20j, and negative 3 times 2i, that's negative 6i, negative 3 times 6j, that's negative 18j.
And now all we need to do is add like terms. 12 minus 6 is 6, and then negative 20 minus 18 is negative 38. And as you can see, when using the standard unit vectors, it's very... easy to perform vector operations.
It reduces to simple algebra. Now you do have to be careful not to make mistakes, so I recommend doing everything one step at a time because it's easy to make a mistake or miss a negative sign. But that's how you could perform vector operations with the standard unit vectors. Now let's talk about the unit circle and how it relates to unit vectors. But let's focus on the first quadrant of the circle.
Now, the unit circle is a circle with a radius of 1. So, the distance between the center and any point on a circle will always be 1, which is the same as the length of a unit vector. So basically, the radius of the unit circle, we could say, is the unit vector. Now, what is the x component of the unit vector, and what is the y component of it? Now this unit vector will have an angle with the x-axis and if you recall from trigonometry the x component of the unit circle is cosine and the y component is sine.
So we can say that ux is equal to cosine theta and uy is equal to sine theta. So as we represent the unit vector u, as being in its component form. We can also say that it's basically cosine theta and sine theta.
Now earlier we said that we can represent any vector as being the product of its magnitude times the unit vector. And so the unit vector is basically just cosine and sine. So the vector V is going to be the magnitude of V times cosine and since that's the x component we're going to attach one of the standard unit vectors associated with the x component that's i and then plus we multiply V by sine and that's the y component so we're going to attach a j to it.
So we could find any vector if we know the magnitude and the angle. So we can express it as the sum of the standard unit vectors i, j, and if we want to, k as well. Now let's work on number five. Write vector v as a linear combination of the unit vectors i and j given that vector v has a magnitude of 16. and an angle of 30 degrees with the positive x-axis.
So let's say this is the y-axis, here is the x-axis, and here's vector v with a magnitude of 16, and it makes a 30-degree angle with the positive x-axis. Let me make this longer. So what we're looking for are the x and y components. We need to determine vx and vy.
But let's use the formula that we wrote earlier. So vector v is going to be the magnitude times cosine. That's going to give us the x component. And to get the y component, it's going to be the magnitude times sine.
So make sure you understand this. Vx, it's going to be V times cosine theta. And Vy is V times sine theta.
So we have V, that's 16. We're going to multiply it by cosine 30. And to get the y component, we're going to multiply sine 30 by 16. And this should be a j. So what is cosine 30? Cosine 30 is the square root of 3 over 2. Let's not forget to put the unit vector i.
And sine 30 is 1 half. So now 16 divided by 2 is 8. So I'm going to have 8 square root 3 times i. And half of 16 is 8 as well. So that's going to be plus 8j. This is the answer.
So that's how you can express vector v in its component form using the unit vectors i and j, if you're given the magnitude and the angle. Now let's work on another problem. So let's say that vector v is 3i minus 8j. Find the magnitude of the vector shown below and determine the angle that it makes with the positive x-axis. So this problem is the reverse of number 5. We're given the vector and we need to determine the magnitude and the angle.
Well we know how to find the magnitude. It's simply going to be the square root of 3 squared plus negative 8 squared. 3 squared is 9, negative 8 squared is 64. 9 plus 64, that's going to be 73. So the magnitude is 73. Now how do we find the angle?
Well, if we were to plot this vector, As a position vector, starting from the origin, we would have to travel 3 units to the right, and then we would travel down 8 units. Now, this graph is not drawn to scale, but vector v would be going in quadrant 4. So how do we find the angle that... the vector makes with the positive x-axis.
And also let's find this angle going counterclockwise from the x-axis. But before we find that angle, let's find what is known as the reference angle, that is the angle inside the triangle. The reference angle will always be a positive angle between 0 and 90. And what we can use is the tangent function.
If you recall from SOHCAHTOA and Trig, tangent theta is equal to the opposite side divided by the hypotenuse. In this case, opposite to theta is vy, and adjacent to it is vx. So tan theta is vy over vx. Now to find a reference angle, it's going to be the arc tangent of the y component. divided by the x component.
Now when looking for the reference angle, even though vy is negative, I recommend using the positive value or the absolute value of vy. It's going to make your life a lot easier. So we're going to take the arc tangent of 8 and divide it by 3. And make sure your calculator is in degree mode Mine was in rating mode just now.
So you should get this angle. Now this is the reference angle. It's 69.4 degrees. Now sometimes your teacher simply wants this angle between the vector and the x-axis.
And that would be the answer. However, if you want the angle measured from the x-axis but in a counterclockwise direction as opposed to a clockwise direction, If you want to go counterclockwise, we need to do some extra work here. But all we need to do, if the angle is in quadrant 4, is subtract it by 360. Or more specifically, subtract the angle from 360. So it's going to be 360 minus 69.4. And so the angle measured from the positive x-axis going in a clockwise, I mean a counterclockwise direction, it's 290.6.
This is the angle that we want. That's the standard angle. And so we have a magnitude of 73 and an angle of 290.6.
So if your vector is in quadrant 1, the angle that you want is the same as the reference angle. In quadrant 2, the angle that you're looking for is going to be 180 minus the reference angle. In quadrant 3, it's going to be the reference angle minus 180. And in quadrant 4... The angle measured from the positive x-axis going in a counterclockwise direction will be 360 minus the reference angle. Now this is going to be the last problem of the video.
Go ahead and work on it. Find the resultant force of two vectors. Take a minute, pause the video, and see if you can get the answer.
Now for those of you who wish to subscribe to this channel, that is of course if you like this video, don't forget to click the notification bell if you wish to receive any updates on new videos that I'm going to be posting in the future. Now let's begin. Let's graph the vectors, or at least a rough sketch of it.
So the first force vector, F1, It's going to be going in this general direction. It has an x component of 5 and a y component of 2. Now F2 will be going towards quadrant 3. We need to travel two units to the left and down eight units. So we could say that For the most part it's going in this general direction. Now the resultant force is the sum of these two force vectors. In which quadrant do you think the resultant force, which we'll call FR, is located in?
So if we take F1 and graphically add it to F2, Where is FR located? So FR is going to be going in this direction. So if we move that vector to the origin of this graph, we can say that approximately it's going in quadrant 4. That means that the x component should be positive and the y component should be negative.
So what we're going to do is we're going to find... the magnitude of the resultant force vector, and also its angle. So let's go ahead and begin.
The resultant force vector is simply the sum of F1 and F2. And F1 is 5i plus 2j. F2 is negative 2i minus 8j.
So what we need to do is add the x components together. So 5i plus negative 2i, that's 3i. And then add the y components together. 2j plus negative 8j is negative 6j.
So that's the resultant force vector in terms of the standard unit vectors i and j. Now once we have that, our next step... is to find the magnitude of the resulting force vector.
And so it's going to be the square root of 3 squared plus negative 6 squared. 3 squared is 9, 6 squared is 36, 9 plus 36 is 45. Now if you want to, you can simplify the square root of 45. 45 is 5 times 9, and the square root of 9 is 3. So you get 3 square root 5. which I'm going to write here. So that's the magnitude of the resultant force vector in this problem.
Now let's go ahead and find the angle that it makes with the positive x-axis going in the counterclockwise direction. But let's graph it first. So to graph this vector, we need to travel three units to the right and then down six units.
So as we can see, This force vector is indeed in quadrant 4. So let's find a reference angle first. So the reference angle is going to be the arc tangent of the absolute value of vy over vx. So vy, it's absolute value, it's positive 6. We're going to get rid of the negative sign. vx is 3, so we need to take the arc tangent of 2. And so it will give us this angle, which is 63.4 degrees. So that's the reference angle inside here.
Let me erase this. So this is 63.4 degrees. Now we need to find this angle, the angle that is counterclockwise measured from the positive x-axis.
Now because our resultant force vector is in quadrant 4, the angle is going to be 360 minus the reference angle. And the reference angle is always an acute angle between 0 and 90. So our reference angle in this example is 63.4. So if we take 360 and subtract it by 63.4, we get this angle which is 296.6 degrees.
And that is the answer. So now you know how to add two vectors to get the resultant force vector and you could describe it using unit vectors or using the magnitude and the angle. So that's it for this video.
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