WEBVTT
00:00:01.720 --> 00:00:05.200
In this video, weβre talking about finding the components of a vector.
00:00:05.640 --> 00:00:13.280
Weβre going to learn how to find these components graphically using a grid as well as using trigonometry.
00:00:14.760 --> 00:00:18.880
Now, as we get started, letβs point out that all vectors are made up of components.
00:00:19.520 --> 00:00:25.480
We could say that components are the pieces we add together to get a vector.
00:00:26.160 --> 00:00:36.240
Given a vector, say that we start with this vector weβve called π, we can see its components by drawing it on a grid.
00:00:37.120 --> 00:00:43.080
First, we sketch in an π₯π¦-coordinate frame and then add in grid lines.
00:00:44.320 --> 00:00:53.760
We can see that our vector extends one, two, three grid spaces along this π₯-axis.
00:00:54.480 --> 00:01:00.800
We would say then that the vectorβs π₯-component, we can call it π sub π₯, equals three.
00:01:01.320 --> 00:01:10.080
The π¦-component of π goes one, two, three, four, five, six units up the π¦-axis.
00:01:10.640 --> 00:01:14.920
So we can say that π sub π¦ is six.
00:01:15.560 --> 00:01:17.760
These are the components of the vector π.
00:01:18.720 --> 00:01:29.440
Now, in general, if we have a two-dimensional vector, we can call that vector π, then we can write the vector π like this.
00:01:30.200 --> 00:01:35.480
Here, π΄ sub π₯ and π΄ sub π¦ are its π₯- and π¦-components.
00:01:36.280 --> 00:01:42.000
And notice that we donβt simply add these components together to get the vector π.
00:01:42.760 --> 00:01:46.800
First, each one has to be multiplied by the correct unit vector.
00:01:47.640 --> 00:01:51.760
In the case of π΄ sub π₯, this is the π’ hat unit vector here.
00:01:52.360 --> 00:01:55.800
And for π΄ sub π¦, itβs the π£ hat unit vector.
00:01:56.280 --> 00:02:03.560
Recall that a unit vector is a vector with a magnitude or a length of one.
00:02:04.680 --> 00:02:13.920
If we drew in our π’ hat unit vector on our grid, it would look like this, while the π£ hat unit vector would look like this.
00:02:15.040 --> 00:02:23.840
These vectors are important because by themselves the π₯- and π¦-components of a given vector are scalar quantities.
00:02:25.080 --> 00:02:28.520
So they canβt give us a vector all by themselves.
00:02:29.160 --> 00:02:33.680
That is, they have a magnitude or a length, but they donβt have any direction.
00:02:34.400 --> 00:02:36.840
Thatβs where the unit vectors come in.
00:02:38.360 --> 00:02:47.880
This first term on the right in our equation says that the vector π has a length π΄ sub π₯, and itβs in this particular direction, the π₯-direction.
00:02:48.640 --> 00:02:52.400
Likewise, it extends a length π΄ sub π¦ in the π¦-direction.
00:02:53.160 --> 00:02:57.680
All this to say, the components of a vector are not themselves vectors.
00:02:58.280 --> 00:02:59.680
Actually, theyβre scalar quantities.
00:03:00.880 --> 00:03:13.200
We saw that here with our vector π, where we found out its π₯-component is three, not a vector but a scalar, and its π¦-component is a scalar too.
00:03:13.920 --> 00:03:24.800
So to write out the vector π in terms of its components, we would say that it has a length of π sub π₯, its π₯-component, in the π’ hat direction.
00:03:25.480 --> 00:03:33.640
And then we add to that another vector, its π¦-component π sub π¦ times the unit vector π£ hat.
00:03:34.160 --> 00:03:40.000
Filling in for our known values, we could write this as three π’ plus six π£.
00:03:40.640 --> 00:03:43.680
And this is called the component form of our vector π.
00:03:44.560 --> 00:03:51.880
Itβs expressed in terms of a horizontal or π₯-component and a vertical or π¦-component.
00:03:53.200 --> 00:03:57.960
Now, we mentioned earlier that thereβs more than one way to solve for the components of a vector.
00:03:58.800 --> 00:04:03.600
In the case of vector π, weβve used grid spacings to calculate these components.
00:04:04.160 --> 00:04:14.840
But letβs imagine that instead weβre given a vector, weβll call it vector π, on an unmarked π₯π¦-plane.
00:04:15.840 --> 00:04:19.880
In this case, weβre not given any grid marks to work off of.
00:04:20.480 --> 00:04:27.480
But say that we are given the magnitude of our vector as well as the angle that it makes with the horizontal.
00:04:28.400 --> 00:04:32.360
Itβs possible to solve for the components of π using just this information.
00:04:33.600 --> 00:04:40.000
Letβs remember that any two-dimensional vector has a horizontal as well as a vertical component.
00:04:41.040 --> 00:04:50.640
If we were to sketch in that horizontal component, it would look like this and the vertical component like this, which now that we think about it is equal in length to this dashed line.
00:04:51.920 --> 00:05:01.160
Since the vertical and horizontal components of any vector are perpendicular to one another, we know that this angle is a right angle.
00:05:02.360 --> 00:05:09.920
And now we have a right triangle where this side is the hypotenuse and these lengths are the other two sides.
00:05:10.840 --> 00:05:17.040
And notice this: the shorter two sides in this triangle are the π₯- and π¦-components of our vector π.
00:05:18.320 --> 00:05:30.840
Whenever we have a right triangle and one of the other interior angles is also known, there are specific trigonometric relationships between the lengths of the three sides of this triangle.
00:05:32.160 --> 00:05:37.320
In a right triangle, the side opposite the right angle is always the hypotenuse.
00:05:37.960 --> 00:05:55.400
Then, because this angle, weβve called it π, is the other one we know, we call this side of the triangle the opposite side, that is, opposite from π, and this the adjacent side, because itβs adjacent or next to π.
00:05:56.840 --> 00:06:08.840
Connecting this triangle to the one created by our vector π, notice that the adjacent and opposite sides of the known angle of 37 degrees are what we want to solve for.
00:06:09.680 --> 00:06:14.640
Those are the horizontal and vertical components of π, respectively.
00:06:16.280 --> 00:06:22.240
Now, these different sides of our right triangle relate to one another through trigonometric functions.
00:06:23.400 --> 00:06:24.640
Hereβs what we mean by that.
00:06:25.640 --> 00:06:32.360
If we take the sin of the angle π, then thatβs equal to the length of the side opposite the angle π divided by the hypotenuse length.
00:06:33.200 --> 00:06:41.160
If we multiply both sides of this equation by the length of the hypotenuse, then we end up with an equation where the opposite side length is the subject.
00:06:42.040 --> 00:06:47.320
In other words, the hypotenuse length times the sin of π will give us the vertical component of our vector.
00:06:48.480 --> 00:06:57.400
When it comes to the vector π, we see that its hypotenuse is nine and π is 37 degrees.
00:06:58.280 --> 00:07:04.040
So if we multiply nine by the sin of 37 degrees, then weβll get the vertical component of π.
00:07:04.560 --> 00:07:09.240
To two significant figures, this is 5.4.
00:07:10.480 --> 00:07:21.320
When it comes to solving for the horizontal component of π, we can note that the cos of the angle π is equal to the adjacent side length over the hypotenuse.
00:07:22.240 --> 00:07:29.880
Rearranging, this gives us that the hypotenuse of our triangle times the cos of π equals that adjacent side length.
00:07:30.520 --> 00:07:33.400
And that in the case of our vector π is the horizontal component.
00:07:34.240 --> 00:07:43.680
π
sub π₯ is equal to nine times the cos of 37 degrees, which rounded to two significant figures is 7.2.
00:07:44.840 --> 00:07:49.000
Weβve solved then for the vertical and horizontal components of π.
00:07:49.880 --> 00:07:54.240
So we could write it out in component form like this.
00:07:54.880 --> 00:08:10.040
We say then that vector π has a magnitude of nine, a direction of 37 degrees above the horizontal, a horizontal component of 7.2, and a vertical component of 5.4.
00:08:11.640 --> 00:08:16.840
Knowing all this about the components of vectors, letβs get a bit of practice through an example.
00:08:18.040 --> 00:08:25.280
The vector π can be written in the form π π₯ times π’ hat plus π π¦ times π£ hat.
00:08:25.880 --> 00:08:27.560
What is the value of π π₯?
00:08:28.240 --> 00:08:29.880
What is the value of π π¦?
00:08:31.000 --> 00:08:38.160
Alright, in this description of the vector π, π π₯ and π π¦ are its π₯- and π¦-components.
00:08:38.840 --> 00:08:44.960
These are scalar quantities that show the length of vector π in the horizontal and vertical directions.
00:08:46.160 --> 00:08:57.080
If we look at this sketch of vector π with the grid around it, we see that thereβs a horizontal axis, weβll call that the π₯-axis, and a vertical one weβll call π¦.
00:08:58.400 --> 00:09:05.560
So looking at this first part of our question, π sub π₯ will be the amount that vector π lies along this π₯-axis.
00:09:06.280 --> 00:09:08.640
This is called the horizontal component of π.
00:09:09.400 --> 00:09:14.520
And we find its value by projecting π downward perpendicularly onto this π₯-axis.
00:09:15.480 --> 00:09:18.560
On this graph then, π sub π₯ would look like this.
00:09:19.280 --> 00:09:28.280
Itβs a length that covers one, two, three, four, five, six, seven grid spaces.
00:09:29.240 --> 00:09:31.800
And so thatβs the value of π sub π₯.
00:09:32.840 --> 00:09:37.480
Knowing this, we then want to figure out what the π¦-component of our vector π is.
00:09:38.040 --> 00:09:42.560
This time, weβll project our vector perpendicularly onto the vertical axis.
00:09:43.480 --> 00:09:44.920
π sub π¦ then would look like this.
00:09:45.880 --> 00:09:48.520
We can count squares to figure out the length of this line.
00:09:49.000 --> 00:09:53.600
Itβs one, two, three, four units long.
00:09:54.120 --> 00:09:56.120
So π sub π¦ is four.
00:09:56.920 --> 00:10:01.400
Weβve solved then for both the horizontal as well as the vertical components of vector π.
00:10:03.040 --> 00:10:04.640
Letβs look now at another example.
00:10:05.360 --> 00:10:07.200
Write π in component form.
00:10:08.040 --> 00:10:11.160
Here, we see this vector π drawn on a grid.
00:10:11.720 --> 00:10:16.160
And we can see the vector starts at the origin of a coordinate frame.
00:10:17.160 --> 00:10:21.400
Letβs call the horizontal axis the π₯-axis and the vertical one the π¦.
00:10:22.400 --> 00:10:34.520
Now, when we go to write this vector π in component form, that means weβll write it in terms of an π₯- and a π¦-component, also called a horizontal and vertical component.
00:10:35.600 --> 00:10:47.600
If we call the π₯-component of vector π π΄ sub π₯ and the π¦-component π΄ sub π¦, then we can multiply each one of these components by the appropriate unit vector.
00:10:48.800 --> 00:10:58.440
The unit vector for the π₯- or horizontal direction is π’ hat, and the unit vector for the vertical or π¦-direction is π£ hat.
00:10:59.560 --> 00:11:07.680
By themselves, the π₯- and π¦-components of vector π are not vectors; theyβre scalar quantities.
00:11:08.440 --> 00:11:16.320
But when we multiply these scalars by a vector, the unit vectors, the result is a vector.
00:11:17.720 --> 00:11:22.320
Finally, adding these vector components together, weβll get the vector π.
00:11:23.240 --> 00:11:28.400
Expressing π this way is known as writing it in component form.
00:11:29.440 --> 00:11:31.520
So then, what are π΄ sub π₯ and π΄ sub π¦?
00:11:32.720 --> 00:11:36.120
To figure that out, weβll need to look at our grid.
00:11:36.920 --> 00:11:41.920
Starting with π΄ sub π₯, thatβs equal to the horizontal component of this vector π.
00:11:42.840 --> 00:11:52.920
In other words, if we project this vector perpendicularly onto the π₯-axis, then the length of that line segment, this length here, is π΄ sub π₯.
00:11:53.840 --> 00:11:59.440
In terms of the units of this grid, that length is one, two, three units long.
00:12:00.480 --> 00:12:05.840
And notice that we moved to the left of the origin, that is, into negative π₯-values.
00:12:06.480 --> 00:12:16.440
So even though this horizontal orange line is three units long, we say that the π₯-component of π is negative three.
00:12:17.360 --> 00:12:25.880
This is because the projection of vector π onto the horizontal axis goes negative three units in the π₯-direction.
00:12:27.000 --> 00:12:31.800
To find the vertical component of π, weβll follow a similar process.
00:12:32.560 --> 00:12:38.960
Once again, we project vector π perpendicularly, this time onto the vertical axis.
00:12:39.560 --> 00:12:44.880
And itβs the length of this line that tells us the vertical or π¦-component of π.
00:12:45.880 --> 00:12:52.400
We see that this is one, two units long and that this is in the positive π¦-direction.
00:12:53.000 --> 00:12:55.680
π΄ sub π¦ then is equal to positive two.
00:12:56.560 --> 00:13:00.040
And now we can write out π in its component form.
00:13:00.520 --> 00:13:09.360
Vector π is equal to negative three times the π’ hat unit vector plus two times the π£ hat unit vector.
00:13:11.360 --> 00:13:14.440
Letβs look now at one last example exercise.
00:13:15.520 --> 00:13:19.400
The diagram shows a vector π that has a magnitude of 22.
00:13:20.240 --> 00:13:26.000
The angle between the vector and the π₯-axis is 36 degrees.
00:13:26.800 --> 00:13:28.880
Work out the horizontal component of the vector.
00:13:29.560 --> 00:13:32.120
Give your answer to two significant figures.
00:13:33.640 --> 00:13:37.240
Alright, we see this vector π sketched out.
00:13:37.920 --> 00:13:41.520
And weβre told it has a length or magnitude of 22.
00:13:42.600 --> 00:13:50.640
Along with this, we know that the vector forms an angle of 36 degrees with the positive π₯-axis.
00:13:51.640 --> 00:13:53.880
Our goal is to solve for its horizontal component.
00:13:54.320 --> 00:13:57.680
This is equal to the horizontal projection of our vector onto this axis.
00:13:59.040 --> 00:14:08.080
As we go about solving for the length of this orange line, letβs note that our dashed line intersects our horizontal axis at a right angle.
00:14:09.000 --> 00:14:12.280
In other words, we have here a right triangle.
00:14:12.920 --> 00:14:17.120
Hereβs the hypotenuse, hereβs another side, and hereβs the third.
00:14:18.520 --> 00:14:25.600
In solving then for the horizontal component of our vector, weβre solving for one of the sides of this right triangle.
00:14:26.280 --> 00:14:29.360
We can do this using trigonometry.
00:14:30.480 --> 00:14:49.240
Letβs remember that, given a right triangle, if we know one of the other interior angles, then we can define the sides of this right triangle as hypotenuse β, the side opposite our angle π π, and the side adjacent to that angle π.
00:14:50.440 --> 00:14:58.480
Set up this way, itβs the adjacent side, what weβve called π over here, that we want to solve for to figure out our horizontal component.
00:14:59.560 --> 00:15:07.320
Now, if we were to take the cos of this angle π, then that would equal the ratio of our adjacent side length to our hypotenuse.
00:15:08.280 --> 00:15:19.480
Or multiplying both sides of this equation by the hypotenuse, canceling that factor out on the right, we have that the adjacent side of our right triangle equals the cos of π times β.
00:15:20.760 --> 00:15:28.640
This relates to our situation with vector π because in this case we know the length of our hypotenuse and we also know this angle.
00:15:29.480 --> 00:15:43.280
So we can actually say that the length of our triangleβs hypotenuse, 22, multiplied by the cos of our angle of 36 degrees is equal to what weβll call π΄ sub π₯, the horizontal component of the vector π.
00:15:44.080 --> 00:15:52.880
When we enter this expression on our calculator and keep two significant figures, our answer is 18.
00:15:53.600 --> 00:15:57.040
This is the horizontal component of the vector π.
00:15:59.200 --> 00:16:03.040
Letβs finish up our lesson now by reviewing a few key points.
00:16:04.080 --> 00:16:13.240
In this lesson, we saw that a vector, say we have a vector π, can be written in terms of components π΄ π₯ and π΄ π¦.
00:16:14.240 --> 00:16:20.200
Here, π΄ π₯ represents the horizontal component of our vector and π΄ π¦ the vertical component.
00:16:21.160 --> 00:16:31.440
We also learn that a vectorβs components are scalar values that can be determined in one of two ways: first from a grid or second using trigonometry.
00:16:32.480 --> 00:16:53.440
And lastly, considering the trigonometry of a right triangle, we saw that, given an interior angle π, which is different from the right angle in the triangle, the sin of π is equal to the ratio of the opposite side length to the hypotenuse, while the cos of π is equal to the adjacent side length to the hypotenuse.
00:16:54.520 --> 00:17:03.680
Here, the opposite and adjacent side lengths often represent the vertical and horizontal components, respectively, of a vector.