### Video Transcript

In this video, weβre talking about
finding the components of a vector. Weβre going to learn how to find
these components graphically using a grid as well as using trigonometry. Now, as we get started, letβs point
out that all vectors are made up of components. We could say that components are
the pieces we add together to get a vector.

Given a vector, say that we start
with this vector weβve called π, we can see its components by drawing it on a
grid. First, we sketch in an
π₯π¦-coordinate frame and then add in grid lines. We can see that our vector extends
one, two, three grid spaces along this π₯-axis. We would say then that the vectorβs
π₯-component, we can call it π sub π₯, equals three. The π¦-component of π goes one,
two, three, four, five, six units up the π¦-axis. So we can say that π sub π¦ is
six. These are the components of the
vector π.

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. Here, π΄ sub π₯ and π΄ sub π¦ are
its π₯- and π¦-components. And notice that we donβt simply add
these components together to get the vector π. First, each one has to be
multiplied by the correct unit vector. In the case of π΄ sub π₯, this is
the π’ hat unit vector here. And for π΄ sub π¦, itβs the π£ hat
unit vector.

Recall that a unit vector is a
vector with a magnitude or a length of one. 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. These vectors are important because
by themselves the π₯- and π¦-components of a given vector are scalar quantities. So they canβt give us a vector all
by themselves. That is, they have a magnitude or a
length, but they donβt have any direction. Thatβs where the unit vectors come
in.

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. Likewise, it extends a length π΄
sub π¦ in the π¦-direction. All this to say, the components of
a vector are not themselves vectors. Actually, theyβre scalar
quantities. 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.

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. And then we add to that another
vector, its π¦-component π sub π¦ times the unit vector π£ hat. Filling in for our known values, we
could write this as three π’ plus six π£. And this is called the component
form of our vector π. Itβs expressed in terms of a
horizontal or π₯-component and a vertical or π¦-component.

Now, we mentioned earlier that
thereβs more than one way to solve for the components of a vector. In the case of vector π, weβve
used grid spacings to calculate these components. But letβs imagine that instead
weβre given a vector, weβll call it vector π, on an unmarked π₯π¦-plane. In this case, weβre not given any
grid marks to work off of. But say that we are given the
magnitude of our vector as well as the angle that it makes with the horizontal. Itβs possible to solve for the
components of π using just this information.

Letβs remember that any
two-dimensional vector has a horizontal as well as a vertical component. 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. Since the vertical and horizontal
components of any vector are perpendicular to one another, we know that this angle
is a right angle. And now we have a right triangle
where this side is the hypotenuse and these lengths are the other two sides. And notice this: the shorter two
sides in this triangle are the π₯- and π¦-components of our vector π.

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. In a right triangle, the side
opposite the right angle is always the hypotenuse. 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. 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. Those are the horizontal and
vertical components of π, respectively.

Now, these different sides of our
right triangle relate to one another through trigonometric functions. Hereβs what we mean by that. 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. 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. In other words, the hypotenuse
length times the sin of π will give us the vertical component of our vector.

When it comes to the vector π, we
see that its hypotenuse is nine and π is 37 degrees. So if we multiply nine by the sin
of 37 degrees, then weβll get the vertical component of π. To two significant figures, this is
5.4. 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. Rearranging, this gives us that the
hypotenuse of our triangle times the cos of π equals that adjacent side length. And that in the case of our vector
π is the horizontal component. π
sub π₯ is equal to nine times
the cos of 37 degrees, which rounded to two significant figures is 7.2. Weβve solved then for the vertical
and horizontal components of π. So we could write it out in
component form like this. 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.

Knowing all this about the
components of vectors, letβs get a bit of practice through an example.

The vector π can be written in the
form π π₯ times π’ hat plus π π¦ times π£ hat. What is the value of π π₯? What is the value of π π¦?

Alright, in this description of the
vector π, π π₯ and π π¦ are its π₯- and π¦-components. These are scalar quantities that
show the length of vector π in the horizontal and vertical directions. 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¦.

So looking at this first part of
our question, π sub π₯ will be the amount that vector π lies along this
π₯-axis. This is called the horizontal
component of π. And we find its value by projecting
π downward perpendicularly onto this π₯-axis. On this graph then, π sub π₯ would
look like this. Itβs a length that covers one, two,
three, four, five, six, seven grid spaces. And so thatβs the value of π sub
π₯. Knowing this, we then want to
figure out what the π¦-component of our vector π is. This time, weβll project our vector
perpendicularly onto the vertical axis. π sub π¦ then would look like
this. We can count squares to figure out
the length of this line. Itβs one, two, three, four units
long. So π sub π¦ is four. Weβve solved then for both the
horizontal as well as the vertical components of vector π.

Letβs look now at another
example.

Write π in component form.

Here, we see this vector π drawn
on a grid. And we can see the vector starts at
the origin of a coordinate frame. Letβs call the horizontal axis the
π₯-axis and the vertical one the π¦. 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. 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. 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. By themselves, the π₯- and
π¦-components of vector π are not vectors; theyβre scalar quantities. But when we multiply these scalars
by a vector, the unit vectors, the result is a vector.

Finally, adding these vector
components together, weβll get the vector π. Expressing π this way is known as
writing it in component form. So then, what are π΄ sub π₯ and π΄
sub π¦? To figure that out, weβll need to
look at our grid. Starting with π΄ sub π₯, thatβs
equal to the horizontal component of this vector π. 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₯. In terms of the units of this grid,
that length is one, two, three units long. And notice that we moved to the
left of the origin, that is, into negative π₯-values. So even though this horizontal
orange line is three units long, we say that the π₯-component of π is negative
three. This is because the projection of
vector π onto the horizontal axis goes negative three units in the
π₯-direction.

To find the vertical component of
π, weβll follow a similar process. Once again, we project vector π
perpendicularly, this time onto the vertical axis. And itβs the length of this line
that tells us the vertical or π¦-component of π. We see that this is one, two units
long and that this is in the positive π¦-direction. π΄ sub π¦ then is equal to positive
two. And now we can write out π in its
component form. Vector π is equal to negative
three times the π’ hat unit vector plus two times the π£ hat unit vector.

Letβs look now at one last example
exercise.

The diagram shows a vector π
that has a magnitude of 22. The angle between the vector
and the π₯-axis is 36 degrees. Work out the horizontal
component of the vector. Give your answer to two
significant figures.

Alright, we see this vector π
sketched out. And weβre told it has a length
or magnitude of 22. Along with this, we know that
the vector forms an angle of 36 degrees with the positive π₯-axis. Our goal is to solve for its
horizontal component. This is equal to the horizontal
projection of our vector onto this axis.

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. In other words, we have here a
right triangle. Hereβs the hypotenuse, hereβs
another side, and hereβs the third. In solving then for the
horizontal component of our vector, weβre solving for one of the sides of this
right triangle. We can do this using
trigonometry.

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. 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.

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. 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. 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. 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. When we enter this expression
on our calculator and keep two significant figures, our answer is 18. This is the horizontal
component of the vector π.

Letβs finish up our lesson now by
reviewing a few key points. 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¦. Here, π΄ π₯ represents the
horizontal component of our vector and π΄ π¦ the vertical component. 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. 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. Here, the opposite and adjacent
side lengths often represent the vertical and horizontal components, respectively,
of a vector.