### Video Transcript

In this lesson, weβre going to
learn how to find the components of a vector. We will start with how to find the
components of a vector that is drawn graphically and how to represent those
components in unit-vector notation. We will also go over how to find
the components of a vector using trigonometry.

Before we learn how to find
components, letβs refresh our memory on what a vector is and how to represent a
vector in graphical form, as well as how to use unit-vector notation. We need to remember that a vector
is a quantity that has both magnitude and direction. We also need to recall that a unit
vector is a vector of length one. Letβs draw an example of two
vectors, π and π. Letβs make vector π have a
magnitude of three units and be oriented along the horizontal axis to the right of
our screen.

When we label vector π, note that
we put a half arrow over the top of the letter. This is a common convention to show
that it is a vector. In text form, it is common for the
variable to be bolded. We will see this form when we do
example problems at the end of our lesson. If we were to write the expression
for vector π using unit vector notation, it would read π is equal to three π’. The π’ is the unit vector that
points in the horizontal direction. The hat over the π’ represents that
itβs a unit vector. In text form, we could bold the π’
to show that it is a unit vector.

We can draw vector π to have the
same magnitude of three units, but this time we can orient the vector vertically to
the top of the screen. Once again, we label our vector
with a half arrow over the top of the letter. The expression for vector π using
unit-vector notation would be vector π is equal to three π£. π£ with a little hat over it is the
unit vector that points in the vertical direction. Now that we have done a quick recap
on what vectors are, how to draw them graphically, and how to write an expression
using unit-vector notation, letβs dive into breaking down a vector into its
components.

A vector can be broken down into
two parts called components. The magnitude of the π₯-component
is represented as π subscript π₯, and the magnitude of the π¦-component is
represented as π subscript π¦. The π₯-component is how much of the
vector points in the horizontal direction and the π¦-component is how much of the
vector points in the vertical direction. We have drawn vector π in on our
coordinate system as shown on the grid. From looking at the grid, letβs
determine what the π₯-component and π¦-component are for our vector.

From the grid, we can see that the
π₯-component of vector π is one, two, three, four, five units. Therefore, the magnitude of π
subscript π₯ is equal to five. The π¦-component of vector π is
equal to one, two units. Therefore, we can say π subscript
π¦ has a magnitude of two. In unit vector notation, the
magnitude of the π₯-component of the vector is what we multiply π’ hat by and the
magnitude of the π¦-component is what we multiply π£ hat by. If we were to write our vector π
in unit-vector notation, we would say that it was equal to five π’ plus two π£,
where the five π’ is the π₯-component of vector π and the two π£ is the
π¦-component of vector π.

Letβs look at one more vector,
vector π. Weβll break down vector π into
both its π₯- and π¦-component and then put it into unit-vector notation. Vector π has a magnitude of two
units to the left of the screen. Therefore, the magnitude, the
π₯-component for vector π, π π₯ would be equal to negative two. Vector π also has a magnitude of
three units along the vertical to the bottom of the screen. So, the magnitude of vector πβs
π¦-component, π π¦, is equal to negative three. Writing out the unit-vector
notation for vector π, we get that vector π is equal to negative two π’ minus
three π£, where the negative two π’ is the π₯-component of vector π and negative
three π£ is the π¦-component of vector π.

Another way to break down a vector
into its components is to use trigonometry. Before we apply trigonometry to our
vectors, letβs refresh our memory on how do you use sine, cosine, and tangent with a
right triangle. When we have a right angle, we can
use trigonometry to solve for an unknown side as long as we know the measure of one
angle and the length of one side. If we label this angle π, then the
side that touches the angle or thatβs next to it is called the adjacent side. The side thatβs opposite to the
angle is called the opposite side, and the side thatβs opposite to the 90-degree
angle is called the hypotenuse.

The sin of the angle π, the cos of
the angle π, and the tan of the angle π are simply the ratios of the sides of
similar triangles. The sin of π is equal to the ratio
of the opposite side of the triangle to the hypotenuse of the triangle. The cos of π is equal to the ratio
of the adjacent side of the triangle to the hypotenuse of the triangle. And the tan of π is the ratio of
the opposite side of the triangle to the adjacent side of the triangle. A common acronym to remember the
ratios is SOH-CAH-TOA. SOH for sin is equal to opposite
over hypotenuse, CAH for cos is equal to adjacent over hypotenuse, and TOA for
tangent is equal to opposite over adjacent.

Letβs apply this bit of
trigonometry to our vectors. If we need to break down our vector
which has a length of eight units and an angle of 35 degrees above the horizontal
into its two components, where do we start? We can start by drawing in the
components so that the π₯-component is along the horizontal axis and the
π¦-component is along the vertical axis to form a right triangle. We have labeled the π₯-component π
subscript π₯ and the π¦-component π subscript π¦.

To find the π₯-component of our
vector π, we need to identify where π π₯ is with respect to the angle. We can see that π π₯ is next to
the angle of 35 degrees or is the adjacent side. We have been given the hypotenuse
of the triangle to be eight, a labeled angle of 35 degrees, and we now know that
weβre solving for the π₯-component or adjacent side of the triangle. Which of our three trigonometric
functions sine, cosine, or tangent should we use? We should use cosine because the
cos of π is equal to the adjacent side divided by the hypotenuse.

We need to isolate the adjacent
side of the triangle, since that is what weβre solving for. We multiply both sides of the
equation by the hypotenuse. The hypotenuse cancels out on the
right side of the equation. The left side of the equation
becomes the hypotenuse times the cos of π. When we plug in the values of our
variables, we put in eight for the hypotenuse, cos of 35 degrees for π, and π π₯
for the adjacent side of the triangle as weβre solving for the π₯-component of our
vector. When we multiply our eight by cos
of 35 degrees, we get an π₯-component of 6.55.

We need to make sure that our
calculators in degrees and not radians as the cos of 35 radians is not the same
thing as the cos of 35 degrees. We will round our answer to two
significant figures because our angle was given to two significant figures. This gives us an π₯-component for
our vector of 6.5.

Moving on to the π¦-component of
the vector, which is the opposite side of our triangle, which trigonometric function
we will use if we were also given the hypotenuse and the angle? We would use the sin of π is equal
to the opposite side of the triangle divided by the hypotenuse. To isolate the opposite side, we
once again multiply both sides by the hypotenuse. Canceling out the hypotenuse on the
right side of the equation and leaving the left side equation to be the hypotenuse
times the sin of π. Substituting the values into our
equation, we would have a hypotenuse of eight, a sin of 35 degrees, and our
π¦-component, π π¦, would represent the opposite side of the triangle.

Multiplying eight by sin of 35
degrees, we get 4.59 as the magnitude of our π¦-component. Again, we need to round our
component to two significant figures in order to match our problem. 4.59 becomes 4.6. So, the π¦-component of our vector
is 4.6. Problems can get tricky by which
axis the angle is drawn to, so we need to pay careful attention to this. If our angle was 35 degrees to the
vertical instead of to the horizontal, what would happen to our components?

In this scenario, the π₯-component
of our vector is the opposite side of the triangle. So, we do π subscript π₯ is equal
to the sin of π times the hypotenuse, just as we did for the π¦-component up
above. We would once again multiply the
sin of 35 degrees by eight. And our π₯-component to two
significant figures would be 4.6, just as the π¦-component was up above. The π¦-component of our vector is
the adjacent side of the angle. Therefore, weβd use the cos of the
angle and multiply it by the hypotenuse to find the π¦-component, just as we did for
the π₯-component up above. We would multiply cos of 35 degrees
by eight, which would give us a π¦-component of our vector of 6.5 when we round to
two significant figures, just as it did up above for our π₯-component.

Letβs try two examples of finding
the components of vectors, one graphically with unit-vector notation and one with
trigonometry.

Write π in component form.

The diagram shows vector π to be
drawn onto a grid. We are asked to write vector π in
component form. The component form of a vector is
typically written as π sub π₯ π’ plus π sub π¦ π£, where π sub π₯ is the
magnitude of the π₯-component of the vector and π sub π¦ is the magnitude of the
π¦-component of the vector. π’ and π£ represent unit vectors,
with π’ being in the direction of the horizontal axis and π£ being in the direction
of the vertical axis. Because weβre handwriting our
answers, we use a half arrow over the letter π to represent that itβs a vector. In text form, we bolded it. In the π’ and π£, we see little
hats to represent that theyβre unit vectors. In text form, these can also be
bolded.

To find the component form of
vector π, we need to determine how much of vector π is in the horizontal direction
and how much of vector π is in the vertical direction. Looking at the grid, we can see
that vector π is one, two, three units to the left of our screen. This is considered negative
three. We could then say that the
π’-component of our vector has a magnitude of negative three. Looking back at the grid, we can
see that the π¦-component or the vertical direction has a magnitude of one, two
units pointing towards the top of the screen, which would be a positive two. We could, therefore, say that the
π£-component of our vector has a magnitude of two. Based on the diagram, the component
form of vector π is equal to negative three π’ plus two π£.

In our next example, we will apply
trigonometry to solve for the components of a vector.

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.

In the problem, weβre being asked
to solve for the π₯-component of our vector, or π sub π₯. With the components drawn in on our
diagram, we can see that the triangle is a right triangle. And therefore, we can use
trigonometry to solve for our unknown side. Letβs recall that we can use
SOH-CAH-TOA. SOH, the sin of the angle is equal
to the opposite side of the triangle divided by the hypotenuse. CAH, the cos of the angle is equal
to the adjacent side of the triangle divided by the hypotenuse. And TOA, the tan of the angle is
equal to the opposite side of the triangle divided by the adjacent side of the
triangle.

Looking back at our diagram, we are
solving for the π₯-component of our vector, weβre given the angle π, and weβre
given the length of the hypotenuse. Which one of our trigonometric
functions would we choose, sine, cosine, or tangent? The best one to choose is cosine
because our vector component is adjacent to the angle. Now, we need to isolate the
adjacent side of the triangle, as thatβs what weβre solving for. To do this, we multiply both sides
by the hypotenuse. This will cancel out the hypotenuse
on the right side of the equation and leave the left side of the equation as the
hypotenuse times the cos of π.

We can now use the values from the
problem to substitute in for our variables. The hypotenuse had a magnitude of
22, π was 36 degrees, and π sub π₯ represents the horizontal component of our
vector. When we multiply 22 by the cos of
36 degrees, we get 17.8. In the problem, we were instructed
to give our answers to two significant figures. Looking at our answer right now,
itβs given to three significant figures. Therefore, we can round up 17.8 to
18. The horizontal component of a
vector that has a magnitude of 22 and makes an angle with the π₯-axis of 36 degrees
is 18.

Key Points of the Lesson

To find the π₯- and π¦-components
of a vector represented graphically on a grid, count how many units along the
π₯-axis and π¦-axis the vector goes. In unit-vector notation, a vector
can be represented as π equals π sub π₯ π’ plus π sub π¦ π£. Trigonometry can be used to
determine an unknown component of a vector.