I'm going to find
the exact components of the vector
using the angle shown below.
Notice how we're given the magnitude of the vector c as 18,
we're also assuming the initial point of the vector
is at the origin on the coordinate plane.
Let's go ahead and sketch the x and y axes.
This would be the x-axis,
and this would be the y-axis.
Notice how the vector is in the second quadrant,
where x is negative and y is positive.
When we have a vector v
with a known magnitude on the coordinate plane,
and makes an angle theta with the positive x-axis,
then the vector has an x-component
of the magnitude times cosine theta,
and a y-component of the magnitude times sine theta.
We can also say in our case
vector c would be equal to the magnitude of vector c,
times the vector with an x-component of cosine theta,
and a y-component of sine theta.
So we know the magnitude is 18,
but angle theta must be formed with the positive x-axis,
so theta is not going to be 60 degrees.
The most obvious positive angle for theta
would be this angle here,
which would be 90 plus 30, or 120 degrees.
If we wanted to use a negative angle
we could use this angle here,
which would be negative 180 minus 60, or negative 240.
Of course we could also use any coterminal angle.
Let's go ahead and use 120 degrees for angle theta,
which means vector c is equal to the magnitude which is 18,
times the vector with an x-component
of cosine 120 degrees comma.
The y-component would be sine 120 degrees.
So notice we have 18 the magnitude times,
this would actually be a unit vector
with a direction of 120 degrees from the positive x-axis
To find these trig function values,
we can use a reference triangle or the unit circle.
We'll go ahead and show both.
Notice how if we were to sketch a reference triangle,
that would be this triangle here,
we'd have a 30-60-90 reference triangle
in the second quadrant which I've also sketched here.
Notice how this 60 degrees is actually the reference angle
for the angle that we're using at 120 degrees.
But because we have a 30-60-90 reference triangle,
we know we can label the short leg one,
the hypotenuse two, and the longer leg square root of three.
But because we are in the second quadrant
where x is negative,
we would change this one to negative one.
Now we can use this reference triangle
to find our cosine and sine function values.
Cosine of 120 degrees would be equal to
the ratio of the adjacent side of the hypotenuse,
which would be negative one divided by two, or negative 1/2.
Sine of 120 degrees would be equal to
the ratio of the opposite side of the hypotenuse,
which would be square root three divided by 2.
So this would give us vector c
is equal to 18 times the vector
with an x-component of cosine of 120 degrees,
is equal to negative 1/2,
and sine 120 degrees is equal to
square root three divided by two.
And I'll perform the scalar multiplication.
So vector c is equal to
the vector with an x-component of 18,
or 18 over one times negative 1/2,
and the y-component would be 18 over one,
times square root of three over two.
I'll go ahead and simplify.
Common factor of two here, one two in two, nine twos in 18,
and the same here,
So we have an x-component of
nine times negative one, or negative nine,
and a y-component of nine times square root of three,
or nine square root three.
So this would be the component form of the given vector c.
And just to show that we could find the trig function
lies on the unit circle,
we can see the angle of 120 degrees would terminate here,
where the unit circle, x is equal to cosine theta
and y equals sine theta.
So cosine 120 degrees is equal to negative 1/2,
and sine 120 degrees is equal to
square root of three divided by two.
The same trig function values we found
using our reference triangle.
I hope you've found this helpful.