### Video Transcript

In this video, weβre talking about
the polar form of a vector. We may be more used to writing
vectors in what is called rectangular form. In this lesson, not only will we
learn what the polar form of a vector is, weβll also learn how to convert from polar
form to rectangular and back.

To start off, say that we have some
vector, weβll call it π, thatβs plotted on a Cartesian plane. This means that π can be divided
up into its π₯- and π¦-components. We could write that π is equal to
π₯ in the π’ hat direction plus π¦ in the π£ hat direction. Another way to write this is using
these angle braces. And we say that this is the
rectangular form of the vector π. Weβre describing the vector using
these rectangular variables π₯ and π¦. Going back to our sketch, we can
start to see, though, that thereβs another way of describing this same vector. We could describe it in terms of
its overall radial distance from the origin, weβll call that distance π, as well as
the angle this vector forms with the positive π₯-axis, and weβll call that π.

And this second way of describing
the vector π, in terms of π and π, is related to the variables π₯ and π¦. For example, since π is the
hypotenuse of a right triangle where the other side lengths are π₯ and π¦, then we
can use the Pythagorean theorem to say that π is equal to the square root of π₯
squared plus π¦ squared. And then as far as the angle π
goes, notice that the tan of this angle π is equal to π¦ divided by π₯. This implies that the variable π
is equal to the inverse or arctan of π¦ over π₯.

And so now, using our rectangular
components, weβre able to represent the vector π in a different way. We can say that itβs defined by a
radial distance π from the origin pointed in a certain direction π relative to the
positive π₯-axis. This is called the polar form of
our vector, and itβs entirely equivalent to the rectangular form, just expressed a
different way. Expressing a vector in polar form
can be helpful in describing the motion of certain objects. For example, if we have an object
moving in a circle centered on the origin of the π₯π¦-plane, then a vector defining
the position of that object at various points in time would be more conveniently
written in polar form than rectangular.

That said, if we are given a vector
in polar form and want to express it in rectangular, we can do that using the
relationships we see in our sketch between π, π, π₯, and π¦. The variable π₯ is equal to π
times the cos of π, while π¦ is equal to π times the sin of π. Given π and π then, we can solve
for π₯ and π¦. Using these starred relationships
then, weβre able to go back and forth between rectangular and polar forms of a
vector. Letβs now get some practice doing
this through an example.

Consider the vector negative two,
three. Calculate the direction of the
vector, giving your solution as an angle to the nearest degree measured
counterclockwise from the positive π₯-axis.

Since weβre given the components of
our vector, letβs start out by plotting it on this π₯π¦-plane. Weβll say that each of these tick
marks represents a distance of one unit. And weβre told that our vector has
an π₯-component of negative two and a π¦-component of positive three. That tells us that with its tail at
the origin our vector will look like this. We want to calculate the direction
of this vector, and weβll do it in terms of an angle that begins at the positive
π₯-axis and goes to our vector. We can give this angle a name. Weβll call it π. At this point, letβs recall that in
general when a vector, we can call it π, is given in polar form, then that vector
is defined by a radial distance from the origin and an angle from the positive
π₯-axis.

In our exercise, we want to solve
for this angle π. And to do this, weβre given a
vector not in polar form but in rectangular form. In using this π₯ and π¦ information
then to solve for π, weβre effectively converting from a rectangular form of a
vector to part of its polar form. We say all that because the tan of
π, where π is the angle of a vector in its polar form, is equal to the ratio of
the rectangular components of that vector, π¦ to π₯. This tells us that the tan of our
angle of interest, π, is equal to three divided by negative two. And then if we take the inverse or
arctan of both sides, the left-hand side simplifies to the angle we want to solve
for.

Now, if we go and evaluate this
expression on our calculator to the nearest hundredth of a degree, we find a result
of negative 56.31 degrees. Looking at our sketch though, we
see that this canβt be the angle π. Whatβs happening here is because
the π₯-value in this fraction is negative β in other words, the vector weβre working
with occupies either the second or the third quadrant β to correctly compute our
angle, weβll need to take our calculated result and add 180 degrees to it. For any vector with a negative
π₯-component like we have here, this is a standard practice for accurately
calculating its direction.

And so now if we add these two
angles together, we get a result of 123.69 degrees. And recalling that we give our
answer rounded to the nearest degree, we can see that π equals 124 degrees. This is the direction of our vector
to the nearest degree measured counterclockwise from the positive π₯-axis.

Now letβs look at an example where
we fully convert the rectangular form of a vector to the polar form.

If π equals negative π’ minus π£,
then the polar form of π is blank. (A) Root two, π over four, (B)
root two, three π over four, (C) root two, five π over four, (D) root two, seven
π over four.

All right, so given this vector π
in rectangular form, we want to solve for its polar form. Another way we can write π in
rectangular form is to express it in terms of its π₯- and π¦-components like
this. And now letβs recall that for a
vector expressed in polar form, we give it not by its π₯-, π¦-components, but rather
by π and π. Here, π is equal to the square
root of π₯ squared plus π¦ squared, and the tan of π is equal to π¦ divided by
π₯. Using these relationships, we then
have the ability to convert π from its rectangular form as given to its polar
form.

That polar form, as weβve seen, is
defined by a radial distance and an angle. And π, we know, is equal to the
square root of the π₯-component squared plus the π¦-component squared. This is equal to the square root of
one plus one or the square root of two, so weβll substitute this result in for π in
our polar form of π. But as we look again at our answer
options, notice that this doesnβt narrow down the list. All four choices had the same
π-value of the square root of two. So letβs move on to calculating the
angle π of our vector. As we do this, it will be helpful
to sketch our vector on a coordinate plane.

Letβs say that each of these tick
marks represents one unit of distance. And since our vector has
rectangular components of negative one and negative one, if the tail of the vector
was at the origin, then the vector would look like this. And the angle π, defining its
direction, would be measured from the positive π₯-axis to the vector. We see then that π will be greater
than π radians but less than three-halves π. We can now use this relationship
here to solve for it. We know that the components of our
vector π₯ and π¦ are both negative one. Here weβve taken the inverse tan of
both sides of our tan of π equation, meaning that π equals the arc or inverse tan
of negative one divided by negative one.

Hereβs whatβs interesting,
though. If we evaluate this inverse tangent
on our calculator, the result we get is exactly π divided by four. But looking at our sketch of our
vector, we know that that canβt be the correct angle for π. Here, we need to recall the rule
that when we calculate an inverse tangent with a negative π₯-value in the fraction,
then to correctly solve for the angle π measured relative to the positive π₯-axis,
we need to add π radians to our result. Itβs by doing this that we avoid a
possible error in calculating π. This potential error can be traced
back to the tangent function.

But suffice to say whenever we
calculate the inverse tangent of a fraction with a negative π₯-value, that is,
whenever our vectorβs in the second or third quadrants, weβll need to add π radians
or 180 degrees, as the case may be, to properly solve for the angle relative to the
positive π₯-axis. π over four plus π is equal to
five over four times π. And substituting this value in for
π in our polar form of π, we see that in polar form π is equal to the square root
of two, five π over four. We find that listed as option (C)
among our answer choices. So, to complete our sentence, if π
equals negative π’ minus π£, then the polar form of π is square root of two, five
π over four.

Now letβs look at an example where
we convert from polar form to rectangular form.

If π equals seven, five π over
three, then vector π, in terms of the fundamental unit vectors, equals blank. (A) Seven-halves π’ plus seven root
three over two π£. (B) Negative seven root three over
two π’ plus seven-halves π£. (C) Seven-halves π’ minus seven
root three over two π£. (D) Seven root three over two π’
plus seven-halves π£.

Okay, so in this case, we have a
vector π given in polar form, and we want to express it in terms of the fundamental
unit vectors. Those vectors are π’ hat and π£
hat. To do this, weβll need to convert
vector π from polar to rectangular form. We can start off by recalling that
for a vector given in polar form, weβre given a radial distance from the origin of a
coordinate plane as well as the direction π in which this vector points relative to
the positive π₯-axis of that plane. So if our vector π, for example,
looked like this, then π would be the length of the vector and π this angle
shown.

Knowing this, we can solve for the
corresponding π₯- and π¦-components of this vector. The π₯-component is equal to π
times the cos of π, and the π¦-component is equal to π times the sin of π. When it comes to our given vector
π then, we can say that in terms of the fundamental unit vectors π’ hat and π£ hat,
π is equal to π₯ times π’ hat plus π¦ times π£ hat. And we see from our sketch that
this equals π times the cos of π π’ hat plus π times the sin of π π£ hat, where
π is equal to seven and π five π over three. We know this because of the vector
π given in our problem statement.

So now, if we substitute in for our
known values of π and π, we find that π₯ equals seven times the cos of five π
over three and π¦ equals seven times the sin of that angle. The cos of five π over three is
equal to exactly one-half, while the sin of five π over three equals negative root
three over two. In total, then, π₯ is equal to
seven-halves and π¦ equals negative seven root three over two. Therefore, our vector π is equal
to seven-halves π’ minus seven root three over two π£. And if we review our answer
options, we see that this is one of the choices. The vector π in terms of the
fundamental unit vectors equals seven-halves π’ minus seven root three over two
π£.

Letβs now look at an example
involving vector magnitude.

If π equals three π’ plus four π£,
π equals four π£, and π equals six π over 10, then the magnitude of π plus the
magnitude of π plus the magnitude of π equals blank. (A) 15, (B) six, (C) 11, (D)
10.

Okay, so here we have these three
vectors π, π, and π. And we want to find the sum of
their magnitudes. In general, for a vector given in
terms of its π₯- and π¦-components, its magnitude is equal to the square root of the
sum of the squares of those components. We can apply this rule to vectors
π and π to solve for their respective magnitudes. The magnitude of π is equal to the
square root of three squared plus four squared. Thatβs equal to the square root of
25 or simply five. And then we could apply the same
rule to vector π. But notice that since this just has
one component, its magnitude is equal to the magnitude of that component. Vector π has a magnitude of
four.

Then lastly, we want to calculate
the magnitude of vector π, which we see is not given in its Cartesian components,
but rather in polar form. When a vector is given in this
form, weβre being told the radial distance of the vector from some origin, in other
words, the vectorβs length, along with the direction that the vector points. So the nice thing is that for a
vector given in this form, we already know its magnitude. Itβs the radial distance π. Looking at vector π then, we can
simply read off its magnitude. It has a magnitude of six. And so when we add these three
magnitudes together, we get five plus four plus six, which is 15. Given these three vectors π, π,
and π, the sum of their magnitudes is 15.

Letβs now finish up our lesson by
summarizing a few key points. Weβve seen that a two-dimensional
vector can be expressed in rectangular or polar form. For a vector π, its rectangular
form involves π₯- and π¦-components, while its polar form involves a radial distance
π and an angle π. Graphically, these four variables
relate like this, where π is the magnitude of the vector π, π is its angle from
the positive π₯-axis, and π₯ and π¦ are its respective components. To convert then from the
rectangular to the polar form of a vector, we can use these relationships for π and
π. And to go the opposite way from
polar to rectangular, we recognize that π₯ equals π times the cos of π and π¦
equals π times the sin of π.