# Lesson Video: Polar Form of a Vector Mathematics

In this video, we will learn how to convert between rectangular and polar forms of a vector.

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### Video Transcript

In this video, weβre 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, weβll also learn how to convert from polar form to rectangular and back.

To start off, say that we have some vector, weβll call it π, thatβs 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. Weβre describing the vector using these rectangular variables π₯ and π¦. Going back to our sketch, we can start to see, though, that thereβs another way of describing this same vector. We could describe it in terms of its overall radial distance from the origin, weβll call that distance π, as well as the angle this vector forms with the positive π₯-axis, and weβll 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, weβre able to represent the vector π in a different way. We can say that itβs 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 itβs 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, weβre able to go back and forth between rectangular and polar forms of a vector. Letβs 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 weβre given the components of our vector, letβs start out by plotting it on this π₯π¦-plane. Weβll say that each of these tick marks represents a distance of one unit. And weβre 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 weβll 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. Weβll call it π. At this point, letβs 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, weβre given a vector not in polar form but in rectangular form. In using this π₯ and π¦ information then to solve for π, weβre 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 canβt be the angle π. Whatβs happening here is because the π₯-value in this fraction is negative β in other words, the vector weβre working with occupies either the second or the third quadrant β to correctly compute our angle, weβll 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 letβs 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 letβs 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 weβve 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 weβll substitute this result in for π in our polar form of π. But as we look again at our answer options, notice that this doesnβt narrow down the list. All four choices had the same π-value of the square root of two. So letβs 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.

Letβs 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 weβve 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.

Hereβs whatβs 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 canβt 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. Itβs 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 vectorβs in the second or third quadrants, weβll 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 letβs 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, weβll need to convert vector π from polar to rectangular form. We can start off by recalling that for a vector given in polar form, weβre 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£.

Letβs 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. Thatβs 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, weβre being told the radial distance of the vector from some origin, in other words, the vectorβs 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. Itβs 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.

Letβs now finish up our lesson by summarizing a few key points. Weβve 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.