Lesson Explainer: Polar Form of a Vector Mathematics

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

When we think about vectors in the plane, we usually think of Cartesian coordinates as this is the most prevalent coordinate system, which leads to the rectangular form of a vector. In particular, the rectangular forms of a vector are used in linear motion where specifying the motion of an axis is simple and where the motion will take a linear path to a particular location.

Rectangular forms of a vector define a position as the linear distance from the origin in two or more mutually perpendicular directions. The standard unit vectors in a coordinate plane are ⃑𝑖=(1,0),⃑𝑗=(0,1).

The origin is the point where the axes intersect, and the vectors on the coordinate plane are specified by a linear combination of the unit vectors using the notation ⃑𝑣=π‘₯⃑𝑖+𝑦⃑𝑗=(π‘₯,𝑦).

For the rectangular form of a vector, any vector can be defined by a unique set of components given as a linear combination of the unit vectors π‘₯⃑𝑖+𝑦⃑𝑗; this form allows both positive and negative components, relative to the origin.

However, there are other ways of representing the a vectorβ€”we will explore one such way known as the polar form of a vector. This polar form defines a vector in space using a combination of radial and angular units, and a vector is specified by a straight-line distance from the origin and the angle from the positive π‘₯-axis.

These are known as radial and angular components of a vector, and the polar form of the vector as shown in the diagram above is ⃑𝑣=(π‘Ÿ,πœƒ).

The polar forms of a vector are often used in nonlinear motion, for example, if the motion involves a circular path. This makes the polar form useful in calculating the equations of motion for a lot of mechanical systems. It also has other real-world applications such as in radars using a plan position indicator, describing the characteristics of a microphone, and guiding industrial robots in various production applications and gravitational fields, just to name a few.

For the polar form of a vector ⃑𝑣, we mark a vector by its linear distance or length from the origin, denoted by π‘Ÿ, and its angle from the positive π‘₯-axis, denoted by πœƒ. In other words, the radial component π‘Ÿ is defined as the length of modulus of the vector, π‘Ÿβ‰‘β€–β€–βƒ‘π‘£β€–β€–.

Since we can form a right triangle using π‘Ÿ as the hypotenuse, we express the sides of the triangle in terms of sinπœƒ and cosπœƒ.

This also allows us to express the components of the rectangular form of a vector ⃑𝑣 in terms of the components of its polar form.

Definition: Converting from Polar Form to Rectangular Form of a Vector

The components of the polar form (π‘Ÿ,πœƒ) of a vector can be converted to rectangular form π‘₯⃑𝑖+𝑦⃑𝑗 as π‘₯=π‘Ÿπœƒ,𝑦=π‘Ÿπœƒ.cossin

So, if we are given the components of the polar form of a vector, the modulus or length π‘Ÿ and the angle πœƒ, we can determine the components of the rectangular form, π‘₯ and 𝑦, using these equations.

As an example, let’s convert a vector from polar form, in degrees, to rectangular form by using the modulus of a vector and an acute angle, represented geometrically.

Example 1: Representing Vectors Geometrically

Consider the vector ⃑𝑣 with modulus 3 and an angle of 45∘ measured counterclockwise from the positive π‘₯-axis. Using trigonometry, calculate the π‘₯- and 𝑦-components of the vector and, hence, write ⃑𝑣 in the form (π‘₯,𝑦). Round your answer to two decimal places.

Answer

In this example, we want to find the rectangular form of a vector using a graphical representation and a given length of the vector.

Recall that the polar form defines a vector according to the distance from the origin, denoted by π‘Ÿ, and the angular direction from the positive π‘₯-axis, denoted by πœƒ.

The radial component is equal to the length or modulus of vector ⃑𝑣, π‘Ÿ=‖‖⃑𝑣‖‖, and is given as π‘Ÿ=3, and we are told that the vector is at an angle of 45∘ above the positive π‘₯-axis.

Substituting the radial and angular components for vector ⃑𝐴, π‘Ÿ=3 and πœƒ=45∘, the components of the rectangular form are given by π‘₯=345=3Γ—1√2=2.121320343…cos∘ and 𝑦=345=3Γ—1√2=2.121320343….sin∘

Therefore, to two decimal places, we have ⃑𝑣=(2.12,2.12).

Now, let’s consider another example where we convert a vector from polar form, in degrees, to rectangular form using a graphical representation of the vector.

Example 2: Converting a Vector from Polar Form to Vector Form Using a Graphical Representation of the Vector

If ‖‖⃑𝐴‖‖=6cm, then ⃑𝐴=.

  1. ο€»6,6√3
  2. ο€»βˆš3,3
  3. ο€»3√3,3
  4. ο€»3,3√3

Answer

In this example, we want to find the rectangular form of a vector using a graphical representation and a given length of the vector.

Recall that the polar form defines a vector according to the distance from the origin, denoted by π‘Ÿ, and the angular direction from the positive π‘₯-axis, denoted by πœƒ.

The components of the rectangular form of a vector ⃑𝑣=(π‘₯,𝑦) can be expressed in terms of the components of the polar form ⃑𝑣=(π‘Ÿ,πœƒ) as π‘₯=π‘Ÿπœƒ,𝑦=π‘Ÿπœƒ.cossin

The radial component is equal to the length or modulus of vector ⃑𝑣: π‘Ÿ=‖‖⃑𝑣‖‖.

Substituting the radial and angular components for vector ⃑𝐴, π‘Ÿ=6cm and πœƒ=60∘, the components of the rectangular form are given by π‘₯=660=6Γ—12=3cos∘ and 𝑦=660=6Γ—βˆš32=3√3.sin∘

Therefore, we have ⃑𝐴=ο€»3,3√3.

This is option D.

Now, let’s convert a vector from polar form, in degrees, to rectangular form to solve a particular word problem involving a force.

Example 3: Converting a Vector from Polar Form to Rectangular Form in a Word Problem

If the force 𝐹=6newtons acts in the direction 60∘ east of north, then ⃑𝐹=.

  1. 3√3βƒ‘π‘–βˆ’3⃑𝑗
  2. 3√3⃑𝑖+3⃑𝑗
  3. 3βƒ‘π‘–βˆ’3√3⃑𝑗
  4. 3⃑𝑖+3√3⃑𝑗

Answer

In this example, we want to find the rectangular form of a force vector from a word problem.

Recall that the polar form defines a vector according to the distance from the origin, denoted by π‘Ÿ, and the angular direction from the positive π‘₯-axis, denoted by πœƒ.

The components of the rectangular form of a vector in terms of fundamental units ⃑𝑣=π‘₯⃑𝑖+𝑦⃑𝑗 can be expressed in terms of the components of the polar form ⃑𝑣=(π‘Ÿ,πœƒ) as π‘₯=π‘Ÿπœƒ,𝑦=π‘Ÿπœƒ.cossin

The radial component is equivalent to the length or modulus of vector ⃑𝑣: π‘Ÿ=‖‖⃑𝑣‖‖.

Since 𝐹=6 acts in the direction 60∘ east of north, the radial component of the polar form is π‘Ÿ=|𝐹|=6 and the angular component, the counterclockwise angle from the positive π‘₯-axis, is πœƒ=90βˆ’60=30∘∘∘.

This is because the given angle is 60∘ east of north, and angles are typically measured from east to north, in the counterclockwise direction from the positive π‘₯-axis, as shown in the diagram above.

Substituting the radial and angular components for vector ⃑𝐹, π‘Ÿ=6 and πœƒ=30∘, the components of the rectangular form are given by π‘₯=630=6Γ—βˆš32=3√3cos∘ and 𝑦=630=6Γ—12=3.sin∘

Therefore, we have ⃑𝐹=3√3⃑𝑖+3⃑𝑗.

This is option B.

The counterclockwise angle is taken to be positive, while the clockwise angle is negative. In previous examples and diagrams, the angular components have been acute, since the vectors in the examples were in the first quadrant. If a vector lies in another quadrant, as shown in the diagram below, its angle is not acute.

In fact, although we derived the equations for the components of the rectangular form, π‘₯=π‘Ÿπœƒ,𝑦=π‘Ÿπœƒ,cossin from acute angles, 0β‰€πœƒ<πœ‹2, we know that they remain true for any angle πœƒ.

Let’s consider an example where we convert a vector that lies in the second quadrant with a nonacute angle from polar form, in radians, to rectangular form, in order to find the difference of two vectors.

Example 4: Solving a Problem Involving Vectors in Rectangular Form and Polar Form

If ⃑𝐴=(8,πœ‹) and ⃑𝐡=βˆ’4⃑𝑖+5⃑𝑗, then 𝐴𝐡=.

  1. (4,5)
  2. (βˆ’4,5)
  3. (4,βˆ’5)
  4. (βˆ’4,βˆ’5)

Answer

In this example, we want to find the directional vector 𝐴𝐡 in rectangular form, where vector ⃑𝐴 is given in polar form and ⃑𝐡 is given in rectangular form. We begin by graphing vectors ⃑𝐴 and ⃑𝐡,

Recall that the polar form defines a vector according to the distance from the origin, denoted by π‘Ÿ, and the angular direction from the positive π‘₯-axis, denoted by πœƒ.

The components of the rectangular form of a vector in terms of fundamental units ⃑𝑣=π‘₯⃑𝑖+𝑦⃑𝑗 can be expressed in terms of the components of the polar form ⃑𝑣=(π‘Ÿ,πœƒ) as π‘₯=π‘Ÿπœƒ,𝑦=π‘Ÿπœƒ.cossin

Substituting the radial and angular components for vector ⃑𝐴, π‘Ÿ=8 and πœƒ=πœ‹, the components of the rectangular form are given by π‘₯=8(πœ‹)=8Γ—βˆ’1=βˆ’8cos and 𝑦=8(πœ‹)=8Γ—0=0.sin

Thus, ⃑𝐴 in rectangular form is ⃑𝐴=βˆ’8⃑𝑖+0⃑𝑗=βˆ’8⃑𝑖.

Now we can determine 𝐴𝐡: 𝐴𝐡=βƒ‘π΅βˆ’βƒ‘π΄=βˆ’4⃑𝑖+5βƒ‘π‘—βˆ’ο€Ίβˆ’8⃑𝑖=4⃑𝑖+5⃑𝑗.

Therefore, we have 𝐴𝐡=(4,5).

This is option A.

In the next example, let’s convert a vector in the fourth quadrant, 3πœ‹2<πœƒ<2πœ‹, from polar form, in radians, to rectangular form.

Example 5: Converting a Vector from Polar Form to Rectangular Form

If ⃑𝐴=ο€Ό7,5πœ‹3, then vector ⃑𝐴, in terms of the fundamental unit vectors, equals .

  1. 72⃑𝑖+7√32⃑𝑗
  2. 72βƒ‘π‘–βˆ’7√32⃑𝑗
  3. βˆ’7√32⃑𝑖+72⃑𝑗
  4. 7√32⃑𝑖+72⃑𝑗

Answer

In this example, we want to determine the rectangular form of a vector from the polar form ο€Ό7,5πœ‹3.

Recall that the polar form defines a vector according to the distance from the origin, denoted by π‘Ÿ, and the angular direction from the positive π‘₯-axis, denoted by πœƒ.

The components of the rectangular form of a vector in terms of fundamental units ⃑𝑣=π‘₯⃑𝑖+𝑦⃑𝑗 can be expressed in terms of the components of the polar form ⃑𝑣=(π‘Ÿ,πœƒ) as π‘₯=π‘Ÿπœƒ,𝑦=π‘Ÿπœƒ.cossin

We know that the distance from the origin is 7, and that the standard angle of the vector is 5πœ‹3. Since this angle is between 3πœ‹2 and 2πœ‹, we know that the vector is located in the fourth quadrant, as shown in the diagram below.

Substituting the radial and angular components for vector ⃑𝐴, π‘Ÿ=7 and πœƒ=5πœ‹3, the components of the rectangular form are given by π‘₯=7ο€Ό5πœ‹3=7Γ—12=72cos and 𝑦=7ο€Ό5πœ‹3=7Γ—ο€Ώβˆ’βˆš32=βˆ’7√32.sin

We note that π‘₯>0 and 𝑦<0, which indicates that this vector lies in the fourth quadrant, as expected.

Thus, vector ⃑𝐴 in terms of fundamental unit vectors equals 72βƒ‘π‘–βˆ’7√32⃑𝑗.

This is option B.

So far, we have seen examples of how to convert a vector from polar form to rectangular form using trigonometry. But what if we wanted to do the reverse, that is, convert a vector from rectangular form to polar form?

Let’s begin by recalling the equations expressing the components of the rectangular form, π‘₯ and 𝑦, in terms of the components of the polar form, π‘Ÿ and πœƒ: π‘₯=π‘Ÿπœƒ,𝑦=π‘Ÿπœƒ.cossin

Using these, we want to write the components of the polar form in terms of π‘₯ and 𝑦, no matter which quadrant the vector lies in.

If we take the square of each of these and add them up, by using the Pythagorean identity, we can eliminate πœƒ and show that these satisfy π‘₯+𝑦=π‘Ÿπœƒ+π‘Ÿπœƒ=π‘Ÿο€Ίπœƒ+πœƒο†=π‘Ÿ.cossincossin

Equivalently, by the Pythagorean theorem, the length or modulus of a vector in rectangular coordinates ⃑𝑣=π‘₯⃑𝑖+𝑦⃑𝑗, which we define to be π‘Ÿ, is given by π‘Ÿ=‖‖⃑𝑣‖‖=√π‘₯+𝑦.

Using this, we can solve a problem involving adding the lengths of three vectors.

Example 6: Solving a Problem Involving Vectors in Rectangular Form and Polar Form

If ⃑𝐴=3⃑𝑖+4⃑𝑗, ⃑𝐡=4⃑𝑗 and ⃑𝐢=ο€»6,πœ‹10, then ‖‖⃑𝐴‖‖+‖‖⃑𝐡‖‖+‖‖⃑𝐢‖‖=.

  1. 6
  2. 11
  3. 10
  4. 15

Answer

In this example, we want to find the sum of the lengths of three vectors, where ⃑𝐴 and ⃑𝐡 are in rectangular form and ⃑𝐢 is in polar form. We begin by graphing the three given vectors.

Recall that the polar form defines a vector according to the displacement from the origin, denoted by π‘Ÿ, and the angular direction from the positive π‘₯-axis, denoted by πœƒ.

The modulus of a vector ⃑𝑣 in rectangular form in terms of fundamental units ⃑𝑣=π‘₯⃑𝑖+𝑦⃑𝑗 is given by ‖‖⃑𝑣‖‖=√π‘₯+𝑦.

Therefore, for the given vectors, ⃑𝐴 and ⃑𝐡, we have ‖‖⃑𝐴‖‖=√3+4=5 and ‖‖⃑𝐡‖‖=√0+4=4.

We can also see the length of vector ⃑𝐡 from the graph.

Vector ⃑𝐢 is given in a polar form (π‘Ÿ,πœƒ), where π‘Ÿ is the radial component equal to the length or modulus of vector ‖‖⃑𝐢‖‖, and πœƒ is the angular component. The modulus of the vector is, therefore, ‖‖⃑𝐢‖‖=6.

Therefore, we have ‖‖⃑𝐴‖‖+‖‖⃑𝐡‖‖+‖‖⃑𝐢‖‖=5+4+6=15.

This is option D.

Also, when we divide the equation for 𝑦 by the equation for π‘₯, we can cancel the π‘Ÿ that appears to obtain 𝑦π‘₯=π‘Ÿπœƒπ‘Ÿπœƒ=πœƒπœƒ=πœƒ.sincossincostan

We note that this only holds true for π‘₯β‰ 0. We have a special case when π‘₯=0,𝑦=π‘Žβˆˆβ„ or (0,π‘Ž) in the rectangular form. For this, we have π‘Ÿπœƒ=0,cos which leads to πœƒ=πœ‹2, πœƒ=βˆ’πœ‹2, or π‘Ÿ=0. We can ignore the case π‘Ÿ=0 since this would imply 𝑦=0, which corresponds to the zero vector, which in polar form is denoted by (0,πœƒ), for any angle πœƒ.

Thus, πœƒ=πœ‹2 or πœƒ=βˆ’πœ‹2, which correspond to the 𝑦-axis. These angles place the vector in the 𝑦-axis, so the modulus of the vector is equal to the absolute value of the 𝑦-coordinate, π‘Ÿ=|π‘Ž|. A representation of the polar form of this vector is ο€»|π‘Ž|,πœ‹2, for π‘Ž>0, and ο€»|π‘Ž|,βˆ’πœ‹2, for π‘Ž<0.

So, if π‘₯β‰ 0, we have the following equation to determine angle πœƒ: tanπœƒ=𝑦π‘₯.

The range for the inverse tangent function is οŸβˆ’πœ‹2,πœ‹2 when the domain of the tangent function is restricted to the same interval, known as the principal branch. This is to ensure that the tangent function is one to one, so that the inverse tangent function evaluates to a single value, known as the principle value.

Thus, as long as πœƒβˆˆοŸβˆ’πœ‹2,πœ‹2, we can take the inverse tangent of both sides of the equation to obtain πœƒ=𝑦π‘₯.tan

The angular coordinates πœƒβˆˆοŸβˆ’πœ‹2,πœ‹2 correspond to the first and fourth quadrants, or the quadrants where π‘₯>0.

In the next example, let’s convert a vector from rectangular form to polar form, in radians.

Example 7: Converting a Vector from Vector Form to Polar Form

If 𝑃𝑀=ο€»5,5√3, then the polar form of vector 𝑃𝑀 is .

  1. ο€»10,πœ‹6
  2. ο€»10,πœ‹2
  3. (10,πœ‹)
  4. ο€»10,πœ‹3

Answer

In this example, we want to determine the polar form (π‘Ÿ,πœƒ), in radians, for the given vector in rectangular coordinates 𝑃𝑀=ο€»5,5√3. We begin by graphing this vector on the coordinate plane.

Recall that the polar form defines a vector according to the distance from the origin, denoted by π‘Ÿ, and the angular direction from the positive π‘₯-axis, denoted by πœƒ. We use the convention where angle πœƒ is the positive counterclockwise angle.

The components of the rectangular form of a vector in terms of fundamental units ⃑𝑣=π‘₯⃑𝑖+𝑦⃑𝑗 can be expressed in terms of the components of the polar form ⃑𝑣=(π‘Ÿ,πœƒ) as π‘₯=π‘Ÿπœƒ,𝑦=π‘Ÿπœƒ.cossin

Now, let’s find the polar form of the vector 𝑃𝑀 by using the graphical representation directly with the definition. The radial coordinate is the distance from the origin to the vector 𝑃𝑀, which we can find from the Pythagorean theorem on the right triangle, as shown the diagram. Let’s first determine the hypotenuse of this triangle, π‘Ÿ=ο„ž5+ο€»5√3=√100=10.

Since the vector 𝑃𝑀 is located in the first quadrant, the angular component πœƒ for the polar form will be the positive counterclockwise angle from the positive π‘₯-axis, and we can form a right triangle with angle πœƒ and sides of lengths 5 and 5√3, as shown in the diagram. Since πœƒ is an acute angle, using right triangle trigonometry we can write this in terms of the inverse tangent as πœƒ=ο€Ώ5√35=ο€»βˆš3=πœ‹3.tantan

We could have also arrived at this answer by using the fact that we can convert a vector from rectangular form, ⃑𝑣=(π‘₯,𝑦), located in the first quadrant into polar form, ⃑𝑣=(π‘Ÿ,πœƒ), by using π‘Ÿ=√π‘₯+𝑦,πœƒ=𝑦π‘₯.tan

This gives the same radial and angular components of the polar form after substituting π‘₯=5 and 𝑦=5√3.

Therefore, the polar form of the vector 𝑃𝑀 is ο€»10,πœ‹3.

This is option D.

Now, let’s consider an example where we identify the graphical polar representation of a given vector.

Example 8: Identifying the Graphical Polar Representation of a Vector

Which of the following is the polar representation of ⃑𝐴=ο€»2√3,2?

Answer

In this example, we want to determine the graphical polar representation, in degrees, for a particular vector in rectangular form, ⃑𝐴=ο€»2√3,2.

Recall that the polar form defines a vector according to the distance from the origin, denoted by π‘Ÿ, and the angular direction from the positive π‘₯-axis, denoted by πœƒ. We use the convention where angle πœƒ is the positive counterclockwise angle. We also restrict the angles to 0β‰€πœƒ<2πœ‹ in order to write them in a standardized form.

The components of the rectangular form of a vector ⃑𝑣=(π‘₯,𝑦) can be expressed in terms of the components of polar form ⃑𝑣=(π‘Ÿ,πœƒ) as π‘₯=π‘Ÿπœƒ,𝑦=π‘Ÿπœƒ.cossin

Since the vector lies in the first quadrant, we can determine the polar form using π‘Ÿ=√π‘₯+𝑦,πœƒ=𝑦π‘₯.tan

For the given vector ⃑𝐴, we can determine the radial component as π‘Ÿ=ο„žο€»2√3+2=√12+4=√16=4 and the angular component as πœƒ=22√3=1√3=30.tantan∘

Thus, the graphical polar representation is the one where the line segment has a length of 4 and is 30∘ from the positive π‘₯-axis in the counterclockwise direction.

This is option B.

In the previous example, we converted the rectangular form of a vector in the first quadrant into polar form. As seen in this example, we can compute the angular coordinate of a vector by using πœƒ=𝑦π‘₯tan in the first and fourth quadrants. However, this is no longer the case if the vector lies in the second or third quadrant. For the second and third quadrants, a value of πœ‹, in radians, or 180∘, in degrees, must be added to angle πœƒ to adjust the angular coordinate so that the vector lies in the correct quadrant. This does not affect the tangent function itself since we have the identity tantanπœƒ=(πœƒ+180),∘ or more generally tantanπœƒ=(πœƒ+180π‘˜),π‘˜βˆˆβ„€.∘

To see this, consider a vector in rectangular coordinates that lies in the second quadrant, (π‘₯,𝑦)=(βˆ’π‘Ž,𝑏) with π‘Ž>0 and 𝑏>0.

The angular component πœƒ for this vector in polar form will be the positive counterclockwise angle from the positive π‘₯-axis, and the angle 𝛼 is measured from the negative π‘₯-axis, as shown in the diagram. We can form a right triangle with angle 𝛼 and sides of lengths π‘Ž and 𝑏. Since 𝛼 is an acute angle, we can write this in terms of the sides, using the inverse tangent, as 𝛼=ο€½π‘π‘Žο‰.tan

We also have 𝛼+πœƒ=180∘. Substituting angle 𝛼, we can rearrange this to find πœƒ=βˆ’π›Ό+180=βˆ’ο€½π‘π‘Žο‰+180=ο€½π‘βˆ’π‘Žο‰+180,∘∘∘tantan where for the last equality we used the fact that the tangent function, and hence the inverse tangent function, is an odd function.

Since we have (π‘₯,𝑦)=(βˆ’π‘Ž,𝑏), this is equivalent to πœƒ=𝑦π‘₯+180.tan∘

In the next example, we will find the angular coordinate of a particular vector, in degrees, that lies in the second quadrant.

Example 9: Finding the Direction Angle of a Given Vector

Consider the vector (βˆ’2,3). Calculate the direction of the vector, giving your solution as an angle to the nearest degree measured counterclockwise from the positive π‘₯-axis.

Answer

In this example, we want to determine the direction of a vector, in degrees, for a particular vector in rectangular form (βˆ’2,3). We want to find the angle measured counterclockwise from the positive π‘₯-axis. We begin by graphing this vector on the coordinate plane.

Since the vector (βˆ’2,3) is located in the second quadrant, the direction of the vector is the angle πœƒ measured counterclockwise from the positive π‘₯-axis. We consider the angle 𝛼, which is measured in the same direction as πœƒ, as shown in the diagram. We can form a right triangle with angle 𝛼 and sides of lengths 2 and 3. Since 𝛼 is an acute angle, we can use right-triangle trigonometry to write the sides in terms of the inverse tangent as 𝛼=ο€Ό32.tan

Since the angles on a straight line add up to 180∘, we also have 𝛼+πœƒ=180∘. Substituting angle 𝛼, we can rearrange this to find πœƒ=βˆ’π›Ό+180=βˆ’ο€Ό32+180=βˆ’56.3099324…+180=123.6900675….∘∘∘∘∘tan

We could have also arrived at this answer by using the fact that a vector in rectangular form, (π‘₯,𝑦), located in the second quadrant, has an angle πœƒ=𝑦π‘₯+πœ‹,tan where πœƒ is measured counterclockwise from the positive π‘₯-axis. This gives the same angle after substituting π‘₯=βˆ’2 and 𝑦=3. This angle is equivalent to the angular component of the polar form of (βˆ’2,3).

Therefore, to the nearest degree, the direction of the vector measured counterclockwise from the positive π‘₯-axis is 124∘.

For the third quadrant, we can show in a similar way that we also have to add 180∘ to tanοŠ±οŠ§ο€»π‘¦π‘₯ to get the angular coordinate πœƒ in the correct quadrant.

Polar forms of a vector are not unique, unless we restrict the angle to a particular range, and there are many ways of representing the same vector. As an example, let’s determine the polar form of the vector ⃑𝑖+⃑𝑗 in rectangular coordinates.

The radial component π‘Ÿ is the distance from the origin, which we can determine using the Pythagorean theorem on the right triangle with sides of length 1 and angle πœƒ. In particular, π‘Ÿ=√1+1=√2.

There are many ways to express the angular component πœƒ. One is the positive counterclockwise angle from the positive π‘₯-axis, which from the right triangle gives us the tangent in terms of the ratio of the opposite and adjacent sides: tanπœƒ=11=1.

Since the vector lies in the first quadrant and πœƒ is an acute angle in the diagram, we can find the angle directly from the inverse tangent, πœƒ=(1)=45.tan∘

Thus, a polar form of the vector ⃑𝑖+⃑𝑗 is ο€»βˆš2,45ο‡βˆ˜.

Another polar form can be found if we use the negative clockwise angle from the positive π‘₯-axis, which would give an equivalent polar form as ο€»βˆš2,45βˆ’360=ο€»βˆš2,βˆ’315ο‡βˆ˜βˆ˜βˆ˜. In fact, if we make a full revolution from this vector, in either the clockwise or counterclockwise direction, we return back to the same vector. So, another representation would be ο€»βˆš2,45+360=ο€»βˆš2,405ο‡βˆ˜βˆ˜βˆ˜.

This shows a key difference when using polar forms, compared to the rectangular form, as it allows an infinite number of sets to describe any given vector. This is because we can add any integer multiple of a full revolution (360∘ or 2πœ‹) to the angular coordinate πœƒ to get an equivalent point in polar coordinates. This follows because the trigonometric functions, which are used to define polar forms, are themselves periodic.

This equivalence condition can be summarized as follows.

Definition: Periodicity Condition for Polar Forms

If (π‘Ÿ,πœƒ) describes the polar form of a vector, then we can express the equivalent polar forms as (π‘Ÿ,πœƒ)=(π‘Ÿ,πœƒ+2πœ‹π‘›)(),=(π‘Ÿ,πœƒ+360𝑛)(),radiansdegrees∘ for any π‘›βˆˆβ„€.

Thus, in order to express the polar form in a standardized way with 0β‰€πœƒ<2πœ‹ in radians or 0β‰€πœƒ<360∘∘ in degrees, we may have to adjust the value of the angular coordinate πœƒ. In particular, for the fourth quadrant, we may have to add a full revolution (2πœ‹ or 360∘) to get the equivalent angle in the standard range, as the inverse tangent gives the negative clockwise angle rather than the positive counterclockwise angle from the positive π‘₯-axis.

The conventions we use take the counterclockwise angle as positive and the clockwise angle as negative. The angle is measured in counterclockwise direction from the positive π‘₯-axis.

We can summarize what we have covered so far in a definition, which can be used to convert the rectangular form of a vector to polar form and vice versa.

Definition: Converting from Rectangular Form to Polar Form of a Vector

One representation of the polar form of a vector ⃑𝑣=(π‘Ÿ,πœƒ) with π‘Ÿβ‰₯0 can be expressed in terms of rectangular form ⃑𝑣=π‘₯⃑𝑖+𝑦⃑𝑗 as π‘Ÿ=√π‘₯+𝑦,πœƒ=⎧βŽͺβŽͺβŽͺβŽͺ⎨βŽͺβŽͺβŽͺβŽͺβŽ©ο€»π‘¦π‘₯π‘₯>0,𝑦>0;𝑦π‘₯+(3602πœ‹)π‘₯>0,𝑦<0;𝑦π‘₯+(180πœ‹)π‘₯<0;90πœ‹2π‘₯=0,𝑦>0;2703πœ‹2π‘₯=0,𝑦<0;∘∘∘∘tanfortanorfortanorfororfororfor for 0β‰€πœƒ<2πœ‹ in radians or 0β‰€πœƒ<360∘∘ in degrees.

This information can be communicated efficiently using the following diagram.

As an example, let’s consider the vectors ⃑𝑣=(4,3),⃑𝑣=(βˆ’4,3),⃑𝑣=(βˆ’4,βˆ’3), and ⃑𝑣=(4,βˆ’3)οŠͺ in rectangular form, where each vector is located in a different quadrant, as shown in the graph. We want to determine the polar forms of these vectors in a standardized way, accurate to two decimal places and in degrees with 0β‰€πœƒ<360∘∘.

The radial component π‘Ÿ for these vectors in polar form will be the same since π‘Ÿ=√4+3=√(βˆ’4)+3=√(βˆ’4)+(βˆ’3)=√4+(βˆ’3)=5.

The difference will be with the angular component πœƒ, since this will determine the direction and, hence, which quadrant the vector in polar form will lie in.

The vector ⃑𝑣=4⃑𝑖+3βƒ‘π‘—οŠ§ is in the first quadrant, and we can determine the angular component from the general formula as πœƒ=ο€Ό34=ο€Ό34=36.86989765….tantan∘

As expected, this angle is acute, as 0β‰€πœƒ<90∘∘, which places the vector in the first quadrant.

The vector ⃑𝑣=βˆ’4⃑𝑖+3βƒ‘π‘—οŠ¨ is in the second quadrant and the angular component is πœƒ=ο€Ό3βˆ’4+180=βˆ’ο€Ό34+180=βˆ’36.86989765…+180=143.1301024….tantan∘∘∘∘∘

As expected, we have 90<πœƒ<180∘∘, which places the vector in the second quadrant.

The vector ⃑𝑣=βˆ’4βƒ‘π‘–βˆ’3βƒ‘π‘—οŠ© is in the third quadrant and the angular component is πœƒ=ο€Όβˆ’3βˆ’4+180=ο€Ό34+180=36.86989765…+180=216.8698976….tantan∘∘∘∘∘

As expected, we have 180<πœƒ<270∘∘, which places the vector in the third quadrant.

Finally, the vector ⃑𝑣=4βƒ‘π‘–βˆ’3⃑𝑗οŠͺ is in the fourth quadrant and the angular component is πœƒ=ο€Όβˆ’34+360=βˆ’ο€Ό34+360=βˆ’36.86989765…+360=323.1301024….tantan∘∘∘∘∘

As expected, we have 270<πœƒ<360∘∘, which places the vector in the fourth quadrant.

Thus, a representation of the polar form of the vectors, accurate to two decimal places, is given by ⃑𝑣=(5,36.87),⃑𝑣=(5,143.13),⃑𝑣=(5,216.87),⃑𝑣=(5,323.13).∘∘∘οŠͺ∘

Finally, let’s consider an example where we convert a vector in the third quadrant from rectangular form to polar form, in radians.

Example 10: Converting a Vector from Rectangular Form to Polar Form

If ⃑𝐴=βˆ’βƒ‘π‘–βˆ’βƒ‘π‘—, then the polar form of ⃑𝐴 is .

  1. ο€Όβˆš2,5πœ‹4
  2. ο€»βˆš2,πœ‹4
  3. ο€Όβˆš2,7πœ‹4
  4. ο€Όβˆš2,3πœ‹4

Answer

In this example, we want to determine the polar form (π‘Ÿ,πœƒ), in radians, for a particular vector in rectangular form ⃑𝐴=βˆ’βƒ‘π‘–βˆ’βƒ‘π‘—.

Recall that the polar form defines a vector according to the distance from the origin, denoted by π‘Ÿ, and the angular direction from the positive π‘₯-axis, denoted by πœƒ. We use the convention where the angle πœƒ is the positive counterclockwise angle. We also restrict the angles to 0β‰€πœƒ<2πœ‹ in order to write them in a standardized form.

The components of the rectangular form of a vector in terms of fundamental units ⃑𝑣=π‘₯⃑𝑖+𝑦⃑𝑗 can be expressed in terms of the components of polar form ⃑𝑣=(π‘Ÿ,πœƒ) as π‘₯=π‘Ÿπœƒ,𝑦=π‘Ÿπœƒ.cossin

Now, let’s find the polar form of the vector ⃑𝐴=βˆ’βƒ‘π‘–βˆ’βƒ‘π‘— by using the graphical representation directly with the definition. The radial component is the distance from the origin to vector ⃑𝐴, and we can find this by applying the Pythagorean theorem on the right triangle, as shown the diagram below.

We can find the hypotenuse of this triangle from π‘Ÿ=√1+1=√2.

Since vector ⃑𝐴 is located in the third quadrant, the angular component πœƒ for the polar form will be the positive counterclockwise angle from the positive π‘₯-axis, and angle 𝛼 is measured from the negative π‘₯-axis. We can form a right triangle with angle 𝛼 and sides of lengths 2 and 5, as shown in the diagram above. Since 𝛼 is an acute angle, we can write this in terms of the sides, using the inverse tangent, as 𝛼=ο€Ό11.tan

We also have πœƒ=𝛼+πœ‹. Substituting angle 𝛼, we can rearrange this to find πœƒ=𝛼+πœ‹=ο€Ό11+πœ‹=πœ‹4+πœ‹=5πœ‹4.tan

We could have also arrived at this answer by using the fact that we can convert a vector from rectangular form ⃑𝑣=π‘₯⃑𝑖+𝑦⃑𝑗 located in the third quadrant into polar form ⃑𝑣=(π‘Ÿ,πœƒ) by using π‘Ÿ=√π‘₯+𝑦,πœƒ=𝑦π‘₯+πœ‹.tan

This gives the same radial and angular components of the polar form after substituting π‘₯=βˆ’1 and 𝑦=βˆ’1.

Therefore, the polar form of ⃑𝐴 is ο€Όβˆš2,5πœ‹4.

This is option A.

Key Points

  • A polar form of a vector is denoted by (π‘Ÿ,πœƒ), where π‘Ÿ represents the distance from the origin and πœƒ represents the angle measured from the π‘₯-axis.
  • The components of the rectangular form of a vector ⃑𝑣=π‘₯⃑𝑖+𝑦⃑𝑗 can be obtained from the components of the polar form ⃑𝑣=(π‘Ÿ,πœƒ) from π‘₯=π‘Ÿπœƒ,𝑦=π‘Ÿπœƒ.cossin The polar form of a vector ⃑𝑣=(π‘Ÿ,πœƒ) is not unique, and there are equivalent ways of describing the same vector, since the trigonometric functions used to define them are periodic.
  • The conventions we use take the counterclockwise angle as positive and the clockwise angle as negative, and we use the standardized form with 0β‰€πœƒ<2πœ‹ in radians or 0β‰€πœƒ<360∘∘ in degrees.
  • We can find an equivalent polar form of a vector by adding or subtracting any integer multiple of a full revolution (360∘ or 2πœ‹): (π‘Ÿ,πœƒ)=(π‘Ÿ,πœƒ+2πœ‹π‘›).
  • The components of the polar form of a vector ⃑𝑣=(π‘Ÿ,πœƒ) can be expressed in terms of the components of the rectangular form ⃑𝑣=π‘₯⃑𝑖+𝑦⃑𝑗 as π‘Ÿ=√π‘₯+𝑦, and the value of πœƒ will depend on the quadrant in which vector ⃑𝑣 lies, for 0β‰€πœƒ<2πœ‹ in radians or 0β‰€πœƒ<360∘∘ in degrees.

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