Lesson Explainer: Vector Applicationsβ€Ž Mathematics

In this explainer, we will learn how to use vector operations to solve various real-world problems.

In the real world, there are some quantities that are scalars, such as mass, length, or age, represented by a single number that we call its magnitude, and others that are vectors, such as displacement, velocity, or force, represented by both a magnitude and particular direction. Vectors are used extensively in mathematics, engineering, physics, and computing. For example, in classical mechanics, if we consider the forces or momentum acting on a body, we have to take both the magnitude and direction into account to know where the body will move at any time; if a body moves in free fall, we have to take into account not only the direction of gravity acting on the body, but also the direction and magnitude of other forces such as air resistance or wind.

Definition: Vector

A vector is a quantity that has both a magnitude and a direction. This can be represented geometrically as a directed line segment whose length is the magnitude of the vector and whose arrow indicates its direction.

This direction can be given in terms of the north, east, west, and south directions, or a combination of them, or more generally as an angle or bearing. For example, an object could be moving at a speed of 25 mph in the eastward direction.

As an example, let’s consider a situation involving velocity and identify the magnitude and direction from the given information.

Example 1: The Magnitude and Direction of Velocity

Velocity is a vector quantity that combines speed and direction. For example, the velocity at a section of the Mississippi river near New Orleans is 3 miles per hour east.

  1. What is the magnitude of this velocity?
  2. What is the direction of this velocity?

Answer

Part 1

Let’s first identify the magnitude of the velocity.

Mississippi river

The magnitude of a velocity is its speed, which in this case, for a section of the Mississippi river near New Orleans, is 3 miles per hour.

Part 2

As stated in the question, a velocity is a vector quantity that combines speed and direction. We are told that the velocity of the water is 3 mph in an easterly direction. Therefore, the direction of the velocity is east.

Another way to represent a vector, instead of using a magnitude and direction, is in rectangular form. In general, vectors can be represented in any number of dimensions, also known as the degrees of freedom, which is the number of components needed to fully determine a vector. In this explainer, we will consider vectors in one and two dimensions, respectively, but the same principles hold for three or higher dimensions.

In one dimension, vectors are represented by a single number, 𝑣, which can be either positive or negative depending on the direction of the vector. In two dimensions, a vector, ⃑𝑣, is represented by two numbers in rectangular form. The rectangular form of a vector defines 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) and ⃑𝑗=(0,1) in two dimensions, as shown in the diagrams.

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).

The direction is encoded into this form of a vector: west to east means we move in the ⃑𝑖 direction, while east to west means we move in the βˆ’βƒ‘π‘– direction. Similarly, south to north means we move in the ⃑𝑗 direction and north to south means we move in the βˆ’βƒ‘π‘— direction. Moving in any other direction, say, southeast, will be a combination of these directions.

For example, if we want to determine the displacement vector of a person who moves 2 m to the west and then 4 m to the south, we can do this by using these mutually perpendicular directions; 2 m to the west represents a displacement of βˆ’2⃑𝑖 and 4 m to the south represents a displacement of βˆ’4⃑𝑗, since we take the north and east directions as positive. Putting this together, the displacement vector would be ⃑𝑠=βˆ’2βƒ‘π‘–βˆ’4⃑𝑗.

Adding or subtracting vectors in one dimension is simple, as we just add or subtract the components or numbers represented by each vector. Recall that adding or subtracting two vectors, ⃑𝑒 and ⃑𝑣, in two dimensions can be represented in a vector diagram as follows.

Adding or subtracting two vectors is equivalent to adding or subtracting their components. For example, if ⃑𝑒=(π‘₯,𝑦) and ⃑𝑣=(π‘₯,𝑦), then ⃑𝑒+⃑𝑣=(π‘₯,𝑦)+(π‘₯,𝑦)=(π‘₯+π‘₯,𝑦+𝑦)=(π‘₯+π‘₯)⃑𝑖+(𝑦+𝑦)⃑𝑗,βƒ‘π‘’βˆ’βƒ‘π‘£=(π‘₯,𝑦)βˆ’(π‘₯,𝑦)=(π‘₯βˆ’π‘₯,π‘¦βˆ’π‘¦)=(π‘₯βˆ’π‘₯)⃑𝑖+(π‘¦βˆ’π‘¦)⃑𝑗.

We can also have a vector relative to another, for example, when we consider the motion of two bodies.

Definition: Relative Velocity Vector

Consider the motion of two bodies, body 𝐴 and body 𝐡.

If these bodies are moving with velocities βƒ‘π‘£οŒ  and βƒ‘π‘£οŒ‘, respectively, the relative velocity of 𝐴 with respect to 𝐡, denoted by βƒ‘π‘£οŒ οŒ‘, is the difference between these two velocities: ⃑𝑣=βƒ‘π‘£βˆ’βƒ‘π‘£.

Similarly, the relative velocity of 𝐡 with respect to 𝐴 can be denoted as βƒ‘π‘£οŒ‘οŒ  and is given by ⃑𝑣=βƒ‘π‘£βˆ’βƒ‘π‘£.

Note that we have ⃑𝑣=βˆ’βƒ‘π‘£οŒ οŒ‘οŒ‘οŒ , which is an intuitive result since the negative of a vector represents the opposite direction.

In simple situations, we can assume that different bodies move along a straight line in two directions, left or right, and that they move either in the same direction or in opposite directions. If they are moving in the same direction, the velocities will have the same sign, while in opposite directions they will have different signs. The speed is just the magnitude of the velocity, that is, ‖‖⃑𝑣‖‖.

Let’s consider an example where we have to determine the relative velocities of two motorcycles moving in the same direction, along a straight line. The velocities of each motorcycle, βƒ‘π‘£οŒ  and βƒ‘π‘£οŒ‘, will be given as a single number, representing a vector in one dimension.

Example 2: Finding the Relative Velocity between Two Bodies Moving in the Same Direction

Two motorcycles, 𝐴 and 𝐡, are moving in the same direction. If the velocity of 𝐴 is 30 km/h and the velocity of 𝐡 is 15 km/h, find the relative velocity of 𝐴 with respect to 𝐡.

Answer

In this example, we want to find the relative velocity of motorcycle 𝐴 and that of motorcycle 𝐡, which are moving in the same direction.

Let the velocity of motorcycle 𝐴 be ⃑𝑣=30/kmh and the velocity of motorcycle 𝐡 be ⃑𝑣=15/kmh, since the motorcycles move in the same direction.

The relative velocity of motorcycle 𝐴 with respect to 𝐡 is ⃑𝑣=βƒ‘π‘£βˆ’βƒ‘π‘£=30βˆ’15=15.

Thus, the relative velocity is 15 km/h.

Now, let’s consider an example where we have to determine the actual speed of a truck, given information about the speed of a police car and relative velocities.

Example 3: Finding the Velocity of a Moving Body Using Its Relative Velocity

A police car was moving along a horizontal highway at 47 km/h. It used a radar to measure the speed of a truck moving in the same direction. Given that the reading on the radar was 50 km/h, determine the actual speed of the truck.

Answer

In this example, we want to determine the actual speed of the truck by using the information given about the velocity of the police car and the relative velocity of the truck to the police car, as measured by the radar.

Let ⃑𝑣=47/kmh be the velocity of the police car and ⃑𝑣=50/kmh be the velocity of the truck as measured by the radar, which is the velocity of the truck relative to the velocity of the police car. Both velocities have the same sign since the police car and truck move in the same direction. We can find the actual velocity of the truck by rearranging ⃑𝑣=βƒ‘π‘£βˆ’βƒ‘π‘£οοŒοοŒ as ⃑𝑣=⃑𝑣+⃑𝑣=50+47=97.

Since this is positive, the actual speed of the truck is 97 km/h.

We can also have a relative velocity vector for the rectangular form in two dimensions. For example, suppose two bodies, 𝐴 and 𝐡, are in motion with velocities βƒ‘π‘£οŒ  and βƒ‘π‘£οŒ‘, respectively, expressed in rectangular form or in terms of two mutually perpendicular unit vectors. This means we take the difference of each component in vector βƒ‘π‘£οŒ  with the corresponding component in vector βƒ‘π‘£οŒ‘.

We do not have to write the unit vectors in terms of ⃑𝑖 and ⃑𝑗, which are the standard unit vectors. We could also have them in terms of other vectors, say, ⃑𝑒 and ⃑𝑓, as long as they are perpendicular to each other and have unit length.

For example, if a body 𝐴 moves at a velocity ⃑𝑣=π‘₯⃑𝑒+π‘¦βƒ‘π‘“οŒ οŠ§οŠ§ and a body 𝐡 moves at a velocity ⃑𝑣=π‘₯⃑𝑒+π‘¦βƒ‘π‘“οŒ‘οŠ¨οŠ¨, the velocity of 𝐴 relative to 𝐡 can be denoted as βƒ‘π‘£οŒ οŒ‘ and is given by ⃑𝑣=βƒ‘π‘£βˆ’βƒ‘π‘£=ο€»π‘₯⃑𝑒+π‘¦βƒ‘π‘“ο‡βˆ’ο€»π‘₯⃑𝑒+𝑦⃑𝑓=(π‘₯βˆ’π‘₯)⃑𝑒+(π‘¦βˆ’π‘¦)⃑𝑓.

Now, let’s look at an example where we have to determine the relative velocity in terms of a given unit vector of two cars moving in opposite directions.

Example 4: Finding the Relative Velocity between Two Bodies Moving in Opposite Directions in Vector Form

Two cars 𝐴 and 𝐡 are moving in opposite directions on the same road at 62 km/h and 31 km/h respectively. Given that ⃑𝑒 is a unit vector in the direction of movement of car 𝐴, determine the velocity of car 𝐴 relative to car 𝐡.

Answer

In this example, we want to determine the velocity of car 𝐴 relative to car 𝐡 in terms of a unit vector ⃑𝑒, the direction of movement of car 𝐴, and the cars move in opposite directions.

Since ⃑𝑒 is a given unit vector in the direction of movement of car 𝐴 and we are told that this car moves at a speed of 62 km/h, the velocity vector of car 𝐴 will be ⃑𝑣=62⃑𝑒.

Similarly, since car 𝐡 moves in the opposite direction to the movement of car 𝐴, at a speed of 31 km/h, its velocity vector will be negative: ⃑𝑣=βˆ’31⃑𝑒.

The velocity of car 𝐴 relative to car 𝐡 is ⃑𝑣=βƒ‘π‘£βˆ’βƒ‘π‘£=ο€Ή62βƒ‘π‘’ο…βˆ’ο€Ήβˆ’31⃑𝑒=93⃑𝑒.

Thus, the relative velocity is ο€Ή93⃑𝑒/.kmh

When considering two or more forces acting on a body, the resultant force, ⃑𝑅, is the vector addition of the various forces.

Definition: Resultant Force

If there are 𝑛 forces, ⃑𝐹,⃑𝐹,…,βƒ‘πΉοŠ§οŠ¨οŠ acting on a single body, then the resultant force ⃑𝑅 is given by ⃑𝑅=⃑𝐹+⃑𝐹+β‹―+⃑𝐹.

A system with such forces in equilibrium occurs when the resultant force is equal to zero, ⃑𝑅=0.

For example, if there are two forces, βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨, acting on a body then the resultant force is given by ⃑𝑅=⃑𝐹+⃑𝐹.

Now, let’s consider an example where we have to determine unknown values appearing in the vectors of forces, given in terms of the standard unit vectors ⃑𝑖 and ⃑𝑗, acting on a particle in equilibrium.

Example 5: Finding the Unknown Components of Two Forces given Equilibrium

A particle is in equilibrium under two forces ⃑𝐹=3⃑𝑖+π‘₯βƒ‘π‘—οŠ§N and ⃑𝐹=𝑦⃑𝑖+2βƒ‘π‘—οŠ¨N. Find the values of π‘₯ and 𝑦.

Answer

In this example, we want to determine the values of π‘₯ and 𝑦, the forces in the directions of ⃑𝑗 and ⃑𝑖 for βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨, respectively, for a particle which is in equilibrium under the action of two forces.

Recall that the resultant force acting on any body is the sum of all the forces. Thus, the resultant force of two forces acting on the particle is ⃑𝑅=⃑𝐹+⃑𝐹=ο€Ί3⃑𝑖+π‘₯⃑𝑗+𝑦⃑𝑖+2⃑𝑗=(3+𝑦)⃑𝑖+(π‘₯+2)⃑𝑗.

Forces are in equilibrium if the resultant force, ⃑𝑅, is equal to zero. If a vector is equal to zero, its individual components must be equal to zero; hence, 3+𝑦=0𝑦=βˆ’3 and π‘₯+2=0π‘₯=βˆ’2.

Thus, the values are π‘₯=βˆ’2,𝑦=βˆ’3.

In the next example, we will again determine unknown values appearing in the vectors of forces, given in terms of the standard unit vectors ⃑𝑖 and ⃑𝑗, this time with three forces acting on a particle with a given resultant force.

Example 6: Finding the Unknown Components of Three Forces given the Components of the Resultant

The forces ⃑𝐹=βˆ’10βƒ‘π‘–βˆ’7βƒ‘π‘—οŠ§, ⃑𝐹=π‘Žβƒ‘π‘–βˆ’βƒ‘π‘—οŠ¨, and ⃑𝐹=5⃑𝑖+(π‘βˆ’10)βƒ‘π‘—οŠ© act on a particle, where ⃑𝑖 and ⃑𝑗 are two perpendicular unit vectors. Given that the forces’ resultant ⃑𝑅=βˆ’13βƒ‘π‘–βˆ’3⃑𝑗, determine the values of π‘Ž and 𝑏.

Answer

In this example, we want to determine the unknown values of π‘Ž and 𝑏, appearing in the components of the forces βƒ‘πΉοŠ¨ and βƒ‘πΉοŠ©, respectively, which, along with βƒ‘πΉοŠ§, act on a particle with a given resultant force.

Recall that the resultant force acting on any body is the sum of all the forces. Thus, the resultant force of three forces acting on the particle is ⃑𝑅=⃑𝐹+⃑𝐹+⃑𝐹=ο€Ίβˆ’10βƒ‘π‘–βˆ’7⃑𝑗+ο€Ίπ‘Žβƒ‘π‘–βˆ’βƒ‘π‘—ο†+ο€Ί5⃑𝑖+(π‘βˆ’10)⃑𝑗=(π‘Žβˆ’5)⃑𝑖+(π‘βˆ’18)⃑𝑗.

Since we are told that the resultant force vector is ⃑𝑅=βˆ’13βƒ‘π‘–βˆ’3⃑𝑗, we equate the ⃑𝑖 and ⃑𝑗 components separately to obtain π‘Žβˆ’5=βˆ’13π‘Ž=βˆ’8 and π‘βˆ’18=βˆ’3𝑏=15.

Thus, the values are π‘Ž=βˆ’8,𝑏=15.

For a vector in rectangular form, ⃑𝑣=π‘₯⃑𝑖+𝑦⃑𝑗, the magnitude is ‖‖⃑𝑣‖‖=√π‘₯+𝑦.

This is the length of the vector or the distance from the origin, which we can determine using the Pythagorean theorem on a right triangle with sides π‘₯ and 𝑦.

The magnitudes of the unit vectors are ‖‖⃑𝑖‖‖=√1+0=1,‖‖⃑𝑗‖‖=√0+1=1, which is the reason they are called unit vectors, since they have unit length. The magnitude of a displacement or velocity vector would give us the distance or speed respectively. For example, if the velocity vector of an object is ⃑𝑣=3⃑𝑖+4⃑𝑗/,ms then the speed of the object is ‖‖⃑𝑣‖‖=√3+4=√25=5/.ms

We can also write the direction in terms of a counterclockwise angle from the positive ⃑𝑖 direction. By using standard trigonometry, we have tanπœƒ=𝑦π‘₯, where πœƒ is known as the direction of the vector. Using the magnitude and direction, we can translate a vector given in terms of unit vectors in rectangular form to a vector represented by its magnitude and direction. In fact, this is equivalent to the polar form of a vector, with radial component π‘Ÿβ‰‘β€–β€–βƒ‘π‘£β€–β€– and angular component πœƒ, where the angle πœƒ is measured counterclockwise from the positive π‘₯-axis.

We may need to sketch the vector first to ensure that we obtain the correct angle, as taking the inverse tangent may not be sufficient. The range for the inverse tangent function is ο€»βˆ’πœ‹2,πœ‹2 when the domain of the tangent function is restricted to the same interval, known as the principal branch.

Thus, as long as πœƒβˆˆο€»βˆ’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€»βˆ’πœ‹2,πœ‹2 correspond to the first and fourth quadrant, or the quadrants where π‘₯>0. However, this is no longer the case if the vector lies in the second or third quadrant. For the second and third quadrant, a value of πœ‹, in radians, or 180∘, in degrees, must be added to the angle πœƒ, to adjust the angle 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).∘

For example, let’s consider a vector in the second quadrant.

From the sketch, we can see that 𝛼=βˆ’ο€»π‘¦π‘₯.tan

All the angles on a straight line add up to 180∘ and hence 𝛼+πœƒ=180∘. Thus, for the second or third quadrant where π‘₯<0, the angle will be πœƒ=βˆ’π›Ό+180=𝑦π‘₯+180.∘∘tan

Finally, let’s consider an example where we determine the displacement, and hence the magnitude and direction, of a particular body from the information given. We can do this by first writing the displacement vector in terms of the the unit vectors and then using the Pythagorean theorem and standard trigonometry.

Example 7: Solving Word Problems by Adding Two Vectors in Magnitude and Direction Form

A body moved 28 m due east and then 14 m due north. Determine the body’s displacement, stating its direction to the nearest minute.

Answer

In this example, we want to determine the displacement, and hence the distance and direction, of a body which moves in particular directions perpendicular to each other.

Let’s begin by determining the displacement vector from the information given. Recall that we can represent a vector in rectangular form in terms of two mutually perpendicular directions, ⃑𝑖 and ⃑𝑗.

We take the direction from west to east (due east) as positive in the ⃑𝑖 direction and from south to north (due north) as positive in the ⃑𝑗 direction.

Since the body moves 28 m, its displacement in that direction will be 28⃑𝑖 and then 14 m due north will have displacement of 14⃑𝑗. Putting this together, the displacement vector is ⃑𝑠=28⃑𝑖+14⃑𝑗.

The distance is the magnitude of the displacement vector, which we can determine by using the Pythagorean theorem as 𝑑=‖‖⃑𝑠‖‖=√28+14=14√5.

The direction is given as the angle πœƒ, which can be found from standard right angle trigonometry as tantantanπœƒ=1428πœƒ=12πœƒ=ο€Ό12=2634β€².∘

Thus, the displacement is 14√5 m, 2634β€²βˆ˜ north of east.

Key Points

  • Vectors are represented by a magnitude and a direction, or in terms of unit vectors in a coordinate plane in two mutually perpendicular directions. The standard unit vectors for these directions are ⃑𝑖 and ⃑𝑗 and the rectangular form of a vector can be written as ⃑𝑣=π‘₯⃑𝑖+𝑦⃑𝑗.
  • We can solve real-world problems by translating the information into vectors and applying operations where we add or subtract different vectors.
  • The relative velocity of two bodies, 𝐴 and 𝐡, with velocities βƒ‘π‘£οŒ  and βƒ‘π‘£οŒ‘, respectively, denoted by βƒ‘π‘£οŒ οŒ‘, is the difference between the two velocities: ⃑𝑣=βƒ‘π‘£βˆ’βƒ‘π‘£.
  • The resultant force on a particle, ⃑𝑅, is the vector addition of all the forces acting on the particle.
  • The magnitude, ‖‖⃑𝑣‖‖, and direction, πœƒ, of a vector in rectangular form can be found by the Pythagorean theorem and standard trigonometry as ‖‖⃑𝑣‖‖=√π‘₯+π‘¦οŠ¨οŠ¨ and tanπœƒ=𝑦π‘₯. Using these, we can translate a vector given in terms of unit vectors in rectangular form to a vector represented by its magnitude and direction. The angle πœƒ will depend on the quadrant in which the vector lies, which can be communicated efficiently using the following diagram.

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