Lesson Explainer: Magnitude of a Vector in 3D Mathematics

In this explainer, we will learn how to find the magnitude of a position vector in space.

There are a number of ways in which a 3D vector can be expressed, however, we are going to focus on component form and unit vector form.

Definition: Component and Unit Vector Representations

Consider the following point 𝑃 in 3D space.

The coordinates of point 𝑃 are 𝑃(π‘₯,𝑦,𝑧), and οƒŸπ‘‚π‘ƒ describes the vector from the origin 𝑂 to point 𝑃.

This vector can be represented in 2 ways: Componentform:UnitVectorform:οƒŸπ‘‚π‘ƒ=(π‘₯,𝑦,𝑧)οƒŸπ‘‚π‘ƒ=π‘₯⃑𝑖+𝑦⃑𝑗+π‘§βƒ‘π‘˜.

Note that these are simply different forms of notation. The last form uses the unit vectors ⃑𝑖, ⃑𝑗, and βƒ‘π‘˜, which are vectors of magnitude 1 in the π‘₯-, 𝑦-, and 𝑧-directions respectively: ⃑𝑖=(1,0,0),⃑𝑗=(0,1,0),βƒ‘π‘˜=(0,0,1).

We should be familiar with the fact that a unit vector has a magnitude (or length) of 1. We should also be familiar with finding the magnitude of a vector in 2D by leveraging the Pythagorean theorem stated below:

π‘Ž+𝑏=𝑐𝑐=βˆšπ‘Ž+𝑏.

Note that this is essentially the same problem as finding the magnitude of a vector in 2D. Recall that, given some vector ⃑𝐴, its magnitude is often represented using the notation |⃑𝐴| or sometimes ‖⃑𝐴‖: ⃑𝐴=(π‘₯,𝑦)‖‖⃑𝐴‖‖=π‘₯+𝑦.

Perhaps a lesser known fact about the Pythagorean theorem is that it can be generalized up to as many dimensions as we require. We can leverage this fact to extend our 2D method to cover 3D.

Instead of finding the long diagonal of a rectangle, as shown in our 2D example, we can use the Pythagorean theorem in 3D to find the long diagonal of a cuboid:

π‘Ž+𝑏+𝑐=𝑑𝑑=βˆšπ‘Ž+𝑏+𝑐.

One might initially assume that since the 2-dimensional theorem uses squares (or square roots), then the 3-dimensional theorem would require cubes (or cube roots). Fortunately, this is not the case!

For this explainer, we will not be going into the proof of the theorem, but it is enough to recognize that finding the length of the long diagonal of a cuboid is essentially the same problem as finding the magnitude of a vector in 3D.

This is the key insight we need for our definition.

Definition: The Magnitude of a Vector in 3D

Consider vector ⃑𝐴 in 3D space, where the vector can be expressed as: ⃑𝐴=(π‘₯,𝑦,𝑧)⃑𝐴=π‘₯⃑𝑖+𝑦⃑𝑗+π‘§βƒ‘π‘˜.or

The magnitude of ⃑𝐴 is given by ‖‖⃑𝐴‖‖=π‘₯+𝑦+𝑧.

Note that it is common to simply use the variables π‘₯, 𝑦, and 𝑧 in magnitude calculations: ‖‖⃑𝐴‖‖=√π‘₯+𝑦+𝑧.

In our magnitude calculation, each component of the vector is squared before the sum is square rooted. This means we will always end up taking the square root of a positive number, which is well defined but has two solutions. For example, √25 may be 5 or βˆ’5.

It is worth noting that the magnitude of a vector is defined to be nonnegative. This means that we can safely ignore the negative solution for all magnitudes in this explainer.

Now that we understand how to find the magnitude of a general vector in 3D, let’s practice applying this knowledge in some examples.

Example 1: Finding the Magnitude of a 3D Vector

If ⃑𝐴=(2,βˆ’5,2), find ‖‖⃑𝐴‖‖.

Answer

This question gives us vector ⃑𝐴 in component form. Since the vector has 3 components, we recognize that it exists in 3D space.

An alternative way to express ⃑𝐴 using unit vectors is ⃑𝐴=2βƒ‘π‘–βˆ’5⃑𝑗+2βƒ‘π‘˜.

We have been asked to find ‖‖⃑𝐴‖‖, which represents the magnitude (or length) of the vector. In order to solve this question, we recall that the magnitude of a vector in 3D space is given by ‖‖⃑𝐴‖‖=√π‘₯+𝑦+𝑧, where π‘₯, 𝑦, and 𝑧 represent the components of the vector in the respective cardinal directions.

Our vector has the following components: π‘₯=2,𝑦=βˆ’5,𝑧=2.

To find its magnitude, we substitute these values into the formula: ‖‖⃑𝐴‖‖=2+(βˆ’5)+2‖‖⃑𝐴‖‖=√4+25+4‖‖⃑𝐴‖‖=√33.

With a small amount of simplification, we reach our answer.

With this type of question, it may occasionally be helpful to find a decimal approximation for your answer (perhaps if needed for comparisons). In this case, it is not necessary and since there is no way to helpfully reduce √33, we can leave our answer expressed as a radical.

Example 2: Finding the Magnitude of a 3D Vector Expressed in terms of Unit Vectors

If ⃑𝐴=2⃑𝑖+3βƒ‘π‘—βˆ’βƒ‘π‘˜, find ‖‖⃑𝐴‖‖.

Answer

This question takes a very similar form to our previous example; however, this time we are working with a 3D vector, ⃑𝐴, which has been given in terms of unit vectors.

Again, we have been asked to find the magnitude of this vector, ‖‖⃑𝐴‖‖ and so we can use the formula for the magnitude of a vector in 3D: ‖‖⃑𝐴‖‖=√π‘₯+𝑦+𝑧.

Our vector has the following components: π‘₯=2,𝑦=3,𝑧=βˆ’1.

It is worth noting that even though the unit vector βƒ‘π‘˜ does not appear to have a coefficient, the fact that it is present in our vector suggests otherwise. We should always be careful not to ignore coefficients of 1, or in this case βˆ’1, when working with vectors expressed as the sum of unit vectors. Again, we substitute our values into the formula and simplify to find ‖‖⃑𝐴‖‖: ‖‖⃑𝐴‖‖=2+3+(βˆ’1)‖‖⃑𝐴‖‖=√4+9+1‖‖⃑𝐴‖‖=√14.

Much like our previous example, it is perfectly fine to leave our magnitude in the form of a radical.

The formula that we have been using not only helps us to find the magnitude of a vector, but also can be used to find a missing component of a vector if we are given the magnitude.

Let’s take a look at an example of this.

Example 3: Finding the Value of an Unknown Component of a Vector Using Its Magnitude

If ⃑𝐴=π‘Žβƒ‘π‘–+βƒ‘π‘—βˆ’βƒ‘π‘˜ and ‖‖⃑𝐴‖‖=√6, find all the possible values of π‘Ž.

Answer

For this example, we have been given the magnitude of a 3D vector and we must use this information to find an unknown component. The coefficient for the unit vector ⃑𝑖 is given by the parameter π‘Ž. This is the unknown that we must work toward finding.

Our vector has the following components: π‘₯=π‘Ž,𝑦=1,𝑧=βˆ’1.

We can substitute these values into the formula for the magnitude of a vector in 3D: ‖‖⃑𝐴‖‖=√π‘₯+𝑦+𝑧‖‖⃑𝐴‖‖=ο„π‘Ž+(1)+(βˆ’1).

In this case, we have another piece of information. The magnitude of the vector ‖‖⃑𝐴‖‖=√6. In order to solve for our unknown, we will need to substitute this into our equation too: √6=ο„π‘Ž+(1)+(βˆ’1).

To proceed, we can square both sides then simplify: 6=π‘Ž+(1)+(βˆ’1)6=π‘Ž+1+14=π‘Ž.

At this stage, we have an equation for π‘ŽοŠ¨. Here, we can take the square root of both sides of the equation: π‘Ž=√4π‘Ž=Β±2.

We should remember that taking the square root of a number has both a positive and a negative solution. Since we are finding the coefficient of a vector (and not a magnitude), we cannot ignore the negative solution.

Our answer is that the possible values for π‘Ž are 2 and βˆ’2.

As a footnote to the previous example, consider that when given the magnitude of a vector, we are only able to find the value of a single unknown component of that same vector. Had there been more than one unknown component, our equation might have looked something like this: √6=ο„π‘Ž+𝑏+(βˆ’1)6=π‘Ž+𝑏+15=π‘Ž+𝑏.

Since the above equation has two unknowns, there would be an infinite number of solutions, and so we would not be able to find a unique pair of values for π‘Ž and 𝑏.

Moving into to our next example, we remember that when working with vectors, operations such as addition and subtraction are important tools. In order to find the magnitude of a vector, we may first need to use these operations to find its components.

Example 4: Solving Problems Involving the Magnitude of a Vector

Given that ⃑𝐴+⃑𝐡=(βˆ’2,4,3) and ⃑𝐴=(3,5,3), determine ‖‖⃑𝐡‖‖.

Answer

Recall that when adding two vectors, the individual components simply add together. The given expression ⃑𝐴+⃑𝐡 is itself a vector, and we have been given its components, along with the components of the lone vector ⃑𝐴. Since the components of ⃑𝐡 are unknown, we will represent them with using π‘₯, 𝑦, and 𝑧: ⃑𝐡=(π‘₯,𝑦,𝑧).

This allows us to form the equation (⃑𝐴+⃑𝐡)=⃑𝐴+⃑𝐡(βˆ’2,4,3)=(3,5,3)+(π‘₯,𝑦,𝑧).

In order to isolate our unknowns, we can subtract vector ⃑𝐴, or (3,5,3), from both sides of this equation, leaving us with just vector ⃑𝐡, or (π‘₯,𝑦,𝑧), on the right-hand side: (βˆ’2,4,3)βˆ’(3,5,3)=(π‘₯,𝑦,𝑧).

At this point, we could set up three separate equations for the ⃑𝑖, ⃑𝑗, and βƒ‘π‘˜ directions; however, this would be unnecessary. Recalling the properties of vector addition and subtraction, we can simplify the left-hand side directly by treating each component separately: ((βˆ’2βˆ’3),(4βˆ’5),(3βˆ’3))=(π‘₯,𝑦,𝑧)(βˆ’5,βˆ’1,0)=(π‘₯,𝑦,𝑧).

Since we defined (π‘₯,𝑦,𝑧) to represent vector ⃑𝐡, we have found that ⃑𝐡=(βˆ’5,βˆ’1,0).

Now that we have found the components of vector ⃑𝐡, we can find its magnitude: ‖‖⃑𝐡‖‖=(βˆ’5)+(βˆ’1)+0‖‖⃑𝐡‖‖=√25+1+0‖‖⃑𝐡‖‖=√26.

We have now completed the question. Since there are no useful simplifications to perform, we can leave our answer in surd form.

Note that the properties of vector addition and subtraction can be used when dealing with systems involving coordinate points.

Consider a system with two points 𝐴(π‘₯,𝑦,𝑧) and 𝐡(π‘₯,𝑦,𝑧). Imagine we wanted to find the distance between these two points.

This problem is the same as finding the magnitude of the vector between points 𝐴 and 𝐡, in other words, ‖‖𝐴𝐡‖‖.

We know that 𝑂𝐴=(π‘₯,𝑦,𝑧),οƒŸπ‘‚π΅=(π‘₯,𝑦,𝑧).

Vector 𝐴𝐡 can easily be found by recalling that 𝐴𝐡=οƒŸπ‘‚π΅βˆ’οƒ π‘‚π΄.

We could then find the distance between the two points by applying our magnitude formula: ‖‖𝐴𝐡‖‖=β€–β€–οƒŸπ‘‚π΅βˆ’οƒ π‘‚π΄β€–β€–.

One very important distinction we will see is that the magnitude of a sum or difference is not necessarily the same as the sum or difference of two magnitudes: Magnitudeofadifference:Differenceofmagnitudes:β€–β€–οƒŸπ‘‚π΅βˆ’οƒ π‘‚π΄β€–β€–β€–β€–οƒŸπ‘‚π΅β€–β€–βˆ’β€–β€–οƒ π‘‚π΄β€–β€–

We will see why this is the case in the next example.

Example 5: Solving Problems Involving the Magnitude of a Vector and Vector Addition

If ⃑𝐴=⃑𝑖+3⃑𝑗+4βƒ‘π‘˜ and ⃑𝐡=2βƒ‘π‘—βˆ’βƒ‘π‘˜, determine ‖‖⃑𝐴+⃑𝐡‖‖ and ‖‖⃑𝐴‖‖+‖‖⃑𝐡‖‖.

Answer

The question has given us two 3D vectors and asked us to find the magnitude of the sum of the vectors, ‖‖⃑𝐴+⃑𝐡‖‖, along with the sum of the magnitude of the vectors, ‖‖⃑𝐴‖‖+‖‖⃑𝐡‖‖. While these two things may look very similar, we should be careful not to assume they are equivalent.

Let’s start with ‖‖⃑𝐴+⃑𝐡‖‖. We will first need to find the components of the vector ⃑𝐴+⃑𝐡 by adding our two given vectors: ⃑𝐴+⃑𝐡=⃑𝑖+3⃑𝑗+4βƒ‘π‘˜ο‡+ο€»2βƒ‘π‘—βˆ’βƒ‘π‘˜ο‡=⃑𝑖+(3+2)⃑𝑗+(4βˆ’1)βƒ‘π‘˜=⃑𝑖+5⃑𝑗+3βƒ‘π‘˜.

With some simplification, we find our components. The coefficients for each of our unit vectors can now be substituted into the formula for the magnitude of a vector: ‖‖⃑𝐴+⃑𝐡‖‖=√1+5+3=√1+25+9=√35.

Now what about ‖‖⃑𝐴‖‖+‖‖⃑𝐡‖‖? At this point, finding the magnitude of a known vector should be familiar. We start by finding ‖‖⃑𝐴‖‖: ‖‖⃑𝐴‖‖=√1+3+4=√1+9+16=√26.

Then, we find ‖‖⃑𝐡‖‖: ‖‖⃑𝐡‖‖=√2+(βˆ’1)=√4+1=√5.

Putting these two pieces of information together gives us our answer: ‖‖⃑𝐴‖‖+‖‖⃑𝐡‖‖=√26+√5.

For the sake of comparison, we may choose to find a decimal approximation for these two quantities: ‖‖⃑𝐴+⃑𝐡‖‖=√35β‰ˆ5.91…,‖‖⃑𝐴‖‖+‖‖⃑𝐡‖‖=√26+√5β‰ˆ7.33….

Certainly, in this case, we have shown that ‖‖⃑𝐴+⃑𝐡‖‖≠‖‖⃑𝐴‖‖+‖‖⃑𝐡‖‖.

If we choose to examine ‖‖⃑𝐴+⃑𝐡‖‖ and ‖‖⃑𝐴‖‖+‖‖⃑𝐡‖‖ more closely, we may realize that this is essentially the triangle inequality, as shown below!

The magnitude of the sum of two vectors can never be greater than the sum of the magnitudes: ‖‖⃑𝐴+⃑𝐡‖‖≀‖‖⃑𝐴‖‖+‖‖⃑𝐡‖‖.

Note that there is a single case when these quantities are equal, which occurs when the vectors ⃑𝐴 and ⃑𝐡 point in the same direction. In other words, ⃑𝐴 and ⃑𝐡 are parallel.

This logic can also be applied to compare β€–β€–βƒ‘π΄βˆ’βƒ‘π΅β€–β€– with β€–β€–βƒ‘π΄β€–β€–βˆ’β€–β€–βƒ‘π΅β€–β€–. Doing so leads to a similar result, which is summarized in the key points below.

Key Points

  • The magnitude of a vector represents its length and is defined to always be a positive number.
  • The term ‖‖⃑𝐴‖‖ represents the magnitude of vector ⃑𝐴.
  • Given that ⃑𝐴=π‘₯⃑𝑖+𝑦⃑𝑗+π‘§βƒ‘π‘˜, its magnitude can be found using the following formula: ‖‖⃑𝐴‖‖=√π‘₯+𝑦+𝑧.
  • If two vectors ⃑𝐴 and ⃑𝐡 are parallel, then ‖‖⃑𝐴+⃑𝐡‖‖=‖‖⃑𝐴‖‖+‖‖⃑𝐡‖‖,β€–β€–βƒ‘π΄βˆ’βƒ‘π΅β€–β€–=β€–β€–βƒ‘π΄β€–β€–βˆ’β€–β€–βƒ‘π΅β€–β€–.
  • If two vectors ⃑𝐴 and ⃑𝐡 are not parallel, then ‖‖⃑𝐴+⃑𝐡‖‖≠‖‖⃑𝐴‖‖+‖‖⃑𝐡‖‖,β€–β€–βƒ‘π΄βˆ’βƒ‘π΅β€–β€–β‰ β€–β€–βƒ‘π΄β€–β€–βˆ’β€–β€–βƒ‘π΅β€–β€–.

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