Lesson Video: Magnitude of a Vector in 3D Mathematics

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

15:28

Video Transcript

In this video, we will learn how to find the magnitude of a position vector in space. We will begin by recalling what we mean by a 3D vector.

A three-dimensional vector has an 𝐒-, 𝐣-, and 𝐀-component. If we consider the point 𝑃 with coordinates two, three, five, we can write the vector 𝐎𝐏 in numerous ways. Firstly, we can consider the 𝐒-, 𝐣-, and 𝐀-components. The vector 𝐎𝐏 is equal to two 𝐒 plus three 𝐣 plus five 𝐀. Vectors are also sometimes written similar to coordinates. With triangular brackets, we have two, three, five. A third way of writing the same vector is as a column inside parentheses, with the 𝐒-component followed by the 𝐣-component and then the 𝐀-component.

The magnitude of any vector is the distance between two points in three-dimensional space. If we consider a vector 𝐀 written in the general form π‘₯𝐒 plus 𝑦𝐣 plus 𝑧𝐀, then the magnitude of vector 𝐀 can be calculated using an application of the Pythagorean theorem. The magnitude of vector 𝐀 denoted by two vertical lines is equal to the square root of π‘₯ squared plus 𝑦 squared plus 𝑧 squared. We find the sum of the squares of the 𝐒-, 𝐣-, and 𝐀-components and then square root the answer.

As well as seeing our unit vectors 𝐒, 𝐣, and 𝐀 written with a hat, you may also have seen them underlined. These notations for handwritten vectors will vary from location to location. Generally when typed, either in a text book or on the internet, vectors will appear in bold. We will now look at some questions where we need to calculate the magnitude of a vector.

If vector 𝐀 is equal to two, negative five, two, find the magnitude of vector 𝐀.

The 𝐒-, 𝐣-, and 𝐀-components of vector 𝐀 are two, negative five, and two, respectively. Therefore, vector 𝐀 could be rewritten as two 𝐒 minus five 𝐣 plus two 𝐀. We recall that the magnitude of any vector could be calculated by square rooting π‘₯ squared plus 𝑦 squared plus 𝑧 squared, where π‘₯, 𝑦, and 𝑧 are the 𝐒-, 𝐣-, and 𝐀-components, respectively. The magnitude of vector 𝐀 is, therefore, equal to the square root of two squared plus negative five squared plus two squared.

Two squared is equal to four. Squaring a negative number gives a positive answer. Therefore, negative five squared is 25. The magnitude of 𝐀 is equal to the square root of four plus 25 plus four. This is equal to the square root of 33. Whilst we could work out this answer on the calculator, as a general rule, we will leave our answers as radicals or surds. The magnitude of vector 𝐀 is the square root of 33.

In our next question, the vector will be written in a different format.

If vector 𝐀 is equal to two 𝐒 plus three 𝐣 minus 𝐀, find the magnitude of vector 𝐀.

For any vector written in the form π‘₯𝐒 plus 𝑦𝐣 plus 𝑧𝐀, the magnitude of the vector is equal to the square root of π‘₯ squared plus 𝑦 squared plus 𝑧 squared. The 𝐒-component of our vector is equal to two, the 𝐣-component is equal to three, and the 𝐀-component is equal to negative one. This means that the magnitude of vector 𝐀 is equal to the square root of two squared plus three squared plus negative one squared.

Two squared is equal to four. Three squared is equal to nine. Squaring a negative number gives us a positive answer. Therefore, negative one squared is one. As four plus nine plus one equals 14, the magnitude of vector 𝐀 is the square root of 14.

In our next question, we will be given the magnitude and need to calculate one of the components of the vector.

If vector 𝐀 is equal to π‘Žπ’ plus 𝐣 minus 𝐀 and the magnitude of vector 𝐀 is equal to the square root of six, find all the possible values of π‘Ž.

Before starting this question, it is worth noting that the wording says find all possible values of π‘Ž. This suggests there will be more than one correct answer. We are given two pieces of information. We are told vector 𝐀 is equal to π‘Žπ’ plus 𝐣 minus 𝐀 and the magnitude of vector 𝐀 is the square root of six. We know that for any vector written in the form π‘₯𝐒 plus 𝑦𝐣 plus 𝑧𝐀, then its magnitude is equal to the square root of π‘₯ squared plus 𝑦 squared plus 𝑧 squared. In this question, the square root of six is equal to the square root of π‘Ž squared plus one squared plus negative one squared. This is because the 𝐒-, 𝐣-, and 𝐀-components are π‘Ž, one, and negative one, respectively.

We can begin to solve this equation by squaring both sides. As squaring is the inverse or opposite of square rooting, the square root of six squared is equal to six. In the same way, the right-hand side becomes π‘Ž squared plus one squared plus negative one squared. Both one squared and negative one squared are equal to one. Therefore, this simplifies to six is equal to π‘Ž squared minus two. We can then subtract two from both sides of this equation so that π‘Ž squared is equal to four.

Our final step is to square root both sides. The square root of π‘Ž squared is π‘Ž. The square root of four is equal to two. But we must take the positive or negative of this. Therefore, π‘Ž is equal to positive or negative two. The possible values of π‘Ž such that the magnitude of vector 𝐀 is the square root of six are two and negative two. This is because when we square both of these, we get an answer of four.

Our next few questions will also involve the addition and subtraction of vectors.

Given that vector 𝐀 plus vector 𝐁 is equal to negative two, four, three and vector 𝐀 is equal to three, five, three, determine the magnitude of vector 𝐁.

We recall that when adding two vectors, we simply add the 𝐒-, 𝐣-, and 𝐀-components separately. The vector 𝐀 plus 𝐁 is equal to the vector 𝐀 plus the vector 𝐁. If we let vector 𝐁 have 𝐒-, 𝐣-, and 𝐀-components π‘₯, 𝑦, and 𝑧, respectively, then negative two, four, three is equal to three, five, three plus π‘₯, 𝑦, 𝑧. We can then subtract vector 𝐀 from both sides of this equation. The left-hand side becomes negative two, four, three minus three, five, three.

Negative two minus three is equal to negative five. Therefore, our 𝐒-component of vector 𝐁 is negative five. Four minus five is equal to negative one, so the 𝐣-component is negative one. Finally, three minus three is equal to zero. Vector 𝐁 is, therefore, equal to negative five, negative one, zero.

We can calculate the magnitude of this vector by squaring each of the components, finding their sum, and then square rooting. The magnitude of vector 𝐁 is equal to the square root of negative five squared plus negative one squared plus zero squared. Negative five squared is 25, negative one squared is one, and zero squared is equal to zero. The magnitude of vector 𝐁 is, therefore, equal to the square root of 26.

In our next question, we’ll find the magnitude of a vector joining the endpoints of two other vectors.

Given that vector 𝐀𝐁 is equal to negative five 𝐒 plus two 𝐣 minus four 𝐀 and vector 𝐁𝐂 is equal to four 𝐒 plus four 𝐣 plus six 𝐀, determine the magnitude of vector 𝐀𝐂.

In this type of question, it is worth drawing a diagram first. This will hopefully ensure that our direction and signs are correct. We are given three points 𝐴, 𝐡, and 𝐢. We can join these to form a triangle. In the question, we are given the value of vector 𝐀𝐁. We are also given the value of vector 𝐁𝐂. Our aim is to calculate the magnitude of vector 𝐀𝐂. Therefore, our first step is to work out vector 𝐀𝐂.

We can see that one way to get from point 𝐴 to point 𝐢 is via point 𝐡. Therefore, vector 𝐀𝐂 is equal to vector 𝐀𝐁 plus vector 𝐁𝐂. Vector 𝐀𝐂 is, therefore, equal to negative five 𝐒 plus two 𝐣 minus four 𝐀 plus four 𝐒 plus four 𝐣 plus six 𝐀.

We can add two vectors by adding the individual components separately. Negative five 𝐒 plus four 𝐒 is equal to negative 𝐒. Two 𝐣 plus four 𝐣 is equal to six 𝐣. Finally, negative four 𝐀 plus six 𝐀 is equal to two 𝐀. Vector 𝐀𝐂 is equal to negative 𝐒 plus six 𝐣 plus two 𝐀.

The magnitude of any vector can be found by squaring the 𝐒-, 𝐣-, and 𝐀-components, finding their sum, and then square rooting the answer. This means that the magnitude of vector 𝐀𝐂 is the square root of negative one squared plus six squared plus two squared. Negative one squared is equal to one, six squared is 36, and two squared is equal to four. One, 36, and four sum to 41. Therefore, the magnitude of vector 𝐀𝐂 is the square root of 41.

In our final question, we will find the magnitude of the difference of two vectors.

If vector 𝐀 is equal to four 𝐒 plus four 𝐣 minus five 𝐀 and vector 𝐁 is equal to three 𝐒 minus 𝐀, determine the magnitude of vector 𝐀 minus vector 𝐁.

Our first step in this question is to calculate vector 𝐀 minus vector 𝐁. This is important as a common mistake here would be to think that the magnitude of vector 𝐀 minus vector 𝐁 is equal to the magnitude of vector 𝐀 minus the magnitude of vector 𝐁. This, however, is not true.

To work out vector 𝐀 minus vector 𝐁, we simply subtract the individual components separately. Four 𝐒 minus three 𝐒 is equal to one 𝐒, or just 𝐒. There is no 𝐣-component in vector 𝐁. Therefore, we are left with four 𝐣. Negative five 𝐀 minus negative 𝐀 is equal to negative four 𝐀. This is because negative five minus negative one is the same as negative five plus one, which is equal to negative four.

We now need to calculate the magnitude of this vector. We know that for any vector written in the form π‘₯𝐒 plus 𝑦𝐣 plus 𝑧𝐀, its magnitude is equal to the square root of π‘₯ squared plus 𝑦 squared plus 𝑧 squared. This means that the magnitude of vector 𝐀 minus vector 𝐁 is the square root of one squared plus four squared plus negative four squared.

Both four squared and negative four squared are equal to 16, so we are left with the square root of one plus 16 plus 16. This is equal to the square root of 33. The magnitude of vector 𝐀 minus vector 𝐁 is the square root of 33.

We will now summarize the key points from this video. For any vector written in the form π‘₯𝐒 plus 𝑦𝐣 plus 𝑧𝐀, the magnitude of the vector is the square root of π‘₯ squared plus 𝑦 squared plus 𝑧 squared. We square the individual components of 𝐒, 𝐣, and 𝐀, find their sum, and then square root the answer. The answer for the magnitude will always be positive.

When adding or subtracting two vectors, we add or subtract the individual components separately. We also saw that the magnitude of the sum or difference of two vectors is not the same as the magnitude of their individual parts. The magnitude of vector 𝐀 plus vector 𝐁 is not equal to the magnitude of vector 𝐀 plus the magnitude of vector 𝐁.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.