Video: Vector Operations in 3D

In this video, we will learn how to perform operations on vectors in 3D, such as addition, subtraction, and scalar multiplication.

17:40

Video Transcript

In this video, we will learn how to perform operations on vectors in three dimensions such as addition, subtraction, and scalar multiplication. We will begin by recalling some key properties of vectors in 3D. We know that a vector in 3D space has magnitude and direction. If we consider the three-dimensional coordinate grid as shown, then the unit vectors in the 𝑥-, 𝑦-, and 𝑧-direction are denoted by 𝐢, 𝐣, and 𝐤, respectively. Any vector can therefore be written in the form 𝑥𝐢 plus 𝑦𝐣 plus 𝑧 or 𝑧𝐤. The values of 𝑥, 𝑦, and 𝑧 will be the number of units traveled in those directions. If the point 𝐴 has coordinates 𝑥, 𝑦, 𝑧, then vector 𝐀 is equal to 𝑥𝐢 plus 𝑦𝐣 plus 𝑧𝐤. This can also be written in component form as shown.

We will now consider how we can add and subtract vectors in three dimensions. If we consider two vectors 𝐀 and 𝐁 with components 𝑥 one, 𝑦 one, 𝑧 one and 𝑥 two, 𝑦 two, 𝑧 two, respectively, then we can add and subtract the two vectors by considering their corresponding components. Vector 𝐀 plus vector 𝐁 will have components 𝑥 one plus 𝑥 two, 𝑦 one plus 𝑦 two, and 𝑧 one plus 𝑧 two. Likewise, 𝐀 minus 𝐁 will have components 𝑥 one minus 𝑥 two, 𝑦 one minus 𝑦 two, and 𝑧 one minus 𝑧 two. This method allows us to add or subtract two or more vectors. We will now look at a couple of examples.

If vector 𝐀 is equal to negative five 𝐢 minus eight 𝐣 plus six 𝐤 and vector 𝐁 is equal to four 𝐢 minus three 𝐣 plus 13𝐤, find vector 𝐀 minus vector 𝐁.

We recall that if vector 𝐀 is equal to 𝑥 one 𝐢 plus 𝑦 one 𝐣 plus 𝑧 one 𝐤 and vector 𝐁 is equal to 𝑥 two 𝐢 plus 𝑦 two 𝐣 plus 𝑧 two 𝐤, then vector 𝐀 minus vector 𝐁 is equal to 𝑥 one minus 𝑥 two 𝐢 plus 𝑦 one minus 𝑦 two 𝐣 plus 𝑧 one minus 𝑧 two 𝐤. In this question, we need to subtract four 𝐢 minus three 𝐣 plus 13𝐤 from negative five 𝐢 minus eight 𝐣 plus six 𝐤. Subtracting four 𝐢 minus three 𝐣 plus 13𝐤 is the same as minus four 𝐢 plus three 𝐣 minus 13𝐤. We can then collect the like terms. Negative five 𝐢 minus four is equal to negative nine 𝐢. Negative eight 𝐣 plus three 𝐣 is equal to negative five 𝐣. Finally, six 𝐤 minus 13𝐤 is equal to negative seven 𝐤.

Vector 𝐀 minus vector 𝐁 is equal to negative nine 𝐢 minus five 𝐣 minus seven 𝐤.

We will now look at an addition problem.

Given the two vectors 𝐀 is equal to negative two, negative three, zero and 𝐁 is equal to negative three, three, negative two, find vector 𝐀 plus vector 𝐁.

In order to add any two vectors in three dimensions, we simply add their corresponding components. This means that if vector 𝐀 has components 𝑥 one, 𝑦 one, 𝑧 one and vector 𝐁 has components 𝑥 two, 𝑦 two, 𝑧 two, then vector 𝐀 plus vector 𝐁 will have components 𝑥 one plus 𝑥 two, 𝑦 one plus 𝑦 two, and 𝑧 one plus 𝑧 two. In this question, we add the 𝑥-components negative two and negative three. We add the 𝑦-components negative three and positive three. Finally, we add 𝑧- or 𝑧-components zero and negative two. Negative two plus negative three is the same as negative two minus three. This is equal to negative five. Negative three plus three is equal to zero. Finally, zero plus negative two is the same as zero minus two, which equals negative two.

If vector 𝐀 is equal to negative two, negative three, zero and vector 𝐁 is equal to negative three, three, negative two, then vector 𝐀 plus vector 𝐁 is equal to negative five, zero, negative two.

We will now look at what happens when we multiply a vector by a scalar. When we multiply any vector by a scalar, we multiply each of the components of the vector by the scalar. This means that if vector 𝐀 has components 𝑥, 𝑦, and 𝑧, then multiplying this by the scalar or constant 𝑘 gives us a vector with components 𝑘𝑥, 𝑘𝑦, and 𝑘𝑧. We will now look at a question involving scalar multiplication.

What is the vector that results from scaling the vector 𝐀 equal to negative six, negative three, negative one by a factor of negative six?

When we multiply any vector by a scalar, we simply multiply each of the components of the vector by the scalar. In this question, we want to multiply the vector negative six, negative three, negative one by the scalar or constant negative six. We recall that when multiplying two negative numbers, we get a positive answer. This means that negative six multiplied by negative six is positive 36. Negative six multiplied by negative three is equal to 18. And negative six multiplied by negative one is equal to six. If vector 𝐀 is equal to negative six, negative three, negative one, multiplying this by the scalar negative six gives us the vector 36, 18, six.

Our next question involves both multiplying by a scalar and subtracting two vectors.

If vector 𝐀 is equal to negative eight, nine, nine and vector 𝐁 is equal to negative six, four, nine, find two-fifths of vector 𝐀 minus four-fifths of vector 𝐁.

We will begin this question by calculating two-fifths of vector 𝐀. When multiplying any vector by a scalar or constant, we simply multiply each of the components by that scalar. In this question, we need to multiply the components negative eight, nine, and nine by two-fifths. Negative eight multiplied by two-fifths is equal to negative sixteen-fifths. Nine multiplied by two-fifths is equal to eighteen-fifths. This means that two-fifths of vector 𝐀 has components negative sixteen-fifths, eighteen-fifths, and eighteen-fifths. We can repeat this process to calculate four-fifths of vector 𝐁. We need to multiply the vector negative six, four, nine by four-fifths. This gives us the vector with components negative twenty-four fifths, sixteen-fifths, and thirty-six fifths.

Our next step is to subtract four-fifths of vector 𝐁 from two-fifths of vector 𝐀. When subtracting two vectors, we simply subtract their corresponding components. We begin by subtracting negative twenty-four fifths from negative sixteen-fifths. Negative 16 minus negative 24 is equal to positive eight. Therefore, the 𝑥-component is eight-fifths. Subtracting sixteen-fifths from eighteen-fifths gives us two-fifths. The 𝑦-component of our vector is two-fifths. Finally, we need to subtract thirty-six fifths from eighteen-fifths. As 18 minus 36 is negative 18, the 𝑧 or 𝑧-component is negative eighteen-fifths. Two-fifths of vector 𝐀 minus four-fifths of actor 𝐁 is equal to eight-fifths, two-fifths, negative eighteen-fifths.

Whilst we don’t need to in this question, we could factor out one-fifth, giving us the scalar one-fifth multiplied by the vector eight, two, negative 18.

In our next question, we will find the missing vector in the vector expression.

If vector 𝐀 is equal to negative one, one, one and vector 𝐁 is equal to one, one, negative two, determine the vector 𝐂 for which two 𝐂 plus five 𝐀 is equal to five 𝐁.

There are lots of ways of approaching this problem. One way would be to begin by rearranging our equation to make vector 𝐂 the subject. We can do this by firstly subtracting five 𝐀 from both sides. This gives us two 𝐂 is equal to five 𝐁 minus five 𝐀. We can then divide both sides of this equation by two such that vector 𝐂 is equal to five 𝐁 minus five 𝐀 divided by two. The right-hand side can be rewritten as one-half of five 𝐁 minus five 𝐀.

We can now calculate five multiplied by vector 𝐁 and five multiplied by vector 𝐀. When multiplying any vector by a scalar, we simply multiply each of the individual components by that scalar. To calculate five 𝐁, we multiply the vector one, one, negative two by five. This gives us the vector five, five, negative 10. In the same way, five multiplied by vector 𝐀 is equal to five multiplied by the vector negative one, one, one. This is equal to negative five, five, five.

We can now subtract these two vectors. And we do this by subtracting the corresponding components. Five minus negative five is equal to 10. Five minus five is equal to zero. And negative 10 minus five is equal to negative 15. Five 𝐁 minus five 𝐀 is equal to 10, zero, negative 15. We know that vector 𝐂 is equal to a half of this. It is equal to a half of the vector 10, zero, negative 15. A half of 10 is equal to five, a half of zero is zero, and a half of negative 15 is negative 15 over two or negative fifteen-halves. The vector 𝐂 for which two 𝐂 plus five 𝐀 is equal to five 𝐁 is five, zero, negative 15 over two.

In our final question, we will also consider the magnitude of vectors in three dimensions.

𝐕 and 𝐖 are two vectors, where vector 𝐕 is equal to negative one, five, negative two and vector 𝐖 is equal to three, one, one. Comparing the magnitude of vector 𝐕 minus vector 𝐖 and the magnitude of vector 𝐕 minus the magnitude of vector 𝐖, which quantity is larger?

We recall that the magnitude of any vector 𝐀 with components 𝑥, 𝑦, and 𝑧 is equal to the square root of 𝑥 squared plus 𝑦 squared plus 𝑧 squared. We find the sum of the squares of the individual components and then square root the answer. This means that the magnitude of vector 𝐕 is equal to the square root of negative one squared plus five squared plus negative two squared. This is equal to the square root of one plus 25 plus four. The magnitude of vector 𝐕 is equal to the square root of 30. We can repeat this process for vector 𝐖. This is equal to the square root of three squared plus one squared plus one squared, which simplifies to the square root of nine plus one plus one, which is the square root of 11. The magnitude of vector 𝐖 is the square root of 11.

Before calculating a numerical value for the magnitude of vector 𝐕 minus the magnitude of vector 𝐖, we will firstly work out vector 𝐕 minus vector 𝐖. We know that to subtract two vectors, we subtract the corresponding components. Negative one minus three is equal to negative four. Five minus one is equal to positive four. And negative two minus one is equal to negative three. Vector 𝐕 minus vector 𝐖 is equal to negative four, four, negative three. The magnitude of this is equal to the square root of negative four squared plus four squared plus negative three squared. This simplifies to the square root of 16 plus 16 plus nine. The magnitude of vector 𝐕 minus vector 𝐖 is therefore equal to the square root of 41.

We now need to calculate the decimal values of the magnitude of vector 𝐕 minus vector 𝐖 and the magnitude of vector 𝐕 minus the magnitude of vector 𝐖. The square root of 41 is equal to 6.4031 and so on. The square root of 30 minus the square root of 11 is equal to 2.1606 and so on. This means that the square root of 41 is greater than the square root of 30 minus the square root of 11. We can therefore conclude that the magnitude of vector 𝐕 minus vector 𝐖 is greater than the magnitude of vector 𝐕 minus the magnitude of vector 𝐖. The larger quantity is the magnitude of vector 𝐕 minus vector 𝐖.

We will now summarize the key points from this video. We saw in this video that to add or subtract vectors in three dimensions, we simply add or subtract the corresponding components. If vector 𝐀 has components 𝑥 one, 𝑦 one, 𝑧 one and vector 𝐁 has components 𝑥 two, 𝑦 two, 𝑧 two, then to calculate vector 𝐀 plus vector 𝐁, we add the components and to calculate vector 𝐀 minus vector 𝐁, we subtract the corresponding components. We also saw that to multiply a vector by a scalar, we multiply each component of the vector by the scalar. If vector 𝐀 has components 𝑥, 𝑦, and 𝑧, then multiplying this by the scalar 𝑘 gives us the vector with components 𝑘𝑥, 𝑘𝑦, and 𝑘𝑧.

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