# Lesson Video: Adding and Subtracting Vectors in 2D Mathematics • 12th Grade

In this video, we will learn how to add and subtract vectors in 2D.

15:32

### Video Transcript

In this video, we will learn how to add and subtract vectors in two dimensions. We will begin by recalling what we mean by a vector and showing how we can draw them on the coordinate plane.

A vector is a quantity that has both a magnitude and direction. A two-dimensional vector can be written in the form 𝑥𝐢 plus 𝑦𝐣; using triangular brackets, 𝑥, 𝑦; or as a column vector, 𝑥 over 𝑦. They can be drawn on the coordinate plane.

Let’s consider the vector four, three or four 𝐢 plus three 𝐣. This vector has an 𝐢-component of four, so it moves four units in the 𝑥-direction. It has a 𝐣-component of three, so it moves three units in the 𝑦-direction. We can then draw the vector on the coordinate plane as shown. Its direction will be going away from the origin, and its magnitude will be the length. Whilst it will not be required in this video, we could calculate the magnitude of the vector using the Pythagorean theorem. Four squared plus three squared is equal to five squared. This means that the magnitude of our vector is five.

We will now consider what happens when we need to add and subtract vectors. Let’s imagine we want to add two vectors 𝐚 and 𝐛. We can show this visually by first drawing vector 𝐚. We can then draw vector 𝐛 from the end of this vector. The vector 𝐚 plus 𝐛 will go from the start of vector 𝐚 to the end of vector 𝐛. If vector 𝐚 was equal to five, two and vector 𝐛 was equal to three, negative three, we could find vector 𝐚 plus 𝐛 by adding the individual components separately. We add five and three and then separately two and negative three. This means that 𝐚 plus 𝐛 would be equal to eight, negative one.

We can follow a similar process when subtracting vectors. To work out vector 𝐚 minus 𝐛, we would subtract three from five and negative three from two. This would give us the vector two, five.

Let’s consider what this would look like in a sketch. Once again, we can begin by drawing vector 𝐚. We know from our top diagram that vector 𝐛 looks as shown. The vector negative 𝐛 would have the same magnitude but go in the opposite direction. This gives us the vector 𝐚 plus negative 𝐛, which is the same as 𝐚 minus 𝐛. We will now look at some specific examples where we need to add and subtract vectors.

Given that vector 𝐀 is equal to negative five 𝐢 plus 10𝐣 and vector 𝐁 is equal to negative four 𝐢 minus five 𝐣, where 𝐢 and 𝐣 are two perpendicular unit vectors, find vector 𝐀 minus vector 𝐁.

We recall that in order to add or subtract two or more vectors, we simply add or subtract the corresponding components. In this question, we need to subtract negative four 𝐢 minus five 𝐣 from negative five 𝐢 plus 10𝐣. We can then subtract the 𝐢- and 𝐣-components separately. Negative five minus negative four is equal to negative one. This is because this is the same as negative five plus four.

Subtracting negative five from 10 gives us 15, as once again this is the same as adding five to 10. Vector 𝐀 minus vector 𝐁 is therefore equal to negative 𝐢 plus 15𝐣.

In our next question, we all need to add and subtract three vectors.

Given that vector 𝐀 is equal to nine, five; vector 𝐁 is equal to negative 10, three; and vector 𝐂 is equal to negative three, six, find vector 𝐀 plus vector 𝐁 minus vector 𝐂.

In order to add or subtract two or more vectors, we need to add or subtract the corresponding components. In this question, we have nine, five plus negative 10, three minus negative three, six. If we consider the 𝑥-components, we have nine plus negative 10 minus negative three, Adding negative 10 to nine is the same as subtracting 10 from nine. We then need to add three to this, as subtracting negative three is the same as adding three. Nine minus 10 equals negative one. And adding three gives us two. The 𝑥-component of our answer is therefore two.

We can then repeat this for the 𝑦-components. We have five plus three minus six. This is also equal to two.

The vector 𝐀 plus 𝐁 minus 𝐂 is equal to two, two.

In our next question, we will also need to multiply vectors by scalars.

Given that vector 𝐀 is equal to negative one, two and vector 𝐁 is equal to three, negative six, find six 𝐀 plus two 𝐁.

In this question, we’re multiplying vectors 𝐀 and 𝐁 by a scalar. We’re multiplying vector 𝐀 by six, and we’re multiplying vector 𝐁 by two. To multiply any vector by a scalar, we simply multiply each of the components by that scalar. Six multiplied by negative one is negative six, and six multiplied by two is equal to 12. Therefore, six 𝐀 is equal to negative six, 12.

We need to multiply vector 𝐁 by two. Two multiplied by three is six, and two multiplied by negative six is 12. Therefore, two 𝐁 is equal to six negative 12.

We need to add these two vectors. When adding or subtracting two vectors, we simply add or subtract the corresponding components. In this case, we need to add negative six and six and then also 12 and negative 12. Negative six plus six is equal to zero, and 12 plus negative 12 is also equal to zero. This means that six 𝐀 plus two 𝐁 is equal to zero, zero. The sum of six multiplied by vector 𝐀 and two multiplied by vector 𝐁 is zero, zero, the origin.

In our next question, we need to determine a vector that satisfies an equation.

Given that vector 𝐁 equals negative nine, negative three; vector 𝐂 equals negative four, negative two; and vector 𝐃 equals negative two, nine, determine the vector 𝐀 that satisfies the equation 𝐀 is equal to negative four 𝐁 plus two 𝐂 minus six 𝐃.

In the equation given, we’ve multiplied vectors 𝐁, 𝐂, and 𝐃 by a scalar. Firstly, vector 𝐁 has been multiplied by negative four. To calculate this, we multiply each of the components by negative four. Multiplying two negative numbers gives a positive answer. Therefore, negative four multiplied by negative nine is 36. And negative four multiplied by negative three is 12.

We can repeat this process by multiplying vector 𝐂 by two. This gives us negative eight, negative four. Finally, we need to multiply vector 𝐃 by six. Six multiplied by negative two is negative 12, and six multiplied by nine is 54. We need to add two 𝐂 to negative four 𝐁 and then subtract six 𝐃 from this. Vector 𝐀 is therefore equal to 36, 12 plus negative eight, negative four minus negative 12, 54.

When adding or subtracting vectors, we simply add or subtract the individual components. For the 𝑥-components in this case, we have 36 plus negative eight minus negative 12. This is the same as 36 minus eight plus 12, which gives us an answer of 40. The 𝑥-component of vector 𝐀 is 40.

We can repeat this for the 𝑦-components. We need to add negative four to 12 and then subtract 54. This is equal to negative 46.

The vector 𝐀 that satisfies the equation negative four 𝐁 plus two 𝐂 minus six 𝐃 is 40, negative 46.

Our next question will briefly deal with what happens when our vectors are in three dimensions.

If vector 𝐀 is equal to six, negative four, seven and vector 𝐁 is equal to five, six, four, determine vector 𝐀 plus vector 𝐁.

We recall that when adding or subtracting vectors in two dimensions, we simply add or subtract the corresponding components. The same is true, as in this case when we have vectors in three dimensions, we simply add the three different components.

Firstly, we add the 𝑥- or 𝐢-component. We need to add six and five. This is equal to 11. Next, we need to add negative four and six. This is equal to two. Finally, we need to add seven and four. This is also equal to 11.

If vector 𝐀 is equal to six, negative four, seven and vector 𝐁 is equal to five, six, four, then vector 𝐀 plus vector 𝐁 is equal to 11, two, 11.

This means that when adding, subtracting, or multiplying vectors in three dimensions by a scalar, we follow the same process as with two dimensions.

In our final question, we will look at a more practical problem involving geometry.

𝐴𝐵𝐶𝐷𝐸𝐹 is a regular hexagon. 𝐴𝐹 is parallel to 𝐺𝐵, and 𝐸𝐹 is parallel to 𝐺𝐷. Express vector 𝐀𝐄 in terms of 𝐮 and 𝐯.

There are some lines missing on our hexagon from 𝐴 to 𝐵, 𝐴 to 𝐺, and 𝐸 to 𝐺. As the hexagon is regular, we have lots of parallel and equal-length lines. The line segments on our diagram from 𝐷 to 𝐶, 𝐺 to 𝐵, and 𝐹 to 𝐴 will all be equal to vector 𝐮. This is because they all have the same direction and magnitude. The same would be true of 𝐸𝐺. 𝐺𝐶, 𝐹𝐺, and 𝐸𝐷 will all be equal to vector 𝐯. Once again, this would also be true for 𝐴𝐵.

We also know that 𝐹𝐸, 𝐺𝐷, and 𝐵𝐶 are all of the same magnitude and direction. Let’s consider what these would be equal to in terms of 𝐮 and 𝐯. We can get from the center of the hexagon 𝐺 to the vertex 𝐷 via vertex 𝐶. This means that 𝐺𝐷 is equal to 𝐺𝐶 plus 𝐶𝐷. We know that 𝐺𝐶 is equal to vector 𝐯. 𝐷𝐶 is equal to vector 𝐮. This means that 𝐶𝐷 is equal to negative 𝐮. It has the same magnitude but is in the opposite direction. We can therefore say that 𝐺𝐷 is equal to 𝐯 minus 𝐮. The same is true for 𝐹𝐸 and 𝐵𝐶.

We are asked to express 𝐴𝐸 in terms of 𝐮 and 𝐯. The quickest way to get from vertex 𝐴 to vertex 𝐸 is via vertex 𝐹. This means that 𝐴𝐸 is equal to 𝐴𝐹 plus 𝐹𝐸. As 𝐹𝐴 is equal to vertex 𝐮, 𝐴𝐹 will be equal to negative 𝐮. We know that 𝐹𝐸 is equal to 𝐯 minus 𝐮. 𝐴𝐸 is therefore equal to negative 𝐮 plus 𝐯 minus 𝐮.

This can be simplified to negative two 𝐮 plus 𝐯. Alternatively, we could write it as 𝐯 minus two 𝐮. We can use this method to work out the vector from any vertex to another vertex on the hexagon.

We will now summarize the key points from this video.

We found out in this video that to add or subtract two or more vectors, we simply add or subtract the corresponding components. To multiply a vector by a scalar, we multiply each component individually. In our final question, we saw that we can use these skills to answer problems involving geometry. We also saw that the above rules apply to vectors in both two and three dimensions.