# Video: Vector Operations in 2D

In this video, we will learn how to perform operations on vectors algebraically such as vector addition, vector subtraction, and scalar multiplication in two dimensions.

15:19

### Video Transcript

In this video, we will learn how to perform operations on vectors algebraically, such as vector addition, vector subtraction, and scalar multiplication, in two dimensions. Let’s begin by recalling what we mean by a vector. A vector has a size, known as its magnitude, and a direction. A two-dimensional vector has a horizontal and vertical component. These are often denoted by 𝑖 and 𝑗 unit vectors.

As with the coordinate grid, moving to the right is the positive direction and moving to the left is the negative horizontal direction. Moving up is the positive vertical direction, and moving down is the negative vertical direction. A vector 𝐮 equal to four 𝑖 plus three 𝑗 will move four units in the horizontal direction and three units up in the vertical direction. This can also be shown as four, three inside triangular brackets, as shown. When adding or subtracting two vectors, we treat the horizontal and vertical components separately. This is also true when multiplying a vector by a scalar.

We will now look at some questions where we need to add and subtract vectors.

Shown on the grid of unit squares are the vectors 𝐮, 𝐯, and 𝐮 plus 𝐯. What are the components of vector 𝐮? What are the components of vector 𝐯? What are the components of vector 𝐮 plus 𝐯?

Let’s begin by considering vector 𝐮. As with any vector, this has a start and end point. The arrow indicates it is moving to the right and up. The horizontal component of this vector will be equal to four, as we move four units to the right. The vertical component will be one, as we move up one square. The components of vector 𝐮 are therefore equal to four, one. We always put the horizontal component first.

We can calculate the components of vector 𝐯 in the same way. This vector moves left and up. This means that its horizontal component will be negative. We move five squares to the left and one square up. Vector 𝐯 is therefore equal to negative five, one.

There are two ways of answering the third part of this question, to calculate the components of the vector 𝐮 plus 𝐯. We could do it directly from the diagram, as the vector 𝐮 plus 𝐯 moves one square to the left and two squares up. This means that it is equal to the vector negative one, two. Alternatively, if we didn’t have the diagram of unit squares as in this case, we could use the fact that vector 𝐮 plus 𝐯 is equal to vector 𝐮 plus vector 𝐯. We need to add the vector four, one to the vector negative five, one.

When adding and subtracting vectors, we treat the horizontal and vertical components separately. In this case, we need to add four and negative five and then separately one and one. Four plus negative five is the same as four minus five, which is equal to negative one. One plus one is equal to two. Therefore, the vertical component of the vector 𝐮 plus 𝐯 is two. This gives us the same answer as from the diagram, the vector negative one, two.

In our next example, we will answer a similar question but without a diagram.

Given that vector 𝐮 is equal to zero, four and vector 𝐯 is equal to zero, negative five, find the components of vector 𝐮 plus 𝐯.

We recall that every two-dimensional vector has a horizontal and vertical component. If our horizontal component is positive, it is moving to the right, whereas if it is negative, it is moving to the left. In a similar way, if the vertical component is positive, our vector is moving upwards, whereas if it is negative, it is moving downwards.

We also recall that when adding two vectors, in this case 𝐮 and 𝐯, we add the horizontal and vertical components separately. The vector 𝐮 plus 𝐯 is equal to the vector 𝐮 zero, four plus the vector 𝐯 zero, negative five. Zero plus zero is equal to zero. Therefore, the horizontal component of vector 𝐮 plus 𝐯 is zero. Four plus negative five is equal to negative one. Therefore, the vertical component is negative one. The vector 𝐮 plus 𝐯 is therefore equal to zero, negative one.

In our next question, we need to express one vector in terms of two other vectors using scalar multiples.

Given that vector 𝐀 is equal to negative four, negative one and vector 𝐁 is equal negative two, negative one, express vector 𝐂 negative eight, negative one in terms of vector 𝐀 and vector 𝐁.

As we want to express vector 𝐂 in terms of vector 𝐀 and 𝐁, we know that 𝐂 is equal to some constant 𝑝 multiplied by vector 𝐀 plus some constant 𝑞 multiplied by vector 𝐁. Substituting in vectors 𝐀, 𝐁, and 𝐂, we have negative eight, negative one is equal to 𝑝 multiplied by negative four, negative one plus 𝑞 multiplied by negative two, negative one.

We recall that when multiplying a vector by any constant or scalar, we need to treat the horizontal and vertical components separately. If we consider the horizontal components first, we have negative eight is equal to negative four 𝑝 plus negative two 𝑞. This is the same as negative eight is equal to negative four 𝑝 minus two 𝑞. We can divide both sides of this equation by negative two. This gives us four is equal to two 𝑝 plus 𝑞. We will call this equation one.

We can now repeat this process with the vertical components. Negative one is equal to negative 𝑝 plus negative 𝑞. We can divide both sides of this equation by negative one. This gives us one is equal to 𝑝 plus 𝑞, which we will call equation two.

We now have a pair of simultaneous equations that we can solve by elimination. We can subtract equation two from equation one. On the left-hand side, four minus one is equal to three, two 𝑝 minus 𝑝 is equal to 𝑝, and 𝑞 minus 𝑞 is equal to zero. Therefore, 𝑝 is equal to three. Substituting this value of 𝑝 back into equation two gives us one is equal to three plus 𝑞. Subtracting three from both sides of this equation gives us 𝑞 is equal to negative two. The vector 𝐂 negative eight, negative one is therefore equal to three multiplied by vector 𝐀 minus two multiplied by vector 𝐁. We have therefore expressed vector 𝐂 in terms of vector 𝐀 and vector 𝐁.

Our next question is a more complicated problem involving scalar multiples and addition of vectors.

On a lattice where vector 𝐀𝐂 is equal to three, three; vector 𝐁𝐂 is equal to 13, negative seven; and two multiplied by vector 𝐂 plus two multiplied by vector 𝐀𝐁 is equal to negative four, negative four; find the coordinates of the point C.

In this question, we are given vector 𝐀𝐂 and vector 𝐁𝐂. We can use these to calculate the vector 𝐀𝐁. If we consider the three points A, B, and C as shown on the diagram, we know that vector 𝐀𝐂 is three, three. Vector 𝐁𝐂 is 13, negative seven. We need to calculate vector 𝐀𝐁. To get from point A to point B via point C, we can add vector 𝐀𝐂 and vector 𝐂𝐁. As vector 𝐂𝐁 is in the opposite direction to vector 𝐁𝐂, vector 𝐀𝐁 is equal to 𝐀𝐂 minus 𝐁𝐂. We need to subtract the vector 13, negative seven from the vector three, three.

When adding and subtracting vectors, we treat the horizontal and vertical components separately. Three minus 13 is equal to negative 10. Three minus negative seven is the same as three plus seven, which is equal to 10. Therefore, the vector 𝐀𝐁 is equal to negative 10, 10. If we let the point C have coordinates 𝑥, 𝑦, we can substitute these values into our equation. Two multiplied by 𝑥, 𝑦 plus two multiplied by negative 10, 10 is equal to negative four, negative four.

Once again, we can treat the horizontal and vertical components separately to create two equations. The horizontal components give us two 𝑥 minus 20 is equal to negative four. The vertical components give us the equation two 𝑦 plus 20 is equal to negative four. In our first equation, we add 20 to both sides, giving us two 𝑥 is equal to 16. Dividing both sides by two gives us a value of 𝑥 equal to eight. In our second equation, we need to subtract 20 from both sides, giving us two 𝑦 is equal to negative 24. Once again, we can divide both sides of this equation by two, giving us 𝑦 is equal to negative 12. As 𝑥 is equal to eight and 𝑦 is equal to negative 12, the coordinates of point C are eight, negative 12.

Our final question involves calculating the magnitude or modulus of a vector.

If vector 𝐀𝐁 is equal to seven 𝑖 plus six 𝑗 and vector 𝐁𝐂 is equal to 𝑖, then the magnitude of vector 𝐀𝐂 is equal to blank.

We recall that the vector seven 𝑖 plus six 𝑗 can be written using triangular brackets as seven, six. The vector 𝑖 can be written one, zero, as there is no vertical component 𝑗. We also recall that when dealing with vectors, the vector 𝐀𝐂 is equal to the vector 𝐀𝐁 plus the vector 𝐁𝐂. In this question, we can calculate the vector 𝐀𝐂 by adding the vector six, one and the vector one, zero.

When adding two vectors, we add the horizontal and vertical components separately. Seven plus one is equal to eight. Six plus zero is equal to six. Therefore, vector 𝐀𝐂 is equal to eight, six. This is not the final answer to this question though, as we need to calculate the magnitude or modulus of vector 𝐀𝐂. In order to do this, we need to know the following rule. If vector 𝐮 is equal to 𝑥, 𝑦, then the magnitude of vector 𝐮 is equal to the square root of 𝑥 squared plus 𝑦 squared. We use the Pythagorean theorem to calculate the magnitude or size of vector 𝐮 by squaring the horizontal and vertical components, finding their sum, and then square rooting the answer.

The magnitude of vector 𝐀𝐂 is equal to the square root of eight squared plus six squared. Eight squared is equal to 64, and six squared is 36. Adding these gives us the square root of 100, which is equal to 10. If vector 𝐀𝐁 is equal to seven 𝑖 plus six 𝑗 and vector 𝐁𝐂 is equal to 𝑖, then the magnitude of vector 𝐀𝐂 is 10.

We will now summarize the key points from this video. We found out in this video that a vector has a horizontal and vertical component. The horizontal component comes first and can be written in triangular brackets — four, negative three — or as four 𝑖 minus three 𝑗, where 𝑖 and 𝑗 are unit vectors. When adding and subtracting vectors, we treat the horizontal and vertical components separately. We also do this when multiplying a vector by a scalar. The magnitude of a vector is its size. And we found out in the last question that we can calculate this using the Pythagorean theorem.