Lesson Explainer: Finding the Components of a Vector Physics

In this explainer, we will learn how to find the 𝑥- and 𝑦-components of a vector given its magnitude and the angle between the vector and one of the axes.

A scalar quantity is a quantity that just has a magnitude; it is just a size or an amount. One example of a scalar quantity in physics is mass.

A vector quantity is a quantity that has both a magnitude and a direction. One example of a vector quantity in physics is velocity. Velocity represents how fast an object is moving and in what direction.

We can represent vectors graphically using arrows. The length of the arrow represents the magnitude of the vector, and the direction of the arrow represents the direction of the vector.

The diagram below shows a vector represented by an arrow.

The magnitude of the vector is 6.4 cm, and the direction of the vector is 51 counterclockwise from the positive 𝑥-axis.

The angle counterclockwise from the positive 𝑥-axis to the vector is called the argument of the vector.

There is another way of representing a vector, however. We can represent a vector in terms of its components. The 𝑥-component, or the horizontal component, of a vector is the size of the vector in the 𝑥-direction. The 𝑦-component, or the vertical component, of a vector is the size of the vector in the 𝑦-direction.

The diagram below shows the same vector as the diagram above, but with a grid drawn on it. The length and width of each grid square are 1 cm.

We can see that the size of the vector in the 𝑥-direction is 4 grid squares, or 4 cm, and the size of the vector in the 𝑦-direction is 5 grid squares, or 5 cm. These are the 𝑥- and 𝑦-components of the vector.

We could simply notate these components as a pair of values, that is, (4,5), but a more useful way of notating them is using unit vectors.

A unit vector is a vector that has a magnitude of 1. Conventionally, we use the symbol 𝑖 to represent a unit vector along the 𝑥-axis and the symbol 𝑗 to represent a unit vector along the 𝑦-axis. These two unit vectors are shown in the diagram below.

We can make the blue vector in the diagram above by adding together multiples of the two unit vectors, 𝑖 and 𝑗, as shown below.

The blue vector is equal to 4 of 𝑖 and 5 of 𝑗 added together. We can write this algebraically as 4𝑖+5𝑗.

In fact, we can represent any vector in this way. For a vector 𝐴, if we say that 𝐴 is the value of the 𝑥-component of the vector and 𝐴 is the value of the 𝑦-component, then 𝐴=𝐴𝑖+𝐴𝑗.

This is how we can work out the components of a vector and write the vector in component form using unit vector notation when we are given a grid. But if we are given the magnitude and argument of a vector, we can instead work out its components using trigonometry.

Any vector makes a right triangle with the 𝑥-axis, as shown below.

If 𝜃 is the angle between the vector and the 𝑥-axis, then the adjacent side of the triangle is the horizontal length of the vector, the opposite side of the triangle is the vertical length of the vector, and the hypotenuse is the magnitude of the vector.

Recall that cosadjacenthypotenuse(𝜃)=.

We can rearrange this to make the adjacent side of the triangle the subject of the equation: adjacenthypotenusecos=×(𝜃).

Since the adjacent side of the triangle is equal to the horizontal component of the vector, 𝐴, and the hypotenuse is equal to the magnitude of the vector, which we will just call 𝐴, this is the same as writing 𝐴=𝐴(𝜃).cos

Similarly, recall that sinoppositehypotenuse(𝜃)=.

We can rearrange this to make the opposite side of the triangle the subject of the equation: oppositehypotenusesin=×(𝜃).

Since the opposite side of the triangle is equal to the vertical component of the vector, 𝐴, and the hypotenuse is equal to the magnitude of the vector, 𝐴, this is the same as writing 𝐴=𝐴(𝜃).sin

We can use these two relations to find the horizontal and vertical components of a vector given its magnitude and argument.

Example 1: Finding the Components of a Vector on a Grid

Write 𝐴 in component form.

Answer

Looking at the grid, we can see that the vector has a horizontal length of 6 grid squares and a vertical length of 3 grid squares.

Recall that we can express any vector as a sum of multiples of unit vectors in the 𝑥- and 𝑦-directions: 𝐴=𝐴𝑖+𝐴𝑗, where 𝐴 is the horizontal component of the vector, 𝐴 is the vertical component of the vector, 𝑖 is a unit vector along the positive 𝑥-axis, and 𝑗 is a unit vector along the positive 𝑦-axis.

Looking at the diagram again, we see that the vector points in the positive 𝑥-direction, so its horizontal component is 6, but in the negative 𝑦-direction, so while its vertical length is 3, its vertical component is 3. Therefore, 𝐴=6𝑖3𝑗.

Example 2: Finding the 𝑥-Component of a Vector given Its Magnitude and Argument

The diagram shows a vector 𝐴 that has a magnitude of 22. The angle between the vector and the 𝑥-axis is 36. Work out the horizontal component of the vector. Give your answer to the nearest integer.

Answer

Recall that we can use the formula 𝐴=𝐴(𝜃),cos where 𝐴 is the magnitude of the vector and 𝜃 is the argument of the vector, to find the horizontal component of the vector, 𝐴.

Substituting in the values given in the question, we get 𝐴=22(36)𝐴=17.798.cos

Rounded to the nearest integer, this is 18.

Example 3: Finding the Components of a Vector given Its Magnitude and Argument

The diagram shows a vector, 𝐴, that has a magnitude of 91. The angle between the vector and the 𝑥-axis is 26. Give this vector in component form. Round all of the numbers in your answer to the nearest whole number.

Answer

Recall that we can use the formulae 𝐴=𝐴(𝜃),𝐴=𝐴(𝜃),cossin where 𝐴 is the magnitude of the vector and 𝜃 is the argument of the vector, to find the horizontal component of the vector, 𝐴, and the vertical component of the vector, 𝐴.

In this question, we are given the angle between the vector and the negative 𝑥-axis, but the argument of a vector is the angle counterclockwise from the positive 𝑥-axis. To find the argument of the vector, we just need to work out 18026, which is 154.

Using this value, as well as the value for the magnitude of the vector given in the question, with the first formula, we get 𝐴=91(154)𝐴=81.790.cos

Rounded to the nearest integer, this is 82.

Doing the same with the second formula, we get 𝐴=91(154)𝐴=39.891.sin

Rounded to the nearest integer, this is 40. The question asks us to give our answer in component form, which is the form 𝐴𝑖+𝐴𝑗, so our final answer is 82𝑖+40𝑗.

Key Points

  • Any vector 𝐴 can be represented as a sum of multiples of unit vectors along the 𝑥- and 𝑦-axes. If 𝐴 is the horizontal component of the vector, 𝐴 is the vertical component of the vector, 𝑖 is a unit vector along the 𝑥-axis, and 𝑗 is a unit vector along the 𝑦-axis, then 𝐴=𝐴𝑖+𝐴𝑗.
  • If a vector has been shown on a grid, we can find the components of the vector by counting the grid squares.
  • If we are given the magnitude and argument of a vector, we can find its components using trigonometry. If 𝐴 is the magnitude of the vector and 𝜃 is the argument, where the argument of a vector is the angle counterclockwise from the positive 𝑥-axis to the vector, then 𝐴=𝐴(𝜃),𝐴=𝐴(𝜃).cossin

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