# Components of a vector

A vector is a quantity that has a length (a non-negative real number) as well as direction (or orientation) Vectors can be represented in one, two or three dimensions.

The components of a vector are the projections of said vector on the coordinate axis; In Figure I we see that vx and Vy are the projections of the vector V on the axes therefore these are the components of V.

## Defining components of a vector

Let's consider a rectangular or Cartesian coordinate system (Figure II). The component "x" (which we will call Ax) vector A  it is the shadow that the latter casts on the x axis; on the other hand, the component "y" (which we will call Ay) vector A is the shadow that the latter casts on the y axis. The vector sum of both components must result in the vector A:

Ax + Ay = A

## Notation

The components of a vector can be enclosed in parentheses and separated with commas:

A = (Ax, TOy)

In the case of three dimensions, it is expressed in this way:

A = (Ax, TOy, TOz)

We can also express them as a combination of unit vectors (i, j, k):

A = Ax î + Ay ĵ and A = Ax î + Ay ĵ + Az k

Other times it can be represented in matrix form as:

A = [Ax, TOy, TOz ]

## Rectangular components of a vector

From Figure II we find that:

• Ax = A cosθ
• Ay = A sinθ

These components are the sides of a right triangle whose hypotenuse has a magnitude A.

The module A and its address is related to its components such as:

|A| = √ [(Ax)2 + (Ay)2 ]

Y

tanθ = Ax / TOy

## Exercises

1. Find the magnitude of the vector A = 5î + 3ĵ - 2k

We see that 5 is the "x" component, 3 the "y" component and -2 the "z" component. Now, from the formula that relates the components to the magnitude of the vector, we have:

|A| = √ [(Ax)2 + (Ay)2 + (Az)2 ] = √ [(5)2 +(3)2 + (-2)2  = √ (25 + 9 +4) = 6.16

1. Find the direction of the vector A = 4î + 5ĵ

We see that 4 is the component "x" and 5 is "y". Now, from the formula that relates the components to the direction of the vector, we have:

tanθ = Ax / TOy = 4/5

θ = arctan (4/5) = 0.67 rad

1. Find the magnitude and direction of the vector A = 9î + 15ĵ

We see that 9 is the component "x" and -15 is "y". From the formula that relates the components to the magnitude of the vector, we have:

|A| = √ [(Ax)2 + (Ay)] = √ [(9)2 +(-15)2 = √ (81 + 225) = 17.49

Now, from the formula that relates the components to the direction of the vector, we have:

tanθ = Ax / TOy = 9/-15

θ = arctan (- 9/15) = - 0.54 rad