## Illustration 3.1: Vector Decomposition

*Please wait for the animation to completely load.*

A red vector is shown on a coordinate grid, and several properties of that
vector are given in a data table **(position is given in meters)**. How can you represent this vector? There are two ways: component form and
magnitude and direction form. Both ways of representing vectors are correct,
but in different circumstances one way may be more convenient than the other.
Restart.

*You can drag the head of the vector by click-dragging the
small circle at the vector's head.*

*Magnitude and Direction Form:* When you think of a vector, such
as the one depicted, you are thinking of the magnitude and direction form. We describe the magnitude as the size of the vector (depicted in the table as
r, which is always a positive number) and the direction as an angle (also depicted in the table and given in
degrees). This angle is measured from the positive x axis to the direction that the
vector is pointing.

*Component Form:* When you are solving problems in two
dimensions, you often need to decompose a vector into the component form.
So how do you do that? Look at the
show components version of the animation. As you drag around the red
vector, the maroon vectors show you the x and y components of the red vector
(these are also shown in the table as x and y). Try to keep the length of
the vector the same and change the angle. How did the components change
with angle? As the angle gets smaller, the x component of the vector gets
larger (it approaches the magnitude of the vector) and the y component of
the vector gets smaller (it approaches zero). As the angle approaches 90°
the x component of the vector gets smaller (it approaches zero) and the y
component of the vector gets larger (it approaches the magnitude of the vector).
Mathematically this is described by the statement that

x = r cos(θ) and y = r
sin(θ).

Once in component form, we can of course, go back to magnitude and direction
form by using the relationships

r = (x^{2} + y^{2})^{1/2} and
θ = tan^{-1}(y/x).

Notice that the magnitude of the vector, here r, must be
positive as stated above.

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