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# 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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