Modelling displacements and velocities

by The Open University

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1.1: Converting to component form

In some applications of vectors there is a need to move backwards and forwards between geometric form and component form; we deal here with how to achieve this.

To start with, we recall definitions of cosine and sine. If P is a point on the unit circle, and the line segment OP makes an angle θ measured anticlockwise from the positive x-axis, then cos θ is the x-coordinate of P and sin θ is the y-coordinate of P (see Figure 1(a)). In other words, P has coordinates (cosθ, sinθ).

Figure 1 Figure 1 (a) Definition of cos θ and sin θ  (b) Coordinates of A Long description

Now suppose that the unit circle is scaled by a factor r in both the x- and y-directions, to produce a circle of radius r, as illustrated in Figure 1(b). If P is moved to the point A by this scaling, then the line segment OA has length r, and the coordinates of A are (r cos θ, r sin θ).

The coordinates of the point A are the components of the position vector of A. Thus the vector

has magnitude

and component form r cos θ i + r sin θ j.

The numbers r, θ are called the polar coordinates of A.

Suppose that a vector a is given in terms of its magnitude |a| and direction θ. Then there is a unique point A in the plane such that the position vector of A is equal to a, namely, the point which is distance |a| from O along a straight line that makes an angle θ with the positive x-axis (see Figure 2).

Figure 2 Figure 2 The vector a defines a point A Long description

But since

for this point A, the quantity |a| takes the place of r in the earlier discussion. We therefore have the following result.

From geometric to component form

Any vector a, with direction θ, has component form

In other words, its i-component is:

a a = |a| cos θ and its j-component is: a b = |a| sin θ.

Example 1 Finding the component form

The vector a has magnitude 4 and direction 120°. Calculate the component form of a, giving the components:

  1. correct to four decimal places;

  2. as exact values.

Answer

Solution

  1. According to the formula in the box above, the component form is

    View larger image 

  2. Exact values for the components can be obtained where one of the ‘special angles? 30°, 45° or 60° is seen to be involved. Here the given direction is 120°, which is 180° − 60°. Using the trigonometric identities

    View larger image 

    we obtain

The outcome is illustrated in Figure 3. A diagram such as this could have been used, instead of the trigonometric identities above, to decide upon the sign and size of each component.

Figure 3 Figure 3 The vector of Example 1Long description

A calculator was used to obtain the 4 d.p. result.

Activity 1 Finding the component form

  1. For each case below, calculate the component form of the vector a whose magnitude |a| and direction θ are given, specifying the components correct to four decimal places.

  2. For each case below, calculate the component form of the vector a whose magnitude |a| and direction θ are given, specifying the components as exact values.

Answer

Solution

In each case, we apply the formula

    1. Here |a| = 3 and θ = 110°, so the component form is

    2. Here |a| = 2.5 and θ = −20°, so the component form is

  2. The results for part (b) are illustrated in Figure 4.

    1. Here |a| = 1 and θ = 45°, so the component form is

      Here |a| = 5 and θ = −150°. Using in turn the trigonometric identities

      we obtain the component form

Figure 4 Figure 4 Long description

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