Dripread - Book

You have seen how any vector given in geometric form, in terms of magnitude and direction, can be written in component form. You will now see how conversion in the opposite sense may be achieved, starting from component form. In other words, given a vector a = a ₁ i + a ₂ j, what are its magnitude |a| and direction θ?

The first part of this question is dealt with using Pythagoras? Theorem: the magnitude of a vector a is given in terms of its components by

As regards finding the direction, we first deal with some special cases, which are illustrated in Figure 5.

Figure 5 Four special cases for the direction of a vectorLong description

These cases may be described as follows:

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Suppose now that none of these special cases applies. It follows from earlier in this subsection that the magnitude |a|, direction θ and components a ₁, a ₂ are related by the pair of equations

Previously, these gave the components in terms of known magnitude and direction, but now the situation is reversed: a ₁ and a ₂ are known, and (with |a| already found as above) these equations must be solved for the remaining unknown, θ. The equations, taken together, have a unique solution for θ within the range −180° < θ ≤180°.

There are various ways of finding this solution. We shall use an approach that involves taking an inverse tangent. If the equation |a| sin θ = a ₂ is divided by the equation |a| cos θ = a ₁, then we obtain

Note that the possibility of division by zero is avoided here, because a ₁ = 0 corresponds to one of the special cases dealt with above and now excluded.

This suggests that the direction θ is found by taking the arctan of both sides of the last equation. However, care is needed at this point, because the equation for tan θ has more than one solution within the desired range −180° < θ ≤ 180°. (For example, we have tan (45°) = 1, but also tan (−135°) = 1.) The following strategy shows how to choose between the possibilities on offer for θ.

The function arctan is the inverse function of the function f(x) = tan x with domain (−90°, 90°). So values of arctan lie in the range from −90° to 90°. In particular, since

|a ₂/a ₁| > 0,

we have

0 < arctan (|a ₂/a ₁|) < 90°.

Evaluate φ = arctan (|a ₂/a ₁|). This gives the angle (between 0 and 90° since |a ₂/a ₁| > 0) that the position vector

makes with the positive or negative x-axis, where the point A has coordinates (a ₁, a ₂). There are then four possibilities, as detailed in Figure 6(a).
The quadrant in which A lies is found from its coordinates, (a ₁, a ₂), and this determines which of the four possibilities applies. The corresponding value of θ is then found from that of φ as indicated in Figure 6(b).

Figure 6 (a) The point A lies in one of the four quadrants (b) Corresponding value of θ Long description

The following box summarises how to find the geometric form of a vector from its component form.

From component to geometric form

A vector a = a ₁ i + a ₂ j has magnitude

If the vector is non-zero, then its direction θ is obtained as follows.

If a = a ₁ i or a = a ₂ j, then θ can be found directly, as shown in Figure 5.
For a case other than those above, find φ = arctan (|a ₂/a ₁|)
Find in which quadrant of the plane the point A (a ₁, a ₂) lies. The value of θ (in terms of φ) is then given according to Figure 6(b).

Example 2 Finding the geometric form

Find the magnitude and direction of each of the following vectors.

Answer

Solution

The magnitude of all these vectors is the same, namely,

It remains to find the direction θ in each case. Arrows to represent the four vectors are shown in Figure 7.

Figure 7 The four vectorsLong description

For

we have

and

Also,

lies in the first quadrant, so the direction of a is θ = φ = 30°.
For

we have

and

Also,

lies in the fourth quadrant, so the direction of b is θ = −φ = −60°.
For

we have

and

Also,

lies in the third quadrant, so the direction of c is θ = −(180° − φ) = −120°.
For

we have

and

Also,

lies in the second quadrant, so the direction of d is θ = 180° − φ = 150°.

All of these cases involved exact values for φ and θ. In general, this will not be so.

Activity 2 Finding the geometric form

Find the magnitude and direction of each of the following vectors. Give answers as exact values where possible; otherwise, give the direction correct to one decimal place. (There is no need to reduce magnitudes to decimal form.)

e = −3i + 3j
f = 4i − 2j
g = −2i − 3j
h = −3.5j

Answer

Solution

Arrows to represent the four vectors are shown in Figure 8.

Figure 8 Long description

The magnitude of the vector e = −3i + 3j is

Since the components of e are e ₁ = −3, e ₂ = 3, we have

Also, (−3, 3) lies in the second quadrant, so the direction of e is θ = 180° − φ = 135°.
The magnitude of the vector f = 4i − 2j is

Since the components of f are f ₁ = 4, f ₂ = −2, we have

Also, (4, −2) lies in the fourth quadrant, so the direction of f is θ = − φ ≈ −26.6°.
The magnitude of the vector g = −2i − 3j is

Since the components of g are g ₁ = −2, g ₂ = −3, we have

Also, (−2, −3) lies in the third quadrant, so the direction of g is θ = −(180° − φ) ≈ −123.7°.
The magnitude of the vector h = −3.5j is

This vector has the form h = h ₂ j, so its direction can be found directly from Figure 5. Since h ₂ = −3.5 < 0, the direction is θ = −90°.