Imagine I am at a position O in a field. I walk 20 metres due east to a position X. My new position is different from my old position. My change of position is obviously 20 m but, in order to be specific, I must say 20 m due east of my former position. This change of position is called my displacement and it is represented by a line bearing an arrow and pointing to the east. This quantity, the displacement, has both magnitude (20 m), and direction (eastward). This is the nature of a vector quantity or a vector (see figure 5.1 (a)). It has both size (‘magnitude’) and direction. It is only completely defined if both its magnitude and its direction are stated. So displacement is a vector quantity.
Let’s now examine three different cases:
(i) I first walk 20 m due east and then I walk 15 m from X in a direction due east (figure 5.1 (a)).
(ii) I first walk 20 m due east and then I walk 15 m from X in a direction due west (figure 5.1 (b)).
(iii) I first walk 20 m due east and then I walk 15 m due south from the point X (figure 5.1 (c)).
The three additional displacements are all different displacements since, although they all have the same magnitude, 15 m, they each have a
different direction. When I add these different displacements to the original displacement, I get three different resultant displacements.
In case (i) (figure 5.1(a)), where I first walk 20 m due east to X and then 15 m due east from X to a point M1, I end up at M1, 35 m from O. My new
displacement ❯ resultant ❯ W E O X M1 M3 M2 X X (a) (b) (c) O O N S Figure 5.1
what direction. Clearly my resultant displacement is now 35 m from O in an easterly direction. It is given by the vector OM1.
In case (ii) (figure 5.1 (b)), I first walk 20 m east and then 15 m due west from X. I end up at a point M2, 5 m to the east of O, and so my new displacement is 5 m due east of O and is given by the vector OM2.
In case (iii) (figure 5.1 (c)), where I first walk 20 m from O to the point X and then 15 m from X due south, I will end up at the point M3, and my overall displacement will be OM3, whose distance from O is, by the geometry of the figure, 25 m, since OM32 = OX2 + XM 3 2 giving OM = √‾(OX2 + M 3X 2) = √‾(202 + 152) = √‾625 = 25
So the resultant obtained by adding vector XM to vector OX (or combining them) depends completely on the direction of vector XM and since this vector XM can take an innumerable number of directions between due east and due west through south and, of course, through north as well, there will be an equally innumerable number of resultants that are all different in magnitude or direction or both magnitude and direction. A careful look at the situation will reveal that the maximum value of the resultant is obtained by adding the first easterly displacement of 15 m, which gives a resultant of 35 m due east and the minimum a resultant of 5 m by subtracting the second westerly displacement of 15 m or, alternatively, by adding –15 m to the original 20 m east. We could summarise all of this by saying, in mathematical language, that the resultant displacement, d, is given by 5 m ≤ d ≤ 35 m.
You can see why I can get a range of very different answers for my distance from my starting point. In figure 5.1 (a) d is clearly equal to 35 m and in (b) it is clearly 5 m. If I walk due south, Pythagoras’ theorem gives d as 25 m.
For any other direction I can use a scale diagram to obtain the answer. Thus, if instead of walking south as in case (c) above, I walk south east from X, I can draw a scale diagram to find my resultant displacement (see figure 5.2). I will use a scale of 2 cm to represent 5 m. I first draw a line OX 8 cm long to represent 20 m of actual distance, then at X construct an angle of 45° with the easterly direction in a clockwise direction and from it cut off a length XM4 of 6 cm to represent 15 m. If we now measure the length OM4 it will represent my displacement from O.
To get the result we have used what is known as the triangle method to find the resultant of two vectors, where the vectors added were OX and XM4. This method is based on the triangle rule for finding the resultant of any two vectors which are in the same plane (we say that they are ‘co-planar’). This rule states that:
The resultant of two co-planar vectors may be represented by the third side of a triangle, two of whose sides are parallel to and proportional to the vectors and are constructed ‘head to tail’.
In using the rule, therefore, to obtain the answer to the case (i) above, you would:
1 Draw one side of the triangle, OX, to a scale of, say, 2 cm to 5 N.
2 From the ‘tail’ of OX (which is the point X) construct the second vector to scale; and so you would draw a line 6 cm from X in the same direction as OX. Denote the end of this vector by M.
Notation for vectors
When using letters (for example, OX) to represent a vector, sometimes we draw an arrow above the letters to indicate the direction of the vector. In this case we could write OX. The direction is in the order of the letters.
MATHEMATICS: algebra: the range of a variable, d
Pythagoras’ theorem ❯
the triangle method ❯
co-planar ❯
O X
M4
45°
Figure 5.2 Using the triangle method to
3 Draw the third ‘side’ of the triangle by joining the origin, O, to the ‘tail’ of the last vector drawn. This vector represents the resultant of the first two vectors.
4 Measure the length of this third ‘side’ OM of the triangle and use the scale to determine the value of the resultant (see figure 5.2).
This is the method by which vectors are added using the triangle of vectors. In this case the vectors we added were displacement vectors, the term displacement being used because it represented the distance we moved from the starting point. So my first displacement from O was OX, then my next displacement from X was XM. My overall displacement, called the resultant displacement, is measured from the original starting point, O, to the point where I ended my journey.
We could represent this movement from O to M using a general vector
equation:
OX + XM = OM
In a vector equation we give a positive sign to vectors directed to the right in accordance with convention. In keeping with this convention, vectors directed to the left are given a negative sign. Using this convention, the equation from figure 5.1 (b) would be:
OX + (–XM2) = OM2 or OX – XM2 = OM2
The treatment applied to displacement, a vector, may be applied to any other vector. Other examples of vectors you will meet in the ‘mechanics’ part of your course are velocity, acceleration, force and momentum. As we would expect for any vector, the effect produced by any one of these depends very much on the direction in which it is acting. Note again that in the examples just discussed, although there can be any number of displacement resultants, there can be only one value
of distance covered, and that is 35 m.
This is the value of the scalar quantity – distance covered.