# MTH5P1: Vectors

##### This unit explains Vectors

We drove into the future looking into a rear view mirror.

What information do you need to decide how close the aircraft
which left these vapour trails
passed to each other?

A quantity which has both size and direction is called a vector. The velocity of an aircraft through the sky is an example of a vector, having size (e.g. 600 mph) and direction (on a course of 254°). By contrast the mass of the aircraft (100 tonnes) is completely described by its size and no direction is associated with it; such a quantity is called a scalar.
Vectors are used extensively in mechanics to represent quantities such as force, velocity and momentum, and in geometry to represent displacements. They are an essential tool in three-dimensional co-ordinate geometry and it is this application of vectors which is the subject of this chapter. However, before coming on to this, you need to be familiar with the associated vocabulary and notation, in two and three dimensions.

Vectors in two dimensions.

Terminology
In two dimensions, it is common to represent a vector by a drawing of a straight line with an arrowhead. The length represents the size, or magnitude, of the vector and the direction is indicated by the line and the arrowhead. Direction is usually given as the angle the vector makes with the positive x axis, with the anticlockwise direction taken to be positive. Note
In the special case when the vector is representing real travel, as in the case of the velocity of an aircraft, the direction may be described by a compass bearing with the angle measured from north, clockwise.

However, this is not done in this chapter, where directions are all taken to be measured anticlockwise from the positive x direction.
An alternative way of describing a vector is in terms of components in given directions. The vector in figure 8.2 is 4 units in the x direction, and 2 in the  The magnitude of a is given by the length a in figure 8.4. The magnitude of a vector is also called its modulus and denoted by the symbols
| | . In the example a = 4i + 2j, the modulus of a, written | a |, is 4.47. Another convention for writing the magnitude of a vector is to use the same letter, but in italics and not bold type; thus the magnitude of a may be written a.

Write the vector (5, 60°) in component form.
SOLUTION
In the right-angled triangle OPX  This technique can be written as a general rule, for all values of θ. Write the vector (10, 290°) in component form.  In Example 8.3 the signs looked after themselves. The component in the i direction came out positive, that in the j direction negative, as must be the case for a direction in the fourth quadrant (270° < θ < 360°). This will always be the case when the conversion is from magnitude−direction form into component form.
The situation is not quite so straightforward when the conversion is carried out the other way, from component form to magnitude−direction form. In that case, it is best to draw a diagram and use it to see the approximate size of the angle required. This is shown in the next example.

Write −5i + 4j in magnitude−direction form. The vector is (6.40, 141.3°) in magnitude−direction form.

Vectors in three dimensions Points
In three dimensions, a point has three co-ordinates, usually called x, y and z. Figure 8.8

The axes are conventionally arranged as shown in figure 8.8, where the point P is (3, 4, 1). Even on correctly drawn three-dimensional grids, it is often hard to see the relationship between the points, lines and planes, so it is seldom worth your while trying to plot points accurately.
The unit vectors i, j and k are used to describe vectors in three dimensions.

Equal vectors
The statement that two vectors a and b are equal means two things.
● The direction of a is the same as the direction of b.
● The magnitude of a is the same as the magnitude of b.
If the vectors are given in component form, each component of a equals the corresponding component of b.
Position vectors
Saying the vector a is given by 3i + 4j + k tells you the components of the vector, or equivalently its magnitude and direction. It does not tell you where the vector is situated; indeed it could be anywhere.
All of the lines in figure 8.9 represent the vector a. There is, however, one special case which is an exception to the rule, that of a vector which starts at the origin. This is called a position vector. Thus the line Points L, M and N have co-ordinates (4, 3), (−2, −1) and (2, 2). Note
A line joining two points, like MN in figure 8.10, is often called a line segment, meaning that it is just that particular part of the infinite straight line that passes through those two points. The length of a vector
In two dimensions, the use of Pythagoras’ theorem leads to the result that a   Express the following vectors in component form. Draw diagrams to show these vectors and then write them in magnitude−direction form. Write, in component form, the vectors represented by the line segments joining the following points.

(i) (2, 3) to (4, 1)                                    (ii) (4, 0) to (6, 0)
(iii) (0, 0) to (0, −4)                             (iv) (0, −4) to (0, 0)
(v) (0, 0, 0) to (0, 0, 5)                        (vi) (0, 0, 0) to (−1, −2, 3)
(vii) (−1, −2 , 3) to (0, 0, 0)                (viii) (0, 2, 0) to (4, 0, 4)
(ix) (1, 2, 3) to (3, 2, 1)                         (x) (4, −5, 0) to (−4, 5, 1)

The points A, B and C have co-ordinates (2, 3), (0, 4) and (−2, 1).
(i) Write down the position vectors of A and C.
(ii) Write down the vectors of the line segments joining AB and CB.
(a) AB and OC
(b) CB and OA?

Vector calculations
Multiplying a vector by a scalar
When a vector is multiplied by a number (a scalar) its length is altered but its direction remains the same.

The vector 2a in figure 8.11 is twice as long as the vector a but in the same direction. When a is given in component form, the components of −a are the same as those for a but with their signs reversed. So; When vectors are given in component form, they can be added component by component. This process can be seen geometrically by drawing them on graph paper, as in the example below.

Add the vectors 2i − 3j and 3i + 5j.
SOLUTION
2i − 3j + 3i + 5j = 5i + 2j The sum of two (or more) vectors is called the resultant and is usually indicated by being marked with two arrowheads.

Adding vectors is like adding the legs of a journey to find its overall outcome (see figure 8.14). When vectors are given in magnitude−direction form, you can find their resultant by making a scale drawing, as in figure 8.14. If, however, you need to calculate their resultant, it is usually easiest to convert the vectors into component form, add component by component, and then convert the answer back to magnitude−direction form.

Subtracting vectors
Subtracting one vector from another is the same as adding the negative of the vector.

Two vectors a and b are given by

a = 2i + 3j b = −i + 2j.
(i) Find a − b.
(ii) Draw diagrams showing a, b, a − b.
SOLUTION

i) a − b = (2i + 3j) − (−i + 2j)
= 3i + j  This is an important result: where p and q are the position vectors of P and Q.
Geometrical figures
It is often useful to be able to express lines in a geometrical figure in terms of given vectors.
The diagram shows a cuboid OABCDEFG. P, Q, R, S and T are the mid-points of the edges they lie on. The origin is at O and the axes lie along OA, OC and OD, as shown in figure 8.17. (i) Name the points with the following co-ordinates.
(a) (6, 5, 4) (b) (0, 5, 0) (c) (6, 2.5, 0)
(d) (0, 2.5, 4) (e) (3, 5, 4)
(ii) Use the letters in the diagram to give displacements which are equal to the following vectors. Give all possible answers; some of them have more than one. Figure 8.18 shows a hexagonal prism.   Unit vectors
A unit vector is a vector with a magnitude of 1, like i and j. To find the unit
vector in the same direction as a given vector, divide that vector by its magnitude. The unit vector in the direction of vector a is written as â and read as ‘a hat’. Relative to an origin O, the position vectors of the points A, B and C are given by     Relative to an origin O, the position vectors of the points A, B and C are given by Find the perimeter of triangle ABC.
Relative to an origin O, the position vectors of the points P and Q are given Find the values of x for which the magnitude of PQ is 7.
Relative to an origin O, the position vectors of the points A and B are given by  As you work through the proof in this section, make a list of all the results that you are assuming.  The next example shows you how to use it to find the angle between two vectors given numerically. Perpendicular vectors
Since cos 90° = 0, it follows that if vectors a and b are perpendicular then a . b = 0.
Conversely, if the scalar product of two non-zero vectors is zero, they are perpendicular. Further points concerning the scalar product
● You will notice that the scalar product of two vectors is an ordinary number. It has size but no direction and so is a scalar, rather than a vector. It is for this reason that it is called the scalar product. There is another way of multiplying vectors that gives a vector as the answer; it is called the vector product. This is beyond the scope of this book.
● The scalar product is calculated in the same way for three-dimensional vectors. For example: The scalar product of two vectors is commutative. It has the same value whichever of them is on the left-hand side or right-hand side. Thus a . b = b . a, as in the following example. How would you prove this result?

The angle between two vectors where a . b is the scalar product of a and b. This result was proved by using the cosine rule on page 271.

Show that the angle between the three-dimensional vectors Working in three dimensions
When working in two dimensions you found the angle between two lines by using the scalar product. As you have just proved, this method can be extended into three dimensions, and its use is shown in the following example.   Three points P, Q and R have position vectors, p, q and r respectively, where
p = 7i + 10j, q = 3i + 12j, r = −i + 4j. Find the angles between these pairs of vectors.  Relative to an origin O, the position vectors of points A and B are 2i + j + 2k and 3i − 2j + pk respectively.
(i) Find the value of p for which OA and OB are perpendicular.
(ii) In the case where p = 6, use a scalar product to find angle AOB, correct to the nearest degree.  Relative to an origin O, the position vectors of the points A and B are given by Relative to an origin O, the position vectors of the points P and Q are given by  The diagram shows a semi-circular prism with a horizontal rectangular base ABCD. The vertical ends AED and BFC are semi-circles of radius 6 cm.
The length of the prism is 20 cm. The mid-point of AD is the origin O, the mid-point of BC is M and the mid-point of DC is N. The points E and F are the highest points of the semi-circular ends of the prism. The point P lies on EF such that EP = 8 cm. Unit vectors i, j and k are parallel to OD, OM and OE respectively. The diagram shows the roof of a house. The base of the roof, OABC, is rectangular and horizontal with OA = CB = 14 m and OC = AB = 8 m. The top of the roof DE is 5 m above the base and DE = 6 m. The sloping edges OD, CD, AE and BE are all equal in length. Unit vectors i and j are parallel to OA and OC respectively and the unit vector k is vertically upwards.   ## STEM Elearning

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