Vectors.

Lesson 1

Addition and subtraction of vectors.

Vectors multiplied by a constant.

Some quantities like temperature and length can be described by using a single number to indicate their magnitude. These are called scalar quantities.
Other quantities such as forces and velocity need direction as well as magnitude. These are called vectors. A vector is not necessarily fixed in space. Moving 5 m to the north on a horizontal plane is the same action wherever it is performed.

In mathematics vectors are drawn as line segments with an arrow. The length of the line represents the magnitude of the vector and the direction of the line represents the direction of the vector.

For example a translation of 5 m to the north could be represented by a vertical line of length 5 cm with the arrow pointing up ( north ). A translation of 2 m east could be drawn as a horizontal line of length 2 cm pointing to the right. (see vectors and in the diagram).

A vector is a directed line segment symbolised by putting an arrow at the end of the line segment in the required direction. The length of the line segment represents the magnitude of the vector.

Vectors are written in texts as small letters with a line or arrow over them. If a vector begins in point A and ends in point B it is often written as ( with a line or an arrow). Vectors that have the same length and direction are equal. A vector is independent of its position in the coordinate system.

In the diagram below the line vectors AE, FK and DH all have the same length and direction and can all be labelled as the vector . If we draw the vectors BF, CG, EJ, GL and NM they can also be labelled as the vector .

In the diagram the vector ( starts in L and ends in G) is parallel to the vector but points in the opposite direction. We can call the vector −.

Now look at further possibilities.

If the directed line AE represents the vector then a line in the same direction but twice as long is the vector 2.
In the same way, if the directed line BD represents the vector then GH, which is half the length of BD, must be ½∙.

Our results can be summarised as follows

Multiplying a vector by a constant number affects only the length not the direction. If the number is negative the direction of the vector is reversed.

These results are shown in the diagram below where we have joined two directed line segments together, both representing the vector .

It is logical to draw the conclusion that + is the same as 2. It is not surprising therefore to see that adding two vectors together is done by joining them together so that the head of one joins the tail of the other.

In the above diagram the directed line AE represents the vector and KM the vector . BK is parallel to AE but twice as long. By joining together BK (2) and KM () we have made the vector 2 + represented by BM.

Adding two vectors involves joining them together so that the second vector begins where the first vector ends. The sum of the two vectors is then a directed line segment that goes from the beginning of the first vector to the end of the second.

Example 1:

In the diagram below, AE represents the vector , AC the vector and AM the vector .

how can we write using and ?
We can do this by doubling (by adding the vector EJ to AE) and then adding 1½ to it. AJ + JM. (See diagram).

The result is = 2 + 1½ .

This operation is called finding the components of the vector in terms of and . The components are 2 and 1½ .

Example 2:

We now use the same coordinate points as in example 1 to find a directed line that can represent − .

We begin by assuming that − is the same vector as + (− ).

We need more room in our diagram so we will draw the vector with G as its starting point. That gives us GL.

Next we add a vector that has the same length as but points in the opposite direction, that’s LJ. Our starting point was G and we finished in J so the resulting translation is from G to J, giving the vector = − .

The above diagram shows that , drawn from the end and to the end of , is the same vector as and therefore also equal to − . This gives us a second method for finding the difference between two vectors.

The difference between two vectors, − can be found by reversing the direction of vector and adding it to vector .

It can also be found by drawing and so that they start in the same point. Then − is the vector that goes from the end of to the end of .

The fact that a vector is represented by lengths and directions ( angles ) means that we often have to use trigonometry to solve vector problems, for example to calculate the length and direction of the sum or difference of two vectors.

It’s customary to use the absolute value symbol when talking about the length of a vector.

The length of vector is written ||.

Example 3:

In the building industry engineers and architects need to be able to calculate the effect of the forces that interact in different parts of the building. Imagine that the vectors in the diagram below represent forces all working on the same object. We need to find out what the total effect will be.

First we use vector addition to add them together.

The added vectors end one unit below the starting point. This means that the effect of all four forces is to pull the object they work on, one unit vertically down. In other words a force of one unit vertically down has exactly the same effect as the other four forces together.

Example 4:

Using the same vectors as in example 3 find the vector − − + .

We reverse the direction of and and then join all the vector together.

The result is the red vector in the diagram, which is drawn from the starting point of the first vector to the end of the fourth one. This emphasises the fact that the position of the vectors does not matter, only the length and direction must remain unchanged. In fact we do not need to use a coordinate system at all at this stage, it is used to make it easier to draw vectors correctly.

Example 5:

A ship sails north at a speed of 24 nautical miles per hour
( knots ). A strong ocean current of 7 knots from the west flows directly across the ship. What is the actual speed and direction of the ship relative to the ocean floor?

First draw the two vectors and add them together. The numbers 24 and 7 form the sides of a right angled triangle where the hypotenuse shows the actual speed and direction of the ship. If we call the starting point A and the finishing point B then | | is the actual speed of the ship and the angle between AB and the vertical, call this A, will give us the angle that the ship has been turned through ( bearings ).

Using Pythagoras to find ||:

| | can we find by using Pythagoras.

| |² = 24² + 7²

= 576 + 49 = 625

| | =

= 25 Knots.

We find the angle A using the rule for tan:

tan A = opposite side/adjacent side.

= 7/24

A = tan⁻¹(7/24) ≈ 16°

Example 6:

A ferry boat has to cross a river that is 100 m wide. The boat can sail at 3 m/s and the river is flowing at a speed of 1 m/s.

a) How long does it take the ferry to cross the river if he makes no attempt to counteract the effect of the rivers flow?

We can ignore the flow of the river as the ferry makes no attempt to work against the stream. We simply need to calculate how long it takes to travel 100m. at a speed of 3 m/s.

Time = 100 m/3 m/s = 33⅓ s

b) Calculate the actual speed of the boat.

The diagram shows the ferry and the river and what happens when we add the two vectors together.

x shows the actual speed of the ferry. using Pythagoras we get :

   x² = 3² + 1² = 9 + 1 = 10

     x = ≈ 3,2 m/s

c)   How far down river has the ferry travelled when it has crossed?

In this case x represents the distance we need to calculate and 100 is the width of the river. They form the short sides of a right angled triangle. This triangle is similar ( in the same proportions) as the previous diagram so we can use ratios to calculate x.

   1/3 = x/100

Multiply through by 100.

      x = 100/ 3 = 33⅓ m

We have used vectors to solve this problem but in fact it is a simple geometric problem that can easily be solved.
We know that the journey takes 33⅓ s and that the river moves the ferry downstream 1 m every second. Therefore it follows that in 33⅓ s the ferry moves 33⅓ m downstream.

d)   The ferry operator wishes to land on the bank exactly opposite to where he started the journey. What direction does he need to take in order to counteract the effect of the rivers flow?

He has to direct the boat upstream so that for every 3 metres that he travels the river takes him 1 m downstream. We call the angle y° and use the rule for sine to find y°.

         y° = sin⁻¹(1/3) ≈ 19,5°

e)   Calculate the actual speed of the boat when he takes the direction found in d).

The actual speed of the boat is labelled x m/s and is found by adding the vectors of length 3 and length 1 together

( see the diagram below). x is one of the short sides in the right angled triangle so we can either use the trig functions or Pythagoras to calculate x.

We will use Pythagoras rule.

   x² + 1² = 3²

           x² = 9 − 1 = 8

            x = Ö8 ≈ 2,8 m/s

Example 7:

A heavy sledge is drawn by using two lengths of rope tied to the front of the sledge and that make an angle of 60° between them. A force of 100 N is used on each of the ropes. Find the combined force that is applied to the sledge to move it straight ahead. First we draw a diagram.

We use vector addition to add the two forces together and then use trig rules in the resulting triangle.

The vector x that we need to calculate divides the angle of 60°in half so that the angle x makes with each force is 30°. The third angle in the triangle is 120°. Using the sine rule we get

The Sine Rule:

Try Quiz 1 on Vectors.
Remember to use the checklist to keep track of your work.