Vector Basics Whereas a coordinate represents a position in space, a vector represents a displacement in space. A D If and are points then is the vector between them. Triangle Law for vector addition: + The vector of multiple vectors is

B If two vectors and have the same magnitude and direction, theyre the same vector and are This might seem parallel. obvious, but and the two vectors are parallel, equal in magnitude but in opposite directions. A vector has 2 properties: Direction Magnitude (i.e. length) C students sometimes think the vector is different because the movement occurred at a different point in

space. Nope! known as the resultant vector. (you will encounter this term in Mechanics) Vector Basics E Vector subtraction is defined using vector addition and negation: G A scalar is a normal number, which can be used to scale a vector. The direction will be the same. But the magnitude will be different (unless the scalar is 1). 2 If scalar is negative,

F 1 opposite direction. 2 The zero vector (a bold 0), represents no movement. H In 2D: Any vector parallel to the vector can be written as , where is a scalar. The implication is that if we can write one vector as a multiple of another, then we can show they are parallel. Show and are parallel. parallel Vectors A You can add and represent vectors using line segments OACB is a parallelogram. The points P, Q, M and N are the midpoints of the sides. a D P

O OB = b Express the following in terms of a and b. b) AB a+b b-a d) CN - /2a 1 c) QC /2b 1 b /2b - /2a 1 B What can you deduce about AB and QN, looking at the vectors? =

1 1 = 2 2 QN is a multiple of AB, so they are parallel! 1 = ( ) 2 e) QN 1 C N M OA = a a) OC Q 6B

Example Edexcel GCSE June 2013 1H Q27 Tip: This ratio wasnt in the original diagram. I like to add the ratio as a visual aid. a 3 : 2 For (b), theres two possible paths to get from to : via or via . But which is best? In (a) we found to rather than to , so it makes sense to go in this direction so that? we can use our result in (a). b Tip: While youre welcome to start your working with the second line, I recommend the first line so that your chosen route is clearer. ? = +

2 ? = + 5 is also because it is exactly the same movement as . Test Your Understanding Edexcel GCSE June 2012 ? ? Vectors You can add and represent vectors using line segments A =2 Express ON in terms of a and b 2 1 =2 + + 3

3 4 1 = + 3 3 a 1 N 2 O OB = b = + a M In triangle OAB, M is the midpoint of OA and N divides AB in the ratio 1:2. OM = a 1 =

3 Sub in values Simplify B b Use the ratio. If N divides AB in the ratio 1:2, show this on the diagram You can see now that AN is one-third of AB We therefore need to know AB To get from A to B, use AO + OB = + =2 + 2 1 = + 3 3 Sub in AO and OB AN = 1/3AB 6B Representing Vectors

3 2 ( ) You should already be familiar that the value of a vector is the displacement in the and direction (if in 2D). 3 0 = ?, = 2? 1 ( ) ( ) ! A unit vector is a vector of magnitude 1. and are unit vectors in the -axis and -axis respectively. e.g. Examples If then: 1) Write in vector form. 2) Find in form. 1

? 2 ? Magnitude of a Vector ! The magnitude of a vector is its length. 3 4 ( ) 2 2 ? | |= 3 + 4 =5 4 3 |( = = If

1 1 )| ? = 2 |( 5 12 4 ||= 4 2+ 12? = 17 1 ( ) 2 ||= 2 22 + 0 =2 ? 0 ( ) ?

)|=13 Unit Vectors ! A unit vector is a vector whose magnitude is 1 Theres certain operations on vectors that require the vectors to be unit vectors. We just scale the vector so that its magnitude is now 1. 3 =? 4 () If is a vector, then the unit vector in the same direction is Test Your Understanding: Convert the following vectors to unit vectors. ? ? Position Vectors Suppose we started at a point and translated by the vector : = 3 2 () 4

0 ( ) ( 3,2 ) ? You might think we can do something like: But only vectors can be added to other vectors. If we treated the point as a vector, then this solves the problem: ( 3 ,2 ) A vector used to represent a position is unsurprisingly known as a position vector. A position can be thought of as a translation from the origin, as per above. It enables us to use positions in all sorts of vector (and matrix!) calculations. ! The position vector of a point is the vector, where is the origin. is usually written as . Example The points and have coordinates and respectively.

Find, in terms of and : a) The position vector of b) The position vector of c) The vector a ? You can see this by inspection of the change in and the change in : b ? +8 c ? More formally: 2 Further Example and . Find: a) The position vector of . b) The exact value of in simplified surd form. a b ? ?

Either a quick sketch will help you see this, or thinking of as the original position and as the translation. Modelling In Mechanics, you will see certain things can be represented as a simple number (without direction), or as a vector (with direction): Vector Quantity Velocity Remember a scalar just means a normal number (in the context of vectors). It can be obtained using the magnitude of the vector. Equivalent Scalar Quantity Speed e.g. This means the position vector of the object changes by each hour. 4 5

? 3 Displacement e.g. Distance ? which is equivalent to moving 5km each hour. Example A girl walks 2 km due east from a fixed point to , and then 3 km due south from to . Find a) the total distance travelled b) the position vector of relative to c) d) The bearing of from . N 2 3 c d

? ? a b (3sf) so bearing = 146o ? ? Further Example In an orienteering exercise, a cadet leaves the starting point and walks 15 km on a bearing of to reach , the first checkpoint. From he walks 9 km on a bearing of to the second checkpoint, at . From he returns directly to . Find: a) the position vector of relative to b) c) the bearing of from d) the position vector of relative . b N N 120 a 30

15? 9 N 240 15 cos 30 13.0 = = 15 sin 30 7.5 ( ? ) ( ) Using cosine rule on , (3sf) N 30

15 60 ? 60 240 9 c Using sine rule on : Bearing (3sf) ? Further Example In an orienteering exercise, a cadet leaves the starting point and walks 15 km on a bearing of to reach , the first checkpoint. From he walks 9 km on a bearing of to the second checkpoint, at . From he returns directly to . Find: a) the position vector of relative to b) c) the bearing of from d) the position vector of relative . d N 120 . 30

. 15 23.41 9 N = ? ( 13.07 sin 23.41 = 5.19 13.07 cos 23.41 12.0 ) ( ) Area of a Triangle and . Determine . Strategy: Find 3 lengths of triangle then?use cosine rule to find angle. 3 2

( ) ? 1 5 ( ) | |= ?26 | |= ? 13 2 3 |= ? 13 ( ?) | A clever student might at this point realise that we can divide all the lengths by without changing , giving a triangle (one of our special triangles!), and thus instantly getting . But lets use a more general method of using the cosine rule: ? Test Your Understanding

In the above diagram, and . We wish to find the ratio . a) If , find an expression for in terms of and . b) If , find an expression for in terms of and . c) By comparing coefficients or otherwise, determine the value of , and hence the ratio . a b ? Comparing coefficients: and , If , then . ? c ? Expand and collect terms and collect terms, so that we can compare coefficients later.

Solving Geometric Problems is a point on such that . is the midpoint of . Show that is parallel to . For any proof question always find the vectors involved first, in this case and . is a multiple of parallel. ? The key is to factor out a scalar such that we see the same vector. The magic words here are is a multiple of. Introducing Scalars and Comparing Coefficients Remember when we had identities like:

we could compare coefficients, so that and . We can do the same with (nonparallel) vectors! is a parallelogram, where and . The diagonals and intersect at a point . Prove that the diagonals bisect each other. (Hint: Perhaps find in two different ways?) By considering the route : We dont know what fraction of the way across the point is, so let be the fraction of the way across. Were hoping that , so that is exactly halfway across and therefore bisects . Similarly, considering the line : Comparing coefficients of : Comparing coefficients of : is the midpoint of each of the diagonals. ?

We need to use a different scalar constant, this time . It is common to use the letters and for scalars. Terminology You can represent a vector by drawing a straight line with an arrowhead. The length of the line represents the magnitude (size) of the vector. The direction of the vector 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. The vector in Figure 12.1 has magnitude 5, direction +30. This is Written (5, 30) and said to be in magnitudedirection form. This is also called polar form. The general form of a vector written in this way is (r, ) where r is its magnitude and its direction. You usually give in the interval [0, 360). The vector a is = 4i + 2j. Write a and 2a in magnitudedirection form. The magnitude of a is given by the length a in Figure 12.4. to 3sf. The direction is given by the angle . to 3sf. So the vector a is (4.47, 26.6o). 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; so the magnitude of a is written a. Write the vectors (i) (5, 60), and (ii) (5, 300) in component form. i) (5, 60o) means r = 5 and .

In the right-angled triangle OPX OX = 5 cos 60 = 2.5 XP = 5 sin 60 = 4.33 to 2 d.p. or 2.5i + 4.33j. ii) In the right-angled triangle OPY YP = 5 cos 60 = 2.5 OY = 5 sin 60 = 4.33 to 2 d.p. or 2.5i 4.33j. In general, for all values of When you convert from component form to magnitudedirection form it is useful to draw a diagram and use it to see which quadrant the angle lies in so that you get the correct signs. This is shown in the next example. Write 5i + 4j in magnitudedirection form. The magnitude to 2 d.p. The direction is given by the angle in the diagram. You need to work out angle first. (to the nearest 0.1o). So The vector is (6.40, 141.3o) in magnitude-direction form. Forces as Vectors Forces have direction, and therefore we can naturally write them as vectors, either in - notation or as column vectors. ! You can find the resultant of two or more forces given as vectors by adding the vectors. The forces , , and act on an object which is in equilibrium. Find the values of and .

? If in equilibrium, resultant force is 0. The vector is due east and due north. A particle begins at rest at the origin. It is acted on by three forces N, N and N. (a) Find the resultant force in the form . (b) Work out the magnitude and bearing of the resultant force. a ? b 4 3 The magnitude of the force is the magnitude of the vector: ? Bearing Test Your Understanding Edexcel M1 May 2009 Q2

Edexcel M1 Jan 2012 Q3 Fro Tip: If a vector is parallel to say , then it could be any multiple of it, i.e. ? ? ? ? ? Motion in 2 dimensions In Chapter 8 we saw that many physical quantities could have both direction and magnitude, and therefore could be represented as a vector: Can be a vector: Force, acceleration, velocity, ? displacement Scalar only: Mass, area, volume ? This naturally means that works with vectors too. Let represent East and North. A resultant force of N acts upon a particle of mass 0.5 kg. (a) Find the acceleration of the particle in the form ms-2. (b) Find the magnitude and bearing of the acceleration of the particle. a ms-2 ? b

6 16 ms-2 (3sf) Bearing: ? Test Your Understanding A boat is modelled as a particle of mass 60 kg being acted on by three forces. Given that the boat is accelerating at a rate of ms-2, find the values of and . Resultant force: N and ? Vectors You can solve problems involving velocity, time, mass and forces (as in earlier chapters) by using vector notation =? =(3 + ) =? =(2 +3 ) =3 =+ =(3 + )+(2 +3 )(3) You can also solve problems involving acceleration by using:

= 3 + +6 +9 =3 +10 v = u + at Where v, u and a are all given in vector form. Particle P has velocity (-3i + j) ms at time t = 0. The particle moves along with constant acceleration a = (2i + 3j) ms-2. Find the speed of the particle after 3 seconds. -1 Sub in values Deal with the brackets Group terms Remember this is the velocity, not the speed! = 32 +10 2 =10.4 Calculate! 1 6F Vectors

You can solve problems involving velocity, time, mass and forces (as in earlier chapters) by using vector notation A force applied to a particle has both a magnitude and a direction, so force is a vector. The force will cause the particle to accelerate. Remember from chapter 3: =? =0 =(10 24 ) =? =10 =+ (10 24 )=0 +(10) 1024 =10 2.4 = Sub in values Tidy up Divide by 10 F = ma A constant force, FN, acts on a particle of mass 2kg for 10 seconds. The particle is initially at rest, and 10 seconds later it has a velocity of (10i 24j) ms-1. Find F. = =(2)( 2.4 ) =( 2 4.8 ) Sub in values

Calculate We need to find a first 6F Vectors You can use vectors to solve problems about forces = 1 + 2+ 3 If several forces are involved in a question a good starting point is to find the resultant force. The following forces: F1 = (2i + 4j) N F2 = (-5i + 4j) N F3 = (6i 5j) N all act on a particle of mass 3kg. Find the acceleration of the particle. ( 2 +4 ) + ( 5 +4 )+(6 5 ) ( 3 +3 ) = (3 +3 )=3 + = Sub in values

Group up Sub in the resultant force, and the mass Divide by 3 The acceleration is (i + j) ms-2 Start by finding the overall resultant force. 6G