Introduction In this chapter you will learn about

Introduction • In this chapter you will learn about Vectors • You will have seen vectors at GCSE level, this chapter focuses on using them to solve problems involving SUVAT equations and forces • Sometimes using vectors offers an easier alternative to regular methods • Vectors are used in video games in the movement of characters and by engineers in the design of buildings, bridges and other structures

Vectors You can use vectors to describe displacements A vector has both direction and magnitude For example: An object is moving north at 20 ms-1 A horizontal force of 7 N An object has moved 5 m to the left These are all vectors. A scalar quantity would be something such as: A force of 10 N Vectors have both direction and magnitude! (It is scalar since it has no direction) 6 A

Vectors N Adj You can use vectors to describe displacements A girl walks 2 km due east from a fixed point O, to A, and then 3 km due south from A to a point B. Describe the displacement of B from O. Start, as always, with a diagram! To describe the displacement you need the distance from O as well as the direction (as a bearing) Remember bearings are always measured from north! “Point B is 3. 61 km from O on a bearing of 146˚” O 2 km θ 56. 3˚ Describing the displacement The distance – use Pythagoras’ Theorem Sub in a and b A 3 km Opp B Calculate The bearing – use Trigonometry to find angle θ Sub in opp and adj Use inverse Tan Bearings are measured from north. Add the north line and add 90˚ 6 A

Vectors N You can use vectors to describe displacements In an orienteering exercise, a cadet leaves the starting point S and walks 15 km on a bearing of 120˚ to reach A, the first checkpoint. From A he then walks 9 km on a bearing of 240˚ to the second checkpoint, and point B. From B he then returns directly to S. Describe the displacement of S from B. Start with a diagram! We need the distance B to S and the bearing from B to S as well. You will need to use angle rules you have learnt pre Alevel. You can use interior angles to find an angle in the triangle Interior angles add up to 180° The missing angle next to 240 is 60° The angle inside the triangle must also be 60° S 120° N b 15 km 60° 13. 1 km a 240° A 9 km B c Finding the distance B to S Sub in values Work out Square root

Vectors N You can use vectors to describe displacements A From B he then returns directly to S. Describe the displacement of S from B. Start with a diagram! We need the distance B to S and the bearing from B to S as well. You will need to use angle rules you have learnt pre Alevel. S is 13. 1 km from B on a bearing of 337° 120° θ 36. 6° In an orienteering exercise, a cadet leaves the starting point S and walks 15 km on a bearing of 120˚ to reach A, the first checkpoint. From A he then walks 9 km on a bearing of 240˚ to the second checkpoint, and point B. S N 15 km 13. 1 km 60° N 60° b Finding the bearing from B to S Show the bearing at B It can be split into 2 sections, one of which is 180° Find angle θ inside the triangle 9 km 156. 6° B 180° 240° A B a You can now use Alternate angles to find the unknown part of the bearing Sub in values Rearrange Calculate θ Add on 180° The bearing is 336. 6°

Vectors You can add and represent vectors using line segments A vector can be represented as a directed line segment Two vectors are equal if they have the same magnitude and direction You can add vectors using the triangle law of addition 3 a a Two vectors are parallel if they have the same direction C A B 6 B

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 -1 / 2 a 1/ 2 b e) QN 1/ 2 b B b What can you deduce about AB and QN, looking at the vectors? c) QC b-a d) CN C N M OA = a a) OC Q - 1/ 2 a QN is a multiple of AB, so they are parallel! 6 B

Vectors You can add and represent vectors using line segments A a M In triangle OAB, M is the midpoint of OA and N divides AB in the ratio 1: 2. OM = a Express ON in terms of a and b a 1 N 2 O OB = b Sub in values Simplify 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 B Sub in AO and OB AN = 1/3 AB 6 B

Vectors You can add and represent vectors using line segments a OABC is a parallelogram. P is the point where OB and AC intersect. The vectors a and c represent OA and OC respectively. Prove that the diagonals bisect each other. If the diagonals bisect each other, then P must be the midpoint of both AC and OB… Try to find a way to represent OP in different ways… (make sure you don’t ‘accidentally’ assume P is the midpoint – this is what we need to prove!) c A O B P C c One way to get from O to P Start with OB OP is parallel to OB so is a multiple of (a + c) We don’t know how much for now, so can use λ (lamda) to represent the unknown quantity 6 B

Vectors A You can add and represent vectors using line segments The vectors a and c represent OA and OC respectively. If the diagonals bisect each other, then P must be the midpoint of both AC and OB… Try to find a way to represent OP in different ways… (make sure you don’t ‘accidentally’ assume P is the midpoint – this is what we need to prove!) c a OABC is a parallelogram. P is the point where OB and AC intersect. Prove that the diagonals bisect each other. O B P -a C c Another way to get from O to P Go from O to A, then A to P We will need AC first… AP is parallel to AC so is a multiple of it. Use a different symbol (usually μ, ‘mew’, for this multiple) Now we have another way to get from O to P Sub in vectors 6 B

Vectors A You can add and represent vectors using line segments The vectors a and c represent OA and OC respectively. (make sure you don’t ‘accidentally’ assume P is the midpoint – this is what we need to prove!) P O C Prove that the diagonals bisect each other. Try to find a way to represent OP in different ways… B a OABC is a parallelogram. P is the point where OB and AC intersect. If the diagonals bisect each other, then P must be the midpoint of both AC and OB… As these represent the same vector, the expressions must be equal! Multiply out brackets Factorise the ‘a’ terms on the right side Now compare sides – there must be the same number of ‘a’s and ‘c’s on each Sub 2 nd equation into the first Rearrange and solve They are equal So P is halfway along OB and AC and hence the lines bisect each other! 6 B

Vectors (0, 1) You can describe vectors using the i, j notation j A unit vector is a vector of length 1. Unit vectors along Cartesian (x, y) axes are usually denoted by i and j respectively. O i C 5 i + 2 j You can write any two-dimensional vector in the form ai + bj Draw a diagram to represent the vector -3 i + j (1, 0) A 2 j B 5 i j -3 i + j -3 i 6 C

Vectors You can solve problems with vectors written using the i, j notation When vectors are written in terms of the unit vectors i and j you can add them together by adding the terms in i and j separately. Subtraction works in a similar way. Add the i terms and j terms separately Given that: p = 2 i + 3 j q = 5 i + j Find p + q in terms of i and j 6 D

Vectors You can solve problems with vectors written using the i, j notation When vectors are written in terms of the unit vectors i and j you can add them together by adding the terms in i and j separately. Subtraction works in a similar way. Multiply out the bracket Careful with the subtraction here! Group terms… Given that: a = 5 i + 2 j b = 3 i - 4 j Find 2 a – b in terms of i and j 6 D

Vectors 3 i You can solve problems with vectors written using the i, j notation -7 j When a vector is given in terms of the unit vectors i and j, you can find its magnitude using Pythagoras’ Theorem. The magnitude of vector a is written as |a| Find the magnitude of the vector: 3 i – 7 j 3 i - 7 j Put in the values from the vectors and calculate Round if necessary 6 D

Vectors You can solve problems with vectors written using the i, j notation Opp y 5 j 51. 3° θ You can also use trigonometry to find an angle between a vector and the axes -4 i x Adj Find the angle between the vector -4 i + 5 j and the positive x-axis Draw a diagram Sub in values Inverse Tan The angle we want is between the vector and the positive x-axis Subtract θ from 180° 6 D

Vectors You can solve problems with vectors written using the i, j notation Given that: Start by calculating a + µb in terms of a, b and µ Move the i and j terms together a = 3 i - j b=i+j Find µ if a + µb is parallel to 3 i + j Multiply out the brackets Factorise the terms in i and j As the vector must be parallel to 3 i + j, the i term must be 3 times the j term! Multiply out the bracket Subtract µ, and add 3 Divide by 2 6 D

Vectors You can solve problems with vectors written using the i, j notation Given that: To show that this works… We now know µ a = 3 i - j b=i+j Find µ if a + µb is parallel to 3 i + j Start by calculating a + µb in terms of a, b and µ Multiply out the brackets Group terms Factorise You can see that using the value of µ = 3, we get a vector which is parallel to 3 i + j 6 D

Vectors 3 i + j You can express the velocity of a particle as a vector The velocity of a particle is a vector in the direction of motion. The magnitude of the vector is its speed. Velocity is usually represented by v. A particle is moving with constant velocity given by: v = (3 i + j) ms-1 Find: a) The speed of the particle b) The distance moved every 4 seconds j 3 i Finding the speed The speed of the particle is the magnitude of the vector Use Pythagoras’ Theorem Calculate Finding the distance travelled every 4 seconds Use GCSE relationships Distance = Speed x Time Sub in values (use the exact speed!) Calculate 6 E

Vectors You can solve problems involving velocity, time, mass and forces (as in earlier chapters) by using vector notation A particle starts from the point with position vector (3 i + 7 j) m and then moves constant velocity (2 i – j) ms -1. Find the position vector of the particle 4 seconds later. If a particle starts from the point with position vector r 0 and moves with constant velocity v, then its displacement from its initial position at time t is given by: (a position vector tells you where a particle is in relation to the origin O) Time Starting position Final position Velocity Sub in values Multiply/remove brackets Simplify 6 F

Vectors You can solve problems involving velocity, time, mass and forces (as in earlier chapters) by using vector notation A particle moving at a constant velocity, ‘v’, and is at the point with position vector (2 i + 4 j) m at time t = 0. Five seconds later the particle is at the point with position vector (12 i + 16 j) m. Find the velocity of the particle. If a particle starts from the point with position vector r 0 and moves with constant velocity v, then its displacement from its initial position at time t is given by: Deal with the brackets! Subtract 2 i and add 4 j Velocity Sub in values Time Starting position Final position Divide by 5 The velocity of the particle is (2 i + 4 j) ms-1 6 F

Vectors 3 i You can solve problems involving velocity, time, mass and forces (as in earlier chapters) by using vector notation At time t = 0, a particle has position vector 4 i + 7 j and is moving at a speed of 15 ms-1 in the direction 3 i – 4 j. Find its position vector after 2 seconds. 3 i – 4 j You have to be careful here, you have been given the speed of the particle. The direction vector does not necessarily have the same speed 5 ms-1 -4 j 9 i – 12 j -12 j 15 ms-1 Multiply all vectors by 3 Find the speed of the direction vector as it is given in the question Then ‘multiply up’ to get the required speed (we need 15 ms-1, not 5 ms-1) Multiplying the vectors will allow you to use the correct velocity 9 i Calculate We can use the vectors as the velocity 6 F

Vectors You can solve problems involving velocity, time, mass and forces (as in earlier chapters) by using vector notation At time t = 0, a particle has position vector 4 i + 7 j and is moving at a speed of 15 ms-1 in the direction 3 i – 4 j. Find its position vector after 2 seconds. You have to be careful here, you have been given the speed of the particle. The direction vector does not necessarily have the same speed Sub in values ‘Deal with’ the brackets Group terms Find the speed of the direction vector as it is given in the question Then ‘multiply up’ to get the required speed (we need 15 ms-1, not 5 ms-1) Multiplying the vectors will allow you to use the correct velocity 6 F

Vectors You can solve problems involving velocity, time, mass and forces (as in earlier chapters) by using vector notation You can also solve problems involving acceleration by using: v = u + at Where v, u and a are all given in vector form. Particle P has velocity (-3 i + j) ms-1 at time t = 0. The particle moves along with constant acceleration a = (2 i + 3 j) ms-2. Find the speed of the particle after 3 seconds. Sub in values ‘Deal with’ the brackets Group terms Remember this is the velocity, not the speed! Calculate! 6 F

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. A constant force, FN, acts on a particle of mass 2 kg for 10 seconds. The particle is initially at rest, and 10 seconds later it has a velocity of (10 i – 24 j) ms-1. Find F. Sub in values ‘Tidy up’ Remember from chapter 3: F = ma Divide by 10 Sub in values Calculate We need to find a first… 6 F

Vectors You can use vectors to solve problems about forces If a particle is resting in equilibrium, then the resultant of all the forces acting on it is zero. The forces (2 i + 3 j), (4 i – j), (-3 i + 2 j) and (ai + bj) are acting on a particle which is in equilibrium. Calculate the values of a and b. Group together the numerical terms The ‘i’ terms must sum to 0 The ‘j’ terms must sum to 0 Set the sum of all the vectors equal to 0 6 G

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

Vectors You can use vectors to solve problems about forces A B 30° A particle P, of weight 7 N is suspended in equilibrium by two light inextensible strings attached to P and to points A and B as shown in the diagram. Line AB is horizontal. Find the tensions in the two strings Draw a sketch of the forces acting on P These can be rearranged into a triangle of forces (the reason being, if the particle is in equilibrium then the overall force is zero – ie) The particle ends up where it started) You will now need to work out the angles in the triangle… 40° P TA TB 7 N TB P 7 N 7 N These are the forces acting on P TA These are the forces rearranged as a triangle 6 G

Vectors You can use vectors to solve problems about forces A B 30° A particle P, of weight 7 N is suspended in equilibrium by two light inextensible strings attached to P and to points A and B as shown in the diagram. Line AB is horizontal. Find the tensions in the two strings You will now need to work out the angles in the triangle… Consider the original diagram, you could work out more angles on it as shown, some of which correspond to our triangle of forces… 40° 60° 50° P TB 50° 7 N 70° 60° TA Now we can calculate the tensions! 7 N The angle between 7 N and TA is 60° The angle between 7 N and TB is 50° (It is vertically opposite on our triangle of forces) The final angle can be worked out from the triangle of forces alone 6 G

Vectors You can use vectors to solve problems about forces A B 30° 40° A particle P, of weight 7 N is suspended in equilibrium by two light inextensible strings attached to P and to points A and B as shown in the diagram. Line AB is horizontal. Find the tensions in the two strings To calculate the tensions you can now use the Sine rule (depending on the information given, you may have to use the Cosine rule instead!) 60° 50° P TB 50° 7 N 70° 60° TA 7 N Multiply by Sin 50 Calculate 6 G

Vectors You can use vectors to solve problems about forces A B 30° 40° A particle P, of weight 7 N is suspended in equilibrium by two light inextensible strings attached to P and to points A and B as shown in the diagram. Line AB is horizontal. Find the tensions in the two strings To calculate the tensions you can now use the Sine rule (depending on the information given, you may have to use the Cosine rule instead!) 60° 50° P TB 50° 7 N 70° 60° TA 7 N Multiply by Sin 60 Calculate 6 G

Vectors You need to be able to solve worded problems in practical contexts The mixed exercise in this chapter is very important as it contains questions in context, the type of which are often on exam papers 6 H

Vectors N You need to be able to solve worded problems in practical contexts A ship S is moving with constant velocity ( – 1 – 2. 5 i + 6 j) kmh. At time 1200, the position vector of S relative to a fixed origin O is (16 i + 5 j) km. Find: (a) the speed of S 6. 5 kmh-1 (b) the bearing on which S is moving. The speed of S 6 j Calculate 67. 4° θ Use Pythagoras’ Theorem 180° -2. 5 i The bearing on which S is travelling 337° The ship is heading directly towards a submerged rock R. A radar tracking station calculates that, if S continues on the same course with the same speed, it will hit R at the time 1500. Find angle θ Use Tan = Opp/ Adj Calculate (c) Find the position vector of R. Consider the north line and read clockwise… 6 H

Vectors You need to be able to solve worded problems in practical contexts A ship S is moving with constant velocity ( – 2. 5 i + 6 j) kmh– 1. At time 1200, the position vector of S relative to a fixed origin O is (16 i + 5 j) km. Find: (a) the speed of S 6. 5 kmh-1 (b) the bearing on which S is moving. 337° Sub in values ‘Deal with’ the brackets Group terms The ship is heading directly towards a submerged rock R. A radar tracking station calculates that, if S continues on the same course with the same speed, it will hit R at the time 1500. (c) Find the position vector of R. 6 H

Vectors Find the position vector of the ship at 1400 You need to be able to solve worded problems in practical contexts A ship S is moving with constant velocity ( – 2. 5 i + 6 j) kmh– 1. At time 1200, the position vector of S relative to a fixed origin O is (16 i + 5 j) km. The tracking station warns the ship’s captain of the situation. The captain maintains S on its course with the same speed until the time is 1400. He then changes course so that S moves due north at a constant speed of 5 km h– 1. Assuming that S continues to move with this new constant velocity. Find: (d) an expression for the position vector of the ship t hours after 1400, (e) the time when S will be due east of R, (f) the distance of S from R at the time 1600 Sub in values ‘Deal with’ the brackets Group terms So at 1400 hours, the ship is at position vector (11 i + 17 j) 6 H

Vectors You need to be able to solve worded problems in practical contexts A ship S is moving with constant velocity ( – 2. 5 i + 6 j) kmh– 1. At time 1200, the position vector of S relative to a fixed origin O is (16 i + 5 j) km. The tracking station warns the ship’s captain of the situation. The captain maintains S on its course with the same speed until the time is 1400. He then changes course so that S moves due north at a constant speed of 5 km h– 1. Assuming that S continues to move with this new constant velocity. At 1400 the ship is at (11 i + 17 j) Find an expression for its position t hours after 1400 Use the same formula, with the updated information Sub in values ‘Deal with’ the brackets Factorise the j terms Find: (d) an expression for the position vector of the ship t hours after 1400, (e) the time when S will be due east of R, (f) the distance of S from R at the time 1600 6 H

Vectors You need to be able to solve worded problems in practical contexts Find the time when S will be due east of R R A ship S is moving with constant velocity ( – 2. 5 i + 6 j) kmh– 1. At time 1200, the position vector of S relative to a fixed origin O is (16 i + 5 j) km. The tracking station warns the ship’s captain of the situation. The captain maintains S on its course with the same speed until the time is 1400. He then changes course so that S moves due north at a constant speed of 5 km h– 1. Assuming that S continues to move with this new constant velocity. Find: (d) an expression for the position vector of the ship t hours after 1400, (e) the time when S will be due east of R, 1512 (f) the distance of S from R at the time 1600 S If S is due east of R, then their j terms must be equal! Subtract 17 Divide by 5 1. 2 hours = 1 hour 12 minutes So S will be due east of R at 1512 hours! 6 H

Vectors You need to be able to solve worded problems in practical contexts A ship S is moving with constant velocity ( – 2. 5 i + 6 j) kmh– 1. At time 1200, the position vector of S relative to a fixed origin O is (16 i + 5 j) km. The tracking station warns the ship’s captain of the situation. The captain maintains S on its course with the same speed until the time is 1400. He then changes course so that S moves due north at a constant speed of 5 km h– 1. Assuming that S continues to move with this new constant velocity. Find: Find the distance of S from R at the time 1600 Find where S is at 1600 hours… Sub in t = 2 (1400 – 1600 hours) Simplify/calculate So the position vectors of the rock and the ship at 1600 hours are: To calculate the vector between them, calculate S - R (d) an expression for the position vector of Now use Pythagoras’ Theorem to work out the distance the ship t hours after 1400, (e) the time when S will be due east of R, 1512 Calculate (f) the distance of S from R at the time 1600 6 H

Summary • We have seen how to use vectors in problems involving forces and SUVAT equations • We have also seen how to answer multipart worded questions • It is essential you practice the mixed exercise in this chapter