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Vectors

Vectors. 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

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Vectors

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  1. Vectors

  2. 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

  3. Teachings for Exercise 6A

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

  5. Vectors N Adj 2km You can use vectors to describe displacements A girl walks 2km due east from a fixed point O, to A, and then 3km 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 3.61km from O on a bearing of 146˚” O A θ 56.3˚ 3km Describing the displacement Opp The distance – use Pythagoras’ Theorem B Sub in a and 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˚ 6A

  6. Vectors N You can use vectors to describe displacements In an orienteering exercise, a cadet leaves the starting point S and walks 15km on a bearing of 120˚ to reach A, the first checkpoint. From A he then walks 9km 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 A-level. • 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° 120° N b S 15km 60° 240° A 60° 13.1km a 9km c B Finding the distance B to S Sub in values Work out Square root

  7. Vectors N You can use vectors to describe displacements In an orienteering exercise, a cadet leaves the starting point S and walks 15km on a bearing of 120˚ to reach A, the first checkpoint. From A he then walks 9km 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 A-level. A 120° N S 15km θ 60° 36.6° 240° A N 60° 13.1km B b 9km a 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 156.6° 180° You can now use Alternate angles to find the unknown part of the bearing Sub in values Add on 180° Rearrange Calculate θ The bearing is 336.6° S is 13.1km from B on a bearing of 337°

  8. Teachings for Exercise 6B

  9. 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 Two vectors are parallel if they have the same direction You can add vectors using the triangle law of addition 3a a C A B 6B

  10. Vectors A Q C 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. OA = a OB = b Express the following in terms of a and b. a) OC b) AB c) QC d) CN e) QN D a N M P O B b What can you deduce about AB and QN, looking at the vectors? a + b b - a 1/2b QN is a multiple of AB, so they are parallel! -1/2a 1/2b - 1/2a 6B

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

  12. Vectors c A B You can add and represent vectors using line segments 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!) a P O 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 6B

  13. Vectors A c B You can add and represent vectors using line segments 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!) a P -a O 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 6B

  14. Vectors A B You can add and represent vectors using line segments 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!) a P O C 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 2nd equation into the first Rearrange and solve So P is halfway along OB and AC and hence the lines bisect each other! They are equal 6B

  15. Teachings for Exercise 6C

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

  17. Teachings for Exercise 6D

  18. 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. Given that: p = 2i + 3j q = 5i + j Find p + q in terms of i and j Add the i terms and j terms separately 6D

  19. 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. Given that: a = 5i + 2j b = 3i- 4j Find 2a – b in terms of i and j 2 Multiply out the bracket 2 Careful with the subtraction here! 2 Group terms… 2 6D

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

  21. Vectors y You can solve problems with vectors written using the i, j notation You can also use trigonometry to find an angle between a vector and the axes Find the angle between the vector -4i + 5j and the positive x-axis  Draw a diagram Opp 5j θ 51.3° x -4i Adj Sub in values Inverse Tan The angle we want is between the vector and the positive x-axis  Subtract θ from 180° 6D

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

  23. Vectors To show that this works… You can solve problems with vectors written using the i, j notation Given that: a = 3i - j b = i + j Find µ if a + µb is parallel to 3i + j  Start by calculating a + µb in terms of a, b and µ We now know 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 3i + j 6D

  24. Teachings for Exercise 6E

  25. Vectors 3i + j 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 = (3i + j) ms-1 Find: • The speed of the particle • The distance moved every 4 seconds 3i 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 6E

  26. Teachings for Exercise 6F

  27. Vectors You can solve problems involving velocity, time, mass and forces (as in earlier chapters) by using vector notation If a particle starts from the point with position vector r0 and moves with constant velocity v, then its displacement from its initial position at time t is given by: A particle starts from the point with position vector (3i + 7j) m and then moves constant velocity (2i – j) ms-1. Find the position vector of the particle 4 seconds later. (a position vector tells you where a particle is in relation to the origin O) Sub in values Time Multiply/remove brackets Final position Starting position Velocity Simplify 6F

  28. Vectors You can solve problems involving velocity, time, mass and forces (as in earlier chapters) by using vector notation If a particle starts from the point with position vector r0 and moves with constant velocity v, then its displacement from its initial position at time t is given by: A particle moving at a constant velocity, ‘v’, and is at the point with position vector (2i + 4j) m at time t = 0. Five seconds later the particle is at the point with position vector (12i + 16j) m. Find the velocity of the particle. Sub in values Deal with the brackets! Subtract 2i and add 4j Time Final position Divide by 5 Starting position Velocity The velocity of the particle is (2i + 4j) ms-1 6F

  29. Vectors 3i 9i 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 4i + 7j and is moving at a speed of 15ms-1 in the direction 3i – 4j. 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 • Find the speed of the direction vector as it is given in the question • Then ‘multiply up’ to get the required speed (we need 15ms-1, not 5ms-1) Multiplying the vectors will allow you to use the correct velocity -4j -12j 3i – 4j 9i – 12j 5ms-1 15ms-1 Multiply all vectors by 3 Calculate We can use the vectors as the velocity 6F

  30. 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 4i + 7j and is moving at a speed of 15ms-1 in the direction 3i – 4j. 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 • Find the speed of the direction vector as it is given in the question • Then ‘multiply up’ to get the required speed (we need 15ms-1, not 5ms-1) Multiplying the vectors will allow you to use the correct velocity Sub in values ‘Deal with’ the brackets Group terms 6F

  31. 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 (-3i + j) ms-1 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. Sub in values ‘Deal with’ the brackets Group terms Remember this is the velocity, not the speed! Calculate! 6F

  32. 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: 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.  We need to find a first… Sub in values ‘Tidy up’ Divide by 10 Sub in values Calculate 6F

  33. Teachings for Exercise 6G

  34. 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 (2i + 3j), (4i – j), (-3i + 2j) and (ai + bj) are acting on a particle which is in equilibrium. Calculate the values of a and b.  Set the sum of all the vectors equal to 0 Group together the numerical terms The ‘i’ terms must sum to 0 The ‘j’ terms must sum to 0 6G

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

  36. Vectors You can use vectors to solve problems about forces A particle P, of weight 7N 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… A B 30° 40° P 7N TA TB TB P 7N 7N TA These are the forces acting on P These are the forces rearranged as a triangle 6G

  37. Vectors You can use vectors to solve problems about forces A particle P, of weight 7N 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… A B 30° 40° 60° 50° P 7N TB 50° The angle between 7N and TA is 60° 7N 70° 60° TA The angle between 7N and TB is 50° (It is vertically opposite on our triangle of forces) Now we can calculate the tensions! The final angle can be worked out from the triangle of forces alone 6G

  38. Vectors You can use vectors to solve problems about forces A particle P, of weight 7N 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!) A B 30° 40° 60° 50° P 7N TB 50° 7N 70° 60° Multiply by Sin50 TA Calculate 6G

  39. Vectors You can use vectors to solve problems about forces A particle P, of weight 7N 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!) A B 30° 40° 60° 50° P 7N TB 50° 7N 70° 60° Multiply by Sin60 TA Calculate 6G

  40. Teachings for Exercise 6H (the mixed exercise – essential!)

  41. 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 6H

  42. Vectors Use Pythagoras’ Theorem N The speed of S You need to be able to solve worded problems in practical contexts A ship S is moving with constant velocity (–2.5i + 6j) kmh–1. At time 1200, the position vector of S relative to a fixed origin O is (16i + 5j) km. Find: • the speed of S (b) the bearing on which S is moving. 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. Calculate 6j 67.4° θ 180° -2.5i 6.5 kmh-1 337° The bearing on which S is travelling  Find angle θ Use Tan = Opp/Adj Calculate Consider the north line and read clockwise… 6H

  43. Vectors You need to be able to solve worded problems in practical contexts A ship S is moving with constant velocity (–2.5i + 6j) kmh–1. At time 1200, the position vector of S relative to a fixed origin O is (16i + 5j) km. Find: • the speed of S (b) the bearing on which S is moving. 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. Sub in values ‘Deal with’ the brackets 6.5 kmh-1 Group terms 337° 6H

  44. Vectors You need to be able to solve worded problems in practical contexts A ship S is moving with constant velocity (–2.5i + 6j) kmh–1. At time 1200, the position vector of S relative to a fixed origin O is (16i + 5j) 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  Find the position vector of the ship at 1400 Sub in values ‘Deal with’ the brackets Group terms So at 1400 hours, the ship is at position vector (11i + 17j) 6H

  45. Vectors You need to be able to solve worded problems in practical contexts A ship S is moving with constant velocity (–2.5i + 6j) kmh–1. At time 1200, the position vector of S relative to a fixed origin O is (16i + 5j) 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  At 1400 the ship is at (11i + 17j) • 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 6H

  46. Vectors You need to be able to solve worded problems in practical contexts A ship S is moving with constant velocity (–2.5i + 6j) kmh–1. At time 1200, the position vector of S relative to a fixed origin O is (16i + 5j) 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  Find the time when S will be due east of R R 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! 1512 6H

  47. Vectors You need to be able to solve worded problems in practical contexts A ship S is moving with constant velocity (–2.5i + 6j) kmh–1. At time 1200, the position vector of S relative to a fixed origin O is (16i + 5j) 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 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 Now use Pythagoras’ Theorem to work out the distance 1512 Calculate 6H

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

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