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 Vectors are used in video games in the movement of characters and by engineers in the design of buildings, bridges and other structures

Teachings for Exercise 6A

Vectors have both direction and magnitude!
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

You can use vectors to describe displacements
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 56.3˚ θ 3km Describing the displacement Opp The distance – use Pythagoras’ Theorem 𝑐= 𝑎 2 + 𝑏 2 B Sub in a and b 𝑐= Calculate 𝑐=3.61𝑘𝑚 The bearing – use Trigonometry to find angle θ 𝑇𝑎𝑛𝜃= 𝑂𝑝𝑝 𝐴𝑑𝑗 Sub in opp and adj 𝑇𝑎𝑛𝜃= 3 2 Use inverse Tan 𝜃=56.3˚ Bearings are measured from north. Add the north line and add 90˚ 𝐵𝑒𝑎𝑟𝑖𝑛𝑔=146˚ 6A

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 𝑎 2 = 𝑏 2 + 𝑐 2 −2𝑏𝑐𝐶𝑜𝑠𝐴 Sub in values 𝑎 2 = −2(15×9)𝐶𝑜𝑠60 Work out 𝑎 2 =171 Square root 𝑎=13.1𝑘𝑚

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 𝑆𝑖𝑛𝐴 𝑎 = 𝑆𝑖𝑛𝐵 𝑏 Sub in values 𝑆𝑖𝑛𝜃 9 = 𝑆𝑖𝑛 Add on 180° Rearrange 𝑆𝑖𝑛θ= 𝑆𝑖𝑛 ×9 Calculate θ 𝜃=36.6° The bearing is 336.6° S is 13.1km from B on a bearing of 337°

Teachings for Exercise 6B

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 𝐴𝐶=𝐴𝐵+𝐵𝐶 6B

Vectors What can you deduce about AB and QN, looking at the vectors?
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 𝐴𝐵 =𝒃−𝒂 𝑄𝑁 = 1 2 𝒃− 1 2 𝒂 QN is a multiple of AB, so they are parallel! -1/2a 1/2b - 1/2a 𝑄𝑁 = 1 2 (𝒃−𝒂) 6B

Vectors 6B 𝐴𝑁 = 1 3 𝐴𝐵 𝑂𝐴 =2𝑎 𝑂𝑁 = 𝑂𝐴 + 𝐴𝑁 𝑂𝑁 =2𝒂+− 2 3 𝒂+ 1 3 𝒃
𝐴𝑁 = 1 3 𝐴𝐵 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 𝑂𝐴 =2𝑎 1 a N M a 2 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 𝑂𝑁 = 𝑂𝐴 + 𝐴𝑁 Sub in values 𝑂𝑁 =2𝒂+− 2 3 𝒂+ 1 3 𝒃 𝐴𝐵 = 𝐴𝑂 + 𝑂𝐵 Sub in AO and OB Simplify 𝐴𝐵 =−2𝒂+𝒃 𝑂𝑁 = 4 3 𝒂+ 1 3 𝒃 AN = 1/3AB 𝐴𝑁 =− 2 3 𝒂+ 1 3 𝒃 6B

Vectors 6B 𝑂𝑃 =λ(𝑎+𝑐) 𝑂𝐵 =𝑎+𝑐 𝑂𝑃 =λ(𝑎+𝑐) 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 𝑂𝐵 =𝑎+𝑐 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 𝑂𝑃 =λ(𝑎+𝑐) 6B

Vectors 6B 𝑂𝑃 =λ(𝑎+𝑐) 𝑂𝑃 =𝑎+𝜇(𝑐−𝑎) 𝐴𝐶 =𝑐−𝑎 𝐴𝑃 =𝜇(𝑐−𝑎) 𝑂𝑃 = 𝑂𝐴 + 𝐴𝑃
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… 𝐴𝐶 =𝑐−𝑎 AP is parallel to AC so is a multiple of it. Use a different symbol (usually μ, ‘mew’, for this multiple) 𝐴𝑃 =𝜇(𝑐−𝑎) Now we have another way to get from O to P 𝑂𝑃 = 𝑂𝐴 + 𝐴𝑃 Sub in vectors 𝑂𝑃 =𝑎+𝜇(𝑐−𝑎) 6B

Vectors 6B 𝑂𝑃 =λ(𝑎+𝑐) 𝑂𝑃 =𝑎+𝜇(𝑐−𝑎) 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 𝑂𝑃 =λ(𝑎+𝑐) 𝑂𝑃 =𝑎+𝜇(𝑐−𝑎) As these represent the same vector, the expressions must be equal! λ 𝑎+𝑐 =𝑎+𝜇(𝑐−𝑎) Multiply out brackets λ𝑎+λ𝑐=𝑎+𝜇𝑐−𝜇𝑎 Factorise the ‘a’ terms on the right side λ𝑎+λ𝑐=(1−𝜇)𝑎+𝜇𝑐 Now compare sides – there must be the same number of ‘a’s and ‘c’s on each λ=1−𝜇 λ=𝜇 Sub 2nd equation into the first λ=1−λ Rearrange and solve So P is halfway along OB and AC and hence the lines bisect each other! λ=0.5 They are equal μ=0.5 6B

Teachings for Exercise 6C

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 𝐴𝐶 = 𝐴𝐵 + 𝐵𝐶 2j 𝐴𝐶 =5𝒊+2𝒋 A B 5i -3i + j j -3i 6C

Teachings for Exercise 6D

Add the i terms and j terms separately
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 𝒑+𝒒= (2𝒊+3𝒋) + (5𝒊+𝒋) Add the i terms and j terms separately 𝒑+𝒒= 7𝒊+4𝒋 6D

Vectors 6D 2𝒂+𝒃 = 2(5𝒊+2𝒋) − (3𝒊−4𝒋) 2𝒂+𝒃 = 10𝒊+4𝒋 − (3𝒊−4𝒋) 2𝒂+𝒃 =
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 2𝒂+𝒃 = 2(5𝒊+2𝒋) − (3𝒊−4𝒋) Multiply out the bracket 2𝒂+𝒃 = 10𝒊+4𝒋 − (3𝒊−4𝒋) Careful with the subtraction here! 2𝒂+𝒃 = 10𝒊+4𝒋 − 3𝒊 + 4𝒋 Group terms… 2𝒂+𝒃 = 7𝒊+8𝒋 6D

Put in the values from the vectors and calculate
You can solve problems with vectors written using the i, j notation 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: 3i – 7j -7j 3i - 7j 𝒗 = (−7) 2 Put in the values from the vectors and calculate 𝒗 = 58 Round if necessary 𝒗 =7.62 (3sf) 6D

The angle we want is between the vector and the positive x-axis
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 51.3° θ x -4i Adj 𝑇𝑎𝑛𝜃= 𝑂𝑝𝑝 𝐴𝑑𝑗 Sub in values 𝑇𝑎𝑛𝜃= 5 4 Inverse Tan 𝜃=51.3° The angle we want is between the vector and the positive x-axis  Subtract θ from 180° 𝐴𝑐𝑡𝑢𝑎𝑙 𝑎𝑛𝑔𝑙𝑒=128.7° 6D

Vectors 𝜇=3 6D 𝒂+𝜇𝒃 = (3𝒊−𝒋)+𝜇(𝒊+𝒋) = 3𝒊−𝒋+𝜇𝒊+𝜇𝒋 = 3𝒊+𝜇𝒊−𝒋+𝜇𝒋 = 3+𝜇 𝒊
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 = 3𝒊−𝒋+𝜇𝒊+𝜇𝒋 Move the i and j terms together = 3𝒊+𝜇𝒊−𝒋+𝜇𝒋 Factorise the terms in i and j = 3+𝜇 𝒊 + −1+𝜇 𝒋 As the vector must be parallel to 3i + j, the i term must be 3 times the j term! 3+𝜇 = 3(−1+𝜇) Multiply out the bracket 3+𝜇 = −3+3𝜇 𝜇=3 Subtract µ, and add 3 6= 2𝜇 Divide by 2 3= 𝜇 6D

Multiply out the brackets
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 𝒂+𝜇𝒃 𝒂+3𝒃 = (3𝒊 −𝒋) + 3(𝒊 + 𝒋) Multiply out the brackets = 3𝒊 − 𝒋 + 3𝒊 +3𝒋 Group terms =6𝒊 + 2𝒋 Factorise =2(3𝒊 + 𝒋) You can see that using the value of µ = 3, we get a vector which is parallel to 3i + j 𝜇=3 6D

Teachings for Exercise 6E

Vectors 6E 𝑣 = 3 2 + 1 2 𝑣 =3.16𝑚 𝑠 −1 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒=𝑠𝑝𝑒𝑒𝑑 ×𝑡𝑖𝑚𝑒
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 𝑣 = Calculate 𝑣 =3.16𝑚 𝑠 −1 Finding the distance travelled every 4 seconds Use GCSE relationships Distance = Speed x Time 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒=𝑠𝑝𝑒𝑒𝑑 ×𝑡𝑖𝑚𝑒 Sub in values (use the exact speed!) 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒=3.16×4 Calculate 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒=12.6𝑚 6E

Teachings for Exercise 6F

Multiply/remove brackets
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) 𝒓 0 =(3𝒊+7𝒋) 𝒗=(2𝒊−𝒋) 𝑡=4 𝒓= 𝒓 0 +𝒗𝑡 𝒓= 𝒓 0 +𝒗𝑡 Sub in values 𝒓= 3𝒊+7𝒋 +(2𝒊−𝒋)(4) Time Multiply/remove brackets 𝒓=3𝒊+7𝒋+8𝒊−4𝒋 Final position Starting position Velocity Simplify 𝒓=11𝒊+3𝒋 6F

The velocity of the particle is (2i + 4j) ms-1
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. 𝒓 0 =(2𝒊−4𝒋) 𝒓=(12𝒊+16𝒋) 𝑡=5 𝒓= 𝒓 0 +𝒗𝑡 Sub in values (12𝒊+16𝒋)= 2𝒊−4𝒋 +(𝒗)(5) Deal with the brackets! 𝒓= 𝒓 0 +𝒗𝑡 12𝒊+16𝒋=2𝒊−4𝐣+5𝒗 Subtract 2i and add 4j Time 10𝒊+20𝒋=5𝒗 Final position Divide by 5 Starting position Velocity 2𝒊+4𝒋=𝒗 The velocity of the particle is (2i + 4j) ms-1 6F

Vectors 𝒓= 𝒓 0 +𝒗𝑡 6F 3i 9i -4j -12j 3i – 4j 9i – 12j 5ms-1 15ms-1
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 𝑆𝑝𝑒𝑒𝑑= (−4) 2 Calculate 𝑆𝑝𝑒𝑒𝑑=5𝑚 𝑠 −1 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦= 9𝒊−12𝒋 𝑚 𝑠 −1 𝑆𝑝𝑒𝑒𝑑=15𝑚 𝑠 −1 We can use the vectors as the velocity 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦= 9𝒊−12𝒋 𝑚 𝑠 −1 6F

Vectors 𝒓= 𝒓 0 +𝒗𝑡 6F 𝒓 0 =(4𝒊+7𝒋) 𝒗=(9𝒊−12𝒋) 𝑡=2 𝒓= 𝒓 0 +𝒗𝑡
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 𝒓 0 =(4𝒊+7𝒋) 𝒗=(9𝒊−12𝒋) 𝑡=2 𝒓= 𝒓 0 +𝒗𝑡 Sub in values 𝒓=(4𝒊+7𝒋)+(9𝒊−12𝒋)(2) ‘Deal with’ the brackets 𝒓=4𝒊+7𝒋+18𝒊−24𝒋 Group terms 𝒓=22𝒊−17𝒋 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦= 9𝒊−12𝒋 𝑚 𝑠 −1 6F

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. 𝑠=? 𝒖=(−3𝒊+𝒋) 𝒗=? 𝒂=(2𝒊+3𝒋) 𝑡=3 𝒗=𝒖+𝒂𝑡 Sub in values 𝒗=(−3𝒊+𝒋)+(2𝒊+3𝒋)(3) ‘Deal with’ the brackets 𝒗=−3𝒊+𝒋+6𝒊+9𝒋 Group terms 𝒗=3𝒊+10𝒋 Remember this is the velocity, not the speed! 𝑆𝑝𝑒𝑒𝑑= Calculate! 𝑆𝑝𝑒𝑒𝑑=10.4𝑚 𝑠 −1 6F

Vectors 6F 𝑠=? 𝒖=0 𝒗=(10𝒊−24𝒋) 𝒂=? 𝑡=10 𝒗=𝒖+𝒂𝑡 (10𝒊−24𝒋)=0+𝒂(10)
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… 𝑠=? 𝒖=0 𝒗=(10𝒊−24𝒋) 𝒂=? 𝑡=10 𝒗=𝒖+𝒂𝑡 Sub in values (10𝒊−24𝒋)=0+𝒂(10) ‘Tidy up’ 10𝒊−24𝒋=10𝒂 Divide by 10 𝒊−2.4𝒋=𝒂 𝑭=𝑚𝒂 Sub in values 𝑭=(2)(𝒊−2.4𝒋) Calculate 𝑭= 2𝒊−4.8𝒋 𝑁 6F

Teachings for Exercise 6G

Vectors 6G 2𝒊+3𝒋 + 4𝒊−𝒋 + −3𝒊+2𝒋 + 𝑎𝒊+𝑏𝒋 =0 3𝒊+𝟒𝒋 + 𝑎𝒊+𝑏𝒋 =0 3+𝑎=0
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 2𝒊+3𝒋 + 4𝒊−𝒋 + −3𝒊+2𝒋 + 𝑎𝒊+𝑏𝒋 =0 Group together the numerical terms 3𝒊+𝟒𝒋 + 𝑎𝒊+𝑏𝒋 =0 The ‘i’ terms must sum to 0 The ‘j’ terms must sum to 0 3+𝑎=0 4+𝑏=0 𝑎=−3 𝑏=−4 6G

Sub in the resultant force, and the mass
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. 𝑅𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 𝐹𝑜𝑟𝑐𝑒= 𝑭 1 + 𝑭 2 + 𝑭 3 Sub in values = 2𝒊+4𝒋 + −5𝒊+4𝒋 +(6𝒊−5𝒋) Group up = 3𝒊+3𝒋 𝑭=𝑚𝒂 Sub in the resultant force, and the mass (3𝒊+3𝒋)=3𝒂 Divide by 3 𝒊+𝒋=𝒂 The acceleration is (i + j) ms-2 6G

Vectors 6G These are the forces acting on P
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

Vectors 6G 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

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° 𝑇 𝐴 𝑆𝑖𝑛50 = 7 𝑆𝑖𝑛70 7N 70° 60° Multiply by Sin50 TA 𝑇 𝐴 = 7 𝑆𝑖𝑛70 ×𝑆𝑖𝑛50 Calculate 𝑇 𝐴 =5.71𝑁 𝑇 𝐴 =5.71𝑁 6G

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° 𝑇 𝐵 𝑆𝑖𝑛60 = 7 𝑆𝑖𝑛70 7N 70° 60° Multiply by Sin60 TA 𝑇 𝐵 = 7 𝑆𝑖𝑛70 ×𝑆𝑖𝑛60 Calculate 𝑇 𝐵 =6.45𝑁 𝑇 𝐴 =5.71𝑁 𝑇 𝐵 =6.45𝑁 6G

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

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

Vectors 6H N The speed of S 𝑆𝑝𝑒𝑒𝑑= (−2.5) 2 + 6 2 6j 𝑆𝑝𝑒𝑒𝑑=6.5 𝑘𝑚 ℎ −1
Use Pythagoras’ Theorem 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. 𝑆𝑝𝑒𝑒𝑑= (−2.5) Calculate 6j 𝑆𝑝𝑒𝑒𝑑=6.5 𝑘𝑚 ℎ −1 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 𝑇𝑎𝑛𝜃= 6 2.5 Calculate 𝜃=67.4° Consider the north line and read clockwise… 𝐵𝑒𝑎𝑟𝑖𝑛𝑔=337° 6H

Vectors 6H 𝒓 0 =(16𝒊+5𝒋) 𝒗=(−2.5𝒊+6𝒋) 𝒕=3 𝒓= 𝒓 0 +𝒗𝑡
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. 𝒓 0 =(16𝒊+5𝒋) 𝒗=(−2.5𝒊+6𝒋) 𝒕=3 𝒓= 𝒓 0 +𝒗𝑡 Sub in values 𝒓=(16𝒊+5𝒋)+(−2.5𝒊+6𝒋)(3) ‘Deal with’ the brackets 𝒓=16𝒊+5𝒋−7.5𝒊+18𝒋 6.5 kmh-1 Group terms 𝒓=8.5𝒊+23𝒋 337° 𝑹=8.5𝒊+23𝒋 6H

Vectors 6H 𝒓 0 =(16𝒊+5𝒋) 𝒗=(−2.5𝒊+6𝒋) 𝒕=2 𝒓= 𝒓 0 +𝒗𝑡
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 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 𝒓 0 =(16𝒊+5𝒋) 𝒗=(−2.5𝒊+6𝒋) 𝒕=2 𝒓= 𝒓 0 +𝒗𝑡 Sub in values 𝑹=8.5𝒊+23𝒋 𝒓=(16𝒊+5𝒋)+(−2.5𝒊+6𝒋)(2) ‘Deal with’ the brackets 𝒓=16𝒊+5𝒋−5𝒊+12𝒋 Group terms 𝒓=11𝒊+17𝒋 So at 1400 hours, the ship is at position vector (11i + 17j) 6H

‘Deal with’ the brackets
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 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 𝒓 0 =(11𝒊+17𝒋) 𝒗=5𝒋 𝑡=𝑡 𝑹=8.5𝒊+23𝒋 𝒓= 𝒓 0 +𝒗𝑡 Sub in values 𝒓=(11𝒊+17𝒋)+(5𝒋)(𝑡) ‘Deal with’ the brackets 𝒓=11𝒊+17𝒋+5𝑡𝒋 Factorise the j terms 𝒓=11𝒊+(17+5𝑡)𝒋 𝒓=11𝒊+(17+5𝑡)𝒋 6H

Vectors 6H 8.5𝒊+23𝒋 11𝒊+(17+5𝑡)𝒋 23=17+5𝑡 6=5𝑡 1.2=𝑡
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 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 8.5𝒊+23𝒋 11𝒊+(17+5𝑡)𝒋 𝑹=8.5𝒊+23𝒋 If S is due east of R, then their j terms must be equal! 23=17+5𝑡 Subtract 17 6=5𝑡 Divide by 5 1.2=𝑡  1.2 hours = 1 hour 12 minutes 𝒓=11𝒊+(17+5𝑡)𝒋  So S will be due east of R at 1512 hours! 1512 6H

So the position vectors of the rock and the ship at 1600 hours are:
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 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… 𝒓=11𝒊+(17+5𝑡)𝒋 Sub in t = 2 (1400 – 1600 hours) 𝒓=11𝒊+(17+5(2))𝒋 Simplify/calculate 𝑹=8.5𝒊+23𝒋 𝒓=11𝒊+27𝒋 So the position vectors of the rock and the ship at 1600 hours are: 𝑹=8.5𝒊+23𝒋 𝑺=11𝒊+27𝒋 To calculate the vector between them, calculate S - R 11𝒊+27𝒋 −(8.5𝒊+23𝒋) =2.5𝒊+4𝒋 𝒓=11𝒊+(17+5𝑡)𝒋 Now use Pythagoras’ Theorem to work out the distance 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒= 1512 Calculate 4.72𝑘𝑚 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒=4.72𝑘𝑚 6H

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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