M1 Vectors for edexcel a level

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Vectors

IntroductionIn 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 structuresTeachings for Exercise 6AVectorsYou 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)6A

Vectors have both direction and magnitude!VectorsYou 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 1466A2km3kmOABNDescribing the displacementThe distance use Pythagoras TheoremThe bearing use Trigonometry to find angle Sub in a and bCalculateSub in opp and adjUse inverse TanBearings are measured from north. Add the north line and add 90OppAdj56.3

VectorsYou 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.NNSAB15km9km120240You can use interior angles to find an angle in the triangleInterior angles add up to 180The missing angle next to 240 is 60The angle inside the triangle must also be 606060Finding the distance B to SSub in valuesabcWork outSquare root13.1km

VectorsYou 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.NNSAB15km9km120240606013.1kmNFinding the bearing from B to S Show the bearing at BIt can be split into 2 sections, one of which is 180Find angle inside the triangleSub in valuesRearrangeCalculate 156.6180You can now use Alternate angles to find the unknown part of the bearingAdd on 180The bearing is 336.6S is 13.1km from B on a bearing of 337AabB36.6

Teachings for Exercise 6BVectorsYou 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 addition6BACBa3a

9VectorsYou 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) OCb) ABc) QC

d) CNe) QN6BMBPONDQACbaa + bb - a1/2b-1/2a1/2b - 1/2aWhat can you deduce about AB and QN, looking at the vectors?QN is a multiple of AB, so they are parallel!


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

6BABOMNaba12Use the ratio. If N divides AB in the ratio 1:2, show this on the diagramYou can see now that AN is one-third of ABWe therefore need to know ABTo get from A to B, use AO + OBSub in AO and OBAN = 1/3ABSub in valuesSimplify


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 dont accidentally assume P is the midpoint this is what we need to prove!)6BPOABCacOne way to get from O to P Start with OBOP is parallel to OB so is a multiple of (a + c)

We dont know how much for now, so can use (lamda) to represent the unknown quantityc







(make sure you dont accidentally assume P is the midpoint this is what we need to prove!)6BPOABCacAnother way to get from O to PGo from O to A, then A to PWe will need AC firstc-aAP is parallel to AC so is a multiple of it. Use a different symbol (usually , mew, for this multiple)Now we have another way to get from O to PSub in vectors







(make sure you dont accidentally assume P is the midpoint this is what we need to prove!)6BPOABCaAs these represent the same vector, the expressions must be equal!Multiply out bracketsFactorise the a terms on the right sideNow compare sides there must be the same number of as and cs on eachSub 2nd equation into the firstThey are equalRearrange and solveSo P is halfway along OB and AC and hence the lines bisect each other!

14Teachings for Exercise 6CVectorsYou 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 + j6CO(0,1)(1,0)jiABC5i2j5i + 2j-3ij-3i + j

16Teachings for Exercise 6DVectorsYou 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 + 3jq = 5i + j

Find p + q in terms of i and j6DAdd the i terms and j terms separately

VectorsYou 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 + 2jb = 3i - 4j

Find 2a b in terms of i and j6DMultiply out the bracketCareful with the subtraction here!Group terms


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 7j6D3i-7j3i - 7jPut in the values from the vectors and calculateRound if necessary


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

6D-4ix5jyOppAdjSub in valuesInverse TanThe angle we want is between the vector and the positive x-axis Subtract from 18051.3


Given that:

a = 3i - jb = i + j

Find if a + b is parallel to 3i + j

Start by calculating a + b in terms of a, b and

6DAs the vector must be parallel to 3i + j, the i term must be 3 times the j term!Multiply out the bracketsDivide by 2Move the i and j terms togetherFactorise the terms in i and jMultiply out the bracketSubtract , and add 3


Given that:

a = 3i - jb = i + j

Find if a + b is parallel to 3i + j

Start by calculating a + b in terms of a, b and

6DTo show that this worksMultiply out the bracketsWe now know Group termsFactoriseYou can see that using the value of = 3, we get a vector which is parallel to 3i + j

Teachings for Exercise 6EVectorsYou 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 particleThe distance moved every 4 seconds6EFinding the speedThe speed of the particle is the magnitude of the vectorUse Pythagoras Theorem3ij3i + jFinding the distance travelled every 4 secondsUse GCSE relationshipsDistance = Speed x TimeSub in values (use the exact speed!)CalculateCalculate

Teachings for Exercise 6FVectorsYou 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:6FFinal positionStarting positionVelocityTimeA 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 valuesMultiply/remove bracketsSimplifyVectorsYou 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:6FFinal positionStarting positionVelocityTimeA 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 valuesThe velocity of the particle is (2i + 4j) ms-1Deal with the brackets!Subtract 2i and add 4jDivide by 5VectorsYou 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 velocity6F

3i-4j3i 4j9i-12j9i 12j5ms-115ms-1Multiply all vectors by 3CalculateWe can use the vectors as the velocityVectorsYou 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 velocity6F

Sub in valuesDeal with the bracketsGroup termsVectorsYou 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.6F

Sub in valuesDeal with the bracketsGroup termsRemember this is the velocity, not the speed!Calculate!VectorsYou 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 first6F

Sub in valuesTidy upDivide by 10Sub in valuesCalculateTeachings for Exercise 6GVectorsYou 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 06GGroup together the numerical termsThe i terms must sum to 0The j terms must sum to 0

VectorsYou 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) NF2 = (-5i + 4j) NF3 = (6i 5j) N

all act on a particle of mass 3kg. Find the acceleration of the particle.

Start by finding the overall resultant force.6G

The acceleration is (i + j) ms-2Sub in valuesGroup upSub in the resultant force, and the massDivide by 3VectorsYou 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 triangle6G

ABP3040PTATB7NTATB7NThese are the forces acting on PThese are the forces rearranged as a triangle7NVectorsYou can use vectors to solve problems about forces



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 forces6G

TATB7NThe angle between 7N and TA is 60ABP30407N5060605070The angle between 7N 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 aloneNow we can calculate the tensions!VectorsYou can use vectors to solve problems about forces



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!)6G

TATB7NABP30407N5060605070Multiply by Sin50CalculateVectorsYou can use vectors to solve problems about forces



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!)6G

TATB7NABP30407N5060605070Multiply by Sin60CalculateTeachings for Exercise 6H(the mixed exercise essential!)VectorsYou 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 papers6H

VectorsYou need to be able to solve worded problems in practical contexts

A ship S is moving with constant velocity (2.5i + 6j) kmh1. 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.6H

-2.5i6jThe speed of SUse Pythagoras TheoremCalculate6.5 kmh-1N180The bearing on which S is travelling Find angle Use Tan = Opp/AdjCalculateConsider the north line and read clockwise33767.4VectorsYou need to be able to solve worded problems in practical contexts


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.6H

6.5 kmh-1337Sub in valuesDeal with the bracketsGroup termsVectorsYou need to be able to solve worded problems in practical contexts


The tracking station warns the ships 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 h1. 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

6H

Find the position vector of the ship at 1400Sub in valuesDeal with the bracketsGroup termsSo at 1400 hours, the ship is at position vector (11i + 17j)VectorsYou need to be able to solve worded problems in practical contexts



Find:


6H

At 1400 the ship is at (11i + 17j)Find an expression for its position t hours after 1400Use the same formula, with the updated informationSub in valuesDeal with the bracketsFactorise the j termsVectorsYou need to be able to solve worded problems in practical contexts



Find:


6H

Find the time when S will be due east of RRSIf S is due east of R, then their j terms must be equal!Subtract 17Divide by 5 1.2 hours = 1 hour 12 minutes So S will be due east of R at 1512 hours!1512VectorsYou need to be able to solve worded problems in practical contexts



Find:


6H

1512Find the distance of S from R at the time 1600 Find where S is at 1600 hoursSub in t = 2 (1400 1600 hours)Simplify/calculateSo the position vectors of the rock and the ship at 1600 hours are:To calculate the vector between them, calculate S - RCalculateNow use Pythagoras Theorem to work out the distanceSummaryWe 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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M1 Vectors for edexcel a level