3DefinitionVectors are quantities that are completely described by a scalar magnitude and a direction. Displacement, velocity, acceleration and force can all be vectors – they can be described fully by their magnitude and direction.Vectors are often represented by a directed line segment.This vector could be represented as: a, a or AB.ABA bold letter is used to represent vectors in text books and on exam papers. When writing by hand, the convention is to underline the letter.
4Column and component form A vector can be represented in component form as ai + bj in two dimensions, or ai + bj + ck in three dimensions.The vectors i, j and k are unit vectors in the positive directions of the x, y and z axes respectively.A vector can also be represented in column formas or÷øöçèæbac
5Position vectorsThe position vector of a point A is the vector OA where O is the origin.r is often used to denote a position vector.For example, if OA = 3i + 2j – k, we might sayr = 3i + 2j – k.Using the notation r to be a position vector is of particular significance when talking about displacement, velocity and acceleration in kinematics.
7Magnitude of a vectorThe magnitude of a vector is the size of the vector. When representing a vector by a directed line segment, the magnitude of the vector is represented by its length.The magnitude of a vector is equal toaThe magnitude of a vector is found using Pythagoras’ Theorem.If x = ai + bj then =xabxSimilarly, in three dimensions,if x = ai + bj + ck then =In 3-D repeated use of Pythagoras’ Theorem will give the magnitude.
8Magnitude questionsFind the magnitude of the following vectors:a) 3i + 2jb) 4i – 6j + kc) -i + 3j – 5k÷øöçèæ-53d)÷øöçèæ-312e)
9Magnitude solutions a) Magnitude = b) Magnitude = c) Magnitude = d) Magnitude =e) Magnitude =
11Unit vectors A unit vector is a vector of magnitude 1. If v is a vector then the corresponding unit vector is represented by .vˆA unit vector in the direction of a given vector is found by dividing the vector by the magnitude:v=ˆ
12Unit vector questionsFind a unit vector in the direction of the following vectors:a) 4i – 2jb) 3i – j + 4kc) –3i + 5j – 2k÷øöçèæ63d)÷øöçèæ-34e)
15Multiplication of a vector by a scalar To multiply a vector by a scalar, multiply each component of the vector by the scalar.3(2i – 5j + k) = 6i – 15j + 3kIf the scalar is greater than 1, the resulting vector is larger than the original vector and parallel to it.If the scale factor is negative then the vector will be in the opposite direction.If the scalar is less than 1, the resulting vector is smaller than the original vector and parallel to it.
16Parallel vectorsFor vectors to be equal, the i, j and k components must be equal.One vector is the negative of another if each of their corresponding components have opposite signs.Two vectors are parallel if one is a scalar multiple of the other:3i – 3j + 4k and –9i + 9j – 12k are parallel vectors since–9i + 9j – 12k = –3(3i – 3j+ 4k)Vectors are still parallel if they are in opposite directions, as in the example given.
18Addition and subtraction Adding and subtracting vectors in component form is simply a case of adding and subtracting the i, j and k components separately.Add the following vectors: 3i + 5j – k, 2i + 3k and –4i + 3j + k.(3i + 5j – 2k) + (2i + 3k) + (–4i + 3j + k) = i + 8j + 2k
19Triangle law of addition To add vectors using the triangle law, one vector is placed on the end of another. The resultant vector is then shown.abDraw the vector representing a + b.
20Parallelogram law of addition To add two vectors, a and b, using the parallelogram law, the vectors are first placed so that they both start from the same fixed point. a is then put on the end of b and b is placed at the end of a. A parallelogram is now formed – the diagonal of this is the vector a + b.abDraw the vector representing a + b.
22Examination-style question 1 A boy swims across a river from a point A on one bank to a point B on the other. A and B are directly opposite each other and the river is 24 metres wide. The river is flowing at 2 ms–1 parallel to the banks. If the boy reaches point B after swimming for 16 seconds, at what speed and in what direction did he swim?Here v is the speed of the swimmer in still water.
23Solution 1If the river is 24 m wide and it takes 16 seconds to cross, the resultant speed is 1.5 ms–1 (24 16).Draw a triangle of velocities:v21.5x°v2 =The angle found was the angle in the triangle. This is subtracted from 90o to give the angle made with the bank. v = 2.5tan x = 2 1.5 x = 53.1°Therefore the boy swims at a speed of 2.5 ms–1 at an angle of 36.9° to the bank.
24Examination-style question 2a Two boats A and B are travelling with constant velocities. A travels with a velocity of (2i + 4j) kmh–1 and B travels with a velocity of (–2i + 5j) kmh–1.a) Find the bearings on which A and B are travelling.A is moving in the first quadrant at an angle of tan–1 2 to the horizontal. This is an angle of 63.4° (to 3 s.f.), so A is travelling on a bearing of 027°.B is moving in the second quadrant at an angle of tan–1 2.5 to the horizontal. This is an angle of 68.2° (to 3 s.f.), so B is travelling on a bearing of 338°.
25Examination-style question 2b and c At 2pm, A is at point O and B is 5 km due east of O. At time t hours after 2pm, the position vectors of A and B relative to O are a and b respectively.b) Find a and b in terms of t, i and j.a = t(2i + 4j) = 2ti + 4tjb = 5i + t(–2i + 5j) = (5 – 2t)i + 5tjc) At what time will A be due north of B?When A is due north of B, the i components are equal. Equating i components gives 2t = 5 – 2t 4t = 5 t = 1.25.Position is found by velocity × time. Initially, A is at O and B is at 5i.Therefore A is due north of B at 3:15pm.
26Examination-style question 2d At time t hours after 2pm the distance between the two boats is d km.d) Show that d2 = 17t2 – 40t + 25The distance between the two boats is AB. d2 is therefore equal to AB2.AB = b – a = (5 – 2t)i – 2ti + (5t – 4t)j= (5 – 4t)i + tjThe vector AB is b – a. AB2 = (5 – 4t)2 + t2 d2 = 25 – 40t + 16t2 + t2= 17t2 – 40t + 25
27Examination-style question 2e e) At 2pm the boats were 5 km apart. Find, correct to the nearest minute, the time when the boats are again 5 km apart.When the boats are 5 km apart, d2 = 25.Therefore, 25 = 17t2 – 40t + 25 17t2 – 40t = 0 t(17t – 40) = 0 t = 0 or t = 40 17When t = 0 this is the initial position of the boats at 2pm.40 17 = 2.35 (to 3 s.f.), which is equivalent to 2 hours 21 minutes (to the nearest minute).Therefore the boats are again 5 km apart at 4:21pm.