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A2-Level Maths: Core 4 for Edexcel C4.7 Vectors 1 This icon indicates the slide contains activities created in Flash. These activities are not editable. For more detailed instructions, see the Getting Started presentation. 1 of 43 © Boardworks Ltd 2006 Vectors in two and three dimensions Contents Vectors in two and three dimensions The magnitude of a vector Multiplying vectors by scalars Adding and subtracting vectors Position vectors and coordinate geometry 2 of 43 © Boardworks Ltd 2006 Vectors and scalars A vector is a quantity that has both size (or magnitude) and direction. Examples of vector quantities include: displacement velocity force A scalar is a quantity that has size (or magnitude) only. Examples of scalar quantities include: length speed mass 3 of 43 © Boardworks Ltd 2006 Representing vectors A vector can be represented using a line segment with an arrow on it. For example, the vector that goes from the point A to the point B can be represented by the following directed line segment. B A The magnitude of the vector is given by the length of the line. The direction of the vector is given by the arrow on the line. 4 of 43 © Boardworks Ltd 2006 Representing vectors We can write this vector as AB. Vectors can also be written using single letters in bold type. For example, we can call this vector a. When this is hand-written, the a is written as a To go from the point A to the point B we must move 6 units to the right and 3 units up. B 3 A 5 of 43 6 © Boardworks Ltd 2006 Representing vectors We can represent this movement using a column vector. 6 AB = 3 This component tells us the number of units moved in the x-direction. This component tells us the number of units moved in the y-direction. We can also represent vectors in three dimensions relative to a three dimensional coordinate grid: A third axis, the z-axis, is added at right angles to the xy-plane. Conventionally, we show the zaxis pointing vertically upwards with the xy-plane horizontal. 6 of 43 © Boardworks Ltd 2006 Representing vectors For example, consider the following three-dimensional vector CD. To go from the point C to the point D 5 C we must move z y x –3 –2 5 units in the x-direction, D –2 units in the z-direction. –3 units in the y-direction This three-dimensional vector can be written in column vector form as: 5 CD = 3 2 7 of 43 This component tells us the number of units moved in the x-direction. This component tells us the number of units moved in the y-direction. This component tells us the number of units moved in the z-direction. © Boardworks Ltd 2006 Equal vectors Two vectors are equal if they have the same magnitude and direction. For example, in the following diagram: B C 4 AB = DC = = 4i + 3 j 3 and A D 5 BC = AD = = 5i j 1 General displacement vectors that are not fixed to any point are often called free vectors. 8 of 43 © Boardworks Ltd 2006 The negative of a vector 5 Here is the vector AB = a = 2 A Suppose the arrow went in the opposite direction, from B to A: A a B B This is the negative (or inverse) of the vector AB. We can describe this new vector as: AB 9 of 43 BA –a or 5 2 © Boardworks Ltd 2006 The negative of a vector In general, a1 a1 If AB = a = then BA = a = a a 2 2 And in three-dimensions, a1 a1 If AB = a = a2 then BA = a = a2 a a 3 3 10 of 43 © Boardworks Ltd 2006 The zero and unit vector A vector with a magnitude of 0 is called the zero vector. The zero vector is written as 0 or hand-written as 0 A vector with a magnitude of 1 is called a unit vector. The most important unit vectors are those that run parallel to the x- and y-axes. These are called unit base vectors. 1 The horizontal unit base vector, , is called i. 0 0 The vertical unit base vector, , is called j. 1 11 of 43 © Boardworks Ltd 2006 The unit base vectors The unit base vectors, i and j, run parallel to the x- and y-axes. y-axis j x-axis i Any column vector can easily be written in terms of i and j. For example, 5 4 = 5i 4 j The number of i’s tells us how many units are moved horizontally, and the number of j’s tells us how many units are moved vertically. 12 of 43 © Boardworks Ltd 2006 The unit base vectors In three dimensions, we introduce a third unit base vector, k, that runs parallel to the z-axis. z-axis 1 0 0 i is 0 , j is 1 and k is 0 . y-axis 0 0 1 k j i x-axis For example, the three-dimensional vector in terms of i, j and k as –i + 6j –3k 1 6 can be written 3 Vectors written in terms of the unit base vectors i, j and k are usually said to be written in component form. 13 of 43 © Boardworks Ltd 2006 Contents The magnitude of a vector Vectors in two and three dimensions The magnitude of a vector Multiplying vectors by scalars Adding and subtracting vectors Position vectors and coordinate geometry 14 of 43 © Boardworks Ltd 2006 Finding the magnitude of a vector The magnitude (or modulus) of a vector is given by the length of the line segment representing it. For example, suppose we have the vector A a 4 AB = a = 2 B The magnitude of this vector is written as AB or a . We can calculate this using Pythagoras’s Theorem. AB = 42 + 22 = 20 =2 5 15 of 43 © Boardworks Ltd 2006 Finding the magnitude of a vector 16 of 43 © Boardworks Ltd 2006 Finding the magnitude of a vector The magnitude of a three-dimensional vector can be found by applying Pythagoras’s Theorem in three dimensions. For example, suppose we have the vector 3 AB = 6 2 The magnitude of this vector is given by AB = 32 + 62 + 22 = 9 + 36 + 4 = 49 =7 17 of 43 © Boardworks Ltd 2006 Finding the magnitude of a vector 18 of 43 © Boardworks Ltd 2006 The distance between two points If we are given the coordinates of two points and we are asked to find the distance between them we use Pythagoras’ Theorem in the same way. For example, Find the distance between the points with coordinates P(–4, 7, –2) and Q(5, 9, –8). If d is the distance between the points then, using Pythagoras’ Theorem in three dimensions gives: d2 = (–4 – 5)2 + (7 – 9)2 + (–2 – –8)2 d2 = 81 + 4 + 36 d2 = 121 d = 11 In general, if d is the distance between the points (x1, y1, z1) and (x2, y2, z2) then d2 = (x1 – x2)2 + (y1 – y2)2 + (z1 – z2)2 19 of 43 © Boardworks Ltd 2006 Unit vectors Remember, if the magnitude of a vector is 1 it is called a unit vector. It is possible to find a unit vector parallel to any given vector, a, by dividing the vector by its magnitude. The unit vector parallel to the vector a is denoted by a. So, in general, a= a a Find a unit vector parallel to b = 4i – j + k 4i j k b= 18 1 = (4i j + k ) 3 2 20 of 43 © Boardworks Ltd 2006 Contents Multiplying vectors by scalars Vectors in two and three dimensions The magnitude of a vector Multiplying vectors by scalars Adding and subtracting vectors Position vectors and coordinate geometry 21 of 43 © Boardworks Ltd 2006 Multiplying vectors by scalars Remember, a scalar quantity can be represented by a single number. It has size but not direction. A vector can be multiplied by a scalar. For example, suppose the vector a is represented as follows: a 2a The vector 2a has the same direction but is twice as long. 3 a= 2 6 2a = 4 22 of 43 © Boardworks Ltd 2006 Multiplying vectors by scalars x y In general, if the vector is multiplied by the scalar k, then: z x kx k y = ky z kz For example, 1 2 2 × 4 = 8 2 4 When a vector is multiplied by a scalar the resulting vector lies either parallel to the original vector or on the same line. 23 of 43 © Boardworks Ltd 2006 Multiplying vectors by scalars 24 of 43 © Boardworks Ltd 2006 Contents Adding and subtracting vectors Vectors in two and three dimensions The magnitude of a vector Multiplying vectors by scalars Adding and subtracting vectors Position vectors and coordinate geometry 25 of 43 © Boardworks Ltd 2006 Adding vectors Adding two vectors is equivalent to applying one vector followed by the other. 5 3 For example, suppose a = and b = . 3 2 We can represent the addition of these two vectors in the following diagram: b a 8 a +b = 1 a+b 26 of 43 © Boardworks Ltd 2006 Adding vectors 27 of 43 © Boardworks Ltd 2006 Adding vectors When two or more vectors are added together the result is called the resultant vector. a1 + b1 a1 b1 In general, if a = a2 and b = b2 , then a + b = a2 + b2 . a +b a b 3 3 3 3 Given that a = 2i + 6j – k and b = –i + 2j + 7k, find a + b. a + b = (2 –1)i + (6 + 2)j + (–1 + 7)k = i + 8j + 6k 28 of 43 © Boardworks Ltd 2006 Subtracting vectors We can think of the subtraction of two vectors, a – b, as a + (–b). 4 2 For example, suppose a = and b = . 4 3 a b –b –b a a–b 6 a b = 1 29 of 43 © Boardworks Ltd 2006 Adding and subtracting vectors 30 of 43 © Boardworks Ltd 2006 The parallelogram law for adding vectors 31 of 43 © Boardworks Ltd 2006 Vector arithmetic We have seen that vectors can be multiplied by scalars, added and subtracted. We have also seen that vector addition is commutative. We can use this to add and subtract any given multiple of a vector given in component or column vector form. For example, Given that a = 2i – 4j + k and b = j + 2k find 3a – 2b. 3a – 2b = 3(2i – 4j + k ) – 2(j + 2k ) = 6i – 12j + 3k – 2j – 4k = 6i – 14j – k 32 of 43 © Boardworks Ltd 2006 Vector arithmetic 6 4 Suppose that a = and b = . 1 7 Find vector c such that 2c + a = b. Start by rearranging the equation to make c the subject. 2c + a = b 2c = b – a c = 21 (b – a) 6 4 = 1 7 5 c= 4 1 2 33 of 43 © Boardworks Ltd 2006 A grid of congruent parallelograms 34 of 43 © Boardworks Ltd 2006 Using vectors to solve problems We can use vectors to solve many problems involving physical quantities such as force and velocity. We can also use vectors to prove geometric results. For example, suppose we have a triangle ABC as follows: The line PQ is such that P is the mid-point of AB and Q is the mid-point of AC. B P A Q 35 of 43 C Use vectors to show that PQ is parallel to BC and that the length of BC is double the length of PQ. © Boardworks Ltd 2006 Using vectors to solve problems Let’s call AP vector a and AQ vector b. PQ = a + b B =b a P BC = 2a + 2b a A = 2b 2a b Therefore, Q C = 2(b a) BC = 2 PQ We can conclude from this that PQ is parallel to BC and that the length of BC is double the length of PQ. 36 of 43 © Boardworks Ltd 2006 Collinear points Three or more points are said to be collinear if they lie on the same line. For example Prove that the three points A(–3, 2, 6), B(1, 4, –2) and C(1, 5, –6) are collinear. 1 3 4 AB = 4 2 = 2 2 6 8 1 1 2 BC = 5 4 = 1 6 2 4 AB = 2BC Since AB is a scalar multiple of BC the two lines must be parallel. They also have the point B in common and so the points A, B and C must be collinear. 37 of 43 © Boardworks Ltd 2006 Contents Position vectors and coordinate geometry Vectors in two and three dimensions The magnitude of a vector Multiplying vectors by scalars Adding and subtracting vectors Position vectors and coordinate geometry 38 of 43 © Boardworks Ltd 2006 Position vectors A position vector is a vector that is fixed relative to a fixed origin O. For example, suppose the point P has coordinates (4, 6). P The position vector of the point P is given by 4 OP = p = p 6 Now, suppose the point Q has coordinates (3, –2). O q Q The position vector of the point Q is given by 3 OQ = q = 2 Write the vector PQ as a column vector. 39 of 43 © Boardworks Ltd 2006 Position vectors To get from the point P to the point Q, we have to go from P to O… … and then from O to Q. P PQ = q p So 3 4 = 2 6 –p p q–p O q Q 40 of 43 1 = 8 We can check this using the vector diagram. © Boardworks Ltd 2006 Position vectors In general, if A is the point with coordinates (x1, y1) and B is the point with coordinate (x2, y2) we can write the position vectors x1 OA = a = y1 and x2 OB = b = y2 The vector AB is given by AB = OB OA =b a x2 x1 = y y 1 2 The vector AB can also be written in terms of i and j as AB = ( x2 x1 )i + ( y2 y1) j 41 of 43 © Boardworks Ltd 2006 The mid-point of a line Returning to our example using the points P(4, 6) and Q(3, –2): P Let M be the mid-point of the line PQ. What is the position vector of the point M? p OM = OP + PM M = OP + 21 PQ = p + 21 (q p) O q Q = p + 21 q 21 p = 21 p + 21 q = 21 (p + q) 42 of 43 © Boardworks Ltd 2006 The mid-point of a line 3 4 p = and q = so 2 6 P 1 2 4+3 OM = 6 + 2 7 = 4 p 1 2 3 21 = 2 M m O q Q In general, if points A and B have position vectors a and b, then the position vector of the mid-point of the line AB is given by: 1 (a + b) 2 43 of 43 © Boardworks Ltd 2006