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Vectors and Matrices – Lecture 2 Dr Mark Kambites School of Mathematics 13/03/2014 Dr Mark Kambites (School of Mathematics) COMP11120 13/03/2014 1 / 20 How do we recover the magnitude of a vector from its component form? The magnitude of a = ax i + ay j + az k is given by |a| = q ax2 + ay2 + az2 Example. The magnitude of −2i + 6j + 3k is given by q √ (−2)2 + 62 + 32 = 49 = 7. Dr Mark Kambites (School of Mathematics) COMP11120 13/03/2014 2 / 20 A vector with magnitude 1 is called a unit vector. Example. − 35 i + 54 j is a unit vector. It has magnitude: s 2 r 9 3 2 4 16 − + = + = 1. 5 5 25 25 Non-Example. 1 2i + 13 j + 14 k is not a unit vector. It has magnitude s 2 2 r 1 2 1 1 61 + + = 6= 1 2 3 4 144 Dr Mark Kambites (School of Mathematics) COMP11120 13/03/2014 3 / 20 Finding Unit Vectors. Given a non-zero vector, we can find a unit vector parallel to it by scaling it to have the right length. Specifically, we scale by 1 m where m is the magnitude of the vector. Example. Find a unit vector parallel to a = i + 8j + 4k. √ Note that |a| = 12 + 82 + 42 = 9 so a unit vector parallel to a is 1 1 8 4 × (i + 8j + 4k) = i + j + k. 9 9 9 9 Dr Mark Kambites (School of Mathematics) COMP11120 13/03/2014 4 / 20 Exercises 1. Give the position vector of the point (−2, 4, 1). 2. Give the displacement vector from (2, 5, 4) to (7, 1, 2). 3. Find the magnitude of the vector a = 3i − 4j + 12k 4. For a = 2i − 3j + 7k, b = −i + 2j + 1k and λ = 4, find 1 2 3 a+b a−b a + λb Dr Mark Kambites (School of Mathematics) COMP11120 13/03/2014 5 / 20 4. Scalar and Vector Products The scalar product (also called the dot product) is an operation which combines two vectors to give a scalar (a number). In terms of arrows, the scalar product is defined by a · b = |a| |b| cos θ where θ is the angle between the directions of vectors a and b. b θ a Notice that the angle is always measured by putting the tails of the two vectors together, and measuring the angle between. Dr Mark Kambites (School of Mathematics) COMP11120 13/03/2014 6 / 20 Example. Find the scalar product of the two vectors a = 4i and b = i + j. Solution. b 45◦ a √ The magnitude of vector a is √42 + 02 = 4. √ The magnitude of vector b is 12 + 12 = 2. The angle between a and b is 45◦ . So, the scalar product a · b is: a · b = |a| |b| cos θ = 4 × Dr Mark Kambites (School of Mathematics) √ 2 × cos 45◦ = 4 × COMP11120 √ 1 2 × √ = 4. 2 13/03/2014 7 / 20 The definition of a · b given above does not make sense if a = 0 or b = 0 (or both). This is because 0 has no direction. Instead we define: a·0=0=0·a for all vectors a. Dr Mark Kambites (School of Mathematics) COMP11120 13/03/2014 8 / 20 The value of the angle θ between two vectors determines the sign of their scalar product: a · b = |a| |b| cos θ. Acute Angles. If θ is an acute angle (between 0 and 90◦ ) the scalar product is positive. Obtuse Angles. If θ is an obtuse angle (between 90◦ and 180◦ ) the scalar product is negative. Right Angles. If θ is exactly 90◦ then the scalar product is 0. Conversely, if the scalar product of two non-zero vectors is 0 then they are perpendicular. Dr Mark Kambites (School of Mathematics) COMP11120 13/03/2014 9 / 20 How do we calculate the scalar product of two vectors from their component forms? Two dimensions. If a = ax i + ay j and b = bx i + by j then a · b = ax bx + ay by . Three dimensions. If a = ax i + ay j + az k and b = bx i + by j + bz k then a · b = ax bx + ay by + az bz . Remark/Exercise. You can check that these formulae work even if a = 0 or b = 0. Dr Mark Kambites (School of Mathematics) COMP11120 13/03/2014 10 / 20 Example. Find the scalar product of the two vectors a = 4i and b = i + j. Solution. Here ax = 4, ay = 0, bx = 1 and by = 1. The scalar product is 4 × 1 + 0 × 1 = 4. Example. Find the scalar product of the two vectors a = 4i − 3j + 2k and b = i + j + 6k. Solution. Here ax = 4, ay = −3, az = 2, bx = 1, by = 1 and bz = 6. The scalar product is 4 × 1 + (−3) × 1 + 2 × 6 = 13. Dr Mark Kambites (School of Mathematics) COMP11120 13/03/2014 11 / 20 The two ways to calculate the scalar product combine to give a way of calculating the angle between two vectors. We know a · b = |a| |b| cos θ. Assuming neither vector is 0 we can divide through by |a| |b|: cos θ = a·b |a| |b| If we know the component forms of a and b, then we can compute a · b, |a| and |b|. So we can compute cos θ and hence θ itself. Dr Mark Kambites (School of Mathematics) COMP11120 13/03/2014 12 / 20 Example. Find the angle between the two vectors a = 2i − j + 3k and b = i + 6j − 2k. Solution. Here a · b = 2 × 1 + (−1) × 6 + 3 × (−2) = −10, |a| = q √ 22 + (−1)2 + 32 = 14, and |b| = q √ 12 + 62 + (−2)2 = 41. Thus cos θ = √ −10 √ = −0.417 14 41 and θ = cos−1 (−0.417) = 115◦ . Dr Mark Kambites (School of Mathematics) COMP11120 13/03/2014 13 / 20 Class Exercises 1. Find the scalar product (i + 6j − 2k) · (−2i − j + 4k). 2. Find the angle between the two vectors a = 2i + 3j and b = −i + 6j. Dr Mark Kambites (School of Mathematics) COMP11120 13/03/2014 14 / 20 The following laws apply to scalar products. a·b=b·a (“commutativity”) a · (λb) = λ(a · b) = (λa) · b (“compatibility with scaling”) a · (b + c) = a · b + a · c (“distributivity over addition”) a · (b − c) = a · b − a · c (“distributivity over subtraction”) Dr Mark Kambites (School of Mathematics) COMP11120 13/03/2014 15 / 20 If a and b are non-zero vectors, then it is possible to express a as the sum of two vectors: one parallel to b; and one perpendicular to b Specifically, we can write a = aparallel + aperp where aparallel = a·b |b|2 b is a vector parallel to b, and aperp = a − aparallel = a − a·b |b|2 b is a vector perpendicular to b. Dr Mark Kambites (School of Mathematics) COMP11120 13/03/2014 16 / 20 Example. Express a = 7i + 4j as the sum of a vector parallel to b = 2i + 3j and a vector perpendicular to b. √ √ Solution. Here a · b = 2 × 7 + 4 × 3 = 26 and |b| = 22 + 32 = 13. So 26 aparallel = √ b = 2(2i + 3j) = 4i + 6j and ( 13)2 aperp = a − aparallel = 7i + 4j − (4i + 6j) = 3i − 2j so that 7i + 4j = (4i + 6j) + (3i − 2j). Dr Mark Kambites (School of Mathematics) COMP11120 13/03/2014 17 / 20 The Vector Product The scalar product gives a way to multiply two vectors and get a scalar. In three dimensions, there is another way to multiply two vectors which gives a vector. The vector product (or cross product) of a and b is written a × b. Provided a 6= 0 and b 6= 0 it has the following properties: it is a vector; it has magnitude |a| |b| sin θ where θ is the angle between a and b; it is perpendicular to both a and b; a, b and a × b form a right-handed system. If a = 0 or b = 0 (or both) then a × b = 0. Dr Mark Kambites (School of Mathematics) COMP11120 13/03/2014 18 / 20 Vector Product in Component Form If a = ax i + ay j + az k and b = bx i + by j + bz k then a × b = (ay bz − az by )i + (az bx − ax bz )j + (ax by − ay bx )k. Example. If a = −2i + 3j + 4k and b = 5i − j − 6k, find the vector product a × b. Solution. Here ax = −2, ay = 3, az = 4, bx = 5, by = −1, bz = −6. So: a × b = (ay bz − az by )i + (az bx − ax bz )j + (ax by − ay bx )k = [3(−6) − 4(−1)]i + [4 × 5 − (−2)(−6)]j + [(−2)(−1) − 3 × 5]k = −14i + 8j − 13k. Dr Mark Kambites (School of Mathematics) COMP11120 13/03/2014 19 / 20 Exercise For c = 2i − 5j − 2k and d = −i + 4j − 3k, evaluate c × d. Dr Mark Kambites (School of Mathematics) COMP11120 13/03/2014 20 / 20