Vectors and Matrices Lecture 2

Transcription

1 Vectors and Matrices Lecture 2 Dr Mark Kambites School of Mathematics 13/03/2014 Dr Mark Kambites (School of Mathematics) COMP /03/ / 20

2 How do we recover the magnitude of a vector from its component form? The magnitude of a = a x i + a y j + a z k is given by a = a 2 x + a 2 y + a 2 z Example. The magnitude of 2i + 6j + 3k is given by ( 2) = 49 = 7. Dr Mark Kambites (School of Mathematics) COMP /03/ / 20

3 A vector with magnitude 1 is called a unit vector. Example. 3 5 i + 4 5j is a unit vector. It has magnitude: ( 3 ) ( ) = = 1. Non-Example. 1 2 i j + 1 4k is not a unit vector. It has magnitude (1 ) ( ) ( ) = Dr Mark Kambites (School of Mathematics) COMP /03/ / 20

4 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 = = 9 so a unit vector parallel to a is 1 9 (i + 8j + 4k) = 1 9 i j k. Dr Mark Kambites (School of Mathematics) COMP /03/ / 20

5 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 a + b 2 a b 3 a + λb Dr Mark Kambites (School of Mathematics) COMP /03/ / 20

6 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) COMP /03/ / 20

7 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 = 4. The magnitude of vector b is = 2. The angle between a and b is 45. So, the scalar product a b is: a b = a b cos θ = 4 2 cos 45 = = 4. Dr Mark Kambites (School of Mathematics) COMP /03/ / 20

8 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: for all vectors a. a 0 = 0 = 0 a Dr Mark Kambites (School of Mathematics) COMP /03/ / 20

9 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) COMP /03/ / 20

10 How do we calculate the scalar product of two vectors from their component forms? Two dimensions. If a = a x i + a y j and b = b x i + b y j then a b = a x b x + a y b y. Three dimensions. If a = a x i + a y j + a z k and b = b x i + b y j + b z k then a b = a x b x + a y b y + a z b z. Remark/Exercise. You can check that these formulae work even if a = 0 or b = 0. Dr Mark Kambites (School of Mathematics) COMP /03/ / 20

11 Example. Find the scalar product of the two vectors a = 4i and b = i + j. Solution. Here a x = 4, a y = 0, b x = 1 and b y = 1. The scalar product is = 4. Example. Find the scalar product of the two vectors a = 4i 3j + 2k and b = i + j + 6k. Solution. Here a x = 4, a y = 3, a z = 2, b x = 1, b y = 1 and b z = 6. The scalar product is ( 3) = 13. Dr Mark Kambites (School of Mathematics) COMP /03/ / 20

12 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) COMP /03/ / 20

13 Example. Find the angle between the two vectors a = 2i j + 3k and b = i + 6j 2k. Solution. Here a b = ( 1) ( 2) = 10, a = ( 1) = 14, and Thus and b = ( 2) 2 = 41. cos θ = = θ = cos 1 ( 0.417) = 115. Dr Mark Kambites (School of Mathematics) COMP /03/ / 20

14 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) COMP /03/ / 20

15 The following laws apply to scalar products. a b = b a a (λb) = λ(a b) = (λa) b a (b + c) = a b + a c a (b c) = a b a c ( commutativity ) ( compatibility with scaling ) ( distributivity over addition ) ( distributivity over subtraction ) Dr Mark Kambites (School of Mathematics) COMP /03/ / 20

16 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 = a parallel + a perp where is a vector parallel to b, and a parallel = a b b 2 b a perp = a a parallel = a a b b 2 b is a vector perpendicular to b. Dr Mark Kambites (School of Mathematics) COMP /03/ / 20

17 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 = = 26 and b = = 13. So a parallel = 26 ( b = 2(2i + 3j) = 4i + 6j and 13) 2 a perp = a a parallel = 7i + 4j (4i + 6j) = 3i 2j so that 7i + 4j = (4i + 6j) + (3i 2j). Dr Mark Kambites (School of Mathematics) COMP /03/ / 20

18 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 0 and b 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) COMP /03/ / 20

19 Vector Product in Component Form If a = a x i + a y j + a z k and b = b x i + b y j + b z k then a b = (a y b z a z b y )i + (a z b x a x b z )j + (a x b y a y b x )k. Example. If a = 2i + 3j + 4k and b = 5i j 6k, find the vector product a b. Solution. Here a x = 2, a y = 3, a z = 4, b x = 5, b y = 1, b z = 6. So: a b = (a y b z a z b y )i + (a z b x a x b z )j + (a x b y a y b x )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) COMP /03/ / 20

20 Exercise For c = 2i 5j 2k and d = i + 4j 3k, evaluate c d. Dr Mark Kambites (School of Mathematics) COMP /03/ / 20