Vectors VECTOR PRODUCT. Graham S McDonald. A Tutorial Module for learning about the vector product of two vectors. Table of contents Begin Tutorial

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1 Vectors VECTOR PRODUCT Graham S McDonald A Tutorial Module for learning about the vector product of two vectors Table of contents Begin Tutorial c 2004

3 Section 1: Theory 3 1. Theory The purpose of this tutorial is to practice working out the vector product of two vectors. It is called the vector product because the result is a vector, i.e. a quantity with both (i) magnitude, and (ii) direction. (i) The MAGNITUDE of the vector product of a and b is a b = a b sin θ a where θ is the angle between a and b. b

4 Section 1: Theory 4 (ii) The DIRECTION of the vector product of a and b is perpendicular to both a and b, such that if we look along a b then a rotates towards b in a clockwise manner. a x b b a C lo c k w is e It can be shown that the above definitions of magnitude and direction of a vector product allow us to calculate the x, y and z components of a b from the individual components of the vectors a and b

5 Section 1: Theory 5 The components of the vector a b are given by the determinant of a matrix with 3 rows. The components of a = a x i + a y j + a z k and b = b x i + b y j + b z k appear in the 2nd and 3rd rows. This 3-row determinant is evaluated by expansion into 2-row determinants, which are themselves then expanded. Matrix theory is itself very useful, and the scheme is shown below i j k a b = a x a y a z b x b y b z = i a y a z j a x b y b z b x a z b z + k a x b x a y b y = i (a y b z a z b y ) j (a x b z a z b x ) + k (a x b y a y b x )

6 Section 2: Exercises 6 2. Exercises Click on Exercise links for full worked solutions (there are 14 exercises in total). Exercise 1. Calculate a b when a = 2, b = 4 and the angle between a and b is θ = 45 Exercise 2. Calculate a b when a = 3, b = 5 and the angle between a and b is θ = 60 Exercise 3. Calculate a b when a = 1, b = 3 and the angle between a and b is θ = 30 Exercise 4. Calculate a b when a = 2, b = 5 and the angle between a and b is θ = 35 Theory Answers Tips Notation

7 Section 2: Exercises 7 Exercise 5. Calculate the magnitude of the torque τ = s F when s = 2 m, F = 4 N and θ = 30 (θ is the angle between the position vector s and the force F ) Exercise 6. Calculate the magnitude of the velocity v = ω s when ω = 3s 1, s = 2m and θ = 45 (θ is the angle between the angular velocity ω and the position vector s) Exercise 7. If a = 4i+2j k and b = 2i 6i 3k then calculate a vector that is perpendicular to both a and b Exercise 8. Calculate the vector a b when a = 2i + j k and b = 3i 6j + 2k Theory Answers Tips Notation

8 Section 2: Exercises 8 Exercise 9. Calculate the vector a b when a = 3i+4j 3k and b = i + 3j + 2k Exercise 10. Calculate the vector a b when a = i + 2j k and b = 3i + 3j + k Exercise 11. Calculate the vector a b when a = 2i+4j +2k and b = i + 5j 2k Exercise 12. Calculate the vector a b when a = 3i 4j + k and b = 2i + 5j k Theory Answers Tips Notation

9 Section 2: Exercises 9 Exercise 13. Calculate the vector a b when a = 2i 3j + k and b = 2i + j + 4k Exercise 14. Calculate the vector a b when a = 2i + 3k and b = i + 2j + 4k Theory Answers Tips Notation

10 Section 3: Answers Answers , , , , 5. 4 J, m s 1, 7. 12i + 10j 28k, 8. 4i 7j 15k, 9. 17i 9j + 5k, 10. 5i 4j 4k, i + 6j + 6k, 12. i + 5j + 23k, i 6j + 8k, 14. 6i 5j + 4k.

11 Section 4: Tips on using solutions Tips on using solutions When looking at the THEORY, ANSWERS, TIPS or NOTATION pages, use the Back button (at the bottom of the page) to return to the exercises Use the solutions intelligently. For example, they can help you get started on an exercise, or they can allow you to check whether your intermediate results are correct Try to make less use of the full solutions as you work your way through the Tutorial

12 Section 5: Alternative notation Alternative notation Here, we use symbols like a to denote a vector. In some texts, symbols for vectors are in bold (eg a instead of a ) In this Tutorial, vectors are given in terms of the unit Cartesian vectors i, j and k. A common alternative notation involves quoting the Cartesian components within brackets. For example, the vector a = 2i + j + 5k can be written as a = (2, 1, 5) The scalar product a b is also called a dot product (reflecting the symbol used to denote this type of multiplication). Likewise, the vector product a b is also called a cross product An alternative notation for the vector product is a b

13 Solutions to exercises 13 Full worked solutions Exercise 1. Here, we have a = 2, b = 4 and θ = 45. a b = a b sin θ = (2)(4) sin 45 = (2)(4) 1 2 = 8 2 = = = 4 2. Return to Exercise 1

14 Solutions to exercises 14 Exercise 2. Here, we have a = 3, b = 5 and θ = 60. a b = a b sin θ = (3)(5) sin 60 3 = (3)(5) 2 = Return to Exercise 2

15 Solutions to exercises 15 Exercise 3. Here, we have a = 1, b = 3 and θ = 30. a b = a b sin θ = (1)(3) sin 30 = (1)(3) 1 2 = 3 2. Return to Exercise 3

16 Solutions to exercises 16 Exercise 4. Here, we have a = 2, b = 5 and θ = 35. a b = a b sin θ = (2)(5) sin 35 (2)(5)(0.5736) = Return to Exercise 4

17 Solutions to exercises 17 Exercise 5. Here, we have s = 2 m, F = 4 N and θ = 30. τ = s F = s F sin θ = (2 m)(4 N) sin 30 = (2 m)(4 N) 1 2 = (8)(N m) 1 2 = (8)(J) 1 2 = 4 J. Return to Exercise 5

18 Solutions to exercises 18 Exercise 6. Here, we have ω = 3 s 1, s = 2 m and θ = 45. τ = ω s = ω s sin θ = (3 s 1 )(2 m) sin 45 = (3 s 1 )(2 m) 1 2 = (6)(m s 1 ) = (6)(m s 1 ) 2 = 3 2 m s 1. Return to Exercise 6

19 Solutions to exercises 19 Exercise 7. a b is a vector that is perpendicular to both a and b. Here, we have a = 4i + 2j k and b = 2i 6j 3k. i j k a b = = i j k = i[(2)( 3) ( 1)( 6)] j[(4)( 3) ( 1)(2)]+k[(4)( 6) (2)(2)] a b = i[ 6 6] j[ ] + k[ 24 4] = 12i + 10j 28k. Return to Exercise 7

20 Solutions to exercises 20 Exercise 8. Here, we have a = 2i + j k and b = 3i 6j + 2k i j k a b = = i j k = i[(1)(2) ( 1)( 6)] j[(2)(2) ( 1)(3)]+k[(2)( 6) (1)(3)] a b = i[2 6] j[4 + 3] + k[ 12 3] = 4i 7j 15k. Return to Exercise 8

21 Solutions to exercises 21 Exercise 9. Here, we have a = 3i + 4j 3k and b = i + 3j + 2k i j k a b = = i j k = i[(4)(2) ( 3)(3)] j[(3)(2) ( 3)(1)]+k[(3)(3) (4)(1)] a b = i[8 + 9] j[6 + 3] + k[9 4] = 17i 9j + 5k. Return to Exercise 9

22 Solutions to exercises 22 Exercise 10. Here, we have a = i + 2j k and b = 3i + 3j + k i j k a b = = i j k = i[(2)(1) ( 1)(3)] j[(1)(1) ( 1)(3)]+k[(1)(3) (2)(3)] i.e. a b = i[2 + 3] j[1 + 3] + k[3 6] = 5i 4j 3k. Return to Exercise 10

23 Solutions to exercises 23 Exercise 11. Here, we have a = 2i + 4j + 2k and b = i + 5j 2k i j k a b = = i j k = i[(4)( 2) (2)(5)] j[(2)( 2) (2)(1)]+k[(2)(5) (4)(1)] a b = i[ 8 10] j[ 4 2] + k[10 4] = 18i + 6j + 6k. Return to Exercise 11

24 Solutions to exercises 24 Exercise 12. Here, we have a = 3i 4j + k and b = 2i + 5j k i j k a b = = i j k = i[( 4)( 1) (1)(5)] j[(3)( 1) (1)(2)]+k[(3)(5) ( 4)(2)] a b = i[4 5] j[ 3 2] + k[15 + 8] = i + 5j + 23k. Return to Exercise 12

25 Solutions to exercises 25 Exercise 13. Here, we have a = 2i 3j + k and b = 2i + j + 4k i j k a b = = i j k = i[( 3)(4) (1)(1)] j[(2)(4) (1)(2)]+k[(2)(1) ( 3)(2)] a b = i[ 12 1] j[8 2] + k[2 + 6] = 13i 6j + 8k. Return to Exercise 13

26 Solutions to exercises 26 Exercise 14. Here, we have a = 2i + 3k = 2i + 0j + 3k and b = i + 2j + 4k i j k a b = = i j k = i[(0)(4) (3)(2)] j[(2)(4) (3)(1)]+k[(2)(2) (0)(1)] a b = i[0 6] j[8 3] + k[4 0] = 6i 5j + 4k. Return to Exercise 14

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