Vector Product of Vectors

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Vectors

VECTOR PRODUCT

Graham S McDonald

A Tutorial Module for learning about thevector product of two vectors

q Table of contentsq Begin Tutorial

c 2004 [email protected]
http://www.cse.salford.ac.uk/mailto:[email protected]:[email protected]://www.cse.salford.ac.uk/


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Table of contents1. Theory

2. Exercises

3. Answers

4. Tips on using solutions

5. Alternative notation

Full worked solutions


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Section 1: Theory 3

1. TheoryThe purpose of this tutorial is to practice working out the vector prod-

uct 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 where is the angle

between a and b.

b

a

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Section 1: Theory 4

(ii) The DIRECTION of the vector product of a and bis perpendicular to both a and b,

such that if we look along a

b

then a rotates towards b

in a clockwise manner.a b

b

a

C lo c k wise

x

It can be shown that the above denitions of magnitude and directionof a vector product allow us to calculate the x , y and z componentsof a b from the individual components of the vectors a and b

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Section 1: Theory 5

The components of the vector a b are given by the determinantof a matrix with 3 rows.The components of a = a x i + a y j + a z k and b = bx i + by j + bz kappear in the 2nd and 3rd rows.

This 3-row determinant is evaluated by expansion into 2-row deter-

minants, which are themselves then expanded. Matrix theory is itself very useful, and the scheme is shown below

a

b =

i j ka x a y a z

bx by bz

= i a y a zby bz ja x a zbx bz

+ k a x a ybx by

= i (a y bz a z by ) j (a x bz a z bx ) + k (a x by a y bx )Toc Back


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Section 2: Exercises 6

2. ExercisesClick 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 anglebetween a and b is = 45

Exercise 2. Calculate |a b| when |a | = 3 , |b| = 5 and the anglebetween a and b is = 60



q Theory q Answers q Tips q Notation

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Exercise 5. Calculate the magnitude of the torque = s F when|s | = 2 m , |F | = 4 N and = 30

( is the angle between the position

vectors

and the forceF

)

Exercise 6. Calculate the magnitude of the velocity v = s when|| = 3s

1 , |s | = 2m and = 45 ( is the angle between the angular

velocity and the position vector s )

Exercise 7. If a = 4 i +2 j k and b = 2 i 6i 3k then calculatea vector that is perpendicular to both a and b

Exercise 8. Calculate the vector a b when a = 2 i + j k andb = 3 i 6 j + 2 k


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Exercise 9. Calculate the vector a b when a = 3 i +4 j 3k andb = i + 3 j + 2 k

Exercise 10. Calculate the vector a b when a = i + 2 j k andb = 3 i + 3 j + k

Exercise 11. Calculate the vector a b when a = 2 i +4 j +2 k andb = i + 5 j 2k

Exercise 12. Calculate the vector a b when a = 3 i 4 j + k andb = 2 i + 5 j k


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Exercise 13. Calculate the vector a b when a = 2 i 3 j + k andb = 2 i + j + 4 k

Exercise 14.Calculate the vector

a

b

whena

= 2i

+ 3k

andb = i + 2 j + 4 k


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Section 3: Answers 10

3. Answers1. 42,2. 152 3,3. 32 ,4. 5.766,5. 4 J,

6. 32 m s 1 ,

7. 12i + 10 j 28k ,8. 4i 7 j 15k ,9. 17i

9 j + 5 k ,

10. 5i 4 j 4k ,11. 18i + 6 j + 6 k ,12. i + 5 j + 23 k ,13.

13i

6 j + 8 k ,

14. 6i 5 j + 4 k .Toc Back


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Section 4: Tips on using solutions 11

4. Tips on using solutions

q When looking at the THEORY, ANSWERS, TIPS or NOTATIONpages, use the Back button (at the bottom of the page) to return tothe exercises

q Use the solutions intelligently. For example, they can help you getstarted on an exercise, or they can allow you to check whether yourintermediate results are correct

q Try to make less use of the full solutions as you work your waythrough the Tutorial

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Section 5: Alternative notation 12

5. Alternative notation

q Here, we use symbols like a to denote a vector.In some texts, symbols for vectors are in bold (eg a instead of a )

q In this Tutorial, vectors are given in terms of the unit Cartesianvectors i , j and k . A common alternative notation involves quot-ing the Cartesian components within brackets. For example, thevector a = 2 i + j + 5 k can be written as a = (2 , 1, 5)

q The scalar product a b is also called a dot product (reectingthe symbol used to denote this type of multiplication). Likewise, thevector product a b is also called a cross product

q An alternative notation for the vector product is ab

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Solutions to exercises 13

Full worked solutionsExercise 1.

Here, we have |a | = 2, |b| = 4 and = 45 .

|a b| = |a ||b|sin = (2)(4) sin 45

= (2)(4)1

2=

82

=8

222

=82

2= 42.

Return to Exercise 1

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Exercise 2.

Here, we have

|a

|= 3,

|b

|= 5 and = 60 .

|a b| = |a ||b|sin = (3)(5) sin 60

= (3)(5) 32

=152

3.


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Exercise 3.

Here, we have

|a

|= 1,

|b

|= 3 and = 30 .

|a b| = |a ||b|sin = (1)(3) sin 30

= (1)(3) 12

=32

.


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Exercise 4.

Here, we have

|a

|= 2,

|b

|= 5 and = 35 .

|a b| = |a ||b|sin = (2)(5) sin 35

(2)(5)(0 .5736)= 5 .736.


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Exercise 5.

Here, we have

|s

|= 2 m,

|F

|= 4 N and = 30 .

| | = |s F |=

|s

||F

|sin

= (2 m)(4 N) sin 30

= (2 m)(4 N)12

= (8)(N m)1

2= (8)(J)

12

= 4 J .


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Exercise 6.

Here, we have || = 3 s 1 , |s | = 2 m and = 45

.

| | = | s |= |||s |sin = (3 s 1 )(2 m) sin 45

= (3 s 1 )(2 m) 12= (6)(m s 1 )

12

22

= (6)(m s 1 ) 22= 32 m s 1 .

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Exercise 7.

a b is a vector that is perpendicular to both a and b.Here, we have a = 4 i + 2 j k and b = 2 i 6 j 3k .a b =

i j k4 2 12 6 3

= i 2 16 3 j 4 12 3

+ k 4 22 6= i[(2)(3)(1)(6)] j [(4)(3)(1)(2)]+ k[(4)(6)(2)(2)]

a b = i[6 6] j [12 + 2] + k[24 4]

= 12i + 10 j 28k.Return to Exercise 7

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S l ti t i 20


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Exercise 8.

Here, we have a = 2 i + j k and b = 3 i 6 j + 2 k

a b =i j k2 1 13 6 2

= i 1 16 2 j 2 13 2 + k 2 13 6

= i[(1)(2)(1)(6)] j [(2)(2)(1)(3)]+ k[(2)(6)(1)(3)]

a b = i[26] j [4 + 3] + k[12 3]

= 4i 7 j 15k.Return to Exercise 8

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Exercise 9.

Here, we have a = 3 i + 4 j 3k and b = i + 3 j + 2 k

a b =i j k3 4 31 3 2

= i 4 33 2 j 3 31 2 + k 3 41 3= i[(4)(2)(3)(3)] j [(3)(2)(3)(1)]+ k[(3)(3)(4)(1)]

a b = i[8 + 9] j [6 + 3] + k[94]

= 17 i 9 j + 5 k.Return to Exercise 9

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Exercise 10.

Here, we have a = i + 2 j k and b = 3 i + 3 j + k

a b =i j k1 2 13 3 1

= i 2 13 1 j 1 13 1 + k 1 23 3= 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[36]

= 5 i 4 j 3k.Return to Exercise 10

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Exercise 11.

Here, we have a = 2 i + 4 j + 2 k and b = i + 5 j 2k

a b =i j k2 4 21 5 2

= i 4 25 2 j 2 21 2

+ k 2 41 5

= i[(4)(2)(2)(5)] j [(2)(2)(2)(1)]+ k[(2)(5)(4)(1)]

a b = i[8 10] j [4 2] + k[104]

= 18i + 6 j + 6 k.Return to Exercise 11

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Exercise 12.

Here, we have a = 3 i 4 j + k and b = 2 i + 5 j k

a b =i j k3 4 12 5 1

= i 4 15 1 j 3 12 1

+ k 3 42 5= i[(4)(1)(1)(5)] j [(3)(1)(1)(2)]+ k[(3)(5)(4)(2)]

a b = i[45] j [3 2] + k[15 + 8]

= i + 5 j + 23 k.Return to Exercise 12

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Exercise 13.

Here, we have a = 2 i 3 j + k and b = 2 i + j + 4 k

a b =i j k2 3 12 1 4

= i 3 11 4 j 2 12 4 + k 2 32 1= i[(3)(4)(1)(1)] j [(2)(4)(1)(2)]+ k[(2)(1)(3)(2)]

a b = i[12 1] j [82] + k[2 + 6]

= 13i 6 j + 8 k.Return to Exercise 13

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Exercise 14.

Here, we have a = 2 i + 3 k = 2 i + 0 j + 3 k and b = i + 2 j + 4 k

a b =i j k2 0 31 2 4

= i 0 32 4 j 2 31 4 + k 2 01 2= i[(0)(4)(3)(2)] j [(2)(4)(3)(1)]+ k[(2)(2)(0)(1)]

a b = i[06] j [83] + k[40]

= 6i 5 j + 4 k.Return to Exercise 14

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Vector Product of Vectors

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