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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/8/8/2019 Vector Product of Vectors
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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
Exercise 3. Calculate |a b| when |a | = 1 , |b| = 3 and the anglebetween a and b is = 30
Exercise 4. Calculate |a b| when |a | = 2 , |b| = 5 and the anglebetween a and b is = 35
q Theory q Answers q Tips q Notation
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Section 2: Exercises 7
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
q Theory q Answers q Tips q Notation
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Section 2: Exercises 8
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
q Theory q Answers q Tips q Notation
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Section 2: Exercises 9
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
q Theory q Answers q Tips q Notation
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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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Solutions to exercises 14
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.
Return to Exercise 2
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Solutions to exercises 15
Exercise 3.
Here, we have
|a
|= 1,
|b
|= 3 and = 30 .
|a b| = |a ||b|sin = (1)(3) sin 30
= (1)(3) 12
=32
.
Return to Exercise 3
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Solutions to exercises 16
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.
Return to Exercise 4
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Solutions to exercises 17
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 .
Return to Exercise 5
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Solutions to exercises 18
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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Solutions to exercises 19
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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Solutions to exercises 20
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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Solutions to exercises 21
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Solutions to exercises 21
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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Solutions to exercises 22
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Solutions to exercises 22
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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Solutions to exercises 23
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Solutions to exercises 23
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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Solutions to exercises 24
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Solutions to exercises 24
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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Solutions to exercises 25
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Solutions to exercises 25
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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Solutions to exercises 26
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Solutions to exercises 26
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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