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University Physics Volume 1

2.4 Products of Vectors

University Physics Volume 12.4 Products of Vectors
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  1. Preface
  2. Unit 1. Mechanics
    1. 1 Units and Measurement
      1. Introduction
      2. 1.1 The Scope and Scale of Physics
      3. 1.2 Units and Standards
      4. 1.3 Unit Conversion
      5. 1.4 Dimensional Analysis
      6. 1.5 Estimates and Fermi Calculations
      7. 1.6 Significant Figures
      8. 1.7 Solving Problems in Physics
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    2. 2 Vectors
      1. Introduction
      2. 2.1 Scalars and Vectors
      3. 2.2 Coordinate Systems and Components of a Vector
      4. 2.3 Algebra of Vectors
      5. 2.4 Products of Vectors
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 3 Motion Along a Straight Line
      1. Introduction
      2. 3.1 Position, Displacement, and Average Velocity
      3. 3.2 Instantaneous Velocity and Speed
      4. 3.3 Average and Instantaneous Acceleration
      5. 3.4 Motion with Constant Acceleration
      6. 3.5 Free Fall
      7. 3.6 Finding Velocity and Displacement from Acceleration
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    4. 4 Motion in Two and Three Dimensions
      1. Introduction
      2. 4.1 Displacement and Velocity Vectors
      3. 4.2 Acceleration Vector
      4. 4.3 Projectile Motion
      5. 4.4 Uniform Circular Motion
      6. 4.5 Relative Motion in One and Two Dimensions
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    5. 5 Newton's Laws of Motion
      1. Introduction
      2. 5.1 Forces
      3. 5.2 Newton's First Law
      4. 5.3 Newton's Second Law
      5. 5.4 Mass and Weight
      6. 5.5 Newton’s Third Law
      7. 5.6 Common Forces
      8. 5.7 Drawing Free-Body Diagrams
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    6. 6 Applications of Newton's Laws
      1. Introduction
      2. 6.1 Solving Problems with Newton’s Laws
      3. 6.2 Friction
      4. 6.3 Centripetal Force
      5. 6.4 Drag Force and Terminal Speed
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    7. 7 Work and Kinetic Energy
      1. Introduction
      2. 7.1 Work
      3. 7.2 Kinetic Energy
      4. 7.3 Work-Energy Theorem
      5. 7.4 Power
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    8. 8 Potential Energy and Conservation of Energy
      1. Introduction
      2. 8.1 Potential Energy of a System
      3. 8.2 Conservative and Non-Conservative Forces
      4. 8.3 Conservation of Energy
      5. 8.4 Potential Energy Diagrams and Stability
      6. 8.5 Sources of Energy
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
    9. 9 Linear Momentum and Collisions
      1. Introduction
      2. 9.1 Linear Momentum
      3. 9.2 Impulse and Collisions
      4. 9.3 Conservation of Linear Momentum
      5. 9.4 Types of Collisions
      6. 9.5 Collisions in Multiple Dimensions
      7. 9.6 Center of Mass
      8. 9.7 Rocket Propulsion
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    10. 10 Fixed-Axis Rotation
      1. Introduction
      2. 10.1 Rotational Variables
      3. 10.2 Rotation with Constant Angular Acceleration
      4. 10.3 Relating Angular and Translational Quantities
      5. 10.4 Moment of Inertia and Rotational Kinetic Energy
      6. 10.5 Calculating Moments of Inertia
      7. 10.6 Torque
      8. 10.7 Newton’s Second Law for Rotation
      9. 10.8 Work and Power for Rotational Motion
      10. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    11. 11 Angular Momentum
      1. Introduction
      2. 11.1 Rolling Motion
      3. 11.2 Angular Momentum
      4. 11.3 Conservation of Angular Momentum
      5. 11.4 Precession of a Gyroscope
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    12. 12 Static Equilibrium and Elasticity
      1. Introduction
      2. 12.1 Conditions for Static Equilibrium
      3. 12.2 Examples of Static Equilibrium
      4. 12.3 Stress, Strain, and Elastic Modulus
      5. 12.4 Elasticity and Plasticity
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    13. 13 Gravitation
      1. Introduction
      2. 13.1 Newton's Law of Universal Gravitation
      3. 13.2 Gravitation Near Earth's Surface
      4. 13.3 Gravitational Potential Energy and Total Energy
      5. 13.4 Satellite Orbits and Energy
      6. 13.5 Kepler's Laws of Planetary Motion
      7. 13.6 Tidal Forces
      8. 13.7 Einstein's Theory of Gravity
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    14. 14 Fluid Mechanics
      1. Introduction
      2. 14.1 Fluids, Density, and Pressure
      3. 14.2 Measuring Pressure
      4. 14.3 Pascal's Principle and Hydraulics
      5. 14.4 Archimedes’ Principle and Buoyancy
      6. 14.5 Fluid Dynamics
      7. 14.6 Bernoulli’s Equation
      8. 14.7 Viscosity and Turbulence
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  3. Unit 2. Waves and Acoustics
    1. 15 Oscillations
      1. Introduction
      2. 15.1 Simple Harmonic Motion
      3. 15.2 Energy in Simple Harmonic Motion
      4. 15.3 Comparing Simple Harmonic Motion and Circular Motion
      5. 15.4 Pendulums
      6. 15.5 Damped Oscillations
      7. 15.6 Forced Oscillations
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    2. 16 Waves
      1. Introduction
      2. 16.1 Traveling Waves
      3. 16.2 Mathematics of Waves
      4. 16.3 Wave Speed on a Stretched String
      5. 16.4 Energy and Power of a Wave
      6. 16.5 Interference of Waves
      7. 16.6 Standing Waves and Resonance
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 17 Sound
      1. Introduction
      2. 17.1 Sound Waves
      3. 17.2 Speed of Sound
      4. 17.3 Sound Intensity
      5. 17.4 Normal Modes of a Standing Sound Wave
      6. 17.5 Sources of Musical Sound
      7. 17.6 Beats
      8. 17.7 The Doppler Effect
      9. 17.8 Shock Waves
      10. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  4. A | Units
  5. B | Conversion Factors
  6. C | Fundamental Constants
  7. D | Astronomical Data
  8. E | Mathematical Formulas
  9. F | Chemistry
  10. G | The Greek Alphabet
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
    8. Chapter 8
    9. Chapter 9
    10. Chapter 10
    11. Chapter 11
    12. Chapter 12
    13. Chapter 13
    14. Chapter 14
    15. Chapter 15
    16. Chapter 16
    17. Chapter 17
  12. Index

Learning Objectives

By the end of this section, you will be able to:
  • Explain the difference between the scalar product and the vector product of two vectors.
  • Determine the scalar product of two vectors.
  • Determine the vector product of two vectors.
  • Describe how the products of vectors are used in physics.

A vector can be multiplied by another vector but may not be divided by another vector. There are two kinds of products of vectors used broadly in physics and engineering. One kind of multiplication is a scalar multiplication of two vectors. Taking a scalar product of two vectors results in a number (a scalar), as its name indicates. Scalar products are used to define work and energy relations. For example, the work that a force (a vector) performs on an object while causing its displacement (a vector) is defined as a scalar product of the force vector with the displacement vector. A quite different kind of multiplication is a vector multiplication of vectors. Taking a vector product of two vectors returns as a result a vector, as its name suggests. Vector products are used to define other derived vector quantities. For example, in describing rotations, a vector quantity called torque is defined as a vector product of an applied force (a vector) and its distance from pivot to force (a vector). It is important to distinguish between these two kinds of vector multiplications because the scalar product is a scalar quantity and a vector product is a vector quantity.

The Scalar Product of Two Vectors (the Dot Product)

Scalar multiplication of two vectors yields a scalar product.

Scalar Product (Dot Product)

The scalar product A·BA·B of two vectors AA and BB is a number defined by the equation

A·B=ABcosφ,A·B=ABcosφ,
(2.27)

where φφ is the angle between the vectors (shown in Figure 2.27). The scalar product is also called the dot product because of the dot notation that indicates it.

In the definition of the dot product, the direction of angle φφ does not matter, and φφ can be measured from either of the two vectors to the other because cosφ=cos(φ)=cos(2πφ)cosφ=cos(φ)=cos(2πφ). The dot product is a negative number when 90°<φ180°90°<φ180° and is a positive number when 0°φ<90°0°φ<90°. Moreover, the dot product of two parallel vectors is A·B=ABcos0°=ABA·B=ABcos0°=AB, and the dot product of two antiparallel vectors is A·B=ABcos180°=ABA·B=ABcos180°=AB. The scalar product of two orthogonal vectors vanishes: A·B=ABcos90°=0A·B=ABcos90°=0. The scalar product of a vector with itself is the square of its magnitude:

A2A·A=AAcos0°=A2.A2A·A=AAcos0°=A2.
(2.28)
Figure a: vectors A and B are shown tail to tail. A is longer than B. The angle between them is phi. Figure b: Vector B is extended using a dashed line and another dashed line is drawn from the head of A to the extension of B, perpendicular to B. A sub perpendicular is equal to A magnitude times cosine phi and is the distance from the vertex where the tails of A and B meet to the location where the perpendicular from A to B meets the extension of B. Figure c: A dashed line is drawn from the head of B to A, perpendicular to A. The distance from the tails of A and B to where the dashed line meets B is B sub perpendicular and is equal to magnitude B times cosine phi.
Figure 2.27 The scalar product of two vectors. (a) The angle between the two vectors. (b) The orthogonal projection AA of vector AA onto the direction of vector BB. (c) The orthogonal projection BB of vector BB onto the direction of vector AA.

Example 2.15

The Scalar Product For the vectors shown in Figure 2.13, find the scalar product A·FA·F.

Strategy From Figure 2.13, the magnitudes of vectors AA and FF are A = 10.0 and F = 20.0. Angle θθ, between them, is the difference: θ=φα=110°35°=75°θ=φα=110°35°=75°. Substituting these values into Equation 2.27 gives the scalar product.

Solution A straightforward calculation gives us

A·F=AFcosθ=(10.0)(20.0)cos75°=51.76.A·F=AFcosθ=(10.0)(20.0)cos75°=51.76.
Check Your Understanding 2.11

For the vectors given in Figure 2.13, find the scalar products A·BA·B and F·CF·C.

In the Cartesian coordinate system, scalar products of the unit vector of an axis with other unit vectors of axes always vanish because these unit vectors are orthogonal:

i^·j^=|i^||j^|cos90°=(1)(1)(0)=0,i^·k^=|i^||k^|cos90°=(1)(1)(0)=0,k^·j^=|k^||j^|cos90°=(1)(1)(0)=0.i^·j^=|i^||j^|cos90°=(1)(1)(0)=0,i^·k^=|i^||k^|cos90°=(1)(1)(0)=0,k^·j^=|k^||j^|cos90°=(1)(1)(0)=0.
(2.29)

In these equations, we use the fact that the magnitudes of all unit vectors are one: |i^|=|j^|=|k^|=1|i^|=|j^|=|k^|=1. For unit vectors of the axes, Equation 2.28 gives the following identities:

i^·i^=i2=j^·j^=j2=k^·k^=k2=1.i^·i^=i2=j^·j^=j2=k^·k^=k2=1.
(2.30)

The scalar product A·BA·B can also be interpreted as either the product of B with the projection AǁAǁ of vector AA onto the direction of vector BB (Figure 2.27(b)) or the product of A with the projection BǁBǁ of vector BB onto the direction of vector AA (Figure 2.27(c)):

A·B=ABcosφ=B(Acosφ)=BAǁ=A(Bcosφ)=ABǁ.A·B=ABcosφ=B(Acosφ)=BAǁ=A(Bcosφ)=ABǁ.

For example, in the rectangular coordinate system in a plane, the scalar x-component of a vector is its dot product with the unit vector i^i^, and the scalar y-component of a vector is its dot product with the unit vector j^j^:

{A·i^=|A||i^|cosθA=AcosθA=AxA·j^=|A||j^|cos(90°θA)=AsinθA=Ay.{A·i^=|A||i^|cosθA=AcosθA=AxA·j^=|A||j^|cos(90°θA)=AsinθA=Ay.

Scalar multiplication of vectors is commutative,

A·B=B·A,A·B=B·A,
(2.31)

and obeys the distributive law:

A·(B+C)=A·B+A·C.A·(B+C)=A·B+A·C.
(2.32)

We can use the commutative and distributive laws to derive various relations for vectors, such as expressing the dot product of two vectors in terms of their scalar components.

Check Your Understanding 2.12

For vector A=Axi^+Ayj^+Azk^A=Axi^+Ayj^+Azk^ in a rectangular coordinate system, use Equation 2.29 through Equation 2.32 to show that A·i^=AxA·i^=Ax A·j^=AyA·j^=Ay and A·k^=AzA·k^=Az.

When the vectors in Equation 2.27 are given in their vector component forms,

A=Axi^+Ayj^+Azk^andB=Bxi^+Byj^+Bzk^,A=Axi^+Ayj^+Azk^andB=Bxi^+Byj^+Bzk^,

we can compute their scalar product as follows:

A·B=(Axi^+Ayj^+Azk^)·(Bxi^+Byj^+Bzk^)=AxBxi^·i^+AxByi^·j^+AxBzi^·k^+AyBxj^·i^+AyByj^·j^+AyBzj^·k^+AzBxk^·i^+AzByk^·j^+AzBzk^·k^.A·B=(Axi^+Ayj^+Azk^)·(Bxi^+Byj^+Bzk^)=AxBxi^·i^+AxByi^·j^+AxBzi^·k^+AyBxj^·i^+AyByj^·j^+AyBzj^·k^+AzBxk^·i^+AzByk^·j^+AzBzk^·k^.

Since scalar products of two different unit vectors of axes give zero, and scalar products of unit vectors with themselves give one (see Equation 2.29 and Equation 2.30), there are only three nonzero terms in this expression. Thus, the scalar product simplifies to

A·B=AxBx+AyBy+AzBz.A·B=AxBx+AyBy+AzBz.
(2.33)

We can use Equation 2.33 for the scalar product in terms of scalar components of vectors to find the angle between two vectors. When we divide Equation 2.27 by AB, we obtain the equation for cosφcosφ, into which we substitute Equation 2.33:

cosφ=A·BAB=AxBx+AyBy+AzBzAB.cosφ=A·BAB=AxBx+AyBy+AzBzAB.
(2.34)

Angle φφ between vectors AA and BB is obtained by taking the inverse cosine of the expression in Equation 2.34.

Example 2.16

Angle between Two Forces Three dogs are pulling on a stick in different directions, as shown in Figure 2.28. The first dog pulls with force F1=(10.0i^20.4j^+2.0k^)NF1=(10.0i^20.4j^+2.0k^)N, the second dog pulls with force F2=(−15.0i^6.2k^)NF2=(−15.0i^6.2k^)N, and the third dog pulls with force F3=(5.0i^+12.5j^)NF3=(5.0i^+12.5j^)N. What is the angle between forces F1F1 and F2F2?

Three dogs pull on a stick.
Figure 2.28 Three dogs are playing with a stick.

Strategy The components of force vector F1F1 are F1x=10.0NF1x=10.0N, F1y=−20.4NF1y=−20.4N, and F1z=2.0NF1z=2.0N, whereas those of force vector F2F2 are F2x=−15.0NF2x=−15.0N, F2y=0.0NF2y=0.0N, and F2z=−6.2NF2z=−6.2N. Computing the scalar product of these vectors and their magnitudes, and substituting into Equation 2.34 gives the angle of interest.

Solution The magnitudes of forces F1F1 and F2F2 are

F1=F1x2+F1y2+F1z2=10.02+20.42+2.02N=22.8NF1=F1x2+F1y2+F1z2=10.02+20.42+2.02N=22.8N

and

F2=F2x2+F2y2+F2z2=15.02+6.22N=16.2N.F2=F2x2+F2y2+F2z2=15.02+6.22N=16.2N.

Substituting the scalar components into Equation 2.33 yields the scalar product

F1·F2=F1xF2x+F1yF2y+F1zF2z=(10.0N)(−15.0N)+(−20.4N)(0.0N)+(2.0N)(−6.2N)=−162.4N2.F1·F2=F1xF2x+F1yF2y+F1zF2z=(10.0N)(−15.0N)+(−20.4N)(0.0N)+(2.0N)(−6.2N)=−162.4N2.

Finally, substituting everything into Equation 2.34 gives the angle

cosφ=F1·F2F1F2=−162.4N2(22.8N)(16.2N)=−0.439φ=cos−1(−0.439)=116.0°.cosφ=F1·F2F1F2=−162.4N2(22.8N)(16.2N)=−0.439φ=cos−1(−0.439)=116.0°.

Significance Notice that when vectors are given in terms of the unit vectors of axes, we can find the angle between them without knowing the specifics about the geographic directions the unit vectors represent. Here, for example, the +x-direction might be to the east and the +y-direction might be to the north. But, the angle between the forces in the problem is the same if the +x-direction is to the west and the +y-direction is to the south.

Check Your Understanding 2.13

Find the angle between forces F1F1 and F3F3 in Example 2.16.

Example 2.17

The Work of a Force When force FF pulls on an object and when it causes its displacement DD, we say the force performs work. The amount of work the force does is the scalar product F·DF·D. If the stick in Example 2.16 moves momentarily and gets displaced by vector D=(−7.9j^4.2k^)cmD=(−7.9j^4.2k^)cm, how much work is done by the third dog in Example 2.16?

Strategy We compute the scalar product of displacement vector DD with force vector F3=(5.0i^+12.5j^)NF3=(5.0i^+12.5j^)N, which is the pull from the third dog. Let’s use W3W3 to denote the work done by force F3F3 on displacement DD.

Solution Calculating the work is a straightforward application of the dot product:

W3=F3·D=F3xDx+F3yDy+F3zDz=(5.0N)(0.0cm)+(12.5N)(−7.9cm)+(0.0N)(−4.2cm)=−98.7N·cm.W3=F3·D=F3xDx+F3yDy+F3zDz=(5.0N)(0.0cm)+(12.5N)(−7.9cm)+(0.0N)(−4.2cm)=−98.7N·cm.

Significance The SI unit of work is called the joule (J)(J), where 1 J = 1 N·mN·m. The unit cm·Ncm·N can be written as 10−2m·N=10−2J10−2m·N=10−2J, so the answer can be expressed as W3=−0.9875J−1.0JW3=−0.9875J−1.0J.

Check Your Understanding 2.14

How much work is done by the first dog and by the second dog in Example 2.16 on the displacement in Example 2.17?

The Vector Product of Two Vectors (the Cross Product)

Vector multiplication of two vectors yields a vector product.

Vector Product (Cross Product)

The vector product of two vectors AA and BB is denoted by A×BA×B and is often referred to as a cross product. The vector product is a vector that has its direction perpendicular to both vectors AA and BB. In other words, vector A×BA×B is perpendicular to the plane that contains vectors AA and BB, as shown in Figure 2.29. The magnitude of the vector product is defined as

|A×B|=ABsinφ,|A×B|=ABsinφ,
(2.35)

where angle φφ, between the two vectors, is measured from vector AA (first vector in the product) to vector BB (second vector in the product), as indicated in Figure 2.29, and is between 0°0° and 180°180°.

According to Equation 2.35, the vector product vanishes for pairs of vectors that are either parallel (φ=0°)(φ=0°) or antiparallel (φ=180°)(φ=180°) because sin0°=sin180°=0sin0°=sin180°=0.

Vector A points out and to the left, and vector B points out and to the right. The angle between them is phi. In figure a we are shown vector C which is the cross product of A cross B. Vector C points up and is perpendicular to both A and B. In figure b we are shown vector minus C which is the cross product of B cross A. Vector minus C points down and is perpendicular to both A and B.
Figure 2.29 The vector product of two vectors is drawn in three-dimensional space. (a) The vector product A×BA×B is a vector perpendicular to the plane that contains vectors AA and BB. Small squares drawn in perspective mark right angles between AA and CC, and between BB and CC so that if AA and BB lie on the floor, vector CC points vertically upward to the ceiling. (b) The vector product B×AB×A is a vector antiparallel to vector A×BA×B.

On the line perpendicular to the plane that contains vectors AA and BB there are two alternative directions—either up or down, as shown in Figure 2.29—and the direction of the vector product may be either one of them. In the standard right-handed orientation, where the angle between vectors is measured counterclockwise from the first vector, vector A×BA×B points upward, as seen in Figure 2.29(a). If we reverse the order of multiplication, so that now BB comes first in the product, then vector B×AB×A must point downward, as seen in Figure 2.29(b). This means that vectors A×BA×B and B×AB×A are antiparallel to each other and that vector multiplication is not commutative but anticommutative. The anticommutative property means the vector product reverses the sign when the order of multiplication is reversed:

A×B=B×A.A×B=B×A.
(2.36)

The corkscrew right-hand rule is a common mnemonic used to determine the direction of the vector product. As shown in Figure 2.30, a corkscrew is placed in a direction perpendicular to the plane that contains vectors AA and BB, and its handle is turned in the direction from the first to the second vector in the product. The direction of the cross product is given by the progression of the corkscrew.

Vector A points out and to the left, and vector B points out and to the right. In figure a we are shown the cross product of A cross B pointing up, perpendicular to both A and B. A screw turning an angle phi from A to B would move up. In figure b we are shown the cross product of B cross A pointing down, perpendicular to both A and B. A screw turning an angle phi from B to A would move down.
Figure 2.30 The corkscrew right-hand rule can be used to determine the direction of the cross product A×BA×B. Place a corkscrew in the direction perpendicular to the plane that contains vectors AA and BB, and turn it in the direction from the first to the second vector in the product. The direction of the cross product is given by the progression of the corkscrew. (a) Upward movement means the cross-product vector points up. (b) Downward movement means the cross-product vector points downward.

Example 2.18

The Torque of a ForceThe mechanical advantage that a familiar tool called a wrench provides (Figure 2.31) depends on magnitude F of the applied force, on its direction with respect to the wrench handle, and on how far from the nut this force is applied. The distance R from the nut to the point where force vector FF is attached is represented by the radial vector RR. The physical vector quantity that makes the nut turn is called torque (denoted by τ)τ), and it is the vector product of the distance between the pivot to force with the force: τ=R×Fτ=R×F.

To loosen a rusty nut, a 20.00-N force is applied to the wrench handle at angle φ=40°φ=40° and at a distance of 0.25 m from the nut, as shown in Figure 2.31(a). Find the magnitude and direction of the torque applied to the nut. What would the magnitude and direction of the torque be if the force were applied at angle φ=45°φ=45°, as shown in Figure 2.31(b)? For what value of angle φφ does the torque have the largest magnitude?

Figure a: a wrench grips a nut. A force F is applied to the wrench at a distance R from the center of the nut. The vector R is the vector from the center of the nut to the location where the force is being applied. The force direction is at an angle phi, measured counterclockwise from the direction of the vector R. Figure b: a wrench grips a nut. A force F is applied to the wrench at a distance R from the center of the nut. The vector R is the vector from the center of the nut to the location where the force is being applied. The force direction is at an angle phi, measured clockwise from the direction of the vector R.
Figure 2.31 A wrench provides grip and mechanical advantage in applying torque to turn a nut. (a) Turn counterclockwise to loosen the nut. (b) Turn clockwise to tighten the nut.

Strategy We adopt the frame of reference shown in Figure 2.31, where vectors RR and FF lie in the xy-plane and the origin is at the position of the nut. The radial direction along vector RR (pointing away from the origin) is the reference direction for measuring the angle φφ because RR is the first vector in the vector product τ=R×Fτ=R×F. Vector ττ must lie along the z-axis because this is the axis that is perpendicular to the xy-plane, where both RR and FF lie. To compute the magnitude ττ, we use Equation 2.35. To find the direction of ττ, we use the corkscrew right-hand rule (Figure 2.30).

SolutionFor the situation in (a), the corkscrew rule gives the direction of R×FR×F in the positive direction of the z-axis. Physically, it means the torque vector ττ points out of the page, perpendicular to the wrench handle. We identify F = 20.00 N and R = 0.25 m, and compute the magnitude using Equation 2.35:

τ=|R×F|=RFsinφ=(0.25m)(20.00N)sin40°=3.21N·m.τ=|R×F|=RFsinφ=(0.25m)(20.00N)sin40°=3.21N·m.

For the situation in (b), the corkscrew rule gives the direction of R×FR×F in the negative direction of the z-axis. Physically, it means the vector ττ points into the page, perpendicular to the wrench handle. The magnitude of this torque is

τ=|R×F|=RFsinφ=(0.25m)(20.00N)sin45°=3.53N·m.τ=|R×F|=RFsinφ=(0.25m)(20.00N)sin45°=3.53N·m.

The torque has the largest value when sinφ=1sinφ=1, which happens when φ=90°φ=90°. Physically, it means the wrench is most effective—giving us the best mechanical advantage—when we apply the force perpendicular to the wrench handle. For the situation in this example, this best-torque value is τbest=RF=(0.25m)(20.00N)=5.00N·mτbest=RF=(0.25m)(20.00N)=5.00N·m.

Significance When solving mechanics problems, we often do not need to use the corkscrew rule at all, as we’ll see now in the following equivalent solution. Notice that once we have identified that vector R×FR×F lies along the z-axis, we can write this vector in terms of the unit vector k^k^ of the z-axis:

R×F=RFsinφk^.R×F=RFsinφk^.

In this equation, the number that multiplies k^k^ is the scalar z-component of the vector R×FR×F. In the computation of this component, care must be taken that the angle φφ is measured counterclockwise from RR (first vector) to FF (second vector). Following this principle for the angles, we obtain RFsin(+40°)=+3.2N·mRFsin(+40°)=+3.2N·m for the situation in (a), and we obtain RFsin(−45°)=−3.5N·mRFsin(−45°)=−3.5N·m for the situation in (b). In the latter case, the angle is negative because the graph in Figure 2.31 indicates the angle is measured clockwise; but, the same result is obtained when this angle is measured counterclockwise because +(360°45°)=+315°+(360°45°)=+315° and sin(+315°)=sin(−45°)sin(+315°)=sin(−45°). In this way, we obtain the solution without reference to the corkscrew rule. For the situation in (a), the solution is R×F=+3.2N·mk^R×F=+3.2N·mk^; for the situation in (b), the solution is R×F=−3.5N·mk^R×F=−3.5N·mk^.

Check Your Understanding 2.15

For the vectors given in Figure 2.13, find the vector products A×BA×B and C×FC×F.

Similar to the dot product (Equation 2.32), the cross product has the following distributive property:

A×(B+C)=A×B+A×C.A×(B+C)=A×B+A×C.
(2.37)

The distributive property is applied frequently when vectors are expressed in their component forms, in terms of unit vectors of Cartesian axes.

When we apply the definition of the cross product, Equation 2.35, to unit vectors i^i^, j^j^, and k^k^ that define the positive x-, y-, and z-directions in space, we find that

i^×i^=j^×j^=k^×k^=0.i^×i^=j^×j^=k^×k^=0.
(2.38)

All other cross products of these three unit vectors must be vectors of unit magnitudes because i^i^, j^j^, and k^k^ are orthogonal. For example, for the pair i^i^ and j^j^, the magnitude is |i^×j^|=ijsin90°=(1)(1)(1)=1|i^×j^|=ijsin90°=(1)(1)(1)=1. The direction of the vector product i^×j^i^×j^ must be orthogonal to the xy-plane, which means it must be along the z-axis. The only unit vectors along the z-axis are k^k^ or +k^+k^. By the corkscrew rule, the direction of vector i^×j^i^×j^ must be parallel to the positive z-axis. Therefore, the result of the multiplication i^×j^i^×j^ is identical to +k^+k^. We can repeat similar reasoning for the remaining pairs of unit vectors. The results of these multiplications are

{i^×j^=+k^,j^×k^=+i^,k^×i^=+j^.{i^×j^=+k^,j^×k^=+i^,k^×i^=+j^.
(2.39)

Notice that in Equation 2.39, the three unit vectors i^i^, j^j^, and k^k^ appear in the cyclic order shown in a diagram in Figure 2.32(a). The cyclic order means that in the product formula, i^i^ follows k^k^ and comes before j^j^, or k^k^ follows j^j^ and comes before i^i^, or j^j^ follows i^i^ and comes before k^k^. The cross product of two different unit vectors is always a third unit vector. When two unit vectors in the cross product appear in the cyclic order, the result of such a multiplication is the remaining unit vector, as illustrated in Figure 2.32(b). When unit vectors in the cross product appear in a different order, the result is a unit vector that is antiparallel to the remaining unit vector (i.e., the result is with the minus sign, as shown by the examples in Figure 2.32(c) and Figure 2.32(d). In practice, when the task is to find cross products of vectors that are given in vector component form, this rule for the cross-multiplication of unit vectors is very useful.

Figure a: The unit vectors, I hat, j hat and k hat of the x y z coordinate system are shown. Arrows indicate the sequence from I hat to j hat to k hat and back to I hat. Figure b: The unit vectors, I hat, j hat and k hat of the x y z coordinate system are shown. I hat equals j hat cross k hat. j hat equals k hat cross i hat. k hat equals i hat cross j hat. Figure c: The unit vectors, I hat and j hat are shown along with minus k hat pointing down. Minus k hat equals j hat cross i hat. Figure d: The unit vectors, I hat and k hat are shown along with minus j hat pointing to the left. Minus j hat equals i hat cross k hat.
Figure 2.32 (a) The diagram of the cyclic order of the unit vectors of the axes. (b) The only cross products where the unit vectors appear in the cyclic order. These products have the positive sign. (c, d) Two examples of cross products where the unit vectors do not appear in the cyclic order. These products have the negative sign.

Suppose we want to find the cross product A×BA×B for vectors A=Axi^+Ayj^+Azk^A=Axi^+Ayj^+Azk^ and B=Bxi^+Byj^+Bzk^B=Bxi^+Byj^+Bzk^. We can use the distributive property (Equation 2.37), the anticommutative property (Equation 2.36), and the results in Equation 2.38 and Equation 2.39 for unit vectors to perform the following algebra:

A×B=(Axi^+Ayj^+Azk^)×(Bxi^+Byj^+Bzk^)=Axi^×(Bxi^+Byj^+Bzk^)+Ayj^×(Bxi^+Byj^+Bzk^)+Azk^×(Bxi^+Byj^+Bzk^)=AxBxi^×i^+AxByi^×j^+AxBzi^×k^+AyBxj^×i^+AyByj^×j^+AyBzj^×k^+AzBxk^×i^+AzByk^×j^+AzBzk^×k^=AxBx(0)+AxBy(+k^)+AxBz(j^)+AyBx(k^)+AyBy(0)+AyBz(+i^)+AzBx(+j^)+AzBy(i^)+AzBz(0).A×B=(Axi^+Ayj^+Azk^)×(Bxi^+Byj^+Bzk^)=Axi^×(Bxi^+Byj^+Bzk^)+Ayj^×(Bxi^+Byj^+Bzk^)+Azk^×(Bxi^+Byj^+Bzk^)=AxBxi^×i^+AxByi^×j^+AxBzi^×k^+AyBxj^×i^+AyByj^×j^+AyBzj^×k^+AzBxk^×i^+AzByk^×j^+AzBzk^×k^=AxBx(0)+AxBy(+k^)+AxBz(j^)+AyBx(k^)+AyBy(0)+AyBz(+i^)+AzBx(+j^)+AzBy(i^)+AzBz(0).

When performing algebraic operations involving the cross product, be very careful about keeping the correct order of multiplication because the cross product is anticommutative. The last two steps that we still have to do to complete our task are, first, grouping the terms that contain a common unit vector and, second, factoring. In this way we obtain the following very useful expression for the computation of the cross product:

C=A×B=(AyBzAzBy)i^+(AzBxAxBz)j^+(AxByAyBx)k^.C=A×B=(AyBzAzBy)i^+(AzBxAxBz)j^+(AxByAyBx)k^.
(2.40)

In this expression, the scalar components of the cross-product vector are

{Cx=AyBzAzBy,Cy=AzBxAxBz,Cz=AxByAyBx.{Cx=AyBzAzBy,Cy=AzBxAxBz,Cz=AxByAyBx.
(2.41)

When finding the cross product, in practice, we can use either Equation 2.35 or Equation 2.40, depending on which one of them seems to be less complex computationally. They both lead to the same final result. One way to make sure if the final result is correct is to use them both.

Example 2.19

A Particle in a Magnetic Field When moving in a magnetic field, some particles may experience a magnetic force. Without going into details—a detailed study of magnetic phenomena comes in later chapters—let’s acknowledge that the magnetic field BB is a vector, the magnetic force FF is a vector, and the velocity uu of the particle is a vector. The magnetic force vector is proportional to the vector product of the velocity vector with the magnetic field vector, which we express as F=ζu×BF=ζu×B. In this equation, a constant ζζ takes care of the consistency in physical units, so we can omit physical units on vectors uu and BB. In this example, let’s assume the constant ζζ is positive.

A particle moving in space with velocity vector u=−5.0i^2.0j^+3.5k^u=−5.0i^2.0j^+3.5k^ enters a region with a magnetic field and experiences a magnetic force. Find the magnetic force FF on this particle at the entry point to the region where the magnetic field vector is (a) B=7.2i^j^2.4k^B=7.2i^j^2.4k^ and (b) B=4.5k^B=4.5k^. In each case, find magnitude F of the magnetic force and angle θθ the force vector FF makes with the given magnetic field vector BB.

Strategy First, we want to find the vector product u×Bu×B, because then we can determine the magnetic force using F=ζu×BF=ζu×B. Magnitude F can be found either by using components, F=Fx2+Fy2+Fz2F=Fx2+Fy2+Fz2, or by computing the magnitude |u×B||u×B| directly using Equation 2.35. In the latter approach, we would have to find the angle between vectors uu and BB. When we have FF, the general method for finding the direction angle θθ involves the computation of the scalar product F·BF·B and substitution into Equation 2.34. To compute the vector product we can either use Equation 2.40 or compute the product directly, whichever way is simpler.

Solution The components of the velocity vector are ux=−5.0ux=−5.0, uy=−2.0uy=−2.0, and uz=3.5uz=3.5.

(a) The components of the magnetic field vector are Bx=7.2Bx=7.2, By=−1.0By=−1.0, and Bz=−2.4Bz=−2.4. Substituting them into Equation 2.41 gives the scalar components of vector F=ζu×BF=ζu×B:

{Fx=ζ(uyBzuzBy)=ζ[(−2.0)(−2.4)(3.5)(−1.0)]=8.3ζFy=ζ(uzBxuxBz)=ζ[(3.5)(7.2)(−5.0)(−2.4)]=13.2ζFz=ζ(uxByuyBx)=ζ[(−5.0)(−1.0)(−2.0)(7.2)]=19.4ζ.{Fx=ζ(uyBzuzBy)=ζ[(−2.0)(−2.4)(3.5)(−1.0)]=8.3ζFy=ζ(uzBxuxBz)=ζ[(3.5)(7.2)(−5.0)(−2.4)]=13.2ζFz=ζ(uxByuyBx)=ζ[(−5.0)(−1.0)(−2.0)(7.2)]=19.4ζ.

Thus, the magnetic force is F=ζ(8.3i^+13.2j^+19.4k^)F=ζ(8.3i^+13.2j^+19.4k^) and its magnitude is

F=Fx2+Fy2+Fz2=ζ(8.3)2+(13.2)2+(19.4)2=24.9ζ.F=Fx2+Fy2+Fz2=ζ(8.3)2+(13.2)2+(19.4)2=24.9ζ.

To compute angle θθ, we may need to find the magnitude of the magnetic field vector,

B=Bx2+By2+Bz2=(7.2)2+(−1.0)2+(−2.4)2=7.6,B=Bx2+By2+Bz2=(7.2)2+(−1.0)2+(−2.4)2=7.6,

and the scalar product F·BF·B:

F·B=FxBx+FyBy+FzBz=(8.3ζ)(7.2)+(13.2ζ)(−1.0)+(19.4ζ)(−2.4)=0.F·B=FxBx+FyBy+FzBz=(8.3ζ)(7.2)+(13.2ζ)(−1.0)+(19.4ζ)(−2.4)=0.

Now, substituting into Equation 2.34 gives angle θθ:

cosθ=F·BFB=0(18.2ζ)(7.6)=0θ=90°.cosθ=F·BFB=0(18.2ζ)(7.6)=0θ=90°.

Hence, the magnetic force vector is perpendicular to the magnetic field vector. (We could have saved some time if we had computed the scalar product earlier.)

(b) Because vector B=4.5k^B=4.5k^ has only one component, we can perform the algebra quickly and find the vector product directly:

F=ζu×B=ζ(−5.0i^2.0j^+3.5k^)×(4.5k^)=ζ[(−5.0)(4.5)i^×k^+(−2.0)(4.5)j^×k^+(3.5)(4.5)k^×k^]=ζ[−22.5(j^)9.0(+i^)+0]=ζ(−9.0i^+22.5j^).F=ζu×B=ζ(−5.0i^2.0j^+3.5k^)×(4.5k^)=ζ[(−5.0)(4.5)i^×k^+(−2.0)(4.5)j^×k^+(3.5)(4.5)k^×k^]=ζ[−22.5(j^)9.0(+i^)+0]=ζ(−9.0i^+22.5j^).

The magnitude of the magnetic force is

F=Fx2+Fy2+Fz2=ζ(−9.0)2+(22.5)2+(0.0)2=24.2ζ.F=Fx2+Fy2+Fz2=ζ(−9.0)2+(22.5)2+(0.0)2=24.2ζ.

Because the scalar product is

F·B=FxBx+FyBy+FzBz=(−9.0ζ)(0)+(22.5ζ)(0)+(0)(4.5)=0,F·B=FxBx+FyBy+FzBz=(−9.0ζ)(0)+(22.5ζ)(0)+(0)(4.5)=0,

the magnetic force vector FF is perpendicular to the magnetic field vector BB.

Significance Even without actually computing the scalar product, we can predict that the magnetic force vector must always be perpendicular to the magnetic field vector because of the way this vector is constructed. Namely, the magnetic force vector is the vector product F=ζu×BF=ζu×B and, by the definition of the vector product (see Figure 2.29), vector FF must be perpendicular to both vectors uu and BB.

Check Your Understanding 2.16

Given two vectors A=i^+j^A=i^+j^ and B=3i^j^B=3i^j^, find (a) A×BA×B, (b) |A×B||A×B|, (c) the angle between AA and BB, and (d) the angle between A×BA×B and vector C=i^+k^C=i^+k^.

In conclusion to this section, we want to stress that “dot product” and “cross product” are entirely different mathematical objects that have different meanings. The dot product is a scalar; the cross product is a vector. Later chapters use the terms dot product and scalar product interchangeably. Similarly, the terms cross product and vector product are used interchangeably.

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