To Choose: The correct option.

Question

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Answer 1

Question

Chapter 10, Problem 2P

To determine

To Choose: The correct option.

Expert Solution & Answer

Answer to Problem 2P

Option (b)

Explanation of Solution

Given:

Two nonzero vectors A→ and B→ .

Formula used:

The formula of vector product:

A→×B→=|A→||B→|sinθ n^

·|A→|= Magnitude of vector B

·|A→|= Magnitude of vector A

·θ= Angle between the vectors A→ and B→ .

·n^= Unit vector perpendicular to the plane A→×B→ .

Calculation:

When A→ and B→ are parallel:

θ=0o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin0o n^=|A→||B→|×0 n^ ∵(sin0o=0)=0 n^

The magnitude of the vector product:

|A→×B→|=02=0

When A→ and B→ are perpendicular:

θ=90o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin90o n^=|A→||B→|×1 n^ ∵(sin 90o=1)=|A→||B→| n^

The magnitude of the vector product:

|A→×B→|= ( coefficient of vector )2= ( | A → || B → | )2=|A→||B→|

When A→ and B→ are antiparallel:

θ=180o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin180o n^=|A→||B→|×0 n^ ∵(sin 180o=0)=0 n^

The magnitude of the vector product:

|A→×B→|=02=0

When A→ and B→ are at 45o :

θ=45o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin45o n^=|A→||B→|×12 n^ ∵(sin 90o=1)=| A →|| B →|2 n^

The magnitude of the vector product:

|A→×B→|= ( coefficient of vector )2= ( | A → || B → | 2 )2=| A →|| B →|2

The greatest magnitude of the vector product is |A→||B→| .

Conclusion:

Hence, when the vectors A→ and B→ are perpendicular to each other, the magnitude of the vector productwould be greatest.

Option (b) is correct.

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Students have asked these similar questions

Sketch the harmonic on graphing paper.

Exercise 1: (a) Using the explicit formulae derived in the lectures for the (2j+1) × (2j + 1) repre- sentation matrices Dm'm, (J/h), derive the 3 × 3 matrices corresponding to the case j = 1. (b) Verify that they satisfy the so(3) Lie algebra commutation relation: [D(Î₁/ħ), D(Î₂/h)]m'm₁ = iƊm'm² (Ĵ3/h). (c) Prove the identity 3 Dm'm,(Î²) = Σ (D(Î¡)D(Î¡))m'¡m; · i=1

Sketch the harmonic.

Answer 2

Question

Chapter 10, Problem 2P

To determine

To Choose: The correct option.

Expert Solution & Answer

Answer to Problem 2P

Option (b)

Explanation of Solution

Given:

Two nonzero vectors A→ and B→ .

Formula used:

The formula of vector product:

A→×B→=|A→||B→|sinθ n^

·|A→|= Magnitude of vector B

·|A→|= Magnitude of vector A

·θ= Angle between the vectors A→ and B→ .

·n^= Unit vector perpendicular to the plane A→×B→ .

Calculation:

When A→ and B→ are parallel:

θ=0o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin0o n^=|A→||B→|×0 n^ ∵(sin0o=0)=0 n^

The magnitude of the vector product:

|A→×B→|=02=0

When A→ and B→ are perpendicular:

θ=90o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin90o n^=|A→||B→|×1 n^ ∵(sin 90o=1)=|A→||B→| n^

The magnitude of the vector product:

|A→×B→|= ( coefficient of vector )2= ( | A → || B → | )2=|A→||B→|

When A→ and B→ are antiparallel:

θ=180o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin180o n^=|A→||B→|×0 n^ ∵(sin 180o=0)=0 n^

The magnitude of the vector product:

|A→×B→|=02=0

When A→ and B→ are at 45o :

θ=45o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin45o n^=|A→||B→|×12 n^ ∵(sin 90o=1)=| A →|| B →|2 n^

The magnitude of the vector product:

|A→×B→|= ( coefficient of vector )2= ( | A → || B → | 2 )2=| A →|| B →|2

The greatest magnitude of the vector product is |A→||B→| .

Conclusion:

Hence, when the vectors A→ and B→ are perpendicular to each other, the magnitude of the vector productwould be greatest.

Option (b) is correct.

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Answer 3

Question

Chapter 10, Problem 2P

To determine

To Choose: The correct option.

Expert Solution & Answer

Answer to Problem 2P

Option (b)

Explanation of Solution

Given:

Two nonzero vectors A→ and B→ .

Formula used:

The formula of vector product:

A→×B→=|A→||B→|sinθ n^

·|A→|= Magnitude of vector B

·|A→|= Magnitude of vector A

·θ= Angle between the vectors A→ and B→ .

·n^= Unit vector perpendicular to the plane A→×B→ .

Calculation:

When A→ and B→ are parallel:

θ=0o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin0o n^=|A→||B→|×0 n^ ∵(sin0o=0)=0 n^

The magnitude of the vector product:

|A→×B→|=02=0

When A→ and B→ are perpendicular:

θ=90o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin90o n^=|A→||B→|×1 n^ ∵(sin 90o=1)=|A→||B→| n^

The magnitude of the vector product:

|A→×B→|= ( coefficient of vector )2= ( | A → || B → | )2=|A→||B→|

When A→ and B→ are antiparallel:

θ=180o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin180o n^=|A→||B→|×0 n^ ∵(sin 180o=0)=0 n^

The magnitude of the vector product:

|A→×B→|=02=0

When A→ and B→ are at 45o :

θ=45o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin45o n^=|A→||B→|×12 n^ ∵(sin 90o=1)=| A →|| B →|2 n^

The magnitude of the vector product:

|A→×B→|= ( coefficient of vector )2= ( | A → || B → | 2 )2=| A →|| B →|2

The greatest magnitude of the vector product is |A→||B→| .

Conclusion:

Hence, when the vectors A→ and B→ are perpendicular to each other, the magnitude of the vector productwould be greatest.

Option (b) is correct.

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Answer 4

Question

Chapter 10, Problem 2P

To determine

To Choose: The correct option.

Expert Solution & Answer

Answer to Problem 2P

Option (b)

Explanation of Solution

Given:

Two nonzero vectors A→ and B→ .

Formula used:

The formula of vector product:

A→×B→=|A→||B→|sinθ n^

·|A→|= Magnitude of vector B

·|A→|= Magnitude of vector A

·θ= Angle between the vectors A→ and B→ .

·n^= Unit vector perpendicular to the plane A→×B→ .

Calculation:

When A→ and B→ are parallel:

θ=0o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin0o n^=|A→||B→|×0 n^ ∵(sin0o=0)=0 n^

The magnitude of the vector product:

|A→×B→|=02=0

When A→ and B→ are perpendicular:

θ=90o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin90o n^=|A→||B→|×1 n^ ∵(sin 90o=1)=|A→||B→| n^

The magnitude of the vector product:

|A→×B→|= ( coefficient of vector )2= ( | A → || B → | )2=|A→||B→|

When A→ and B→ are antiparallel:

θ=180o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin180o n^=|A→||B→|×0 n^ ∵(sin 180o=0)=0 n^

The magnitude of the vector product:

|A→×B→|=02=0

When A→ and B→ are at 45o :

θ=45o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin45o n^=|A→||B→|×12 n^ ∵(sin 90o=1)=| A →|| B →|2 n^

The magnitude of the vector product:

|A→×B→|= ( coefficient of vector )2= ( | A → || B → | 2 )2=| A →|| B →|2

The greatest magnitude of the vector product is |A→||B→| .

Conclusion:

Hence, when the vectors A→ and B→ are perpendicular to each other, the magnitude of the vector productwould be greatest.

Option (b) is correct.

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Answer 5

Chapter 10, Problem 2P

To determine

To Choose: The correct option.

Expert Solution & Answer

Answer to Problem 2P

Option (b)

Explanation of Solution

Given:

Two nonzero vectors A→ and B→ .

Formula used:

The formula of vector product:

A→×B→=|A→||B→|sinθ n^

·|A→|= Magnitude of vector B

·|A→|= Magnitude of vector A

·θ= Angle between the vectors A→ and B→ .

·n^= Unit vector perpendicular to the plane A→×B→ .

Calculation:

When A→ and B→ are parallel:

θ=0o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin0o n^=|A→||B→|×0 n^ ∵(sin0o=0)=0 n^

The magnitude of the vector product:

|A→×B→|=02=0

When A→ and B→ are perpendicular:

θ=90o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin90o n^=|A→||B→|×1 n^ ∵(sin 90o=1)=|A→||B→| n^

The magnitude of the vector product:

|A→×B→|= ( coefficient of vector )2= ( | A → || B → | )2=|A→||B→|

When A→ and B→ are antiparallel:

θ=180o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin180o n^=|A→||B→|×0 n^ ∵(sin 180o=0)=0 n^

The magnitude of the vector product:

|A→×B→|=02=0

When A→ and B→ are at 45o :

θ=45o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin45o n^=|A→||B→|×12 n^ ∵(sin 90o=1)=| A →|| B →|2 n^

The magnitude of the vector product:

|A→×B→|= ( coefficient of vector )2= ( | A → || B → | 2 )2=| A →|| B →|2

The greatest magnitude of the vector product is |A→||B→| .

Conclusion:

Hence, when the vectors A→ and B→ are perpendicular to each other, the magnitude of the vector productwould be greatest.

Option (b) is correct.

Answer 6

Chapter 10, Problem 2P

To determine

To Choose: The correct option.

Expert Solution & Answer

Answer to Problem 2P

Option (b)

Explanation of Solution

Given:

Two nonzero vectors A→ and B→ .

Formula used:

The formula of vector product:

A→×B→=|A→||B→|sinθ n^

·|A→|= Magnitude of vector B

·|A→|= Magnitude of vector A

·θ= Angle between the vectors A→ and B→ .

·n^= Unit vector perpendicular to the plane A→×B→ .

Calculation:

When A→ and B→ are parallel:

θ=0o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin0o n^=|A→||B→|×0 n^ ∵(sin0o=0)=0 n^

The magnitude of the vector product:

|A→×B→|=02=0

When A→ and B→ are perpendicular:

θ=90o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin90o n^=|A→||B→|×1 n^ ∵(sin 90o=1)=|A→||B→| n^

The magnitude of the vector product:

|A→×B→|= ( coefficient of vector )2= ( | A → || B → | )2=|A→||B→|

When A→ and B→ are antiparallel:

θ=180o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin180o n^=|A→||B→|×0 n^ ∵(sin 180o=0)=0 n^

The magnitude of the vector product:

|A→×B→|=02=0

When A→ and B→ are at 45o :

θ=45o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin45o n^=|A→||B→|×12 n^ ∵(sin 90o=1)=| A →|| B →|2 n^

The magnitude of the vector product:

|A→×B→|= ( coefficient of vector )2= ( | A → || B → | 2 )2=| A →|| B →|2

The greatest magnitude of the vector product is |A→||B→| .

Conclusion:

Hence, when the vectors A→ and B→ are perpendicular to each other, the magnitude of the vector productwould be greatest.

Option (b) is correct.

Answer 7

Chapter 10, Problem 2P

To determine

To Choose: The correct option.

Answer 8

Expert Solution & Answer

Answer to Problem 2P

Option (b)

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

Two nonzero vectors A→ and B→ .

Formula used:

The formula of vector product:

A→×B→=|A→||B→|sinθ n^

·|A→|= Magnitude of vector B

·|A→|= Magnitude of vector A

·θ= Angle between the vectors A→ and B→ .

·n^= Unit vector perpendicular to the plane A→×B→ .

Calculation:

When A→ and B→ are parallel:

θ=0o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin0o n^=|A→||B→|×0 n^ ∵(sin0o=0)=0 n^

The magnitude of the vector product:

|A→×B→|=02=0

When A→ and B→ are perpendicular:

θ=90o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin90o n^=|A→||B→|×1 n^ ∵(sin 90o=1)=|A→||B→| n^

The magnitude of the vector product:

|A→×B→|= ( coefficient of vector )2= ( | A → || B → | )2=|A→||B→|

When A→ and B→ are antiparallel:

θ=180o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin180o n^=|A→||B→|×0 n^ ∵(sin 180o=0)=0 n^

The magnitude of the vector product:

|A→×B→|=02=0

When A→ and B→ are at 45o :

θ=45o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin45o n^=|A→||B→|×12 n^ ∵(sin 90o=1)=| A →|| B →|2 n^

The magnitude of the vector product:

|A→×B→|= ( coefficient of vector )2= ( | A → || B → | 2 )2=| A →|| B →|2

The greatest magnitude of the vector product is |A→||B→| .

Conclusion:

Hence, when the vectors A→ and B→ are perpendicular to each other, the magnitude of the vector productwould be greatest.

Option (b) is correct.

Answer 12

Ch. 10 - Prob. 1P Ch. 10 - Prob. 2P Ch. 10 - Prob. 3P Ch. 10 - Prob. 4P Ch. 10 - Prob. 5P Ch. 10 - Prob. 6P Ch. 10 - Prob. 7P Ch. 10 - Prob. 8P Ch. 10 - Prob. 9P Ch. 10 - Prob. 10P

To Choose: The correct option.

To Choose: The correct option.

Answer to Problem 2P

Explanation of Solution

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Chapter 10 Solutions