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

Given the vectors A = (8i + 4j - 2k) N B = (2j + 6k) m C = (3i - 2j + 4k) m Calculate the following: (a) the dot product of vectors A and B; (b) the projected component of B in the direction of C; (c) the angle between B and C.

"If vectors A = 4i + 3j + 2k and B = 2i + 3j + 4k, then the dot product A.B will be equal to"

Consider the vectors a = i + 8j + 4k and b = 7i − 6j − 6k, where i, j, and k are mutuallyperpendicular unit vectors forming a right-handed system. (a) Calculate (v) The vector product a × b. (vi) The direction cosines of a.

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